Version of September 3, 2012

A Term of

Commutative Algebra

By Allen ALTMAN

and Steven KLEIMAN

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . iii 1. Rings and Ideals . . . . . . . . . . . . . . . . . . . 1 2. Prime Ideals . . . . . . . . . . . . . . . . . . . . 6 3. Radicals . . . . . . . . . . . . . . . . . . . . . . 10 4. Modules . . . . . . . . . . . . . . . . . . . . . . 14 5. Exact Sequences . . . . . . . . . . . . . . . . . . . 20 6. Direct Limits . . . . . . . . . . . . . . . . . . . . 26 7. Filtered Direct Limits . . . . . . . . . . . . . . . . . 33 8. Tensor Products . . . . . . . . . . . . . . . . . . . 37 9. Flatness . . . . . . . . . . . . . . . . . . . . . . 43 10. Cayley–Hamilton Theorem . . . . . . . . . . . . . . . 49 11. Localization of Rings . . . . . . . . . . . . . . . . . 55 12. Localization of Modules . . . . . . . . . . . . . . . . 61 13. Support . . . . . . . . . . . . . . . . . . . . . 66 14. Krull–Cohen–Seidenberg Theory . . . . . . . . . . . . . 71 15. Noether Normalization . . . . . . . . . . . . . . . . 75

Appendix: Jacobson Rings . . . . . . . . . . . . . . . 80 16. Chain Conditions . . . . . . . . . . . . . . . . . . 82 17. Associated Primes . . . . . . . . . . . . . . . . . . 87 18. Primary Decomposition . . . . . . . . . . . . . . . . 91 19. Length . . . . . . . . . . . . . . . . . . . . . . 97 20. Hilbert Functions . . . . . . . . . . . . . . . . . . 101

Appendix: Homogeneity . . . . . . . . . . . . . . . . 107 21. Dimension . . . . . . . . . . . . . . . . . . . . 109 22. Completion . . . . . . . . . . . . . . . . . . . . 115 23. Discrete Valuation Rings . . . . . . . . . . . . . . . 122 24. Dedekind Domains . . . . . . . . . . . . . . . . . 127 25. Fractional Ideals . . . . . . . . . . . . . . . . . . 131 26. Arbitrary Valuation Rings . . . . . . . . . . . . . . . 136 Solutions . . . . . . . . . . . . . . . . . . . . . . 141

1. Rings and Ideals . . . . . . . . . . . . . . . . . 141 2. Prime Ideals . . . . . . . . . . . . . . . . . . 143 3. Radicals . . . . . . . . . . . . . . . . . . . . 145 4. Modules . . . . . . . . . . . . . . . . . . . . 148 5. Exact Sequences . . . . . . . . . . . . . . . . . 149 6. Direct Limits . . . . . . . . . . . . . . . . . . 153 7. Filtered direct limits . . . . . . . . . . . . . . . . 156 8. Tensor Products . . . . . . . . . . . . . . . . . 158 9. Flatness . . . . . . . . . . . . . . . . . . . . 159 10. Cayley–Hamilton Theorem . . . . . . . . . . . . . 161 11. Localization of Rings . . . . . . . . . . . . . . . 164 12. Localization of Modules . . . . . . . . . . . . . . 167 13. Support . . . . . . . . . . . . . . . . . . . . 168 14. Krull–Cohen–Seidenberg Theory . . . . . . . . . . . 171 15. Noether Normalization . . . . . . . . . . . . . . 174 16. Chain Conditions . . . . . . . . . . . . . . . . 177 17. Associated Primes . . . . . . . . . . . . . . . . 179 18. Primary Decomposition . . . . . . . . . . . . . . 180 19. Length . . . . . . . . . . . . . . . . . . . . 183 20. Hilbert Functions . . . . . . . . . . . . . . . . 185 21. Dimension . . . . . . . . . . . . . . . . . . . 188 22. Completion . . . . . . . . . . . . . . . . . . 190 23. Discrete Valuation Rings . . . . . . . . . . . . . . 193 24. Dedekind Domains . . . . . . . . . . . . . . . . 197 25. Fractional Ideals . . . . . . . . . . . . . . . . . 199 26. Arbitrary Valuation Rings . . . . . . . . . . . . . 200

References . . . . . . . . . . . . . . . . . . . . . . 202 Index . . . . . . . . . . . . . . . . . . . . . . . . 203

Preface

There is no shortage of books on Commutative Algebra, but the present book is different. Most books are monographs, with extensive coverage. There is one notable exception: Atiyah and Macdonald’s 1969 classic [2]. It is a clear, concise, and efficient textbook, aimed at beginners, with a good selection of topics. So it has remained popular. However, its age and flaws do show. So there is need for an updated and improved version, which the present book aims to be.

Atiyah and Macdonald explain their philosophy in their introduction. They say their book “has the modest aim of providing a rapid introduction to the subject. It is designed to be read by students who have had a first elementary course in general algebra. On the other hand, it is not intended as a substitute for the more voluminous tracts on Commutative Algebra … The lecture-note origin of this book accounts for the rather terse style, with little general padding, and for the condensed account of many proofs.” They “resisted the temptation to expand it in the hope that the brevity of [the] presentation will make clearer the mathematical structure of what is by now an elegant and attractive theory.” They endeavor “to build up to the main theorems in a succession of simple steps and to omit routine verifications.”

Their successful philosophy is wholeheartedly embraced below (it is a feature, not a flaw!), and also refined a bit. The present book also “grew out of a course of lectures.” That course was based primarily on their book, but has been offered a number of times, and has evolved over the years, influenced by other publications and the reactions of the students. Their book comprises eleven chapters, split into forty-two sections. The present book comprises twenty-six sections; each represents a single lecture, and is self-contained.

Atiyah and Macdonald “provided … exercises at the end of each chapter.” They “provided hints, and sometimes complete solutions, to the hard” exercises. Moreover, they developed a significant amount of the main content in the exercises. By contrast, in the present book, the exercises are integrated into the development, and complete solutions are given at the end of the book.

There are well over two hundred exercises below. Included are nearly all the exercises in Atiyah and Macdonald’s book. Included also are many exercises that come from other publications and many that originate here. Here the exercises are tailored to provide a means for students to check, solidify, and expand their understanding of the material. The exercises are intentionally not difficult, tricky, or involved. Rarely do they introduce new techniques, although some introduce new concepts and many statements are used later.

Students are encouraged to try to solve each and every exercise, and to do so before looking up its solution. If they become stuck, then they should review the relevant material; if they remain stuck, then they should study the given solution, possibly discussing it with others, but always making sure they can eventually solve the whole exercise completely on their own. In any event, students should read the given solution, even if they think they already know it, just to make sure; also, some exercises provide enlightening alternative solutions. Finally, instructors are encouraged to examine their students, possibly orally at a blackboard, on a small randomly chosen subset of exercises that have been assigned for the students to write up in their own words over the course of the term.

Atiyah and Macdonald explain that “a proper treatment of Homological Algebra is impossible within the confines of a small book; on the other hand, it is hardly sensible to ignore it completely.” So they “use elementary homological methods — exact sequence, diagrams, etc. — but … stop short of any results requiring a deep study of homology.” Again, their philosophy is embraced and refined in the present book. Notably, below, elementary methods are used, not Tor’s as they do, to prove the Ideal Criterion for flatness, and to relate flat modules and free modules over local rings. Also, projective modules are treated below, but not in their book.

In the present book, Category Theory is a basic tool; in Atiyah and Macdonald’s, it seems like a foreign language. Thus they discuss the universal (mapping) property (UMP) of localization of a ring, but provide an ad hoc characterization. They also prove the UMP of tensor product of modules, but do not use the term this time. Below, the UMP is fundamental: there are many canonical constructions; each has a UMP, which serves to characterize the construction up to unique isomorphism owing to one general observation of Category Theory. For example, the Left Exactness of Hom is viewed simply as expressing in other words that the kernel and the cokernel of a map are characterized by their UMPs; by contrast, Atiyah and Macdonald prove the Left Exactness via a tedious elementary argument.

Atiyah and Macdonald prove the Adjoint-Associativity Formula. They note it says that Tensor Product is the left adjoint of Hom. From it and the Left Exactness of Hom, they deduce the Right Exactness of Tensor Product. They note that this derivation shows that any “left adjoint is right exact.” More generally, as explained below, this derivation shows that any left adjoint preserves arbitrary direct limits, ones indexed by any small category. Atiyah and Macdonald consider only direct limits indexed by a directed set, and sketch an ad hoc argument showing that tensor product preserves direct limit. Also, arbitrary direct sums are direct limits indexed by a discrete category (it is not a directed set); hence, the general result yields that Tensor Product and other left adjoints preserve arbitrary Direct Sum.

Below, left adjoints are proved unique up to unique isomorphism. Therefore, the functor of localization of a module is canonically isomorphic to the functor of tensor product with the localized base ring, as both are left adjoints of the same functor, Restriction of Scalars from the localized ring to the base ring. There is an alternative argument. Since Localization is a left adjoint, it preserves Direct Sum and Cokernel; whence, it is isomorphic to that tensor-product functor by Watts Theorem, which characterizes all tensor-product functors as those linear functors that preserve Direct Sum and Cokernel. Atiyah and Macdonald’s treatment is ad hoc. However, they do use the proof of Watts Theorem directly to show that, under the appropriate conditions, Completion of a module is Tensor Product with the completed base ring.

Below, Direct Limit is also considered as a functor, defined on the appropriate category of functors. As such, Direct Limit is a left adjoint. Hence, direct limits preserve other direct limits. Here the theory briefly reaches a higher level of abstraction. This discussion is completely elementary, but by far the most abstract part of the book. The extra abstraction can be difficult, especially for beginners.

Below, filtered direct limits are treated too. They are closer to the kind of limits treated by Atiyah and Macdonald. In particular, filtered direct limits preserve exactness and flatness. Further, they appear in the following lovely form of Lazard’s Theorem: in a canonical way, every module is the direct limit of free modules of finite rank; moreover, the module is flat if and only if that direct limit is filtered.

Atiyah and Macdonald handle primary decomposition in a somewhat personal and dated fashion. First, they study primary decompositions of ideals in rings. Then, in the exercises, they indicate how to translate the theory to modules. The decompositions need not exist, as the rings and modules need not be Noetherian. Associated primes play a secondary role: they are defined as the radicals of the primary components, and then characterized as the primes that are the radicals of annihilators of elements. Finally, they prove that, when the rings and modules are Noetherian, decompositions exist and the associated primes are annihilators. To prove existence, they study irreducible modules. Nowadays, associated primes are normally defined as prime annihilators of elements, and studied on their own at first; sometimes, as below, irreducible modules are not considered.

There are several other significant differences between Atiyah and Macdonald’s treatment and the one below. First, the Noether Normalization Lemma is proved below in a stronger form for nested sequences of ideals; consequently, for algebras that are finitely generated over a field, dimension theory can be developed directly without treating Noetherian local rings first. Second, in a number of results below, the modules are assumed to be finitely presented, rather than finitely generated over a Noetherian ring. Third, there is a rudimentary treatment of regular sequences below and a proof of Serre’s Criterion for Normality. Fourth, below, the Adjoint-Associativity Formula is proved over a pair of base rings; hence, it yields both a left and a right adjoint to the functor restriction of scalars.

The present book is a beta edition. Please do the community a service by sending the authors lists of comments, corrections, and typos. Thanks!

1. Rings and Ideals

We begin by reviewing basic notions and conventions to set the stage. Throughout this book, we emphasize universal mapping properties (UMPs); they are used to characterize notions and to make constructions. So, although polynomial rings and residue rings should already be familiar in other ways, we present their UMPs immediately, and use them extensively. We close this section with a brief treatment of idempotents and the Chinese Remainder Theorem.

(1.1) (Rings). — Recall that a ring R is an abelian group, written additively, with an associative multiplication that is distributive over the addition.

Throughout this book, every ring has a multiplicative identity, denoted by 1. Further, every ring is commutative (that is, xy = yx in it), with an occasional exception, which is always marked (normally, it’s a ring of matrices).

As usual, the additive identity is denoted by 0. Note that, for any x in R,

x · 0 = 0;

indeed, x · 0 = x(0 + 0) = x · 0 + x · 0, and x · 0 can be canceled by adding −(x · 0). We allow 1 = 0. If 1 = 0, then R = 0; indeed, x = x · 1 = x · 0 = 0 for any x. A unit is an element u with a reciprocal 1/u such that u·1/u = 1. Alternatively,

1/u is denoted u^{−1} and is called the multiplicative inverse of u. The units form a multiplicative group, denoted R^{×}.

For example, the ordinary integers form a ring ℤ, and its units are 1 and −1. A ring homomorphism, or simply a ring map, φ: R → R′ is a map preserving

sums, products, and 1. Clearly, φ(R^{×}) ⊂ R^{′×}. We call φ an isomorphism if it is bijective, and then we write φ : R ⟶^{∼} R′. We call φ an endomorphism if R′ = R. We call φ an automorphism if it is bijective and if R′ = R.

If there is an unspecified isomorphism between rings R and R′, then we write R = R′ when it is canonical; that is, it does not depend on any artificial choices, so that for all practical purposes, R and R′ are the same. (Recognizing that an isomorphism is canonical provides insight and obviates verifications.) Otherwise, we write R ≃ R′.

A subset R′′ ⊂ R is a subring if R′′ is a ring and the inclusion R′′ ↪ R a ring map. For example, given a ring map φ: R → R′, its image Im(φ) := φ(R) is a subring of R′.

An R-algebra is a ring R′ that comes equipped with a ring map φ: R → R′, called the structure map. An R-algebra homomorphism, or R-algebra map, R′ → R′′ is a ring map between R-algebras compatible with their structure maps.

(1.2) (Boolean rings). — The simplest nonzero ring has two elements, 0 and 1. It is unique, and denoted 𝔽_{2}.

Given any ring R and any set X, let R^{X} denote the set of functions f: X → R. Then R^{X} is, clearly, a ring under valuewise addition and multiplication.

For example, take R := 𝔽_{2}. Given f: X → R, put S := f^{−1}{1}. Then f(x) = 1 if x ∈ S, and f(x) = 0 if x ∉ S; in other words, f is the characteristic function χ_{S}. Thus the characteristic functions form a ring, namely, 𝔽_{2}^{X}.

Given T ⊂ X, clearly χ_{S} · χ_{T} = χ_{s∩T}. Further, χ_{S} + χ_{T} = χ_{S△T}, where S△T is

September 3, 2012 11Nts.tex the symmetric difference:

S△T := (S ∪ T) − (S ∩ T) = (S − T) ∪ (T − S),

where S − T denotes, as usual, the set of elements of S not in T. Thus the subsets of X form a ring: sum is symmetric difference, and product is intersection. This ring is canonically isomorphic to 𝔽_{2}^{X}.

A ring B is said to be Boolean if f^{2} = f for all f ∈ B. Clearly, 𝔽_{2}^{X} is Boolean. Suppose X is a topological space, and give 𝔽_{2} the discrete topology; that is,

every subset is both open and closed. Consider the continuous functions f: X → 𝔽_{2}. Clearly, they are just the χ_{S} where S is both open and closed. Clearly, they form a Boolean subring of 𝔽_{2}^{X}. Conversely, Stone’s Theorem (13.7) asserts that every Boolean ring is canonically isomorphic to the ring of continuous functions from a compact Hausdorff topological space to 𝔽_{2}, or equivalently, to the ring of open and closed subsets of that space.

(1.3) (Polynomial rings). — Let R be a ring, P := R[X_{1}, …, X_{n}] the polynomial ring in n variables (see [1, pp. 352–3] or [4, p. 268]). Recall that P has this Universal Mapping Property (UMP): given a ring map φ: R → R′ and given an element x_{i} of R′ for each i, there is a unique ring map π: P → R′ with π|R = φ and π(X_{i}) = x_{i}. In fact, since π is a ring map, necessarily π is given by the formula:

π(∑ a_{(i_{1},...,i_{n})}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}}) = ∑ φ(a_{(i_{1},...,i_{n})})x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}}.

In other words, P is the universal example of an R-algebra equipped with a list of n elements: P is one example, and it maps uniquely to any other [1, (3.4), p. 353].

Similarly, let P′ := R[{X_{λ}}_{λ∈Λ}] be the polynomial ring in an arbitrary set of variables: its elements are the polynomials in any finitely many of the X_{λ}; sum and product are defined as in P. Thus P′ contains as a subring the polynomial ring in any finitely many X_{λ}, and P′ is the union of these subrings. Clearly, P′ has essentially the same UMP as P: given φ: R → R′ and given x_{λ} ∈ R′ for each λ, there is a unique π: P′ → R′ with π|R = φ and π(X_{λ}) = x_{λ}.

(1.4) (Ideals). — Let R be a ring. Recall that a subset 𝔞 is called an ideal if

(1) 0 ∈ 𝔞 (or 𝔞 is nonempty),

(2) whenever a, b ∈ 𝔞, also a + b ∈ 𝔞, and

(3) whenever x ∈ R and a ∈ 𝔞, also xa ∈ 𝔞.

Given elements a_{λ} ∈ R for λ ∈ Λ, by the ideal ⟨a_{λ}⟩_{λ∈Λ} they generate, we mean the smallest ideal containing them all. If Λ = ∅, then this ideal consists just of 0.

Any ideal containing all the a_{λ} contains any (finite) linear combination ∑ x_{λ}a_{λ} with x_{λ} ∈ R and almost all 0. Form the set 𝔞 of all such linear combinations; clearly, 𝔞 is an ideal containing all a_{λ}. Thus 𝔞 is the ideal generated by the a_{λ}.

Given a single element a, we say that the ideal ⟨a⟩ is principal. By the preceding observation, ⟨a⟩ is equal to the set of all multiples xa with x ∈ R.

Similarly, given ideals 𝔞_{λ} of R, by the ideal they generate, we mean the smallest ideal ∑ 𝔞_{λ} that contains them all. Clearly, ∑ 𝔞_{λ} is equal to the set of all finite linear combinations ∑ x_{λ}a_{λ} with x_{λ} ∈ R and a_{λ} ∈ 𝔞_{λ}.

September 3, 2012 11Nts.tex Given two ideals 𝔞 and 𝔟, consider these three nested sets:

𝔞 + 𝔟 := {a + b | a ∈ 𝔞 and b ∈ 𝔟},
𝔞 ∩ 𝔟 := {a | a ∈ 𝔞 and a ∈ 𝔟},
𝔞𝔟 := {∑ a_{i}b_{i} | a_{i} ∈ 𝔞 and b_{i} ∈ 𝔟}.

They are clearly ideals. They are known as the sum, intersection, and product of 𝔞 and 𝔟. Further, for any ideal 𝔠, the distributive law holds: 𝔞(𝔟 + 𝔠) = 𝔞𝔟 + 𝔞𝔠.

Let 𝔞 be an ideal. Then 𝔞 = R if and only if 1 ∈ 𝔞. Indeed, if 1 ∈ 𝔞, then x = x · 1 ∈ 𝔞 for every x ∈ R. It follows that 𝔞 = R if and only if 𝔞 contains a unit. Further, if ⟨x⟩ = R, then x is a unit, since then there is an element y such that xy = 1. If 𝔞 ≠ R, then 𝔞 is said to be proper.

Let φ: R → R′ be a ring map. Let 𝔞R′ denote the ideal of R′ generated by φ(𝔞); we call 𝔞R′ the extension of 𝔞. Let 𝔞′ be an ideal of R′. Clearly, the preimage φ^{−1}(𝔞′) is an ideal of R; we call φ^{−1}(𝔞′) the contraction of 𝔞′.

(1.5) (Residue rings). — Let φ: R → R′ be a ring map. Recall its kernel Ker(φ) is defined to be the ideal φ^{−1}(0) of R. Recall Ker(φ) = 0 if and only if φ is injective.

Conversely, let 𝔞 be an ideal of R. Form the set of cosets of 𝔞:

R/𝔞 := {x + 𝔞 | x ∈ R}.

Recall that R/𝔞 inherits a ring structure, and is called the residue ring (or quotient ring or factor ring) of R modulo 𝔞. Form the quotient map

κ: R → R/𝔞 by κx := x + 𝔞.

The element κx ∈ R/𝔞 is called the residue of x. Clearly, κ is surjective, κ is a ring map, and κ has kernel 𝔞. Thus every ideal is a kernel!

Note that Ker(φ) ⊃ 𝔞 if and only if φ𝔞 = 0.

Recall that, if Ker(φ) ⊃ 𝔞, then there is a ring map ψ: R/𝔞 → R′ with ψκ = φ; that is, the following diagram is commutative:

[figure: R; R/𝔞; R′; κ; φ; ψ]

Conversely, if ψ exists, then Ker(φ) ⊃ 𝔞, or φ𝔞 = 0, or 𝔞R′ = 0, since κ𝔞 = 0. Further, if ψ exists, then ψ is unique as κ is surjective. Finally, as κ is surjective, if ψ exists, then ψ is surjective if and only if φ is so.

In addition, then ψ is injective if and only if 𝔞 = Ker(φ). Hence then ψ is an isomorphism if and only if φ is surjective and 𝔞 = Ker(φ). In particular, always

R/ Ker(φ) ⟶^{∼} Im(φ).  (1.5.1)

In practice, it is usually more convenient to view R/𝔞 not as a set of cosets, but simply as another ring R′ that comes equipped with a surjective ring map φ: R → R′ whose kernel is the given ideal 𝔞.

Finally, R/𝔞 has, as we saw, this UMP: κ(𝔞) = 0 and, given φ: R → R′ such that φ(𝔞) = 0, there is a unique ring map ψ: R/𝔞 → R′ such that ψκ = φ. In other words, R/𝔞 is the universal example of an R-algebra R′ such that 𝔞R′ = 0.

The UMP applies, first of all, to the underlying sets, providing a unique map ψ

September 3, 2012 11Nts.tex of sets. Now, φ and κ are ring maps, and ψκ = φ; so

ψ(κ(a) + κ(b)) = ψκ(a + b) = ψκ(a) + ψκ(b),
ψ(κ(a)κ(b)) = ψκ(ab) = ψκ(a) · ψκ(b), and ψ(1) = ψκ(1) = 1.

But κ is surjective; so κ(a), κ(b) ∈ R/𝔞 are arbitrary. Thus ψ is a ring map. The UMP serves to determine R/𝔞 up to unique isomorphism. Indeed, say R′, equipped with φ: R → R′, has the UMP too. Then φ(𝔞) = 0; so

there is a unique ψ: R/𝔞 → R′ with ψκ = φ. And κ(𝔞) = 0; so there is a unique ψ′: R′ → R/𝔞 with ψ′φ = κ. Then, as shown, (ψ′ψ)κ = κ, but 1 ∘ κ = κ where 1

[figure: R/𝔞; R; R′; R/𝔞; 1; ψ; φ; κ; κ; ψ′]

is the identity map of R/𝔞; hence, ψ′ψ = 1 by uniqueness. Similarly, ψψ′ = 1 where 1 now stands for the identity map of R′. Thus ψ and ψ′ are inverse isomorphisms.

The preceding proof is completely formal, and so works widely. There are many more constructions to come, and each one has an associated UMP, which therefore serves to determine the construction up to unique isomorphism.

Exercise (1.6). — Let R be a ring, 𝔞 an ideal, and P := R[X_{1}, …, X_{n}] the polynomial ring. Construct an isomorphism ψ from P/𝔞P onto (R/𝔞)[X_{1}, …, X_{n}].

Proposition (1.7). — Let R be a ring, P := R[X] the polynomial ring in one variable, a ∈ R, and π: P → R the R-algebra map defined by π(X) := a. Then

Ker(π) = ⟨X − a⟩, and R[X]/⟨X − a⟩ ⟶^{∼} R.

Proof: Given F (X) ∈ P, the Division Algorithm yields F (X) = G(X)(X−a)+b with G(X) ∈ P and b ∈ R. Then π(F (X)) = b. Hence Ker(π) = ⟨X − a⟩. Finally,

(1.5.1) yields R[X]/⟨X − a⟩ ⟶^{∼} R. □

(1.8) (Nested ideals). — Let R be a ring, 𝔞 an ideal, and κ: R → R/𝔞 the quotient map. Given an ideal 𝔟 ⊃ 𝔞, form the corresponding set of cosets of 𝔞:

𝔟/𝔞 := {b + 𝔞 | b ∈ 𝔟} = κ(𝔟).

Clearly, 𝔟/𝔞 is an ideal of R/𝔞. Also 𝔟/𝔞 = 𝔟(R/𝔞).

Clearly, the operations 𝔟 ↦ 𝔟/𝔞 and 𝔟′ ↦ κ^{−1}(𝔟′) are inverse to each other, and establish a bijective correspondence between the set of ideals 𝔟 of R containing 𝔞 and the set of all ideals 𝔟′ of R/𝔞. Moreover, this correspondence preserves inclusions.

Given an ideal 𝔟 ⊃ 𝔞, form the composition of the quotient maps

φ: R → R/𝔞 → (R/𝔞)/(𝔟/𝔞).

Clearly, φ is surjective, and Ker(φ) = 𝔟. Hence, owing to (1.5), φ factors through the canonical isomorphism ψ in this commutative diagram:

R ⟶ R/𝔟

↓  ^{ψ}↓^{≃}
R/𝔞 ⟶ (R/𝔞)/(𝔟/𝔞)

Exercise (1.9). — Let R be ring, and P := R[X_{1}, …, X_{n}] the polynomial ring. Let m ≤ n and a_{1}, …, a_{m} ∈ R. Set 𝔭 := ⟨X_{1} − a_{1}, …, X_{m} − a_{m}⟩. Prove that P/𝔭 = R[X_{m+1}, …, X_{n}].

(1.10) (Idempotents). — Let R be a ring. Let e ∈ R be an idempotent; that is, e^{2} = e. Then Re is a ring with e as 1, because (xe)e = xe. But Re is not a subring of R unless e = 1, although Re is an ideal. Set e′ := 1 − e. Then e′ is idempotent, and e · e′ = 0. We call e and e′ complementary and orthogonal idempotents

Example (1.11). — Let R := R′ × R′′ be a product of two rings: its operations are performed componentwise. The additive identity is (0, 0); the multiplicative identity is (1, 1). Set e := (1, 0) and e′ := (0, 1). Then e and e′ are complementary idempotents. The next proposition shows this example is the only one possible.

Proposition (1.12). — Let R be a ring with complementary idempotents e and e′. Set R′ := Re and R′′ := Re′, and form the map φ: R → R′ × R′′ defined by φ(x) := (xe, xe′). Then φ is a ring isomorphism.

Proof: Define a map φ′: R → R′ by φ′(x) := xe. Then φ′ is a ring map since xye = xye^{2} = (xe)(ye). Hence φ is a ring map. Further, φ is surjective, since (xe, x′e′) = φ(xe + x′e′). Also φ is injective, since if xe = 0 and xe′ = 0, then x = xe + xe′ = 0. Thus φ is an isomorphism. □

Exercise (1.13) (Chinese Remainder Theorem). — Let R be a ring. (1) Let 𝔞 and 𝔟 be comaximal ideals; that is, 𝔞 + 𝔟 = R. Prove

(a) 𝔞𝔟 = 𝔞 ∩ 𝔟 and (b) R/𝔞𝔟 = (R/𝔞) × (R/𝔟).

(2) Let 𝔞 be comaximal to both 𝔟 and 𝔟′. Prove 𝔞 is also comaximal to 𝔟𝔟′.

(3) Let 𝔞, 𝔟 be comaximal, and m, n ≥ 1. Prove 𝔞^{m} and 𝔟^{n} are comaximal.

(4) Let 𝔞_{1}, …, 𝔞_{n} be pairwise comaximal. Prove (a) 𝔞_{1} and 𝔞_{2} ⋯ 𝔞_{n} are comaximal;

(b) 𝔞_{1} ∩ ⋯ ∩ 𝔞_{n} = 𝔞_{1} ⋯ 𝔞_{n};

(c) R/(𝔞_{1} ⋯ 𝔞_{n}) ⟶^{∼} ∏(R/𝔞_{i}).

Exercise (1.14). — First, given a prime number p and a k ≥ 1, find the idempotents in ℤ/⟨p^{k}⟩. Second, find the idempotents in ℤ/⟨12⟩. Third, find the number of idempotents in ℤ/⟨n⟩ where n = ∏_{i=1}^{N} p_{i}^{n_{i}} with p_{i} distinct prime numbers.

Exercise (1.15). — Let R := R′ × R′′ be a product of rings, 𝔞 ⊂ R an ideal. Show 𝔞 = 𝔞′ × 𝔞′′ with 𝔞′ ⊂ R′ and 𝔞′′ ⊂ R′′ ideals. Show R/𝔞 = (R′/𝔞′) × (R′′/𝔞′′).

Exercise (1.16). — Let R be a ring, and e, e′ idempotents. (See (10.6) also.) (1) Set 𝔞 := ⟨e⟩. Show 𝔞 is idempotent; that is, 𝔞^{2} = 𝔞.

(2) Let 𝔞 be a principal idempotent ideal. Show 𝔞 = ⟨f⟩ with f idempotent.

(3) Assume ⟨e⟩ = ⟨e′⟩. Show e = e′.

(4) Set e′′ := e + e′ − ee′. Show ⟨e, e′⟩ = ⟨e′′⟩ and e′′ is idempotent.

(5) Let e_{1}, …, e_{r} be idempotents. Show ⟨e_{1}, …, e_{r}⟩ = ⟨f⟩ with f idempotent.

(6) Assume R is Boolean. Show every finitely generated ideal is principal.

2. Prime Ideals

Prime ideals are the key to the structure of commutative rings. So we review the basic theory. Specifically, we define prime ideals, and show their residue rings are domains. We show maximal ideals are prime, and discuss examples. Then we use Zorn’s Lemma to prove the existence of maximal ideals in every nonzero ring.

Definition (2.1). — Let R be a ring. An element x is called a zerodivisor if there is a nonzero y with xy = 0; otherwise, x is called a nonzerodivisor. Denote the set of zerodivisors by z.div(R).

A subset S is called multiplicative if 1 ∈ S and if x, y ∈ S implies xy ∈ S. An ideal 𝔭 is called prime if its complement R − 𝔭 is multiplicative, or equiva-

lently, if 1 ∉ 𝔭 and if xy ∈ 𝔭 implies x ∈ 𝔭 or y ∈ 𝔭.

Exercise (2.2). — Let 𝔞 and 𝔟 be ideals, and 𝔭 a prime ideal. Prove that these conditions are equivalent: (1) 𝔞 ⊂ 𝔭 or 𝔟 ⊂ 𝔭; and (2) 𝔞 ∩ 𝔟 ⊂ 𝔭; and (3) 𝔞𝔟 ⊂ 𝔭.

(2.3) (Fields, Domains). — A ring is called a field if 1 ≠ 0 and if every nonzero element is a unit. Standard examples include the rational numbers ℚ, the real numbers ℝ, and the complex numbers ℂ.

A ring is called an integral domain, or simply a domain, if ⟨0⟩ is prime, or equivalently, if R is nonzero and has no nonzero zerodivisors.

Every domain R is a subring of its fraction field Frac(R), which consists of the fractions x/y with x, y ∈ R and y ≠ 0. Conversely, any subring R of a field K, including K itself, is a domain; indeed, any nonzero x ∈ R cannot be a zerodivisor, because, if xy = 0, then (1/x)(xy) = 0, so y = 0. Further, Frac(R) has this UMP: the inclusion of R into any field L extends uniquely to an inclusion of Frac(R) into L. For example, the ring of integers ℤ is a domain, and Frac(ℤ) = ℚ ⊂ ℝ ⊂ ℂ.

Let R be a domain, and R[X] the polynomial ring in one variable. Then R[X] is a domain too; in fact, given any two nonzero polynomials f and g, not only is their product fg nonzero, but its leading coefficient is the product of the leading coefficients of f and g.

By induction, the polynomial ring in n variables R[X_{1}, …, X_{n}] is a domain, since

R[X_{1}, …, X_{n}] = R[X_{1}, …, X_{n−1}][X_{n}].

Hence the polynomial ring in an arbitrary set of variables R[{X_{λ}}_{λ∈Λ}] is a domain, since any two elements lie in a polynomial subring in finitely many of the X_{λ}.

A similar argument proves that R^{×} is full group of units in the polynomial ring. This statement can fail if R is not a domain. For example, if a^{2} = 0 in R, then (1 + aX)(1 − aX) = 1 in R[X].

The fraction field Frac(R[{X_{λ}}_{λ∈Λ}]) is called the field of rational functions, and is also denoted by K({X_{λ}}_{λ∈Λ}) where K := Frac(R).

Exercise (2.4). — Given a prime number p and an integer n ≥ 2, prove that the residue ring ℤ/⟨p^{n}⟩ does not contain a domain.

Exercise (2.5). — Let R := R′ × R′′ be a product of two rings. Show that R is a domain if and only if either R′ or R′′ is a domain and the other is 0.

(2.6) (Unique factorization). — Let R be a domain, p a nonzero nonunit. We call p prime if, whenever p | xy (that is, there exists z ∈ R such that pz = xy), either p | x or p | y. Clearly, p is prime if and only if the ideal ⟨p⟩ is prime.

We call p irreducible if, whenever p = yz, either y or z is a unit. We call R a Unique Factorization Domain (UFD) if every nonzero element is a product of irreducible elements in a unique way up to order and units.

In general, prime elements are irreducible; in a UFD, irreducible elements are prime. Standard examples of UFDs include any field, the integers ℤ, and a polynomial ring in n variables over a UFD; see [1, p. 398, p. 401], [4, Cor. 18.23, p. 297].

Lemma (2.7). — Let φ: R → R′ be a ring map, and T ⊂ R′ a subset. If T is multiplicative, then φ^{−1}T is multiplicative; the converse holds if φ is surjective.

Proof: Both assertions are easy to check. □

Proposition (2.8). — Let φ: R → R′ be a ring map, and 𝔮 ⊂ R′ an ideal. If 𝔮 is prime, then φ^{−1}𝔮 is prime; the converse holds if φ is surjective.

Proof: By (2.7), R−𝔭 is multiplicative if and only if R′ −𝔮 is. So the assertion results from Definitions (2.1). □

Proposition (2.9). — Let R be a ring, 𝔭 an ideal. Then 𝔭 is prime if and only if R/𝔭 is a domain.

Proof: By (2.8), 𝔭 is prime if and only if ⟨0⟩ ⊂ R/𝔭 is. So the assertion results from the definition of domain in (2.3). □

Exercise (2.10). — Let R be a ring, 𝔭 a prime ideal, R[X] the polynomial ring. Show that 𝔭R[X] and 𝔭R[X] + ⟨X⟩ are prime ideals of R[X].

Exercise (2.11). — Let R be a domain, and R[X_{1}, …, X_{n}] the polynomial ring in n variables. Let m ≤ n, and set 𝔭 := ⟨X_{1}, …, X_{m}⟩. Prove 𝔭 is a prime ideal.

Exercise (2.12). — Let R := R′ × R′′ be a product of rings. Show every prime ideal of R has the form 𝔭′ × R′′ with 𝔭′ ⊂ R′ prime or R′ × 𝔭′′ with 𝔭′′ ⊂ R′′ prime.

Definition (2.13). — Let R be a ring. An ideal 𝔪 is said to be maximal if 𝔪 is proper and if there is no proper ideal 𝔞 with 𝔪 ⫋ 𝔞.

Example (2.14). — Let R be a domain. In the polynomial ring R[X, Y] in two variables, ⟨X⟩ is prime by (2.11). However, ⟨X⟩ is not maximal since ⟨X⟩ ⫋ ⟨X, Y ⟩.

Proposition (2.15). — A ring R is a field if and only if ⟨0⟩ is a maximal ideal.

Proof: Suppose R is a field. Let 𝔞 be a nonzero ideal, and a a nonzero element of 𝔞. Since R is a field, a ∈ R^{×}. So (1.4) yields 𝔞 = R.

Conversely, suppose ⟨0⟩ is maximal. Take x ≠ 0. Then ⟨x⟩ ≠ ⟨0⟩. So ⟨x⟩ = R. So x is a unit by (1.4). Thus R is a field. □

Exercise (2.16). — Let k be a field, R a nonzero ring, φ: k → R a ring map. Prove φ is injective.

Proposition (2.17). — Let R be a ring, 𝔪 an ideal. Then 𝔪 is maximal if and only if R/𝔪 is a field.

Proof: Clearly, 𝔪 is maximal in R if and only if ⟨0⟩ is maximal in R/𝔪 by (1.8). Hence the assertion results from (2.15). □

Exercise (2.18). — Let B be a Boolean ring. Show that every prime 𝔭 is maximal, and B/𝔭 = 𝔽_{2}.

Example (2.19). — Let k be a field, a_{1}, …, a_{n} ∈ k, and P := k[X_{1}, …, X_{n}] the polynomial ring in n variables. Set 𝔪 := ⟨X_{1} − a_{1}, …, X_{n} − a_{n}⟩. Then P/𝔪 = k by (1.9); so 𝔪 is maximal by (2.17).

Exercise (2.20). — Prove the following statements or give a counterexample. (1) The complement of a multiplicative subset is a prime ideal.

(2) Given two prime ideals, their intersection is prime.

(3) Given two prime ideals, their sum is prime.

(4) Given a ring map φ: R → R′, the operation φ^{−1} carries maximal ideals of R′ to maximal ideals of R.

(5) In (1.8), κ^{−1} takes maximal ideals of R/𝔞 to maximal ideals of R.

Exercise (2.21). — Let k be a field, P := k[X_{1}, …, X_{n}] the polynomial ring, f ∈ P nonzero. Let d be the highest power of any variable appearing in f.

(1) Let S ⊂ k have at least dn + 1 elements. Proceeding by induction on n, find a_{1}, …, a_{n} ∈ S with f(a_{1}, …, a_{n}) ≠ 0.

(2) Using the algebraic closure K of k, find a maximal ideal 𝔪 of P with f ∉ 𝔪.

Corollary (2.22). — In a ring, every maximal ideal is prime.

Proof: A field is a domain by (2.3). So (2.9) and (2.17) yield the result. □

(2.23) (PIDs). — A domain R is called a Principal Ideal Domain (PID) if every ideal is principal. Examples include the polynomial ring k[X] in one variable over a field k, and the ring ℤ of integers. Every PID is a UFD by [1, (2.12), p. 396], [4, Thm. 18.11, p. 291].

Let R be a PID, and p ∈ R irreducible. Then ⟨p⟩ is maximal; indeed, if ⟨p⟩ ⫋ ⟨x⟩, then p = xy for some nonunit y, and so x must be a unit since p is irreducible. So (2.17) implies that R/⟨p⟩ is a field.

Exercise (2.24). — Prove that, in a PID, elements x and y are relatively prime (share no prime factor) if and only if the ideals ⟨x⟩ and ⟨y⟩ are comaximal.

Example (2.25). — Let R be a PID, and p ∈ R a prime. Set k := R/⟨p⟩. Let P := R[X] be the polynomial ring in one variable. Take g ∈ P, let g′ be its image in k[X], and assume g′ is irreducible. Set 𝔪 := ⟨p, g⟩. Then 𝔪 is maximal by (2.17); indeed, P/𝔪 ⟶^{∼} k[X]/⟨g′⟩ by (1.5), and k[X]/⟨g′⟩ is a field by (2.23).

Theorem (2.26). — Let R be a PID. Let P := R[X] be the polynomial ring in one variable, and 𝔭 a prime ideal of P.

(1) Then 𝔭 = ⟨0⟩, or 𝔭 = ⟨f⟩ with f prime, or 𝔭 is maximal.

(2) Assume 𝔭 is maximal. Then either 𝔭 = ⟨f⟩ with f prime, or 𝔭 = ⟨p, g⟩ with p ∈ R prime and g ∈ P with image g′ ∈ (R/⟨p⟩)[X] prime.

Proof: Assume 𝔭 ≠ ⟨0⟩. Take a nonzero f_{1} ∈ 𝔭. Since 𝔭 is prime, 𝔭 contains a prime factor f′_{1} of f_{1}. Replace f_{1} by f′_{1}. Assume 𝔭 ≠ ⟨f_{1}⟩. Then there is an prime f_{2} ∈ 𝔭 − ⟨f_{1}⟩. Set K := Frac(R). Gauss’s Lemma [1, p. 401], [4, Thm. 18.15, p. 295] implies that f_{1} and f_{2} are also prime in K[X]. So f_{1} and f_{2} are relatively prime in K[X]. So (2.23) and (2.24) yield g_{1}, g_{2} ∈ P and c ∈ R with (g_{1}/c)f_{1} + (g_{2}/c)f_{2} = 1. So c = g_{1}f_{1} + g_{2}f_{2} ∈ R ∩ 𝔭. Hence R ∩ 𝔭 ≠ 0. But R ∩ 𝔭

September 3, 2012 11Nts.tex is prime, and R is a PID; so R ∩ 𝔭 = ⟨p⟩ where p is prime by (2.6).

Set k := R/⟨p⟩. Then k is a field by (2.23). Set 𝔮 := 𝔭/⟨p⟩ ⊂ k[X]. Then k[X]/𝔮 = P/𝔭 by (1.6) and (1.8). But P/𝔭 is a domain as 𝔭 is prime. Hence 𝔮 = ⟨g′⟩ where g′ is prime in k[X] by (2.6). Then 𝔮 is maximal by (2.23). So 𝔭 is maximal by (1.7). Take g ∈ 𝔭 with image g′. Then 𝔭 = ⟨p, g⟩ as 𝔭/⟨p⟩ = ⟨g′⟩. □

Exercise (2.27). — Preserve the setup of (2.26). Let f := a_{0}X^{n} + ⋯ + a_{n} be a polynomial of positive degree n. Assume that R has infinitely many prime elements p, or simply that there is a p such that p ∤ a_{0}. Show that ⟨f⟩ is not maximal.

Theorem (2.28). — Every proper ideal 𝔞 is contained in some maximal ideal.

Proof: Set 𝒮 := {ideals 𝔟 | 𝔟 ⊃ 𝔞 and 𝔟 ∌ 1}. Then 𝔞 ∈ 𝒮, and 𝒮 is partially ordered by inclusion. Given a totally ordered subset {𝔟_{λ}} of 𝒮, set 𝔟 := ⋃ 𝔟_{λ}. Then 𝔟 is clearly an ideal, and 1 ∉ 𝔟; so 𝔟 is an upper bound of {𝔟_{λ}} in 𝒮. Hence by Zorn’s Lemma [7, pp. 25, 26], [6, p. 880, p. 884], 𝒮 has a maximal element, and it is the desired maximal ideal. □

Corollary (2.29). — Let R be a ring, x ∈ R. Then x is a unit if and only if x belongs to no maximal ideal.

Proof: By (1.4), x is a unit if and only if ⟨x⟩ is not proper. So (2.28) yields the assertion. □

3. Radicals

Two radicals of a ring are commonly used in Commutative Algebra: the Jacobson radical, which is the intersection of all maximal ideals, and the nilradical, which is the set of all nilpotent elements. Closely related to the nilradical is the radical of a subset. We define these three radicals, and discuss examples. In particular, we study local rings; a local ring has only one maximal ideal, which is then its Jacobson radical. We prove two important general results: Prime Avoidance, which states that, if an ideal lies in a finite union of primes, then it lies in one of them, and the Scheinnullstellensatz, which states that the nilradical of an ideal is equal to the intersection of all the prime ideals containing it.

Definition (3.1). — Let R be a ring. Its (Jacobson) radical rad(R) is defined to be the intersection of all its maximal ideals.

Proposition (3.2). — Let R be a ring, x ∈ R. Then x ∈ rad(R) if and only if 1 − xy is a unit for all y ∈ R.

Proof: Assume x ∈ rad(R). Let 𝔪 be a maximal ideal. Suppose 1 − xy ∈ 𝔪. Since x ∈ 𝔪 too, also 1 ∈ 𝔪, a contradiction. So 1 − xy is a unit by (2.29).

Conversely, assume x ∉ rad(R). Then there is a maximal ideal 𝔪 with x ∉ 𝔪. So ⟨x⟩ + 𝔪 = R. Hence there exist y ∈ R and m ∈ 𝔪 such that xy + m = 1. Then 1 − xy = m ∈ 𝔪. So 1 − xy is not a unit by (2.29), or directly by (1.4). □

Definition (3.3). — A ring A is called local if it has exactly one maximal ideal, and semilocal if it has at least one and at most finitely many.

Lemma (3.4) (Nonunit Criterion). — Let A be a ring, 𝔫 the set of nonunits. Then A is local if and only if 𝔫 is an ideal; if so, then 𝔫 is the maximal ideal.

Proof: Every proper ideal 𝔞 lies in 𝔫 as 𝔞 contains no unit. So, if 𝔫 is an ideal, then it is a maximal ideal, and the only one. Thus A is local.

Conversely, assume A is local with maximal ideal 𝔪. Then A − 𝔫 = A − 𝔪 by (2.29). So 𝔫 = 𝔪. Thus 𝔫 is an ideal. □

Example (3.5). — The product ring R′ × R′′ is not local by (3.4) if both R′ and R′′ are nonzero. Indeed, (1, 0) and (0, 1) are nonunits, but their sum is a unit.

Exercise (3.6). — Let A be a ring, 𝔪 a maximal ideal such that 1 + m is a unit for every m ∈ 𝔪. Prove A is local. Is this assertion still true if 𝔪 is not maximal?

Example (3.7). — Let R be a ring. A formal power series in the n variables X_{1}, …, X_{n} is a formal infinite sum of the form ∑ a_{(i)}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}} where a_{(i)} ∈ R and where (i) = (i_{1}, …, i_{n}) with each i_{j} ≥ 0. Addition and multiplication are performed as for polynomials; with these operations, these series form a ring R[[X_{1}, …, X_{n}]].

Set P := R[[X_{1}, …, X_{n}]] and 𝔞 := ⟨X_{1}, …, X_{n}⟩. Then ∑ a_{(i)}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}} ↦ a_{(0)} is a canonical surjective ring map P → R with kernel 𝔞; hence, P/𝔞 = R.

Given an ideal 𝔪 ⊂ R, set 𝔫 := 𝔞 + 𝔪P. Then (1.8) yields P/𝔫 = R/𝔪.

Suppose R is a local ring with maximal ideal 𝔪. Then any power series f ∉ 𝔫 is of the form f = a(1 − g) with a ∈ R^{×} and g ∈ 𝔞. Set h := a^{−1}(1 + g + g^{2} + ⋯); this sum makes sense as the component of degree d involves only the first d + 1

September 3, 2012 11Nts.tex summands. Clearly f · h = 1. Hence the nonunits constitute 𝔫. Thus P is local with maximal ideal 𝔫 by (3.4).

Example (3.8). — Let k be a ring, and A := k[[X]] the formal power series ring in one variable. A Laurent series is a formal sum of the form ∑_{i=−m}^{∞} a_{i}X^{i} with a_{i} ∈ k and m ∈ ℤ. The Laurent series form a ring k{{X}}. Set K := k{{X}}.

Set f := ∑_{i=−m}^{∞} a_{i}X^{i}. If a_{−m} ∈ k^{×}, then f ∈ K^{×}; indeed, f = a_{−m}X^{−m}(1 − g) where g ∈ A, and f · a_{−m}^{−1}X^{m}(1 + g + g^{2} + ⋯) = 1.

Assume k is a field. If f ≠ 0, then f = X^{−m}u where u ∈ A^{×}. Let 𝔞 ⊂ A be a nonzero ideal. Suppose f ∈ 𝔞. Then X^{−m} ∈ 𝔞. Let n be the smallest integer such that X^{n} ∈ 𝔞. Then −m ≥ n. Set b := X^{−m−n}u. Then b ∈ A and f = bX^{n}. Hence 𝔞 = ⟨X^{n}⟩. Thus A is a PID.

Further, K is a field. In fact, K = Frac(A) as any nonzero f ∈ K is of the form f = u/X^{m} where u, X^{m} ∈ A.

Let A[Y] be the polynomial ring in one variable, and ι: A ↪ K the inclusion. Define φ: A[Y] → K by φ|A = ι and φ(Y) := X^{−1}. Then φ is surjective. Set 𝔪 := Ker(φ). Then 𝔪 is maximal by (2.17) and (1.5). So by (2.26), 𝔪 has the form ⟨f⟩ with f irreducible, or the form ⟨p, g⟩ with p ∈ A irreducible and g ∈ A[Y]. But 𝔪 ∩ A = 0 as ι is injective. So 𝔪 = ⟨f⟩. But XY − 1 belongs to 𝔪, and is clearly irreducible; hence, XY − 1 = fu with u a unit. Thus ⟨XY − 1⟩ is maximal.

In addition, ⟨X, Y ⟩ is maximal. Indeed, A[Y]/⟨Y ⟩ = A by (1.7), and so (3.7) yields A[Y]/⟨X, Y ⟩ = A/⟨X⟩ = k. However, ⟨X, Y ⟩ is not principal, as no nonunit of A[Y] divides both X and Y. Thus A[Y] has both principal and nonprincipal maximal ideals, the two types allowed by (2.26).

Proposition (3.9). — Let R be a ring, S a multiplicative subset, and 𝔞 an ideal with 𝔞 ∩ S = ∅. Set 𝒮 := {ideals 𝔟 | 𝔟 ⊃ 𝔞 and 𝔟 ∩ S = ∅}. Then 𝒮 has a maximal element 𝔭, and every such 𝔭 is prime.

Proof: CLearly, 𝔞 ∈ 𝒮, and 𝒮 is partially ordered by inclusion. Given a totally ordered subset {𝔟_{λ}} of 𝒮, set 𝔟 := ⋃ 𝔟_{λ}. Then 𝔟 is an upper bound for {𝔟_{λ}} in 𝒮. So by Zorn’s Lemma, 𝒮 has a maximal element 𝔭. Let’s show 𝔭 is prime.

Take x, y ∈ R − 𝔭. Then 𝔭 + ⟨x⟩ and 𝔭 + ⟨y⟩ are strictly larger than 𝔭. So there are p, q ∈ 𝔭 and a, b ∈ R with p + ax ∈ S and q + by ∈ S. Since S is multiplicative, pq + pby + qax + abxy ∈ S. But pq + pby + qax ∈ 𝔭, so xy ∉ 𝔭. Thus 𝔭 is prime. □

Exercise (3.10). — Let φ: R → R′ be a ring map, 𝔭 an ideal of R. Prove (1) there is an ideal 𝔮 of R′ with φ^{−1}(𝔮) = 𝔭 if and only if φ^{−1}(𝔭R′) = 𝔭;

(2) if 𝔭 is prime with φ^{−1}(𝔭R′) = 𝔭, then there’s a prime 𝔮 of R′ with φ^{−1}(𝔮) = 𝔭.

Exercise (3.11). — Use Zorn’s lemma to prove that any prime ideal 𝔭 contains a prime ideal 𝔮 that is minimal containing any given subset 𝔰 ⊂ 𝔭.

(3.12) (Saturated multiplicative subsets). — Let R be a ring, and S a multiplicative subset. We say S is saturated if, given x, y ∈ R with xy ∈ S, necessarily x, y ∈ S.

For example, it’s easy to see the following. The group of units R^{×} is a saturated multiplicative subset. Further, let φ: R → R′ be a ring map, T ⊂ R′ a subset. If T is saturated multiplicative, then so is φ^{−1}T. The converse holds if φ is surjective.

Exercise (3.13). — Let R be a ring, S a subset. Show that S is saturated multiplicative if and only if R − S is a union of primes.

Exercise (3.14). — Let R be a ring, and S a multiplicative subset. Define its saturation to be the subset

\overline{S} := \{ x ∈ R | there is y ∈ R with xy ∈ S \}.

(1) Show (a) that \overline{S} ⊃ S, and (b) that \overline{S} is saturated multiplicative, and (c) that any saturated multiplicative subset T containing S also contains \overline{S}.

(2) Show that R − \overline{S} is the union U of all the primes 𝔭 with 𝔭 ∩ S = ∅.

(3) Let 𝔞 be an ideal; assume S = 1 + 𝔞; set W := ⋃_{𝔭∈V(𝔞)} 𝔭. Show R − \overline{S} = W.

Lemma (3.15) (Prime Avoidance). — Let R be a ring, 𝔞 a subset of R that is stable under addition and multiplication, and 𝔭_{1}, …, 𝔭_{n} ideals such that 𝔭_{3}, …, 𝔭_{n} are prime. If 𝔞 ⊄ 𝔭_{j} for all j, then there is an x ∈ 𝔞 such that x ∉ 𝔭_{j} for all j; or equivalently, if 𝔞 ⊂ ⋃_{i=1}^{n} 𝔭_{i}, then 𝔞 ⊂ 𝔭_{i} for some i.

Proof: Proceed by induction on n. If n = 1, the assertion is trivial. Assume that n ≥ 2 and by induction that, for every i, there is an x_{i} ∈ 𝔞 such that x_{i} ∉ 𝔭_{j} for all j ≠ i. We may assume x_{i} ∈ 𝔭_{i} for every i, else we’re done. If n = 2, then clearly x_{1} + x_{2} ∉ 𝔭_{j} for j = 1, 2. If n ≥ 3, then (x_{1} ⋯ x_{n−1}) + x_{n} ∉ 𝔭_{j} for all j as, if j = n, then x_{n} ∈ 𝔭_{n} and 𝔭_{n} is prime, and if j < n, then x_{n} ∉ 𝔭_{j} and x_{j} ∈ 𝔭_{j}. □

Exercise (3.16). — Let k be an infinite field.

(1) Let V be a vector space, W_{1}, …, W_{r} proper subspaces. Show ⋃ W_{i} ≠ V.

(2) In (1), let W ⊂ ⋃ W_{i} be a subspace. Show W ⊂ W_{i} for some i.

(3) Let R a k-algebra, 𝔞, 𝔞_{1}, …, 𝔞_{r} ideals. If 𝔞 ⊂ ⋃ 𝔞_{i}, show 𝔞 ⊂ 𝔞_{i} for some i.

Exercise (3.17). — Let k be a field, R := k[X, Y] the polynomial ring in two variables, 𝔪 := ⟨X, Y ⟩. Show 𝔪 is a union of smaller primes.

(3.18) (Nilradical). — Let R be a ring, 𝔞 a subset. Then the radical of 𝔞 is the set \sqrt{𝔞} defined by the formula \sqrt{𝔞} := \{x ∈ R | x^{n} ∈ 𝔞 for some n = n(x) ≥ 1\}.

Notice \sqrt{\sqrt{𝔞}} = \sqrt{𝔞}. Also, if 𝔞 is an intersection of prime ideals, then \sqrt{𝔞} = 𝔞. We call \sqrt{⟨0⟩} the nilradical, and sometimes denote it by nil(R). We call an

element x ∈ R nilpotent if x belongs to \sqrt{⟨0⟩}, that is, if x^{n} = 0 for some n ≥ 1. We call R reduced if nil(R) = ⟨0⟩, that is, if R has no nonzero nilpotents.

Exercise (3.19). — Find the nilpotents in ℤ/⟨n⟩. In particular, take n = 12.

Exercise (3.20). — Let φ: R → R′ be a ring map, 𝔟 ⊂ R′ a subset. Prove

φ^{−1}\sqrt{𝔟} = \sqrt{φ^{−1}𝔟}.

Exercise (3.21). — Let R be a ring, 𝔞 ⊂ \sqrt{⟨0⟩} an ideal, and P := R[Y] the polynomial ring in one variable. Let u ∈ R be a unit, and x ∈ R a nilpotent.

(1) Prove (a) that u + x is a unit in R and (b) that u + xY is a unit in P.

(2) Suppose w ∈ R maps to a unit of R/𝔞. Prove that w is a unit in R.

Theorem (3.22) (Scheinnullstellensatz). — Let R be a ring, 𝔞 an ideal. Then

\sqrt{𝔞} = ⋂_{𝔭⊃𝔞} 𝔭

where 𝔭 runs through all the prime ideals containing 𝔞. (By convention, the empty intersection is equal to R.)

Proof: Take x ∉ \sqrt{𝔞}. Set S := \{1, x, x^{2}, …\}. Then S is multiplicative, and 𝔞 ∩ S = ∅. By (3.9), there is a 𝔭 ⊃ 𝔞, but x ∉ 𝔭. So x ∉ ⋂_{𝔭⊃𝔞} 𝔭. Thus \sqrt{𝔞} ⊃ ⋂_{𝔭⊃𝔞} 𝔭.

Conversely, take x ∈ \sqrt{𝔞}. Say x^{n} ∈ 𝔞 ⊂ 𝔭. Then x ∈ 𝔭. Thus \sqrt{𝔞} = ⋂_{𝔭⊃𝔞} 𝔭. □

Exercise (3.23). — Let B be a Boolean ring. Show that rad(B) = nil(B) = ⟨0⟩.

Proposition (3.24). — Let R be a ring, 𝔞 an ideal. Then \sqrt{𝔞} is an ideal.

Proof: Take x, y ∈ \sqrt{𝔞}; say x^{n} ∈ 𝔞 and y^{m} ∈ 𝔞. Then

(x + y)^{n+m−1} = ∑_{i+j=m+n−1} \binom{n+m−1}{j}x^{i}y^{j}.

This sum belongs to 𝔞 as, in each summand, either x^{i} or y^{j} does, since, if i ≤ n − 1 and j ≤ m − 1, then i + j ≤ m + n − 2. Thus x + y ∈ \sqrt{𝔞}. So clearly \sqrt{𝔞} is an ideal.

Alternatively, given any collection of ideals 𝔞_{λ}, note that ⋂ 𝔞_{λ} is also an ideal. So \sqrt{𝔞} is an ideal owing to (3.22). □

Exercise (3.25). — Let R be a ring, and 𝔞 an ideal. Assume \sqrt{𝔞} is finitely generated. Show (\sqrt{𝔞})^{n} ⊂ 𝔞 for all large n.

Exercise (3.26). — Let R be a ring, 𝔮 an ideal, 𝔭 a finitely generated prime. Prove that 𝔭 = \sqrt{𝔮} if and only if there is n ≥ 1 such that 𝔭 ⊃ 𝔮 ⊃ 𝔭^{n}.

Proposition (3.27). — Let R be a ring. Assume R is reduced with only one minimal prime 𝔮. Then R is a domain.

Proof: Since R is reduced, ⟨0⟩ = \sqrt{⟨0⟩} by (3.18). Hence ⟨0⟩ is equal to the intersection of all the prime ideals 𝔭 by (3.22). By (3.11), every 𝔭 contains 𝔮. So ⟨0⟩ = 𝔮. Thus R is a domain. □

Exercise (3.28). — Let R be a ring. Assume R is reduced and has finitely many minimal prime ideals 𝔭_{1}, …, 𝔭_{n}. Prove φ: R → ∏(R/𝔭_{i}) is injective, and for each i, there is some (x_{1}, …, x_{n}) ∈ Im(φ) with x_{i} ≠ 0 but x_{j} = 0 for j ≠ i.

4. Modules

In Commutative Algebra, it has proven advantageous to expand the study of rings to include modules. Thus we obtain a richer theory, which is more flexible and more useful. We begin the expansion here by discussing residue modules, kernels, and images. In particular, we identify the universal property of the residue module, and use it to construct the Noether isomorphisms. We also construct free modules, direct sums, and direct products, and we describe their universal properties.

(4.1) (Modules). — Let R be a ring. Recall that an R-module M is an abelian group, written additively, with a scalar multiplication, R × M → M, written (x, m) ↦ xm, which is

(1) distributive, x(m + n) = xm + xn and (x + y)m = xm + xn,

(2) associative, x(ym) = (xy)m, and

(3) unitary, 1 · m = m.

For example, if R is a field, then an R-module is a vector space. Moreover, a ℤ-module is just an abelian group; multiplication is repeated addition.

As in (1.1), for any x ∈ R and m ∈ M, we have x · 0 = 0 and 0 · m = 0.

A submodule N of M is a subgroup that is closed under multiplication; that is, xn ∈ N for all x ∈ R and n ∈ N. For example, the ring R is itself an R-module, and the submodules are just the ideals. Given an ideal 𝔞, let 𝔞N denote the smallest submodule containing all products an with a ∈ 𝔞 and n ∈ N. Similar to (1.4), clearly 𝔞N is equal to the set of finite sums ∑ a_{i}n_{i} with a_{i} ∈ 𝔞 and n_{i} ∈ N.

Given m ∈ M, we call the set of x ∈ R with xm = 0 the annihilator of m, and denote it Ann(m). We call the set of x ∈ R with xm = 0 for all m ∈ M the annihilator of M, and denote it Ann(M). Clearly, Ann(m) and Ann(M) are ideals.

(4.2) (Homomorphisms). — Let R be a ring, M and N modules. Recall that a homomorphism, or module map is a map α: M → N that is R-linear:

α(xm + yn) = x(αm) + y(αn).

Associated to a homomorphism α: M → N are its kernel and its image

Ker(α) := α^{−1}(0) ⊂ M and Im(α) := α(M) ⊂ N.

They are defined as subsets, but are obviously submodules.

A homomorphism α is called an isomorphism if it is bijective. If so, then we write α: M ⟶^{∼} N. Then the set-theoretic inverse α^{−1} : N → M is a homomorphism too. So α is an isomorphism if and only if there is a set map β: N → M such that βα = 1_{M} and αβ = 1_{N}, and then β = α^{−1}. If there is an unspecified isomorphism between M and N, then we write M = N when it is canonical (that is, it does not depend on any artificial choices), and we write M ≃ N otherwise.

The set of homomorphisms α is denoted by Hom_{R}(M, N) or simply Hom(M, N). It is an R-module with addition and scalar multiplication defined by

(α + β)m := αm + βm and (xα)m := x(αm) = α(xm).

Homomorphisms α: L → M and β: N → P induce, via composition, a map

Hom(α, β): Hom(M, N) → Hom(L, P),

which is obviously a homomorphism. When α is the identity map 1_{M}, we write Hom(M, β) for Hom(1_{M}, β); similarly, we write Hom(α, N) for Hom(α, 1_{N}).

Exercise (4.3). — Let R be a ring, M a module. Consider the set map

θ: Hom(R, M) → M defined by θ(ρ) := ρ(1).

Show that θ is an isomorphism, and describe its inverse.

(4.4) (Endomorphisms). — Let R be a ring, M a module. An endomorphism of M is a homomorphism α: M → M. The module of endomorphisms Hom(M, M) is also denoted End_{R}(M).It is a ring, usually noncommutative, with multiplication given by composition. Further, End_{R}(M) is a subring of End_{ℤ}(M).

Given x ∈ R, let μ_{x}: M → M denote the map of multiplication by x, defined by μ_{x}(m) := xm. It is an endomorphism. Further, x ↦ μ_{x} is a ring map

μ_{R}: R → End_{R}(M) ⊂ End_{ℤ}(M).

(Thus we may view μ_{R} as representing R as a ring of operators on the abelian group M.) Note that Ker(μ_{R}) = Ann(M).

Conversely, given an abelian group N and a ring map

ν: R → End_{ℤ}(N),

we obtain a module structure on N by setting xn := (νx)(n). Then μ_{R} = ν.

We call M faithful if μ_{R}: R → End_{R}(M) is injective, or Ann(M) = 0. For example, R is a faithful R-module, as x · 1 = 0 implies x = 0.

(4.5) (Algebras). — Fix two rings R and R′.

Suppose R′ is an R-algebra with structure map φ. Let M′ be an R′-module. Then M′ is also an R-module by restriction of scalars: xm := φ(x)m. In other words, the R-module structure on M′ corresponds to the composition

R ⟶^{φ} R′ ⟶^{μ_{R′}} End_{ℤ}(M′).

In particular, R′ it is an R-module; further, for all x ∈ R and y, z ∈ R′,

(xy)z = x(yz).

Indeed, R′ is an R′-module, so an R-module by restriction of scalars; further, (xy)z = x(yz) since (φ(x)y)z = φ(x)(yz) by associativity in R′.

Conversely, suppose R′ is an R-module such that (xy)z = x(yz). Then R′ has an R-algebra structure that is compatible with the given R-module structure. Indeed, define φ: R → R′ by φ(x) := x · 1. Then φ(x)z = xz as (x · 1)z = x(1 · z). So the composition μ_{R′}φ: R → R′ → End_{ℤ}(R′) is equal to μ_{R}. Hence φ is a ring map, because μ_{R} is one and μ_{R′} is injective by (4.4). Thus R′ is an R-algebra, and restriction of scalars recovers its given R-module structure.

Suppose that R′ = R/𝔞 for some ideal 𝔞. Then an R-module M has a compatible R′-module structure if and only if 𝔞M = 0; if so, then the R′-structure is unique. Indeed, the ring map μ_{R}: R → End_{ℤ}(M) factors through R′ if and only if μ_{R}(𝔞) = 0 by (1.5), so if and only if 𝔞M = 0; as End_{ℤ}(M) may be noncommutative, we must apply (1.5) to μ_{R}(R), which is commutative.

Again suppose R′ is an arbitrary R-algebra with structure map φ. A subalgebra R′′ of R′ is a subring such that φ maps into R′′. The subalgebra generated by

September 3, 2012 11Nts.tex x_{1}, …, x_{n} ∈ R′ is the smallest R-subalgebra that contains them. We denote it by R[x_{1}, …, x_{n}]. It clearly contains all polynomial combinations f(x_{1}, …, x_{n}) with coefficients in R. In fact, the set R′′ of these polynomial combinations is itself clearly an R-subalgebra; hence, R′′ = R[x_{1}, …, x_{n}].

We say R′ is a finitely generated R-algebra or is algebra finite over R if there exist x_{1}, …, x_{n} ∈ R′ such that R′ = R[x_{1}, …, x_{n}].

(4.6) (Residue modules). — Let R be a ring, M a module, M′ ⊂ M a submodule. Form the set of cosets

M/M′ := {m + M′ | m ∈ M}.

Recall that M/M′ inherits a module structure, and is called the residue module or quotient of M modulo M′. Form the quotient map

κ: M → M/M′ by κ(m) := m + M′.

Clearly κ is surjective, κ is linear, and κ has kernel M′.

Let α: M → N be linear. Note that Ker(α) ⊃ M′ if and only if α(M′) = 0. Recall that, if Ker(α) ⊃ M′, then there exists a homomorphism β: M/M′ → N

such that βκ = α; that is, the following diagram is commutative:

[figure: M; M/M′; N; κ; α; β]

Conversely, if β exists, then Ker(α) ⊃ M′, or α(M′) = 0, as κ(M′) = 0. Further, if β exists, then β is unique as κ is surjective. Finally, since κ is surjective, if β exists, then β is surjective if and only if α is

so. In addition, then β is injective if and only if M′ = Ker(α). Hence β is an isomorphism if and only if α is surjective and M′ = Ker(α). In particular, always

M/ Ker(α) ⟶^{∼} Im(α).  (4.6.1)

In practice, it is usually more convenient to view M/M′ not as a set of cosets, but simply another module M′′ that comes equipped with a surjective homomorphism α: M → M′′ whose kernel is the given submodule M′.

Finally, as we have seen, M/M′ has the following UMP: κ(M′) = 0, and given α: M → N such that α(M′) = 0, there is a unique homomorphism β: M/M′ → N such that βκ = α. Formally, the UMP determines M/M′ up to unique isomorphism.

(4.7) (Cyclic modules). — Let R be a ring. A module M is said to be cyclic if there exists m ∈ M such that M = Rm. If so, form α: R → M by x ↦ xm; then α induces an isomorphism R/ Ann(m) ⟶^{∼} M as Ker(α) = Ann(m); see (4.6.1). Note that Ann(m) = Ann(M). Conversely, given any ideal 𝔞, the R-module R/𝔞 is cyclic, generated by the coset of 1, and Ann(R/𝔞) = 𝔞.

(4.8) (Noether Isomorphisms). — Let R be a ring, N a module, and L and M submodules.

First, assume L ⊂ M ⊂ N. Form the following composition of quotient maps:

α: N → N/L → (N/L)/(M/L).

Clearly α is surjective, and Ker(α) = M. Hence owing to (4.6), α factors through

September 3, 2012 11Nts.tex the isomorphism β in this commutative diagram:

\begin{matrix}N ⟶ N/M \\ ↓ & ^{β}↓^{≃} \\ N/L ⟶ (N/L)/(M/L) \end{matrix}  (4.8.1)

Second, let L + M denote the set of all sums ℓ + m with ℓ ∈ L and m ∈ M. Clearly L + M is a submodule of N. It is called the sum of L and M.

Form the composition α′ of the inclusion map L → L + M and the quotient map L + M → (L + M)/M. Clearly α′ is surjective and Ker(α′) = L ∩ M. Hence owing to (4.6), α′ factors through the isomorphism β′ in this commutative diagram:

\begin{matrix}L ⟶ L/(L ∩ M) \\ ↓ & ^{β′}↓^{≃} \\ L + M ⟶ (L + M)/M \end{matrix}  (4.8.2)

The isomorphisms of (4.6.1) and (4.8.1) and (4.8.2) are called Noether’s First, Second, and Third Isomorphisms.

(4.9) (Cokernels, coimages). — Let R be a ring, α: M → N a linear map. Associated to α are its cokernel and its coimage,

Coker(α) := N/ Im(α) and Coim(α) := M/ Ker(α);

they are quotient modules, and their quotient maps are both denoted by κ.

Note (4.6) yields the UMP of the cokernel: κα = 0, and given a map β: N → P with βα = 0, there is a unique map γ: Coker(α) → P with γκ = β as shown below

[figure: M; N; Coker(α); P; α; β; κ; γ]

Further, (4.6.1) becomes Coim(α) ⟶^{∼} Im(α).

(4.10) (Free modules). — Let R be a ring, Λ a set, M a module. Given elements m_{λ} ∈ M for λ ∈ Λ, by the submodule they generate, we mean the smallest submodule that contains them all. Clearly, any submodule that contains them all contains any (finite) linear combination ∑ x_{λ}m_{λ} with x_{λ} ∈ R. On the other hand, consider the set N of all such linear combinations; clearly, N is a submodule containing the m_{λ}. Thus N is the submodule generated by the m_{λ}.

The m_{λ} are said to be free or linearly independent if, whenever ∑ x_{λ}m_{λ} = 0, also x_{λ} = 0 for all λ. Finally, the m_{λ} are said to form a free basis of M if they are free and generate M; if so, then we say M is free on the m_{λ}.

We say M is finitely generated if it has a finite set of generators.

We say M is free if it has a free basis. If so, then by (10.5) below, any two free bases have the same number ℓ of elements, and we say M is free of rank ℓ.

For example, form the set of restricted vectors

R^{⊕Λ} := {(x_{λ}) | x_{λ} ∈ R with x_{λ} = 0 for almost all λ}.

It is a module under componentwise addition and scalar multiplication. It has a standard basis, which consists of the vectors e_{μ} whose λth component is the value

September 3, 2012 11Nts.tex of the Kronecker delta function; that is,

e_{μ} := (δ_{μλ}) where δ_{μλ} := \begin{cases}1, & if λ = μ; \\ 0, & if λ ≠ μ. \end{cases}

Clearly the standard basis is free. If Λ has a finite number ℓ of elements, then R^{⊕Λ} is often written R^{ℓ} and called the direct sum of ℓ copies of R.

The free module R^{⊕Λ} has the following UMP: given a module M and elements m_{λ} ∈ M for λ ∈ Λ, there is a unique homomorphism

α: R^{⊕Λ} → M with α(e_{λ}) = m_{λ} for each λ ∈ Λ;

namely, α((x_{λ})) = α(∑ x_{λ}e_{λ}) = ∑ x_{λ}m_{λ}. Note the following obvious statements: (1) α is surjective if and only if the m_{λ} generate M.

(2) α is injective if and only if the m_{λ} are linearly independent.

(3) α is an isomorphism if and only if the m_{λ} form a free basis. Thus M is free of rank ℓ if and only if M ≃ R^{ℓ}.

Example (4.11). — Take R := ℤ and M := ℚ. Then any two x, y in M are not free; indeed, if x = a/b and y = −c/d, then bcx + ady = 0. So M is not free. Also M is not finitely generated. Indeed, given any m_{1}/n_{1}, …, m_{r}/n_{r} ∈ M, let d be a common multiple of n_{1}, …, n_{r}. Then (1/d)ℤ contains every linear combination x_{1}(m_{1}/n_{1}) + ⋯ + x_{ℓ}(m_{ℓ}/n_{ℓ}), but (1/d)ℤ ≠ M.

Exercise (4.12). — Let R be a domain, and x ∈ R nonzero. Let M be the submodule of Frac(R) generated by 1, x^{−1}, x^{−2}, . . .. Suppose that M is finitely generated. Prove that x^{−1} ∈ R, and conclude that M = R.

(4.13) (Direct Products, Direct Sums). — Let R be a ring, Λ a set, M_{λ} a module for λ ∈ Λ. The direct product of the M_{λ} is the set of arbitrary vectors:

∏ M_{λ} := {(m_{λ}) | m_{λ} ∈ M_{λ}}.

Clearly, ∏ M_{λ} is a module under componentwise addition and scalar multiplication. The direct sum of the M_{λ} is the subset of restricted vectors:

⨁ M_{λ} := {(m_{λ}) | m_{λ} = 0 for almost all λ} ⊂ ∏ M_{λ}.

Clearly, ⨁ M_{λ} is a submodule of ∏ M_{λ}. Clearly, ⨁ M_{λ} = ∏ M_{λ} if Λ is finite. If Λ = {λ_{1}, …, λ_{n}}, then ⨁ M_{λ} is also denoted by M_{λ_{1}} ⊕ ⋯ ⊕ M_{λ_{n}}.

The direct product comes equipped with projections

π_{κ}: ∏ M_{λ} → M_{κ} given by π_{κ}((m_{λ})) := m_{κ}.

It is easy to see that ∏ M_{λ} has this UMP: given homomorphisms α_{κ}: N → M_{κ}, there is a unique homomorphism α: N → ∏ M_{λ} satisfying π_{κ}α = α_{κ} for all κ ∈ Λ; namely, α(n) = (α_{λ}(n)). Often, α is denoted (α_{λ}). In other words, the π_{λ} induce a bijection of sets,

Hom(N, ∏ M_{λ}) ⟶^{∼} ∏ Hom(N, M_{λ}).  (4.13.1)

Clearly, this bijection is an isomorphism of modules.

Similarly, the direct sum comes equipped with injections

ι_{κ}: M_{κ} → ⨁ M_{λ} given by ι_{κ}(m) := (m_{λ}) where m_{λ} := \begin{cases}m, & if λ = κ; \\ 0, & if λ ≠ κ. \end{cases}

It is easy to see that it has this UMP: given homomorphisms β_{κ}: M_{κ} → N, there is

September 3, 2012 11Nts.tex a unique homomorphism β: ⨁ M_{λ} → N satisfying βι_{κ} = β_{κ} for all κ ∈ Λ; namely, β((m_{λ})) = ∑ β_{λ}(m_{λ}). Often, β is denoted ∑ β_{λ}; often, (β_{λ}). In other words, the ι_{κ} induce this bijection of sets:

Hom(⨁ M_{λ}, N) ⟶^{∼} ∏ Hom(M_{λ}, N).  (4.13.2)

Clearly, this bijection is an isomorphism of modules.

For example, if M_{λ} = R for all λ, then ⨁ M_{λ} = R^{⊕Λ} by construction. Further, if N_{λ} := N for all λ, then Hom(R^{⊕Λ}, N) = ∏ N_{λ} by (4.13.2) and (4.3).

Exercise (4.14). — Let Λ be an infinite set, R_{λ} a ring for λ ∈ Λ. Endow ∏ R_{λ} and ⨁ R_{λ} with componentwise addition and multiplication. Show that ∏ R_{λ} has a multiplicative identity (so is a ring), but that ⨁ R_{λ} does not (so is not a ring).

Exercise (4.15). — Let L, M, and N be modules. Consider a diagram

L ⟶^{α} ⟵_{ρ} M ⟶^{β} ⟵_{σ} N

where α, β, ρ, and σ are homomorphisms. Prove that

M = L ⊕ N and α = ι_{L}, β = π_{N}, σ = ι_{N}, ρ = π_{L}

if and only if the following relations hold:

βα = 0, βσ = 1, ρσ = 0, ρα = 1, and αρ + σβ = 1.

Exercise (4.16). — Let N be a module, Λ a nonempty set, M_{λ} a module for λ ∈ Λ. Prove that the injections ι_{κ}: M_{κ} → ⨁ M_{λ} induce an injection

⨁ Hom(N, M_{λ}) ↪ Hom(N, ⨁ M_{λ}),

and that it is an isomorphism if N is finitely generated.

Exercise (4.17). — Let 𝔞 be an ideal, Λ a nonempty set, M_{λ} a module for λ ∈ Λ. Prove 𝔞(⨁ M_{λ}) = ⨁ 𝔞M_{λ}. Prove 𝔞(∏ M_{λ}) = ∏ 𝔞M_{λ} if 𝔞 is finitely generated.

5. Exact Sequences

In the study of modules, the exact sequence plays a central role. We relate it to the kernel and image, the direct sum and direct product. We introduce diagram chasing, and prove the Snake Lemma, which is a fundamental result in homological algebra. We define projective modules, and characterize them in four ways. Finally, we prove Schanuel’s Lemma, which relates two arbitrary presentations of a module.

Definition (5.1). — A (finite or infinite) sequence of module homomorphisms

⋯ → M_{i−1} ⟶^{α_{i−1}} M_{i} ⟶^{α_{i}} M_{i+1} → ⋯

is said to be exact at M_{i} if Ker(α_{i}) = Im(α_{i−1}). The sequence is said to be exact if it is exact at every M_{i}, except an initial source or final target.

Example (5.2). — (1) A sequence 0 → L ⟶^{α} M is exact if and only if α is injective. If so, then we often identify L with its image α(L).

Dually — that is, in the analogous situation with all arrows reversed — a sequence M ⟶^{β} N → 0 is exact if and only if β is surjective.

(2) A sequence 0 → L ⟶^{α} M ⟶^{β} N is exact if and only if L = Ker(β), where ‘=’ means “canonically isomorphic.” Dually, a sequence L ⟶^{α} M ⟶^{β} N → 0 is exact if and only if N = Coker(α) owing to (1) and (4.6.1).

(5.3) (Short exact sequences). — A sequence 0 → L ⟶^{α} M ⟶^{β} N → 0 is exact if and only if α is injective and N = Coker(α), or dually, if and only if β is surjective and L = Ker(β). If so, then the sequence is called short exact, and often we regard L as a submodule of M, and N as the quotient M/L.

For example, the following sequence is clearly short exact:

0 → L ⟶^{ι_{L}} L ⊕ N ⟶^{π_{N}} N → 0.

Often, we identify L with ι_{L}L and N with ι_{N}N.

Proposition (5.4). — For λ ∈ Λ, let M′_{λ} → M_{λ} → M′′_{λ} be a sequence of module homomorphisms. If every sequence is exact, then so are the two induced sequences

⨁ M′_{λ} → ⨁ M_{λ} → ⨁ M′′_{λ} and ∏ M′_{λ} → ∏ M_{λ} → ∏ M′′_{λ}.

Conversely, if either induced sequence is exact then so is every original one.

Proof: The assertions are immediate from (5.1) and (4.13). □

Exercise (5.5). — Let M′ and M′′ be modules, N ⊂ M′ a submodule. Set M := M′ ⊕ M′′. Using (5.2)(1) and (5.3) and (5.4), prove M/N = M′/N ⊕ M′′.

Exercise (5.6). — Let 0 → M′ → M → M′′ → 0 be a short exact sequence. Prove that, if M′ and M′′ are finitely generated, then so is M.

Lemma (5.7). — Let 0 → M′ ⟶^{α} M ⟶^{β} M′′ → 0 be a short exact sequence, and N ⊂ M a submodule. Set N′ := α^{−1}(N) and N′′ := β(N). Then the induced sequence 0 → N′ → N → N′′ → 0 is short exact.

Proof: It is simple and straightforward to verify the asserted exactness. □

Definition (5.8). — We say that a short exact sequence

0 → M′ ⟶^{α} M ⟶^{β} M′′ → 0  (5.8.1)

splits if there is an isomorphism φ: M ⟶^{∼} M′⊕M′′ with φα = ι_{M′} and β = π_{M′′}φ. We call a homomorphism ρ: M → M′ a retraction of α if ρα = 1_{M′}. Dually, we call a homomorphism σ: M′′ → M a section of β if βσ = 1_{M′′}.

Proposition (5.9). — Let 0 → M′ ⟶^{α} M ⟶^{β} M′′ → 0 be a short exact sequence. Then the following conditions are equivalent:

(1) The sequence splits.

(2) There exists a retraction ρ: M → M′ of α.

(3) There exists a section σ: M′′ → M of β.

Proof: Assume (1). Then there exists φ: M ⟶^{∼} M′ ⊕ M′′ such that φα = ι_{M′} and β = π_{M′′}φ. Set ρ := π_{M′}φ and σ := φ^{−1}ι_{M′′}. Then clearly (2) and (3) hold.

Assume (2). Set σ′ := 1_{M} − αρ. Then σ′α = α − αρα = 0. So there exists σ: M′′ → M with σβ = σ′ by (5.2)(2) and the UMP of (4.9). So 1_{M} = αρ + σβ. Since βσβ = β and β is surjective, βσ = 1_{M′′}. Hence αρσ = 0. Since α is injective, ρσ = 0. Thus (4.15) yields (1) and also (3).

Assume (3). Then similarly (1) and (2) hold. □

Exercise (5.10). — Let M′, M′′ be modules, and set M := M′ ⊕ M′′. Let N be a submodule of M containing M′, and set N′′ := N ∩ M′′. Prove N = M′ ⊕ N′′.

Exercise (5.11). — Criticize the following misstatement of (5.9): given a short exact sequence 0 → M′ ⟶^{α} M ⟶^{β} M′′ → 0, there is an isomorphism M ≃ M′ ⊕ M′′ if and only if there is a section σ: M′′ → M of β.

Lemma (5.12) (Snake). — Consider this commutative diagram with exact rows:

M′ ⟶^{α} M ⟶^{β} M′′ ⟶ 0

γ′↓  ^{γ}↓ ^{γ′′}↓

0 ⟶ N′ ⟶^{α′} N ⟶^{β′} N′′

It yields the following exact sequence:

Ker(γ′) ⟶^{φ} Ker(γ) ⟶^{ψ} Ker(γ′′) ⟶^{∂} Coker(γ′) ⟶^{φ′} Coker(γ) ⟶^{ψ′} Coker(γ′′). (5.12.1)

Moreover, if α is injective, then so is φ; dually, if β′ is surjective, then so is ψ′.

Proof: Clearly α yields a unique compatible homomorphism Ker(γ′) → Ker(γ) because γα(Ker(γ′)) = 0. By the UMP discussed in (4.9), α′ yields a unique compatible homomorphism φ′ because M′ goes to 0 in Coker(γ). Similarly, β and β′ induce corresponding homomorphisms ψ and ψ′. Thus all the homomorphisms in (5.12.1) are defined except for ∂.

To define ∂, chase an m′′ ∈ Ker(γ′′) through the diagram. Since β is surjective, there is m ∈ M such that β(m) = m′′. By commutativity, γ′′β(m) = β′γ(m). So β′γ(m) = 0. By exactness of the bottom row, there is a unique n′ ∈ N′ such that α′(n′) = γ(m). Define ∂(m′′) to be the image of n′ in Coker(γ′).

To see ∂ is well defined, choose another m_{1} ∈ M with β(m_{1}) = m′′. Let n′_{1} ∈ N′ be the unique element with α′(n′_{1}) = γ(m_{1}) as above. Since β(m − m_{1}) = 0, there is an m′ ∈ M′ with α(m′) = m − m_{1}. But α′γ′ = γα. So α′γ′(m′) = α′(n′ − n′_{1}). Hence γ′(m′) = n′ − n′_{1} since α′ is injective. So n′ and n′_{1} have the same image in

September 3, 2012 11Nts.tex Coker(γ′). Thus ∂ is well defined.

Let’s show that (5.12.1) is exact at Ker(γ′′). Take m′′ ∈ Ker(γ′′). As in the construction of ∂, take m ∈ M such that β(m) = m′′ and take n′ ∈ N′ such that α′(n′) = γ(m). Suppose m′′ ∈ Ker(∂). Then the image of n′ in Coker(γ′) is equal to 0; so there is m′ ∈ M′ such that γ′(m′) = n′. Clearly γα(m′) = α′γ′(m′). So γα(m′) = α′(n′) = γ(m). Hence m − α(m′) ∈ Ker(γ). Since β(m − α(m′)) = m′′, clearly m′′ = ψ(m − α(m′)); so m′′ ∈ Im(ψ). Hence Ker(∂) ⊂ Im(ψ).

Conversely, suppose m′′ ∈ Im(ψ). We may assume m ∈ Ker(γ). So γ(m) = 0 and α′(n′) = 0. Since α′ is injective, n′ = 0. Thus ∂(m′′) = 0, and so Im(ψ) ⊂ Ker(∂). Thus Ker(∂) is equal to Im(ψ); that is, (5.12.1) is exact at Ker(γ′′).

The other verifications of exactness are similar or easier.

The last two assertions are clearly true. □

Exercise (5.13). — Referring to (4.8), give an alternative proof that β is an isomorphism by applying the Snake Lemma to the diagram

0 ⟶ M ⟶ N ⟶ N/M ⟶ 0

↓  ^{κ}↓  ^{β}↓

0 ⟶ M/L ⟶ N/L ⟶^{λ} (N/L)/(M/L) ⟶ 0

Exercise (5.14) (Five Lemma). — Consider this commutative diagram:

M_{4} ⟶^{α_{4}} M_{3} ⟶^{α_{3}} M_{2} ⟶^{α_{2}} M_{1} ⟶^{α_{1}} M_{0}

γ_{4}↓  ^{γ_{3}}↓  ^{γ_{2}}↓  ^{γ_{1}}↓  ^{γ_{0}}↓

N_{4} ⟶^{β_{4}} N_{3} ⟶^{β_{3}} N_{2} ⟶^{β_{2}} N_{1} ⟶^{β_{1}} N_{0}

Assume it has exact rows. Via a chase, prove these two statements:

(1) If γ_{3} and γ_{1} are surjective and if γ_{0} is injective, then γ_{2} is surjective.

(2) If γ_{3} and γ_{1} are injective and if γ_{4} is surjective, then γ_{2} is injective.

Exercise (5.15) (Nine Lemma). — Consider this commutative diagram:

\begin{matrix}0  0  0 \\ ↓ & ↓ & ↓ \\ 0 ⟶ L′ ⟶ L ⟶ L′′ ⟶ 0 \\ ↓ & ↓ & ↓ \\ 0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0 \\ ↓ & ↓ & ↓ \\ 0 ⟶ N′ ⟶ N ⟶ N′′ ⟶ 0 \\ ↓ & ↓ & ↓ \\ 0  0  0 \end{matrix}  (5.15.1)

Assume all the columns are exact and the middle row is exact. Applying the Snake Lemma, prove that the first row is exact if and only if the third is.

Exercise (5.16). — Consider this commutative diagram with exact rows:

M′ ⟶^{β} M ⟶^{γ} M′′

α′↓  ^{α}↓ _{α′′}↓

N′ ⟶^{β′} N ⟶^{γ′} N′′

September 3, 2012 11Nts.tex Assume α′ and γ are surjective. Given n ∈ N and m′′ ∈ M′′ with α′′(m′′) = γ′(n), show that there is m ∈ M such that α(m) = n and γ(m) = m′′.

Theorem (5.17) (Left exactness of Hom). — (1) Let M′ → M → M′′ → 0 be a sequence of module homomorphisms. Then it is exact if and only if, for all modules N, the following induced sequence is exact:

0 → Hom(M′′, N) → Hom(M, N) → Hom(M′, N).  (5.17.1)

(2) Let 0 → N′ → N → N′′ be a sequence of module homomorphisms. Then it is exact if and only if, for all modules M, the following induced sequence is exact:

0 → Hom(M, N′) → Hom(M, N) → Hom(M, N′′).

Proof: By (5.2)(2), the exactness of M′ ⟶^{α} M ⟶^{β} M′′ → 0 means simply that M′′ = Coker(α). On the other hand, the exactness of (5.17.1) means that a φ ∈ Hom(M, N) maps to 0, or equivalently φα = 0, if and only if there is a unique γ: M′′ → N such that γβ = φ. So (5.17.1) is exact if and only if M′′ has the UMP of Coker(α), discussed in (4.9); that is, M′′ = Coker(α). Thus (1) holds.

The proof of (2) is similar. □

Definition (5.18). — A (free) presentation of a module M is an exact sequence

G → F → M → 0

with G and F free. If G and F are free of finite rank, then the presentation is called finite. If M has a finite presentation, then M is said to be finitely presented.

Proposition (5.19). — Given a module M and a set of generators {m_{λ}}_{λ∈Λ}, there is an exact sequence 0 → K → R^{⊕Λ}⟶^{α}M → 0 with α(e_{λ}) = m_{λ}, where {e_{λ}} is the standard basis; further, there is a presentation R^{⊕Σ} → R^{⊕Λ}⟶^{α}M → 0.

Proof: By (4.10)(1), there is a surjection α: R^{⊕Λ} → → M with α(e_{λ}) = m_{λ}. Set K := Ker(α). Then 0 → K → R^{⊕Λ} → M → 0 is exact by (5.3). Take a set of generators {k_{σ}}_{σ∈Σ} of K, and repeat the process to obtain a surjection R^{⊕Σ} → → K. Then R^{⊕Σ} → R^{⊕Λ} → M → 0 is a presentation. □

Definition (5.20). — A module P is called projective if, given any surjective homomorphism β: M → → N, every homomorphism α: P → N lifts to a homomorphism γ: P → M; that is, α = βγ.

Exercise (5.21). — Show that a free module R^{⊕Λ} is projective.

Theorem (5.22). — The following conditions on a module P are equivalent:

(1) The module P is projective.

(2) Every short exact sequence 0 → K → M → P → 0 splits.

(3) There is a module K such that K ⊕ P is free.

(4) Every exact sequence N′ → N → N′′ induces an exact sequence

Hom(P, N′) → Hom(P, N) → Hom(P, N′′).  (5.22.1)

(5) Every surjective homomorphism β: M → → N induces a surjection

Hom(P, β): Hom(P, M) → Hom(P, N).

Proof: Assume (1). In (2), the surjection M → → P and the identity P → P yield a section P → M. So the sequence splits by (5.9). Thus (2) holds.

Assume (2). By (5.19), there is an exact sequence 0 → K → R^{⊕Λ} → P → 0. Then (2) implies K ⊕ P ≃ R^{⊕Λ}. Thus (3) holds.

Assume (3); say K ⊕ P ≃ R^{⊕Λ}. For each λ ∈ Λ, take a copy N′_{λ} → N_{λ} → N′_{λ} of the exact sequence N′ → N → N′′ of (4). Then the induced sequence

∏ N′_{λ} → ∏ N_{λ} → ∏ N′′_{λ}.

is exact by (5.4). But by the end of (4.13), that sequence is equal to this one:

Hom(R^{⊕Λ}, N′) → Hom(R^{⊕Λ}, N) → Hom(R^{⊕Λ}, N′′).

But K ⊕ P ≃ R^{⊕Λ}. So owing to (4.13.2), the latter sequence is also equal to

Hom(K, N′) ⊕ Hom(P, N′) → Hom(K, N) ⊕ Hom(P, N) → Hom(K, N′′) ⊕ Hom(P, N′′).

Hence (5.22.1) is exact by (5.4). Thus (4) holds.

Assume (4). Then every exact sequence M ⟶^{β} N → 0 induces an exact sequence

Hom(P, M) → Hom(P, N) → 0.

In other words, (5) holds.

Assume (5). By definition, Hom(P, β)(γ) = βγ. Therefore, (1) holds. □

Lemma (5.23) (Schanuel). — Given two short exact sequences

0 → L ⟶^{i} P ⟶^{α} M → 0 and 0 → L′ ⟶^{i′} P′ ⟶^{α′} M → 0

with P and P′ projective, there is an isomorphism of exact sequences:

0 ⟶ L ⊕ P′ ⟶^{i⊕1_{P′}} P ⊕ P′ ⟶^{(α 0)} M ⟶ 0

≅↓^{β}  ≅↓^{γ}  ^{=}↓^{1_{M}}

0 ⟶ P ⊕ L′ ⟶^{1_{P} ⊕i′} P ⊕ P′ ⟶^{(0 α′)} M ⟶ 0

Proof: First, let’s construct an intermediate isomorphism of exact sequences:

0 ⟶ L ⊕ P′ ⟶^{i⊕1_{P′}} P ⊕ P′ ⟶^{(α 0)} M ⟶ 0

≅↑^{λ}  ≅↑^{θ}  ^{=}↑^{1_{M}}

0 ⟶ K ⟶ P ⊕ P′ ⟶^{(α α′)} M ⟶ 0

Take K := Ker(α α′). To form θ, recall that P′ is projective and α is surjective. So there is a map π: P′ → P such that α′ = απ. Take θ := ( \begin{matrix}1 & π \\ 0 & 1 \end{matrix} ).

Then θ has ( \begin{matrix}1 & −π \\ 0 & 1 \end{matrix} ) as inverse. Further, the right-hand square is commutative:

(α 0)θ = (α 0)( \begin{matrix}1 & π \\ 0 & 1 \end{matrix} ) = (α απ) = (α α′).

So θ induces the desired isomorphism λ: K ⟶^{∼} L ⊕ P′.

Symmetrically, form an automorphism θ′ of P ⊕P′, which induces an isomorphism

λ′: K ⟶^{∼} P ⊕ L′. Finally, take γ := θ′θ^{−1} and β := λ′λ^{−1}. □

Exercise (5.24). — Let R be a ring, and 0 → L → R^{n} → M → 0 an exact sequence. Prove M is finitely presented if and only if L is finitely generated.

Exercise (5.25). — Let R be a ring, X_{1}, X_{2}, … infinitely many variables. Set P := R[X_{1}, X_{2}, …] and M := P/⟨X_{1}, X_{2}, … ⟩. Is M finitely presented? Explain.

Proposition (5.26). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence with L finitely generated and M finitely presented. Then N is finitely presented.

Proof: Let R be the ground ring, μ: R^{m} → M any surjection. Set ν := βμ, set K := Ker ν, and set λ := μ|K. Then the following diagram is commutative:

0 ⟶ K ⟶ R^{m} ⟶^{ν} N ⟶ 0

λ↓ ^{μ}↓ ^{1_{N}} ↓

0 ⟶ L ⟶^{α} M ⟶^{β} N ⟶ 0

The Snake Lemma (5.12) yields an isomorphism Ker λ ⟶^{∼} Ker μ. But Ker μ is finitely generated by (5.24). So Ker λ is finitely generated. Also, the Snake Lemma implies Coker λ = 0 as Coker μ = 0; so 0 → Ker λ → K ⟶^{λ} L → 0 is exact. Hence K is finitely generated by (5.6). Thus N is finitely presented by (5.24). □

Exercise (5.27). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence with M finitely generated and N finitely presented. Prove L is finitely generated.

Proposition (5.28). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence with L and N finitely presented. Prove M is finitely presented too.

Proof: Let R be the ground ring, λ: R^{ℓ} → L and ν: R^{n} → → N any surjections. Define γ: R^{ℓ} → M by γ := αλ. Note R^{n} is projective by (5.21), and define δ: R^{n} → M by lifting ν along β. Define μ: R^{ℓ} ⊕ R^{n} → M by μ := γ + δ. Then the following diagram is, plainly, commutative, where ι := ι_{R^{ℓ}} and π := π_{R^{n}}:

0 ⟶ R^{ℓ} ⟶^{ι} R^{ℓ} ⊕ R^{n} ⟶^{π} R^{n} ⟶ 0

λ↓  ^{μ}↓  ^{ν}↓

0 ⟶ L ⟶^{α} M ⟶^{β} N ⟶ 0

Since λ and ν are surjective, the Snake Lemma (5.12) yields an exact sequence

0 → Ker λ → Ker μ → Ker ν → 0,

and implies Coker μ = 0. Also, Ker λ and Ker ν are finitely generated by (5.24). So Ker μ is finitely generated by (5.6). Thus M is finitely presented by (5.24). □

6. Direct Limits

Category theory provides the right abstract setting for certain common concepts, constructions, and proofs. Here we treat adjoints and direct limits. We elaborate on two key special cases of direct limits: coproducts (direct sums) and coequalizers (cokernels). Then we construct arbitrary direct limits of sets and of modules. Further, we prove direct limits are preserved by left adjoints; whence, direct limits commute with each other, and in particular, with coproducts and coequalizers.

Although this section is the most abstract of the entire book, all the material here is elementary, and none of it is very deep. In fact, many statements are just concise restatements in more expressive language; they can be understood through a simple translation of terms. Experience shows that it pays to learn this more abstract language, but that doing so requires determined, yet modest effort.

(6.1) (Categories). — A category 𝒞 is a collection of elements, called objects. Each pair of objects A, B is equipped with a set Hom_{𝒞} (A, B) of elements, called maps or morphisms. We write α: A → B or A ⟶^{α} B to mean α ∈ Hom_{𝒞} (A, B).

Further, given objects A, B, C, there is a composition law

Hom_{𝒞} (A, B) × Hom_{𝒞} (B, C) → Hom_{𝒞} (A, C), written (α, β) ↦ βα,

and there is a distinguished map 1_{B} ∈ Hom_{𝒞} (B, B), called the identity such that (1) composition is associative, or γ(βα) = (γβ)α for γ: C → D, and

(2) 1_{B} is unitary, or 1_{B}α = α and β1_{B} = β. We say α is an isomorphism with inverse β: B → A if αβ = 1_{B} and βα = 1_{A}.

For example, four common categories are those of sets ((Sets)), of rings ((Rings)), of R-modules ((R-mod)), and of R-algebras ((R-alg)); the corresponding maps are the set maps, and the ring, R-module, and R-algebra homomorphisms.

Given categories 𝒞 and 𝒞′, their product 𝒞 × 𝒞′ is the category whose objects are the pairs (A, A′) with A an object of 𝒞 and A′ an object of 𝒞′ and whose maps are the pairs (α, α′) of maps α in 𝒞 and α′ in 𝒞′.

(6.2) (Functors). — A map of categories is known as a functor. Namely, given categories 𝒞 and 𝒞′, a (covariant) functor F: 𝒞 → 𝒞′ is a rule that assigns to each object A of 𝒞 an object F (A) of 𝒞′ and to each map α: A → B of 𝒞 a map F (α): F (A) → F (B) of 𝒞′ preserving composition and identity; that is,

(1) F (βα) = F (β)F (α) for maps α: A → B and β: B → C of 𝒞, and

(2) F (1_{A}) = 1_{F (A)} for any object A of 𝒞.

We also denote a functor F by F (•), by A ↦ F (A), or by A ↦ F_{A}.

Note that a functor F preserves isomorphisms. Indeed, if αβ = 1_{B} and βα = 1_{A}, then F (α)F (β) = 1_{F (B)} and F (β)F (α) = F (1_{A}).

For example, let R be a ring, M a module. Then clearly Hom_{R}(M, •) is a functor from ((R-mod)) to ((R-mod)). A second example is the forgetful functor from ((R-mod)) to ((Sets)); it sends a module to its underlying set and a homomorphism to its underlying set map.

A map of functors is known as a natural transformation. Namely, given two functors F, F′: 𝒞 ⇉ 𝒞′, a natural transformation θ: F → F′ is a collection of maps θ(A): F (A) → F′(A), one for each object A of 𝒞, such that θ(B)F (α) = F′(α)θ(A)

September 3, 2012 11Nts.tex for every map α: A → B of 𝒞; that is, the following diagram is commutative:

F (A) ⟶^{F (α)} F (B)

θ(A)↓  ^{θ(B)}↓

F′(A) ⟶^{F′(α)} F′(B)

For example, the identity maps 1_{F (A)} trivially form a natural transformation 1_{F} from any functor F to itself. We call F and F′ isomorphic if there are natural transformations θ: F → F′ and θ′: F′ → F with θ′θ = 1_{F} and θθ′ = 1_{F′}.

A contravariant functor G from 𝒞 to 𝒞′ is a rule similar to F, but G reverses the direction of maps; that is, G(α) carries G(B) to G(A), and G satisfies the analogues of (1) and (2). For example, fix a module N; then Hom(•, N) is a contravariant functor from ((R-mod)) to ((R-mod)).

Exercise (6.3). — (1) Show that the condition (6.2)(1) is equivalent to the commutativity of the corresponding diagram:

Hom_{𝒞} (B, C) ⟶ Hom_{𝒞′} (F (B), F (C))
↓  ↓
Hom_{𝒞} (A, C) ⟶ Hom_{𝒞′} (F (A), F (C))

(2) Given γ: C → D, show (6.2)(1) yields the commutativity of this diagram:

Hom_{𝒞} (B, C) ⟶ Hom_{𝒞′} (F (B), F (C))
↓  ↓
Hom_{𝒞} (A, D) ⟶ Hom_{𝒞′} (F (A), F (D))

(6.4) (Adjoints). — Let 𝒞 and 𝒞′ be categories, F: 𝒞 → 𝒞′ and F′: 𝒞′ → 𝒞 functors. We call (F, F′) an adjoint pair, F the left adjoint of F′, and F′ the right adjoint of F if, for each object A ∈ 𝒞 and object A′ ∈ 𝒞′, there is a natural bijection

Hom_{𝒞′} (F (A), A′) ≃ Hom_{𝒞} (A, F′(A′)).  (6.4.1)

Here natural means that maps B → A and A′ → B′ induce a commutative diagram:

Hom_{𝒞′} (F (A), A′) ≃ Hom_{𝒞} (A, F′(A′))
↓  ↓
Hom_{𝒞′} (F (B), B′) ≃ Hom_{𝒞} (B, F′(B′))

Naturality serves to determine an adjoint up to canonical isomorphism. Indeed, let F and G be two left adjoints of F′. Given A ∈ 𝒞, define θ(A): G(A) → F (A) to be the image of 1_{F (A)} under the adjoint bijections

Hom_{𝒞′} (F (A), F (A)) ≃ Hom_{𝒞} (A, F′F (A)) ≃ Hom_{𝒞′} (G(A), F (A)).

To see that θ(A) is natural in A, take a map α: A → B. It induces the following

September 3, 2012 11Nts.tex diagram, which is commutative owing to the naturality of the adjoint bijections:

Hom_{𝒞′} (F (A), F (A)) ≃ Hom_{𝒞} (A, F′F (A)) ≃ Hom_{𝒞′} (G(A), F (A))

↓  ↓  ↓
Hom_{𝒞′} (F (A), F (B)) ≃ Hom_{𝒞} (A, F′F (B)) ≃ Hom_{𝒞′} (G(A), F (B))
↑  ↑  ↑
Hom_{𝒞′} (F (B), F (B)) ≃ Hom_{𝒞} (B, F′F (B)) ≃ Hom_{𝒞′} (G(B), F (B))

Chase after 1_{F (A)} and 1_{F (B)}. Both map to F (α) ∈ Hom_{𝒞′} (F (A), F (B)). So both map to the same image in Hom_{𝒞′} (G(A), F (B)). But clockwise, 1_{F (A)} maps to F (α)θ(A); counterclockwise, 1_{F (B)} maps to θ(B)G(α). So θ(B)G(α) = F (α)θ(A). Thus the θ(A) form a natural transformation θ: G → F.

Similarly, there is a natural transformation θ′: F → G. It remains to show θ′θ = 1_{G} and θθ′ = 1_{F}. However, by naturality, this diagram is commutative:

Hom_{𝒞′} (F (A), F (A)) ≃ Hom_{𝒞} (A, F′F (A)) ≃ Hom_{𝒞} (G(A), F (A))

↓  ↓  ↓
Hom_{𝒞′} (F (A), G(A)) ≃ Hom_{𝒞} (A, F′G(A)) ≃ Hom_{𝒞} (G(A), G(A))

Chase after 1_{F (A)}. Clockwise, its image is θ′(A)θ(A) in the lower right corner. Counterclockwise, its image is 1_{G(A)}, owing to the definition of θ′. Thus θ′θ = 1_{G}. Similarly, θθ′ = 1_{F}, as required.

For example, the “free module” functor is the left adjoint of the forgetful functor from ((R-mod)) to ((Sets)), since by (4.10),

Hom_{((R-mod))}(R^{⊕Λ}, M) = Hom_{((Sets))}(Λ, M).

Similarly, the “polynomial ring” functor is the left adjoint of the forgetful functor from ((R-alg)) to ((Sets)), since by (1.3),

Hom_{((R-alg))}(R[X_{1}, …, X_{n}], R′) = Hom_{((Sets))}({X_{1}, …, X_{n}}, R′).

Exercise (6.5). — Let 𝒞 and 𝒞′ be categories, F: 𝒞 → 𝒞′ and F′: 𝒞′ → 𝒞 an adjoint pair. Let φ_{A,A′}: Hom_{𝒞′} (F A, A′) ⟶^{∼} Hom_{𝒞} (A, F′A′) denote the natural bijection, and set η_{A} := φ_{A,F A}(1_{F A}). Do the following:

(1) Prove η_{A} is natural in A; that is, given g: A → B, the induced square

A ⟶^{η_{A}} F′F A

g↓  ↓^{F′F g}

B ⟶^{η_{B}} F′F B

is commutative. We call the natural transformation A ↦ η_{A} the unit of (F, F′). (2) Given f′: F A → A′, prove φ_{A,A′}(f′) = F′f′ ∘ η_{A}.

(3) Prove the natural map η_{A}: A → F′F A is universal from A to F′; that is, given f: A → F′A′, there is a unique map f′: F A → A′ with F′f′ ∘ η_{A} = f.

(4) Conversely, instead of assuming (F, F′) is an adjoint pair, assume given a natural transformation η: 1_{𝒞} → F′F satisfying (1) and (3). Prove the equation in (2) defines a natural bijection making (F, F′) an adjoint pair, whose unit is η.

(5) Identify the units in the two examples in (6.4): the “free module” functor and the “polynomial ring” functor.

(Dually, we can define a counit ε: F F′ → 1_{𝒞′} , and prove similar statements.)

September 3, 2012 11Nts.tex (6.6) (Direct limits). — Let Λ, 𝒞 be categories. Assume Λ is small; that is, its objects form a set. Given a functor λ ↦ M_{λ} from Λ to 𝒞, its direct limit or colimit, denoted \underrightarrow{lim} M_{λ} or \underrightarrow{lim}_{λ∈Λ} M_{λ}, is defined as the universal example of an object P of 𝒞 equipped with maps β_{μ}: M_{μ} → P, called insertions, that are compatible with the transition maps α_{μ}^{κ}: M_{κ} → M_{μ}, which are the images of the maps of Λ. In other words, there is a unique map β such that all these diagrams commute:

M_{κ} ⟶^{α_{μ}^{κ}} M_{μ} ⟶^{α_{μ}} \underrightarrow{lim} M_{λ}

↓^{β_{κ}}  ↓^{β_{μ}}  ↓^{β}

P ⟶^{1_{P}} P ⟶^{1_{P}} P

To indicate this context, the functor λ ↦ M_{λ} is often called a direct system.

As usual, universality implies that, once equipped with its insertions α_{μ}, the limit \underrightarrow{lim} M_{λ} is determined up to unique isomorphism, assuming it exists. In practice, there is usually a canonical choice for \underrightarrow{lim} M_{λ}, given by a construction. In any case, let us use \underrightarrow{lim} M_{λ} to denote a particular choice.

We say that 𝒞 has direct limits indexed by Λ if, for every functor λ ↦ M_{λ} from Λ to 𝒞, the direct limit \underrightarrow{lim} M_{λ} exists. We say that 𝒞 has direct limits if it has direct limits indexed by every small category Λ. We say that a functor F: 𝒞 → 𝒞′ preserves direct limits if, given any direct limit \underrightarrow{lim} M_{λ} in 𝒞, the direct limit \underrightarrow{lim} F (M_{λ}) exists, and is equal to F (\underrightarrow{lim} M_{λ}); more precisely, the maps F (α_{μ}): F (M_{μ}) → F (\underrightarrow{lim} M_{λ}) induce a canonical map

ϕ: \underrightarrow{lim} F (M_{λ}) → F (\underrightarrow{lim} M_{λ}),

and ϕ is an isomorphism. Sometimes, we construct \underrightarrow{lim} F (M_{λ}) by showing that F (\underrightarrow{lim} M_{λ}) has the requisite UMP.

Assume 𝒞 has direct limits indexed by Λ. Then, given a natural transformation from λ ↦ M_{λ} to λ ↦ N_{λ}, universality yields unique commutative diagrams

M_{μ} ⟶ \underrightarrow{lim} M_{λ}

↓  ↓
N_{μ} ⟶ \underrightarrow{lim} N_{λ}

To put it in another way, form the functor category 𝒞^{Λ}: its objects are the functors λ ↦ M_{λ} from Λ to 𝒞; its maps are the natural transformations (they form a set as Λ is one). Then taking direct limits yields a functor \underrightarrow{lim} from 𝒞^{Λ} to 𝒞.

In fact, it is just a restatement of the definitions that the “direct limit” functor \underrightarrow{lim} is the left adjoint of the diagonal functor

Δ: 𝒞 → 𝒞^{Λ}.

By definition, Δ sends each object M to the constant functor ΔM, which has the same value M at every λ ∈ Λ and has the same value 1_{M} at every map of Λ; further, Δ carries a map γ: M → N to the natural transformation Δγ: ΔM → ΔN, which has the same value γ at every λ ∈ Λ.

(6.7) (Coproducts). — Let 𝒞 be a category, Λ a set, and M_{λ} an object of 𝒞 for each λ ∈ Λ. The coproduct ∐_{λ∈Λ} M_{λ}, or simply ∐ M_{λ}, is defined as the universal example of an object P equipped with a map β_{μ}: M_{μ} → P for each μ ∈ Λ. The maps ι_{μ}: M_{μ} → ∐ M_{λ} are called the inclusions. Thus, given an example P, there

September 3, 2012 11Nts.tex exists a unique map β: ∐ M_{λ} → P with βι_{μ} = β_{μ} for all μ ∈ Λ.

If Λ = ∅, then the coproduct is an object B with a unique map β to every other object P. There are no μ in Λ, so no inclusions ι_{μ}: M_{μ} → B, so no equations βι_{μ} = β_{μ} to restrict β. Such a B is called an initial object.

For instance, suppose 𝒞 = ((R-mod)). Then the zero module is an initial object. For any Λ, the coproduct ∐ M_{λ} is just the direct sum ⨁ M_{λ} (a convention if Λ = ∅). Further, suppose 𝒞 = ((Sets)). Then the empty set is an initial object. For any Λ, the coproduct ∐ M_{λ} is the disjoint union ⨆ M_{λ} (a convention if Λ = ∅).

Note that the coproduct is a special case of the direct limit. Indeed, regard Λ as a discrete category: its objects are the λ ∈ Λ, and it has just the required maps, namely, the 1_{λ}. Then \underrightarrow{lim} M_{λ} = ∐ M_{λ} with the insertions equal to the inclusions.

(6.8) (Coequalizers). — Let α, α′: M → N be two maps in a category 𝒞. Their coequalizer is defined as the universal example of an object P equipped with a map η: N → P such that ηα = ηα′.

For instance, if 𝒞 = ((R-mod)), then the coequalizer is Coker(α − α′). In particular, the coequalizer of α and 0 is just Coker(α).

Suppose 𝒞 = ((Sets)). Take the smallest equivalence relation ∼ on N with α(m) ∼ α′(m) for all m ∈ M; explicitly, n ∼ n′ if there are elements m_{1}, …, m_{r} with α(m_{1}) = n, with α′(m_{r}) = n′, and with α(m_{i}) = α′(m_{i+1}) for 1 ≤ i < r. Clearly, the coequalizer is the quotient N/∼ equipped with the quotient map.

Note that the coequalizer is a special case of the direct limit. Indeed, let Λ be the category consisting of two objects κ, μ and two nontrivial maps φ, φ′: κ → μ. Define λ ↦ M_{λ} in the obvious way: set M_{κ} := M and M_{μ} := N; send φ to α and φ′ to α′. Then the coequalizer is \underrightarrow{lim} M_{λ}.

Exercise (6.9). — Let α: L → M and β: L → N be two maps. Their pushout is defined as the universal example of an object P equipped with a pair of maps γ: M → P and δ: N → P such that γα = δβ. Express the pushout as a direct limit. Show that, in ((Sets)), the pushout is the disjoint union M ⊔ N modulo the smallest equivalence relation ∼ with m ∼ n if there is ℓ ∈ L with α(ℓ) = m and β(ℓ) = n. Show that, in ((R-mod)), the pushout is equal to the direct sum M ⊕ N modulo the image of L under the map (α, −β).

Lemma (6.10). — A category 𝒞 has direct limits if and only if 𝒞 has coproducts and coequalizers. If a category 𝒞 has direct limits, then a functor F: 𝒞 → 𝒞′ preserves them if and only if F preserves coproducts and coequalizers.

Proof: If 𝒞 has direct limits, then 𝒞 has coproducts and coequalizers because they are special cases by (6.7) and (6.8). By the same token, if F: 𝒞 → 𝒞′ preserves direct limits, then F preserves coproducts and coequalizers.

Conversely, assume that 𝒞 has coproducts and coequalizers. Let Λ be a small category, and λ ↦ M_{λ} a functor from Λ to 𝒞. Let Σ be the set of transition maps α_{μ}^{λ}: M_{λ} → M_{μ}. For each σ := α_{μ}^{λ} ∈ Σ, set M_{σ} := M_{λ}. Set M := ∐_{σ∈Σ} M_{σ} and N := ∐_{λ∈Λ} M_{λ}. For each σ, there are two maps M_{σ} := M_{λ} → N: the inclusion ι_{λ} and the composition ι_{μ}α_{μ}^{λ}. Correspondingly, there are two maps α, α′: M → N. Let C be their coequalizer, and η: N → C the insertion.

Given maps β_{λ}: M_{λ} → P with β_{μ}α_{μ}^{λ} = β_{λ}, there is a unique map β: N → P with βι_{λ} = β_{λ} by the UMP of the coproduct. Clearly βα = βα′; so β factors uniquely through C by the UMP of the coequalizer. Thus C = \underrightarrow{lim} M_{Λ}, as desired.

Finally, if F: 𝒞 → 𝒞′ preserves coproducts and coequalizers, then F preserves arbitrary direct limits as F preserves the above construction. □

Theorem (6.11). — The categories ((R-mod)) and ((Sets)) have direct limits.

Proof: The assertion follows from (6.10) because ((R-mod)) and ((Sets)) have coproducts by (6.7) and have coequalizers by (6.8). □

Theorem (6.12). — Every left adjoint F: 𝒞 → 𝒞′ preserves direct limits.

Proof: Let Λ be a small category, λ ↦ M_{λ} a functor from Λ to 𝒞 such that \underrightarrow{lim} M_{λ} exists. Given an object P′ of 𝒞′, consider all possible commutative diagrams

\begin{matrix}F (M_{κ}) ⟶^{F (α_{μ}^{κ})} F (M_{μ}) ⟶^{F (α_{μ})} F (\underrightarrow{lim} M_{λ}) \\ ↓^{β′_{κ}} & ↓^{β′_{μ}} & ↓^{β′} \\ P′ ⟶^{1} P′ ⟶^{1} P′ \end{matrix}  (6.12.1)

where α_{μ}^{κ} is any transition map and α_{μ} is the corresponding insertion. Given the β′_{κ}, we must show there is a unique β′.

Say F is the left adjoint of F′: 𝒞′ → 𝒞. Then giving (6.12.1) is equivalent to giving this corresponding commutative diagram:

M_{κ} ⟶^{α_{μ}^{κ}} M_{μ} ⟶^{α_{μ}} \underrightarrow{lim} M_{λ}

↓^{β_{κ}}  ↓^{β_{μ}}  ↓^{β}

F′(P′) ⟶^{1} F′(P′) ⟶^{1} F′(P′)

However, given the β_{κ}, there is a unique β by the UMP of \underrightarrow{lim} M_{λ}. □

Proposition (6.13). — Let 𝒞 be a category, Λ and Σ small categories. Assume 𝒞 has direct limits indexed by Σ. Then the functor category 𝒞^{Λ} does too.

Proof: Let σ ↦ (λ ↦ M_{σλ}) be a functor from Σ to 𝒞^{Λ}. Then a map σ → τ in Σ yields a natural transformation from λ → M_{σλ} to λ ↦ M_{τλ}. So a map λ ↦ μ in Λ yields a commutative square

\begin{matrix}M_{σλ} ⟶ M_{σμ} \\ ↓ & ↓ \\ M_{τλ} ⟶ M_{τμ} \end{matrix}  (6.13.1)

in a manner compatible with composition in Σ. Hence, with λ fixed, the rule σ ↦ M_{σλ} is a functor from Σ to 𝒞.

By hypothesis, \underrightarrow{lim}_{σ∈Σ} M_{σλ} exists. So λ ↦ \underrightarrow{lim}_{σ∈Σ} M_{σλ} is a functor from Λ to 𝒞. Further, as τ ∈ Σ varies, there are compatible natural transformations from the λ ↦ M_{τλ} to λ ↦ \underrightarrow{lim}_{σ∈Σ} M_{σλ}. Finally, the latter is the direct limit of the functor τ ↦ (λ ↦ M_{τλ}) from Σ to 𝒞^{Λ}, because, given any functor λ ↦ P_{λ} from Λ to 𝒞 equipped with, for τ ∈ Σ, compatible natural transformations from the λ ↦ M_{τλ} to λ ↦ P_{λ}, there are, for λ ∈ Λ, compatible unique maps \underrightarrow{lim}_{σ∈Σ} M_{σλ} → P_{λ}. □

Theorem (6.14) (Direct limits commute). — Let 𝒞 be a category with direct limits indexed by small categories Σ and Λ. Let σ ↦ (λ ↦ M_{σλ}) be a functor from Σ to 𝒞^{Λ}. Then

\underrightarrow{lim}_{σ∈Σ} \underrightarrow{lim}_{λ∈Λ} M_{σ,λ} = \underrightarrow{lim}_{λ∈Λ} \underrightarrow{lim}_{σ∈Σ} M_{σ,λ}.

Proof: By (6.6), the functor \underrightarrow{lim}_{λ∈Λ} : 𝒞^{Λ} → 𝒞 is a left adjoint. By (6.13), the category 𝒞^{Λ} has direct limits indexed by Σ. So (6.12) yields the assertion. □

Corollary (6.15). — Let Λ be a small category, R a ring, and 𝒞 either ((Sets)) or ((R-mod)). Then functor \underrightarrow{lim}: 𝒞^{Λ} → 𝒞 preserves coproducts and coequalizers.

Proof: By (6.7) and (6.8), both coproducts and coequalizers are special cases of direct limits, and 𝒞 has them. So (6.14) yields the assertion. □

Exercise (6.16). — Let 𝒞 be a category, Σ and Λ small categories.

(1) Prove 𝒞^{Σ×Λ} = (𝒞^{Λ})^{Σ} with (σ, λ) ↦ M_{σ,λ} corresponding to σ ↦ (λ ↦ M_{σ,λ}).

(2) Assume 𝒞 has direct limits indexed by Σ and by Λ. Prove that 𝒞 has direct limits indexed by Σ × Λ and that \underrightarrow{lim}_{λ∈Λ} \underrightarrow{lim}_{σ∈Σ} = \underrightarrow{lim}_{(σ,λ)∈Σ×Λ}.

Exercise (6.17). — Let λ ↦ M_{λ} and λ ↦ N_{λ} be two functors from a small category Λ to ((R-mod)), and {θ_{λ}: M_{λ} → N_{λ}} a natural transformation. Show

\underrightarrow{lim} Coker(θ_{λ}) = Coker(\underrightarrow{lim} M_{λ} → \underrightarrow{lim} N_{λ}).

Show that the analogous statement for kernels can be false by constructing a counterexample using the following commutative diagram with exact rows:

ℤ ⟶^{μ_{2}} ℤ ⟶ ℤ/⟨2⟩ ⟶ 0

↓^{μ_{2}}  ↓^{μ_{2}}  ↓^{μ_{2}}

ℤ ⟶^{μ_{2}} ℤ ⟶ ℤ/⟨2⟩ ⟶ 0

7. Filtered Direct Limits

Filtered direct limits are direct limits indexed by a filtered category, which is a more traditional sort of index set. We give an alternative construction of these limits for modules. We conclude that forming them preserves exact sequences, and so commutes with forming the module of homomorphisms out of a fixed finitely presented source. We end by proving that every module is a filtered direct limit of finitely presented modules.

(7.1) (Filtered categories). — We call a small category Λ filtered if (1) given objects κ and λ, for some μ there are maps κ → μ and λ → μ,

(2) given two maps σ, τ: η ⇉ κ with the same source and the same target, for

some μ there is a map φ: κ → μ such that φσ = φτ. Given a category 𝒞, we say a functor λ ↦ M_{λ} from Λ to 𝒞 is filtered if Λ is

filtered. If so, then we say the direct limit \underrightarrow{lim} M_{λ} is filtered if it exists.

For example, let Λ be a partially ordered set. Suppose Λ is directed; that is, given κ, λ ∈ Λ, there is a μ with κ ≤ μ and λ ≤ μ. Regard Λ as a category whose objects are its elements and whose sets Hom(κ, λ) consist of a single element if κ ≤ λ, and are empty if not; morphisms can be composed as the ordering is transitive. Clearly, the category Λ is filtered.

Exercise (7.2). — Let R be a ring, M a module, Λ a set, M_{λ} a submodule for each λ ∈ Λ. Assume ⋃ M_{λ} = M. Assume, given λ, μ ∈ Λ, there is ν ∈ Λ such that M_{λ}, M_{μ} ⊂ M_{ν}. Order Λ by inclusion: λ ≤ μ if M_{λ} ⊂ M_{μ}. Prove that M = \underrightarrow{lim} M_{λ}.

Exercise (7.3). — Show that every module M is the filtered direct limit of its finitely generated submodules.

Exercise (7.4). — Show that every direct sum of modules is the filtered direct limit of its finite direct subsums.

Example (7.5). — Let Λ be the set of all positive integers, and for each n ∈ Λ, set M_{n} := {r/n | r ∈ ℤ} ⊂ ℚ. Then ⋃ M_{n} = ℚ and M_{m}, M_{n} ⊂ M_{mn}. Then (7.2) yields ℚ = \underrightarrow{lim} M_{n} where Λ is ordered by inclusion of the M_{n}.

However, M_{m} ⊂ M_{n} if and only if 1/m = s/n for some s, if and only if m | n. Thus we may view Λ as ordered by divisibility of the n ∈ Λ.

For each n ∈ Λ, set R_{n} := ℤ, and define β_{n}: R_{n} → M_{n} by β_{n}(r) := r/n. Clearly, β_{n} is a ℤ-module isomorphism. And if n = ms, then this diagram is commutative:

\begin{matrix}R_{m} ⟶^{μ_{s}} R_{n} \\ ^{β_{m}}↓^{≃} & ^{β_{n}}↓^{≃} \\ M_{m} ⟶^{ι_{n}^{m}} M_{n} \end{matrix}  (7.5.1)

where ι_{n}^{m} is the inclusion. Hence ℚ = \underrightarrow{lim} R_{n} where the transition maps are the multiplication maps μ_{s}.

Exercise (7.6). — Keep the setup of (7.5). For each n ∈ Λ, set N_{n} := ℤ/⟨n⟩; if n = ms, define α_{n}^{m}: N_{m} → N_{n} by α_{n}^{m}(x) := xs (mod n). Show \underrightarrow{lim} N_{n} = ℚ/ℤ.

Proposition (7.7). — Let Λ be a filtered category, R a ring, and 𝒞 either ((Sets)) or ((R-mod)) or ((R-alg)). Let λ ↦ M_{λ} be a functor from Λ to 𝒞. Define a relation ∼ on the disjoint union ⨆ M_{λ} as follows: m_{1} ∼ m_{2} for m_{i} ∈ M_{λ_{i}} if there are transition maps α_{μ}^{λ_{i}}: M_{λ_{i}} → M_{μ} such that α_{μ}^{λ_{1}}m_{1} = α_{μ}^{λ_{2}}m_{2}. Then ∼ is an equivalence relation. Set M := (⨆ M_{λ})/∼. Then M = \underrightarrow{lim} M_{λ}, and for each μ, the canonical map α_{μ}: M_{μ} → M is equal to the insertion map M_{μ} → \underrightarrow{lim} M_{λ}.

Proof: Clearly ∼ is reflexive and symmetric. Let’s show it is transitive. Given m_{i} ∈ M_{λ_{i}} for i = 1, 2, 3 with m_{1} ∼ m_{2} and m_{2} ∼ m_{3}, there are α_{μ}^{λ_{i}} for i = 1, 2 and α_{ν}^{λ_{i}} for i = 2, 3 with α_{μ}^{λ_{1}}m_{1} = α_{μ}^{λ_{2}}m_{2} and α_{ν}^{λ_{2}}m_{2} = α_{ν}^{λ_{3}}m_{3}. Then (7.1)(1) yields α_{ρ}^{μ} and α_{ρ}^{ν}. Possibly, α_{ρ}^{μ}α_{μ}^{λ_{2}} ≠ α_{ρ}^{ν}α_{ν}^{λ_{2}}, but in any case, (7.1)(2) yields α_{σ}^{ρ} with α_{σ}^{ρ}(α_{ρ}^{μ}α_{μ}^{λ_{2}}) = α_{σ}^{ρ}(α_{ρ}^{ν}α_{ν}^{λ_{2}}). Hence, (α_{σ}^{ρ}α_{ρ}^{μ})α_{μ}^{λ_{1}}m_{1} = (α_{σ}^{ρ}α_{ρ}^{ν})α_{ν}^{λ_{3}}m_{3}. Thus m_{1} ∼ m_{3}.

If 𝒞 = ((R-mod)), define addition in M as follows. Given m_{i} ∈ M_{λ_{i}} for i = 1, 2, there are α_{μ}^{λ_{i}} by (7.1)(1). Set

α_{λ_{1}}m_{1} + α_{λ_{2}}m_{2} := α_{μ}(α_{μ}^{λ_{1}}m_{1} + α_{μ}^{λ_{2}}m_{2}).

We must check that this addition is well defined.

First, consider μ. Suppose there are α_{ν}^{λ_{i}} too. Then (7.1)(1) yields α_{ρ}^{μ} and α_{ρ}^{ν}. Possibly, α_{ρ}^{μ}α_{μ}^{λ_{i}} ≠ α_{ρ}^{ν}α_{ν}^{λ_{i}}, but (7.1)(2) yields α_{σ}^{ρ} with α_{σ}^{ρ}(α_{ρ}^{μ}α_{μ}^{λ_{1}}) = α_{σ}^{ρ}(α_{ρ}^{ν}α_{ν}^{λ_{1}}) and then α_{τ}^{σ} with α_{τ}^{σ}(α_{σ}^{ρ}α_{ρ}^{μ}α_{μ}^{λ_{2}}) = α_{τ}^{σ}(α_{σ}^{ρ}α_{ρ}^{ν}α_{ν}^{λ_{2}}). Therefore,

(α_{τ}^{σ}α_{σ}^{ρ}α_{ρ}^{μ})(α_{μ}^{λ_{1}}m_{1} + α_{μ}^{λ_{2}}m_{2}) = (α_{τ}^{σ}α_{σ}^{ρ}α_{ρ}^{ν})(α_{ν}^{λ_{1}}m_{1} + α_{ν}^{λ_{2}}m_{2}).

Thus both μ and ν yield the same value for α_{λ_{1}}m_{1} + α_{λ_{2}}m_{2}.

Second, suppose m_{1} ∼ m′_{1} ∈ M_{λ′_{1}}. Then a similar, but easier, argument yields α_{λ_{1}}m_{1} + α_{λ_{2}}m_{2} = α_{λ′_{1}}m′_{1} + α_{λ_{2}}m_{2}. Thus addition is well defined on M.

Define scalar multiplication on M similarly. Then clearly M is an R-module. If 𝒞 = ((R-alg)), then we can see similarly that M is canonically an R-algebra. Finally, let β_{λ}: M_{λ} → N be maps with β_{λ}α_{λ}^{κ} = β_{κ} for all α_{λ}^{κ}. The β_{λ} induce

a map ⨆ M_{λ} → N. Suppose m_{1} ∼ m_{2} for m_{i} ∈ M_{λ_{i}}; that is, α_{μ}^{λ_{1}}m_{1} = α_{μ}^{λ_{2}}m_{2} for some α_{μ}^{λ_{i}}. Then β_{λ_{1}}m_{1} = β_{λ_{2}}m_{2} as β_{μ}α_{μ}^{λ_{i}} = β_{λ_{i}}. So there is a unique map β: M → N with βα_{λ} = β_{λ} for all λ. Further, if 𝒞 = ((R-mod)) or 𝒞 = ((R-alg)), then clearly β is a homomorphism. The proof is now complete. □

Corollary (7.8). — Preserve the conditions of (7.7).

(1) Given m ∈ \underrightarrow{lim} M_{λ}, for some λ, there is m_{λ} ∈ M_{λ} such that m = α_{λ}m_{λ}.

(2) Given m_{i} ∈ M_{λ_{i}} for i = 1, 2 such that α_{λ_{1}}m_{1} = α_{λ_{2}}m_{2}, there are α_{μ}^{λ_{i}} such that α_{μ}^{λ_{1}}m_{1} = α_{μ}^{λ_{2}}m_{2}.

(3) Suppose 𝒞 = ((R−mod)) or 𝒞 = ((R−alg)). Then given m_{λ} ∈ M_{λ} such that α_{λ}m_{λ} = 0, there is α_{μ}^{λ} such that α_{μ}^{λ}m_{λ} = 0.

Proof: The assertions follow immediately from (7.7). □

Exercise (7.9). — Let R be a filtered direct limit of rings R_{λ}. Show R = 0 if and only if R_{λ} = 0 for some λ. Show R is a domain if R_{λ} is a domain for every λ.

Theorem (7.10) (Exactness of filtered direct limits). — Let R be a ring, Λ a filtered category. Let 𝒞 be the category of 3-term exact sequences of R-modules: its

September 3, 2012 11Nts.tex objects are the 3-term exact sequences, and its maps are the commutative diagrams

L ⟶ M ⟶ N
↓  ↓  ↓
L′ ⟶ M′ ⟶ N′

Then, for any functor λ ↦ (L_{λ} ⟶^{β_{λ}} M_{λ} ⟶^{γ_{λ}} N_{λ}) from Λ to 𝒞, the induced sequence \underrightarrow{lim} L_{λ} ⟶^{β} \underrightarrow{lim} M_{λ} ⟶^{γ} \underrightarrow{lim} N_{λ} is exact.

Proof: Abusing notation, in all three cases, denote by α_{λ}^{κ} the transition maps and by α_{λ} the insertions. Then given ℓ ∈ \underrightarrow{lim} L_{λ}, there is ℓ_{λ} ∈ L_{λ} with α_{λ}ℓ_{λ} = ℓ by (7.8)(1). By hypothesis, γ_{λ}β_{λ}ℓ_{λ} = 0; so γβℓ = 0. Thus Im(β) ⊂ Ker(γ).

For the opposite inclusion, take m ∈ \underrightarrow{lim} M_{λ} with γm = 0. By (7.8)(1), there is m_{λ} ∈ M_{λ} with α_{λ}m_{λ} = m. Now, α_{λ}γ_{λ}m_{λ} = 0 by commutativity. So by (7.8)(3), there is α_{μ}^{λ} with α_{μ}^{λ}γ_{λ}m_{λ} = 0. So γ_{μ}α_{μ}^{λ}m_{λ} = 0 by commutativity. Hence there is ℓ_{μ} ∈ L_{μ} with β_{μ}ℓ_{μ} = α_{μ}^{λ}m_{λ} by exactness. Apply α_{μ} to get

βα_{μ}ℓ_{μ} = α_{μ}β_{μ}ℓ_{μ} = α_{μ}α_{μ}^{λ}m_{λ} = m.

Thus Ker(γ) ⊂ Im(β). So Ker(γ) = Im(β) as asserted. □

Exercise (7.11). — Let M := \underrightarrow{lim} M_{λ} be a filtered direct limit of modules, and N ⊂ M a submodule. For each λ, let α_{λ}: M_{λ} → M be the insertion, and set N_{λ} := α_{λ}^{−1}N ⊂ M_{λ}. Prove that N = \underrightarrow{lim} N_{λ}.

Proposition (7.12). — Let Λ a filtered category, R a ring, λ ↦ M_{λ} a functor from Λ to ((R-mod)), and N an R-module. Consider the canonical homomorphism

θ(N): \underrightarrow{lim} Hom(N, M_{λ}) → Hom(N, \underrightarrow{lim} M_{λ}),

which is induced by the insertions M_{λ} → \underrightarrow{lim} M_{λ}. Then θ(N) is injective if N is finitely generated; further, θ(N) is bijective if N is finitely presented.

Proof: If N := R, then θ(N) is bijective by (4.3). Assume N is finitely generated, and take a presentation R^{⊕Σ} → R^{n} → N → 0 with Σ finite if N is finitely presented. It induces the following commutative diagram:

0 ⟶ \underrightarrow{lim} Hom(N, M_{λ}) ⟶ \underrightarrow{lim} Hom(R^{n}, M_{λ}) ⟶ \underrightarrow{lim} Hom(R^{⊕Σ}, M_{λ})
θ(N)↓  ^{θ(R^{n})}↓^{≃}  ^{θ(R^{⊕Σ})}↓
0 ⟶ Hom(N, \underrightarrow{lim} M_{λ}) ⟶ Hom(R^{n}, \underrightarrow{lim} M_{λ}) ⟶ Hom(R^{⊕Σ}, \underrightarrow{lim} M_{λ})

The rows are exact owing to (5.17), the left exactness of Hom, and to (7.10), the exactness of filtered direct limits. Now, Hom preserves finite direct sums by (4.13), and direct limit does so by (6.15) and (6.7); hence, θ(R^{n}) is bijective, and θ(R^{⊕Σ}) is bijective if Σ is finite. A diagram chase yields the assertion. □

Exercise (7.13). — Let Λ and Λ′ be small categories, C: Λ′ → Λ a functor. Assume Λ′ is filtered. Assume C is cofinal; that is,

(1) given λ ∈ Λ, there is a map λ → Cλ′ for some λ′ ∈ Λ′, and

(2) given ψ, φ: λ ⇉ Cλ′, there is χ: λ′ → λ′_{1} with (Cχ)ψ = (Cχ)φ.

September 3, 2012 11Nts.tex 36 7. Filtered Direct Limits Let λ ↦ M_{λ} be a functor from Λ to 𝒞 whose direct limit exists. Show that

\underrightarrow{lim}_{λ′∈Λ′} M_{Cλ′} = \underrightarrow{lim}_{λ∈Λ} M_{λ};

more precisely, show that the right side has the UMP characterizing the left. Exercise (7.14). — Show that every R-module M is the filtered direct limit over a directed set of finitely presented modules.

8. Tensor Products

Given two modules, their tensor product is the target of the universal bilinear map. We construct the product, and establish various properties: bifunctoriality, commutativity, associativity, cancellation, and most importantly, adjoint associativity; the latter relates the product to the module of homomorphisms. With one factor fixed, the product becomes a linear functor. We prove Watt’s Theorem; it characterizes “tensor-product” functors as those linear functors that commute with direct sums and cokernels. Lastly, we discuss the tensor product of algebras.

(8.1) (Bilinear maps). — Let R a ring, and M, N, P modules. We call a map

α: M × N → P

bilinear if it is linear in each variable; that is, given m ∈ M and n ∈ N, the maps

m′ ↦ α(m′, n) and n′ ↦ α(m, n′)

are R-linear. Denote the set of all these maps by Bil_{R}(M, N; P). It is clearly an R-module, with sum and scalar multiplication performed valuewise.

(8.2) (Tensor product). — Let R be a ring, and M, N modules. Their tensor product, denoted M ⊗_{R} N or simply M ⊗N, is constructed as the quotient of the free module R^{⊕(M×N)} modulo the submodule generated by the following elements, where (m, n) stands for the standard basis element e_{(m,n)}:

\begin{matrix}(m + m′, n) − (m, n) − (m′, n) & and (m, n + n′) − (m, n) − (m, n′), \\ (xm, n) − x(m, n) & and (m, xn) − x(m, n) \end{matrix} (8.2.1)

for all m, m′ ∈ M and n, n′ ∈ N and x ∈ R.

Note that M ⊗ N is the target of the canonical map with source M × N

β: M × N → M ⊗ N,

which sends each (m, n) to its residue class m ⊗ n. By construction, β is bilinear.

Theorem (8.3) (UMP of tensor product). — Let R be a ring, M, N modules. Then β: M × N → M ⊗ N is the universal example of a bilinear map with source M × N; in fact, β induces, not simply a bijection, but a module isomorphism,

θ: Hom_{R}(M ⊗_{R} N, P) ⟶^{∼} Bil_{R}(M, N; P).  (8.3.1)

Proof: Note that, if we follow any bilinear map with any linear map, then the result is bilinear; hence, θ is well defined. Clearly, θ is a module homomorphism. Further, θ is injective since M ⊗_{R} N is generated by the image of β. Finally, given any bilinear map α: M ×N → P, by (4.10) it extends to a map α′: R^{⊕(M×N)} → P, and α′ carries all the elements in (8.2.1) to 0; hence, α′ factors through β. Thus θ is also surjective, so an isomorphism, as asserted. □

(8.4) (Bifunctoriality). — Let R be a ring, α: M → M′ and α′: N → N′ module homomorphisms. Then there is a canonical commutative diagram:

M × N ⟶^{α×α′} M′ × N′

↓^{β}  ↓^{β′}

M ⊗ N ⟶^{α⊗α′} M′ ⊗ N′

Indeed, β′ ∘ (α × α′) is clearly bilinear; so the UMP (8.3) yields α ⊗ α′. Thus • ⊗ N and M ⊗ • are commuting linear functors, that is, linear on maps (see (9.2)).

Proposition (8.5). — Let R be a ring, M and N modules.

(1) Then the switch map M × N → N × M induces an isomorphism

M ⊗_{R} N = N ⊗_{R} M.  (commutative law)

(2) Then multiplication of R on M induces an isomorphism

R ⊗_{R} M = M.  (unitary law)

Proof: The switch map induces an isomorphism R^{⊕(M×N)} ⟶^{∼} R^{⊕(N×M)}, and it preserves the elements of (8.2.1). Thus (1) holds.

Define β: R × M → M by β(x, m) := xm. Clearly β is bilinear. Let’s check β has the requisite UMP. Given a bilinear map α: R × M → P, define γ: M → P by γ(m) := α(1, m). Then γ is linear as α is bilinear. Also, α = γβ as

α(x, m) = xα(1, m) = α(1, xm) = γ(xm) = γβ(x, m).

Further, γ is unique as β is surjective. Thus the UMP holds, so (2) does too. □

Exercise (8.6). — Let R be a domain, 𝔞 a nonzero ideal. Set K := Frac(R). Show that 𝔞 ⊗_{R} K = K.

(8.7) (Bimodules). — Let R and R′ be rings. An abelian group N is an (R, R′)- bimodule if it is both an R-module and an R′-module and if x(x′n) = x′(xn) for all x ∈ R, all x′ ∈ R′, and all n ∈ N. At times, we think of N as a left Rmodule, with multiplication xn, and as a right R′-module, with multiplication nx′. Then the compatibility condition becomes the associative law: x(nx′) = (xn)x′. A (R, R′)-homomorphism of bimodules is a map that is both R-linear and R′-linear.

Let M be an R-module, and let N be an (R, R′)-bimodule. Then M ⊗_{R} N is an (R, R′)-bimodule with R-structure as usual and with R′-structure defined by x′(m ⊗ n) := m ⊗ (x′n) for all x′ ∈ R′, all m ∈ M, and all n ∈ N. The latter multiplication is well defined and the two multiplications commute because of bifunctoriality (8.4) with α := μ_{x} and α′ := μ_{x′}.

For instance, suppose R′ is an R-algebra. Then R′ is an (R, R′)-bimodule. So M ⊗_{R} R′ is an R′-module. It is said to be obtained by extension of scalars.

Exercise (8.8). — Let R be a ring, R′ an R-algebra, M, N two R′-modules. Show there is a canonical R-linear map τ: M ⊗_{R} N → M ⊗_{R′} N.

Let K ⊂ M ⊗_{R} N denote the R-submodule generated by all the differences (x′m) ⊗ n − m ⊗ (x′n) for x′ ∈ R′ and m ∈ M and n ∈ N. Show K = Ker(τ). Show τ is surjective, and is an isomorphism if R′ is a quotient of R.

Theorem (8.9). — Let R and R′ be rings, M an R-module, P an R′-module, N an (R, R′)-bimodule, . Then there are two canonical (R, R′)-isomorphisms:

M ⊗_{R} (N ⊗_{R′} P) = (M ⊗_{R} N) ⊗_{R′} P,  (associative law)
Hom_{R′}(M ⊗_{R} N, P) = Hom_{R}(M, Hom_{R′}(N, P)). (adjoint associativity)

Proof: Note that M ⊗_{R} (N ⊗_{R′} P) and (M ⊗_{R} N)⊗_{R′} P are (R, R′)-bimodules. For each (R, R′)-bimodule Q, call a map τ: M × N × P → Q trilinear if it is R-bilinear in M × N and R′-bilinear in N × P. Denote the set of all these τ by Tril(M, N, P; Q). It is, clearly, an (R, R′)-bimodule.

September 3, 2012 11Nts.tex

A trilinear map τ yields an R-bilinear map M × (N ⊗_{R′} P) → Q, whence a map M ⊗_{R} (N ⊗_{R′} P) → Q, which is both R-linear and R′-linear, and vice versa. Thus

Tril_{(R,R′)}(M, N, P; Q) = Hom(M ⊗_{R} (N ⊗_{R′} P), Q).

Similarly, there is a canonical isomorphism of (R, R′)-bimodules

Tril_{(R,R′)}(M, N, P; Q) = Hom((M ⊗_{R} N) ⊗_{R′} P, Q).

Hence both M ⊗_{R} (N ⊗_{R′} P) and (M ⊗_{R} N) ⊗_{R′} P are universal examples of a target of a trilinear map with source M × N × P. Thus they are equal, as asserted.

To establish the isomorphism of adjoint associativity, define a map

α: Hom_{R′}(M ⊗_{R} N, P) → Hom_{R}(M, Hom_{R′}(N, P)) by
(α(γ)(m))(n) := γ(m ⊗ n).

Let’s check α is well defined. First, α(γ)(m) is R′-linear, because given x′ ∈ R′,

γ(m ⊗ (x′n)) = γ(x′(m ⊗ n)) = x′γ(m ⊗ n)

since γ is R′-linear. Further, α(γ) is R-linear, because given x ∈ R,

(xm) ⊗ n = m ⊗ (xn) and so (α(γ)(xm))(n) = (α(γ)(m))(xn).

Thus α(γ) ∈ Hom_{R}(M, Hom_{R′}(N, P)). Clearly, α is an (R, R′)-homomorphism.

To obtain an inverse to α, given η ∈ Hom_{R}(M, Hom_{R′}(N, P)), define a map ζ: M × N → P by ζ(m, n) := (η(m))(n). Clearly, ζ is ℤ-bilinear, so ζ induces a ℤ-linear map δ: M ⊗_{ℤ} N → P. Given x ∈ R, clearly (η(xm))(n) = (η(m))(xn); so δ((xm) ⊗ n) = δ(m ⊗ (xn)). Hence, δ induces a ℤ-linear map β(η): M ⊗_{R} N → P owing to (8.8) with ℤ for R and with R for R′. Clearly, β(η) is R′-linear as η(m) is so. Finally, it is easy to verify that α(β(η)) = η and β(α(γ)) = γ, as desired. □

Corollary (8.10). — Let R and R′ be rings, M an R-module, P an R′-module. If R′ is an R-algebra, then there are two canonical (R, R′)-isomorphisms:

(M ⊗_{R} R′) ⊗_{R′} P = M ⊗_{R} P,  (cancellation law)
Hom_{R′}(M ⊗_{R} R′, P) = Hom_{R}(M, P).  (left adjoint)

Instead, if R is an R′-algebra, then there is another canonical (R, R′)-isomorphism:

Hom_{R′}(M, P) = Hom_{R}(M, Hom_{R′}(R, P)).  (right adjoint)

In other words, • ⊗_{R} R′ is the left adjoint of restriction of scalars from R′ to R, and Hom_{R′}(R, •) is the right adjoint of restriction of scalars from R to R′.

Proof: The cancellation law results from the associative and unitary laws; the adjoint isomorphisms, from adjoint associativity, (4.3) and the unitary law. □

Corollary (8.11). — Let R, R′ be rings, N a bimodule. Then the functor •⊗_{R}N preserves direct limits, or equivalently, direct sums and cokernels.

Proof: By adjoint associativity, •⊗_{R}N is the left adjoint of Hom_{R′}(N, •). Thus the assertion results from (6.12) and from (6.7) and (6.8). □

Example (8.12). — Tensor product does not preserve kernels, nor even injections. Indeed, consider the injection μ_{2} : ℤ → ℤ. Tensor it with N := ℤ/⟨2⟩, obtaining μ_{2} : N → N. This map is zero, but not injective as N ≠ 0.

Exercise (8.13). — Let R be a ring, 𝔞 and 𝔟 ideals, and M a module. (1) Use (8.11) to show that (R/𝔞) ⊗ M = M/𝔞M.

(2) Use (1) to show that (R/𝔞) ⊗ (R/𝔟) = R/(𝔞 + 𝔟).

Exercise (8.14). — Let k be a field, M and N nonzero vector spaces. Prove that M ⊗ N ≠ 0.

Theorem (8.15) (Watts). — Let F: ((R-mod)) → ((R-mod)) be a linear functor. Then there is a natural transformation θ(•): • ⊗F (R) → F (•) with θ(R) = 1, and θ(•) is an isomorphism if and only if F preserves direct sums and cokernels.

Proof: As F is a linear functor, there is, by definition, a natural R-linear map θ(M): Hom(R, M) → Hom(F (R), F (M)). But Hom(R, M) = M by (4.3). Set N := F (R). Then, with P := F (M), adjoint associativity yields the desired map

θ(M) ∈ Hom(M, Hom(N, F (M))) = Hom(M ⊗ N, F (M)).

Explicitly, θ(M)(m ⊗ n) = F (ρ)(n) where ρ: R → M is defined by ρ(1) = m. Alternatively, this formula can be used to construct θ(M), as (m, n) ↦ F (ρ)(n) is clearly bilinear. Either way, it is not hard to see that θ(M) is natural in M.

If θ(•) is an isomorphism, then F preserves direct sums and cokernels by (8.11). To prove the converse, take a presentation R^{⊕Λ} ⟶^{β} R^{⊕Σ} ⟶^{α} M → 0; one exists

by (5.19). Applying θ, we get this commutative diagram:

\begin{matrix}R^{⊕Λ} ⊗ N ⟶ R^{⊕Σ} ⊗ N ⟶ M ⊗ N ⟶ 0 \\ ↓^{θ(R^{⊕Λ})} & ↓^{θ(R^{⊕Σ})} & ↓^{θ(M)} \\ F (R^{⊕Λ}) ⟶ F (R^{⊕Σ}) ⟶ F (M) ⟶ 0 \end{matrix}  (8.15.1)

By construction, θ(R) = 1_{N}. If F preserves direct sums, then θ(R^{⊕Λ}) = 1_{N^{⊕Λ}} and θ(R^{⊕Σ}) = 1_{N^{⊕Σ}}; in fact, given any natural transformation θ: T → U, let’s show that, if T and U preserve direct sums, then so does θ.

Given a collection of modules M_{λ}, each inclusion ι_{λ}: M_{λ} → ⨁ M_{λ} yields, because of naturality, the following commutative diagram:

T (M_{λ}) ⟶^{T (ι_{λ})} ⨁ T (M_{λ})

↓^{θ(M_{λ})}  ↓^{θ(⨁ M_{λ})}

U(M_{λ}) ⟶^{U(ι_{λ})} ⨁ U(M_{λ})

Hence θ(⨁ M_{λ})T (ι_{λ}) = ⨁ θ(M_{λ})T (ι_{λ}). But the UMP of direct sum says that, given any N, a map ⨁ T (M_{λ}) → N is determined by its compositions with the inclusions T (ι_{λ}). Thus θ(⨁ M_{λ}) = ⨁ θ(M_{λ}), as desired.

Suppose F preserves cokernels. Since • ⊗ N does too, the rows of (8.15.1) are exact by (5.2). Therefore, θ(M) is an isomorphism. □

Exercise (8.16). — Let F: ((R-mod)) → ((R-mod)) be a linear functor. Show that F always preserves finite direct sums. Show that θ(M): M ⊗ F (R) → F (M) is surjective if F preserves surjections and M is finitely generated, and that θ(M) is an isomorphism if F preserves cokernels and M is finitely presented.

(8.17) (Additive functors). — Let R be a ring, M a module, and form the diagram

M ⟶^{δ_{M}} M ⊕ M ⟶^{σ_{M}} M

where δ_{M} := (1_{M}, 1_{M}) and σ_{M} := 1_{M} + 1_{M}.

September 3, 2012 11Nts.tex Let α, β: M → N be two maps of modules. Then

σ_{N}(α ⊕ β)δ_{M} = α + β,  (8.17.1)

because, for any m ∈ M, we have

(σ_{N}(α ⊕ β)δ_{M})(m) = σ_{N}(α ⊕ β)(m, m) = σ_{N}(α(m), β(m)) = α(m) + β(m).

Let F: ((R-mod)) → ((R-mod)) be a functor that preserves finite direct sums. Then F (α ⊕ β) = F (α) ⊕ F (β). Also, F (δ_{M}) = δ_{F (M)} and F (σ_{M}) = σ_{F (M)} as F (1_{M}) = 1_{F (M)}. Hence F (α + β) = F (α) + F (β) by (8.17.1). Thus F is additive, that is, ℤ-linear.

Conversely, every additive functor preserves finite direct sums owing to (8.16). However, not every additive functor is R-linear. For example, take R := ℂ.

Define F (M) to be M, but with the scalar product of x ∈ ℂ and m ∈ M to be \overline{x}m where \overline{x} is the conjugate. Define F (α) to be α. Then F is additive, but not linear.

Lemma (8.18) (Equational Criterion for Vanishing). — Let R be a ring, M and N modules, and {n_{λ}}_{λ∈Λ} a set of generators of N. Then any element of M ⊗ N can be written as a finite sum ∑ m_{λ} ⊗ n_{λ} with m_{λ} ∈ M. Further, ∑ m_{λ} ⊗ n_{λ} = 0 if and only if there are m_{σ} ∈ M and x_{λσ} ∈ R for σ ∈ Σ for some Σ such that

∑_{σ} x_{λσ}m_{σ} = m_{λ} for all λ and ∑_{λ} x_{λσ}n_{λ} = 0 for all σ.

Proof: By (8.2), M ⊗ N is generated by elements of the form m ⊗ n with m ∈ M and n ∈ N, and if n = ∑ x_{λ}n_{λ} with x_{λ} ∈ R, then m ⊗ n = ∑(x_{λ}m) ⊗ n_{λ}. Thus any element of M ⊗ N has the asserted form ∑ m_{λ} ⊗ n_{λ}.

Assume the m_{σ} and the x_{λσ} exist. Then

∑ m_{λ} ⊗ n_{λ} = ∑_{λ} ( ∑_{σ} x_{λσ}m_{σ}) ⊗ n_{λ} = ∑_{σ} (m_{σ} ⊗ ∑_{λ} x_{λσ}n_{λ}) = 0.

Conversely, by (5.19), there is a presentation R^{⊕Σ} ⟶^{β} R^{⊕Λ} ⟶^{α} N → 0 with α(e_{λ}) = n_{λ} for all λ where {e_{λ}} is the standard basis of R^{⊕Λ}. Then by (8.11) the following sequence is exact:

M ⊗ R^{⊕Σ} ⟶^{1⊗β} M ⊗ R^{⊕Λ} ⟶^{1⊗α} M ⊗ N → 0.

Further, (1 ⊗ α)(∑ m_{λ} ⊗ e_{λ}) = 0. So the exactness implies there is an element s ∈ M ⊗ R^{⊕Σ} such that (1 ⊗ β)(s) = ∑ m_{λ} ⊗ e_{λ}. Let {e_{σ}} be the standard basis of R^{⊕Σ}, and write s = ∑ m_{σ} ⊗ e_{σ} with m_{σ} ∈ M. Write β(e_{σ}) = ∑_{λ} x_{λσ}e_{λ}. Then clearly 0 = αβ(e_{σ}) = ∑_{λ} x_{λσ}n_{λ}, and

0 = ∑_{λ} m_{λ} ⊗ e_{λ} − ∑_{σ} m_{σ} ⊗ (∑_{λ} x_{λσ}e_{λ}) = ∑_{λ}(m_{λ} − ∑_{σ} x_{λσ}m_{σ}) ⊗ e_{λ}.

Since the e_{λ} are independent, m_{λ} = ∑_{σ} x_{λσ}m_{σ}, as asserted. □

(8.19) (Algebras). — Let R be a ring, S and T algebras with structure maps σ: R → S and τ: R → T. Set U := S ⊗_{R} T; it is an R-module. Now, define S × T × S × T → U by (s, t, s′, t′) ↦ ss′ ⊗ tt′. This map is clearly linear in each factor. So it induces a bilinear map

μ: U × U → U with μ(s ⊗ t, s′ ⊗ t′) = (ss′ ⊗ tt′).

It is easy to check that U is a ring with μ as product. In fact, U is an R-algebra with structure map ω given by ω(r) := σ(r) ⊗ 1 = 1 ⊗ τ(r), called the tensor product of S and T over R.

September 3, 2012 11Nts.tex

Define ι_{S}: S → S ⊗_{R} T by ι_{S}(s) := s ⊗ 1. Clearly ι_{S} is an R-algebra homomorphism. Define ι_{T} : T → S ⊗ T similarly. Given an R-algebra V, define a map

γ: Hom_{((R-alg))}(S ⊗_{R} T, V) → Hom_{((R-alg))}(S, V) × Hom_{((R-alg))}(T, V).

by γ(ψ) := (ψι_{S}, ψι_{T}). Conversely, given R-algebra homomorphisms θ: S → V and ζ: T → V, define η: S × T → V by η(s, t) := θ(s) · ζ(t). Then η is clearly bilinear, so it defines a linear map ψ: S ⊗_{R} T → V. It is easy to see that the map (θ, ζ) ↦ ψ is an inverse to γ. Thus γ is bijective.

In other words, S ⊗_{R} T is the coproduct of S and T in ((R-alg)):

[figure: S; R; S ⊗_{R} T; V; T; ι_{S}; θ; σ; τ; ψ; ι_{T}; ζ]

Example (8.20). — Let R be a ring, S an algebra, and X_{1}, …, X_{n} variables. Then there is a canonical S-algebra isomorphism

S ⊗_{R} R[X_{1}, …, X_{n}] = S[X_{1}, …, X_{n}].

Indeed, given an S-algebra homomorphism S → T and elements x_{1}, …, x_{n} of T, there is an R-algebra homomorphism R[X_{1}, …, X_{n}] → T by (1.3). So by (8.19), there is a unique S-algebra homomorphism S ⊗_{R} R[X_{1}, …, X_{n}] → T. Thus both S ⊗_{R} R[X_{1}, …, X_{n}] → T and S[X_{1}, … X_{n}] possess the same UMP.

In particular, for variables Y_{1}, …, Y_{m}, we obtain

R[X_{1}, …, X_{n}] ⊗_{R} R[Y_{1}, …, Y_{m}] = R[X_{1}, …, X_{n}, Y_{1}, …, Y_{m}].

Exercise (8.21). — Let X be a variable, ω a complex cubic root of 1, and \sqrt[3]{2} the real cube root of 2. Set k := ℚ(ω) and K := k[\sqrt[3]{2}]. Show K = k[X]/⟨X^{3} − 2⟩ and then K ⊗_{k} K = K × K × K.

9. Flatness

A module is called flat if tensor product with it is an exact functor. First, we study exact functors in general. Then we prove various properties of flat modules. Notably, we prove Lazard’s Theorem, which characterizes the flat modules as the filtered direct limits of free modules of finite rank. Lazard’s Theorem yields the Ideal Criterion for Flatness, which characterizes the flat modules as those whose tensor product with any finitely generated ideal is equal to the ordinary product.

Lemma (9.1). — Let R be a ring, α: M → N a homomorphism of modules. Then there is a diagram with two short exact sequences involving N′

[figure: 0; M′; M; N; P; 0; 0; N′; 0; α; α′; α′′]  (9.1.1)

if and only if M′ = Ker(α) and N′ = Im(α) and P = Coker(α).

Proof: The equations yield the diagram since Coim(α) ⟶^{∼} Im(α) by (4.9). Conversely, given the diagram, note that Ker(α) = Ker(α′) since α′′ is injective.

So M′ = Ker(α). So N′ = Coim(α) since α′ is surjective. Hence N′ = Im(α). Therefore, P = Coker(α). Thus the equations hold. □

(9.2) (Exact Functors). — Let R be a ring, R′ an algebra, F a functor from ((R-mod)) to ((R′-mod)). Assume F is R-linear; that is, the associated map

Hom_{R}(M, N) → Hom_{R′}(F M, F N)

is R-linear. Then, if a map α: M → N is 0, so is F α: F M → F N. But M = 0 if and only if 1_{M} = 0. Further, F (1_{M}) = 1_{F M}. Thus if M = 0, then F M = 0.

We call F exact if it preserves exact sequences. For example, Hom(P, •) is exact if and only if P is projective by (5.22).

We call F left exact if it preserves kernels. When F is contravariant, we call F left exact if it takes cokernels to kernels. For example, Hom(N, •) and Hom(•, N) are left exact covariant and contravariant functors.

We call F right exact if it preserves cokernels. For example, M ⊗ • is right exact.

Proposition (9.3). — Let R be a ring, R′ an algebra, F an R-linear functor. Then the following conditions are equivalent:

(1) F preserves exact sequences; that is, F is exact.

(2) F preserves short exact sequences.

(3) F preserves kernels and surjections.

(4) F preserves cokernels and injections.

(5) F preserves kernels and images.

Proof: Trivially, (1) implies (2). In view of (5.2), clearly (1) yields (3) and (4). Assume (3). Let 0 → M′ → M → M′′ → 0 be a short exact sequence. Since F

preserves kernels, 0 → F M′ → F M → F M′′ is exact; since F preserves surjections, F M → F M′′ → 0 is also exact. Thus (2) holds. Similarly, (4) implies (2).

Assume (2). Given α: M → N, form the diagram (9.1.1). Applying F to it and

September 3, 2012 11Nts.tex using (2), we obtain a similar diagram for F (α). Hence (9.1) yields (5).

Finally, assume (5). Let M′ ⟶^{α} M ⟶^{β} M′′ be exact; that is, Ker(β) = Im(α). Now, (5) yields Ker(F (β)) = F (Ker(β)) and Im(F (α)) = F (Im(α)). Therefore, Ker(F (β)) = Im(F (α)). Thus (1) holds. □

(9.4) (Flatness). — An R-module M is said to be flat over R or R-flat if the functor M ⊗_{R} • preserves injections. It is equivalent by (9.3) that M ⊗_{R} • be exact since it is right exact.

An R-algebra R′ and its structure map are said to be flat if R′ is flat as an R-module.

Lemma (9.5). — A direct sum ⨁ M_{λ} is flat if and only if each summand is flat.

Proof: Let β: N′ → N be an injective map. Then (8.11) yields

(⨁ M_{λ}) ⊗ β = ⨁(M_{λ} ⊗ β);

see the end of the proof of (8.15), taking T (M) := M ⊗ N′ and U(M) := M ⊗ N. Now, the map on the right is injective if and only if each summand M_{λ} ⊗ β is injective by (5.4). The assertion follows. □

Proposition (9.6). — A free module is flat; in fact, a projective module is flat.

Proof: The unitary law implies that R is flat over R. Hence a free module is flat by (9.5). Finally, a projective module is a direct summand of a free module by (5.22), and therefore flat by (9.5). □

Exercise (9.7). — Let R be a ring, R′ a flat algebra, and P a flat R′-module. Show that P is a flat R-module.

Exercise (9.8). — Let R be a ring, M a flat module, and R′ an algebra. Show that M ⊗_{R} R′ is a flat R′-module.

Exercise (9.9). — Let R be a ring, 𝔞 an ideal. Assume that R/𝔞 is R-flat. Show that 𝔞 = 𝔞^{2}.

Exercise (9.10). — Let R be a ring, R′ a flat algebra with structure map φ. Then R′ is said to be faithfully flat if for every R-module M, the map M → M ⊗ R′ given by x ↦ x ⊗ 1 is injective. Show that the following conditions are equivalent:

(1) R′ is faithfully flat.

(2) φ^{−1}(𝔞R′) = 𝔞 for every ideal 𝔞 of R.

(3) Spec(R′) → Spec(R) is surjective.

(4) For every maximal ideal 𝔪 of R, the ideal 𝔪R′ ≠ R′.

(5) For any nonzero R-module M, the module M ⊗_{R} R′ ≠ 0.

Exercise (9.11). — Let A and B be local rings, 𝔪 and 𝔫 their maximal ideals. Let φ: A → B be a local homomorphism; that is, φ(𝔪) ⊂ 𝔫. Assume φ is flat. Show that φ is faithfully flat.

Proposition (9.12). — Let R be a ring, 0 → M′ → M → M′′ → 0 an exact sequence of modules. Assume M′′ is flat.

(1) Then 0 → M′ ⊗ N → M ⊗ N → M′′ ⊗ N → 0 is exact for any module N.

(2) Then M is flat if and only if M′ is flat.

Proof: By (5.19), there is an exact sequence 0 → K → R^{⊕Λ} → N → 0. Tensor it with the given sequence to obtain the following commutative diagram:

0 ↓

M′ ⊗ K ⟶ M ⊗ K ⟶ M′′ ⊗ K ⟶ 0

↓  ↓  ↓^{α}

0 ⟶ M′ ⊗ R^{⊕Λ} ⟶^{β} M ⊗ R^{⊕Λ} ⟶ M′′ ⊗ R^{⊕Λ}

↓  ↓

M′ ⊗ N ⟶^{γ} M ⊗ N

↓  ↓

0 0

Here α and β are injective by Definition (9.4), as M′′ and R^{⊕Λ} are flat by hypothesis and by (9.6). So the rows and columns are exact, as tensor product is right exact. Finally, the Snake Lemma, (5.12), implies γ is injective. Thus (1) holds.

To prove (2), take an injection β: N′ → N, and form this commutative diagram:

0 ⟶ M′ ⊗ N′ ⟶ M ⊗ N′ ⟶ M′′ ⊗ N′ ⟶ 0
α′↓  ^{α}↓  _{α′′}↓
0 ⟶ M′ ⊗ N ⟶ M ⊗ N ⟶ M′′ ⊗ N ⟶ 0

Its rows are exact by (1).

Assume M is flat. Then α is injective. Hence α′ is too. Thus M′ is flat.

Conversely, assume M′ is flat. Then α′ is injective. But α′′ is injective as M′′ is flat. Hence α is injective by the Snake lemma. Thus M is flat. Thus (2) holds. □

Proposition (9.13). — A filtered direct limit of flat modules \underrightarrow{lim} M_{λ} is flat.

Proof: Let β: N′ → N be injective. Then M_{λ} ⊗ β is injective for each λ since M_{λ} is flat. So \underrightarrow{lim}(M_{λ} ⊗ β) is injective by the exactness of filtered direct limits, (7.10). So (\underrightarrow{lim} M_{λ}) ⊗ β is injective by (8.11). Thus \underrightarrow{lim} M_{λ} is flat. □

Proposition (9.14). — Let R and R′ be rings, M an R-module, N an (R, R′)- bimodule, and P an R′-module. Then there is a canonical homomorphism

θ: Hom_{R}(M, N) ⊗_{R′} P → Hom_{R}(M, N ⊗_{R′} P).  (9.14.1)

Assume P is flat. If M is finitely generated, then θ is injective; if M is finitely presented, then θ is an isomorphism.

Proof: The map θ exists by Watts’s Theorem, (8.15), with R′ for R, applied to Hom_{R}(M, N ⊗_{R′} •). Explicitly, θ(φ ⊗ p)(m) = φ(m) ⊗ p. Alternatively, this formula can be used to construct θ, as (φ, n) ↦ ψ, where ψ(m) := φ(m) ⊗ p, is clearly bilinear.

Clearly, θ is bijective if M = R. So θ is bijective if M = R^{n} for any n, as Hom_{R}(•, Q) preserves finite direct sums for any Q by (4.13).

Assume that M is finitely generated. Then from (5.19), we obtain a presentation R^{⊕Λ} → R^{n} → M → 0, with Λ finite if P is finitely presented. Since θ is natural, it

September 3, 2012 11Nts.tex yields this commutative diagram:

0 ⟶ Hom_{R}(M, N) ⊗_{R′} P ⟶ Hom_{R}(R^{n}, N) ⊗_{R′} P ⟶ Hom_{R}(R^{⊕Λ}, N) ⊗_{R′} P

θ↓  ^{≃}↓  ↓
0 ⟶ Hom_{R}(M, N ⊗_{R′} P) ⟶ Hom_{R}(R^{n}, N ⊗_{R′} P) ⟶ Hom_{R}(R^{⊕Λ}, N ⊗_{R′} P)

Its rows are exact owing to the left exactness of Hom and to the flatness of P. The right-hand vertical map is bijective if Λ is finite. The assertion follows. □

Exercise (9.15). — Let R be a ring, R′ an algebra, M and N modules. Show that there is a canonical map

σ: Hom_{R}(M, N) ⊗_{R} R′ → Hom_{R′}(M ⊗_{R} R′, N ⊗_{R} R′).

Assume R′ is flat over R. Show that if M is finitely generated, then σ is injective, and that if M is finitely presented, then σ is an isomorphism.

Definition (9.16). — Let R be a ring, M a module. Let Λ_{M} be the category whose objects are the pairs (R^{m}, α) where α: R^{m} → M is a homomorphism, and whose maps (R^{m}, α) → (R^{n}, β) are the homomorphisms φ: R^{m} → R^{n} with βφ = α.

Proposition (9.17). — Let R be a ring, M a module, and (R^{m}, α) ↦ R^{m} the forgetful functor from Λ_{M} to ((R-mod)). Then M = \underrightarrow{lim}_{(R^{m},α)∈Λ_{M}} R^{m}.

Proof: By the UMP, the α: R^{m} → M induce a map ζ: \underrightarrow{lim} R^{m} → M. Let’s show ζ is bijective. First, ζ is surjective, because each x ∈ M is in the image of (R, α_{x}) where α_{x}(r) := rx.

For injectivity, let y ∈ Ker(ζ). By construction, ⨁_{(R^{m},α)} R^{m} → \underrightarrow{lim} R^{m} is surjective; see the proof of (6.10). So y is in the image of some finite sum ⨁_{(R^{m_{i}} , α_{i})} R^{m_{i}}. Set m := ∑ m_{i}. Then ⨁ R^{m_{i}} = R^{m}. Set α := ∑ α_{i}. Then y is the image of some y′ ∈ R^{m} under the insertion ι_{m}: R^{m} → \underrightarrow{lim} R^{m}. But y ∈ Ker(ζ). So α(y′) = 0.

Let θ, φ: R ⇉ R^{m} be the homomorphisms with θ(1) := y′ and φ(1) := 0. They yield maps in Λ_{M}. So, by definition of direct limit, they have the same compositions with the insertion ι_{m}. Hence y = ι_{m}(y′) = 0. Thus ζ is injective, so bijective. □

Theorem (9.18) (Lazard). — Let R be a ring, M a module. Then the following conditions are equivalent:

(1) M is flat.

(2) Given a finitely presented module P, this version of (9.14.1) is surjective:

Hom_{R}(P, R) ⊗_{R} M → Hom_{R}(P, M).

(3) Given a finitely presented module P and a map β: P → M, there exists a factorization β: P ⟶^{γ} R^{n} ⟶^{α} M;

(4) Given an α: R^{m} → M and a k ∈ Ker(α), there exists a factorization α: R^{m} ⟶^{φ} R^{n} → M such that φ(k) = 0.

(5) Given an α: R^{m} → M and k_{1}, …, k_{r} ∈ Ker(α) there exists a factorization

α: R^{m} ⟶^{φ} R^{n} → M such that φ(k_{i}) = 0 for i = 1, …, r.

(6) Given R^{r} ⟶^{ρ} R^{m} ⟶^{α} M such that αρ = 0, there exists a factorization α: R^{m} ⟶^{φ} R^{n} → M such that φρ = 0.

(7) Λ_{M} is filtered.

(8) M is a filtered direct limit of free modules of finite rank.

Proof: Assume (1). Then (9.14) yields (2).

Assume (2). Consider (3). There are γ_{1}, …, γ_{n} ∈ Hom(P, R) and x_{1}, …, x_{n} ∈ M such that β(p) = ∑ γ_{i}(p)x_{i}. Let γ: P → R^{n} be (γ_{1}, …, γ_{n}), and let α: R^{n} → M be given by α(r_{1}, …, r_{n}) = ∑ r_{i}x_{i}. Then β = αγ, as (3) requires.

Assume (3), and consider (4). Set P := R^{m}/Rk, and let κ: R^{m} → P denote the quotient map. Then P is finitely presented, and there is β: P → M such that βκ = α. By (3), there is a factorization β: P ⟶^{γ} R^{n} → M. Set φ := γκ. Then β: R^{m} ⟶^{φ} R^{n} → M is a factorization of β and φ(k) = 0.

Assume (4), and consider (5). Set m_{0} := m and α_{0} = α. Inductively, (4) yields

α_{i−1} : R^{m_{i−1}} ⟶^{φ_{i}} R^{m_{i}} ⟶^{α_{i}} M for i = 1, …, r

such that φ_{i} ⋯ φ_{1}(k_{i}) = 0. Set φ := φ_{r} ⋯ φ_{1} and n := m_{r}. Then (5) holds.

Assume (5), and consider (6). Let e_{1}, …, e_{r} be the standard basis of R^{r}, and set k_{i} := ρ(e_{i}). Then α(k_{i}) = 0. So (5) yields a factorization α: R^{m} ⟶^{φ} R^{n} → M such that φ(k_{i}) = 0. Then φρ = 0, as required by (6).

Assume (6). Given (R^{m_{1}}, α_{1}) and (R^{m_{2}}, α_{2}) in Λ_{M}, set m := m_{1} + m_{2} and α := α_{1} + α_{2}. Then the inclusions R^{m_{i}} → R^{m} induce maps in Λ_{M}. Thus the first condition of (7.1) is satisfied.

Given σ, τ: (R^{r}, ω) ⇉ (R^{m}, α) in Λ_{M}, set ρ := σ − τ. Then αρ = 0. So (6) yields a factorization α: R^{m} ⟶^{φ} R^{n} → M with φρ = 0. Then φ is a map of Λ_{M}, and φσ = φτ. Hence the second condition of (7.1) is satisfied. Thus (7) holds.

If (7) holds, then (8) does too, since M = \underrightarrow{lim}_{(R^{m},α)∈Λ_{M}} R^{m} by (9.17).

Assume (8). Say M = \underrightarrow{lim} M_{λ} with the M_{λ} free. Each M_{λ} is flat by (9.4), and a filtered direct limit of flat modules is flat by (9.13). Thus M is flat □

Exercise (9.19) (Equational Criterion for Flatness). — Prove that the Condition (9.18)(4) can be reformulated as follows: For every relation ∑_{i} x_{i}y_{i} = 0 with x_{i} ∈ R and y_{i} ∈ M, there are x_{ij} ∈ R and y′_{j} ∈ M such that

∑_{j} x_{ij}y′_{j} = y_{i} for all i and ∑_{i} x_{ij}x_{i} = 0 for all j.  (9.19.1)

Lemma (9.20) (Ideal Criterion for Flatness). — A module N is flat if and only if, for every finitely generated ideal 𝔞, the natural map is an isomorphism:

𝔞 ⊗ N ⟶^{∼} 𝔞N.

Proof: In any case, (8.5)(2) implies R⊗N ⟶^{∼} N with a⊗x ↦ ax. If N is flat, then the inclusion 𝔞 ↪ R yields an injection 𝔞⊗N ↪ R⊗N, and so 𝔞⊗N ⟶^{∼} 𝔞N.

To prove the converse, let’s check the criterion (9.19). Given ∑_{i=1}^{n} x_{i}y_{i} = 0 with

x_{i} ∈ R and y_{i} ∈ N, set 𝔞 := ⟨x_{1}, …, x_{n}⟩. If 𝔞 ⊗ N ⟶^{∼} 𝔞N, then ∑_{i} x_{i} ⊗ y_{i} = 0; so

the Equational Criterion for Vanishing (8.18) yields (9.19.1). Thus N is flat. □

Example (9.21). — Let R be a domain, and set K := Frac(R). Then K is flat, but K is not projective unless R = K. Indeed, (8.6) says 𝔞 ⊗_{R} K = K, with a ⊗ x = ax, for any ideal 𝔞 of R. So K is flat by (9.20).

Suppose K is projective. Then K ↪ R^{Λ} for some Λ by (5.22). So there is a nonzero map α: K → R. So there is an x ∈ K with α(x) ≠ 0. Set a := α(x). Take any nonzero b ∈ R. Then ab · α(x/ab) = α(x) = a. Since R is a domain, b · α(x/ab) = 1. Hence b ∈ R^{×}. Thus R is a field. So (2.3) yields R = K.

Exercise (9.22). — Let R be a domain, M a module. Prove that, if M is flat, then M is torsion free; that is, μ_{x}: M → M is injective for all nonzero x ∈ R. Prove that, conversely, if R is a PID and M is torsion free, then M is flat.

10. Cayley–Hamilton Theorem

The Cayley–Hamilton Theorem says that a matrix satisfies its own characteristic polynomial. We prove an equivalent form, known as the “Determinant Trick.” Using the Trick, we obtain various results, including the uniqueness of the rank of a finitely generated free module. We also obtain Nakayama’s Lemma, and use it to study finitely generated modules further. Then we turn to the important notions of integrality and module finiteness for an algebra. Using the Trick, we relate these notions to each other, and study their properties. We end with a discussion of integral extensions and normal rings.

(10.1) (Cayley–Hamilton Theorem). — Let R be a ring, and M := (a_{ij}) an n × n matrix with a_{ij} ∈ R. Let I_{n} be the n × n identity matrix, and T a variable. The characteristic polynomial of M is the following polynomial:

p_{M}(T) := T^{n} + a_{1}T^{n−1} + ⋯ + a_{n} := det(T I_{n} − M).

Let 𝔞 be an ideal. If a_{ij} ∈ 𝔞 for all i, j, then clearly a_{k} ∈ 𝔞^{k} for all k.

The Cayley–Hamilton Theorem asserts that, in the ring of matrices,

p_{M}(M) = 0.

It is a special case of (10.2) below; indeed, take M := R^{n}, take m_{1}, …, m_{n} to be the standard basis, and take φ to be the endomorphism defined by M.

Conversely, given the setup of (10.2), form the surjection α: R^{n} → → M taking the ith standard basis element to m_{i}, and form the map Φ: R^{n} → R^{n} associated to the matrix M. Then φα = αΦ. Hence, given any polynomial p(T), we have p(φ)α = αp(Φ). Hence, if p(Φ) = 0, then p(φ) = 0 as α is surjective. Thus the Cayley–Hamilton Theorem and the Determinant Trick (10.2) are equivalent.

Theorem (10.2) (Determinant Trick). — Let M be an R-module generated by m_{1}, …, m_{n}, and φ: M → M an endomorphism. Say φ(m_{i}) =: ∑_{j=1}^{n} a_{ij}m_{j} with a_{ij} ∈ R, and form the matrix M := (a_{ij}). Then p_{M}(φ) = 0 in End(M).

Proof: Let δ_{ij} be the Kronecker delta function, μ_{a_{ij}} the multiplication map. Let Δ stand for the matrix (δ_{ij}φ − μ_{a_{ij}}) with entries in End(M), and X for the column vector (m_{j}). Then clearly ΔX = 0. Multiply on the left by the matrix of cofactors Γ of Δ: the (i, j)th entry of Γ is (−1)^{i+j} times the determinant of the matrix obtained by deleting the jth row and the ith column of Δ. Then ΓΔX = 0. Now, ΓΔ = det(Δ)I_{n}. Hence det(Δ)m_{j} = 0 for all j. Thus p_{M}(φ) = 0. □

Proposition (10.3). — Let M be a finitely generated module, 𝔞 an ideal. Then M = 𝔞M if and only if there exists a ∈ 𝔞 such that (1 + a)M = 0.

Proof: Assume M = 𝔞M. Say m_{1}, …, m_{n} generate M, and m_{i} = ∑_{j=1}^{n} a_{ij}m_{j} with a_{ij} ∈ 𝔞. Set M := (a_{ij}). Say p_{M}(T) = T^{n} + a_{1}T^{n−1} + ⋯ + a_{n}. Set a := a_{1} + ⋯ + a_{n} ∈ 𝔞. Then (1 + a)M = 0 by (10.2) with φ := 1_{M}.

Conversely, if there exists a ∈ 𝔞 such that (1 + a)M = 0, then m = −am for all m ∈ M. So M ⊂ 𝔞M ⊂ M. Thus M = 𝔞M. □

Corollary (10.4). — Let R be a ring, M a finitely generated module, and φ an endomorphism of M. If φ is surjective, then φ is an isomorphism.

Proof: Let P := R[X] be the polynomial ring in one variable. By the UMP of P, there is an R-algebra homomorphism μ: P → End(M) with μ(X) = φ. So M is an P -module such that p(X)M = p(φ)M for any p(X) ∈ P by (4.4). Set 𝔞 := ⟨X⟩. Since φ is surjective, M = 𝔞M. By (10.3), there is a ∈ 𝔞 with (1 + a)M = 0. Say a = Xr for some polynomial r. Then 1_{M} + φr(φ) = 0. Thus φ is invertible. □

Corollary (10.5). — Let R be a nonzero ring, m and n positive integers. (1) Then any n generators v_{1}, …, v_{n} of the free module R^{n} form a free basis.

(2) If R^{m} ≃ R^{n}, then m = n.

Proof: Form the surjection α: R^{n} → → R^{n} taking the ith standard basis element to v_{i}. Then φ is an isomorphism by (10.4). So the v_{i} form a free basis by (4.10)(3).

To prove (2), say m ≤ n. Then R^{n} has m generators. Add to them n − m zeros. The result is a free basis by (1), so can contain no zeros. Thus n − m = 0. □

Exercise (10.6). — Let R be a ring, 𝔞 an ideal. Assume 𝔞 is finitely generated and idempotent (or 𝔞 = 𝔞^{2}). Prove there is a unique idempotent e with ⟨e⟩ = 𝔞.

Proposition (10.7). — Let R be a ring, 𝔞 an ideal. Then these conditions are equivalent:

(1) R/𝔞 is projective over R.

(2) R/𝔞 is flat over R, and 𝔞 is finitely generated.

(3) 𝔞 is finitely generated and idempotent.

(4) 𝔞 is generated by an idempotent.

(5) 𝔞 is a direct summand of R.

Proof: Suppose (1) holds. Then R/𝔞 is flat by (9.6). Further, the sequence 0 → 𝔞 → R → R/𝔞 → 0 splits by (5.22), and so 𝔞 is principal. Thus (2) holds.

If (2) holds, then (3) holds by (9.9). If (3) holds, then (4) holds by (10.6). If (4) holds, then (5) holds by (1.12). If (5) holds, then (1) holds by by (5.22). □

Exercise (10.8). — Prove the following conditions on a ring R are equivalent: (1) R is absolutely flat; that is, every module is flat.

(2) Every finitely generated ideal is a direct summand of R.

(3) Every finitely generated ideal is idempotent.

(4) Every principal ideal is idempotent.

Exercise (10.9). — Let R be a ring.

(1) Assume R is Boolean. Prove R is absolutely flat.

(2) Assume R is absolutely flat. Prove any quotient ring R′ is absolutely flat.

(3) Assume R is absolutely flat. Prove every nonunit x is a zerodivisor.

(4) Assume R is absolutely flat and local. Prove R is a field.

Lemma (10.10) (Nakayama). — Let R be a ring, 𝔪 ⊂ rad(R) an ideal, M a finitely generated module. Assume M = 𝔪M. Then M = 0.

Proof: By (10.3), there is a ∈ 𝔪 with (1 + a)M = 0. By (3.2), 1 + a is a unit. Thus M = (1 + a)^{−1}(1 + a)M = 0.

Alternatively, suppose M ≠ 0. Say m_{1}, …, m_{n} generate M with n minimal. Then m_{1} = a_{1}m_{1}+⋯+a_{n}m_{n} with a_{i} ∈ 𝔪. By (3.2), we may set x_{i} := (1−a_{1})^{−1}a_{i}. Then m_{1} = x_{2}m_{2} + ⋯ + x_{n}m_{n}, contradicting minimality of n. Thus M = 0. □

Proposition (10.11). — Let R be a ring, 𝔪 ⊂ rad(R) an ideal, N ⊂ M modules. (1) If M/N is finitely generated and if N + 𝔪M = M, then N = M.

(2) Assume M is finitely generated. Then elements m_{1}, …, m_{n} generate M if and only if their images m′_{1}, …, m′_{n} generate M′ := M/𝔪M.

Proof: In (1), the second hypothesis holds if and only if 𝔪(M/N) = M/N. Hence (1) holds by (10.10) applied with M/N for M.

In (2), let N be the submodule generated by m_{1}, …, m_{n}. Since M is finitely generated, so is M/N. Hence N = M if the m′_{i} generate M/𝔪M by (1). The converse is obvious. □

Exercise (10.12). — Let R be a ring, 𝔪 ⊂ rad(R) an ideal. Let α, β: M → N be two maps of finitely generated modules. Assume α is surjective and β(M) ⊂ 𝔪N. Set γ := α + β. Show that γ is an isomorphism.

Exercise (10.13). — Let A be a local ring, 𝔪 the maximal ideal, M a finitely generated A-module, and m_{1}, …, m_{n} ∈ M. Set k := A/𝔪 and M′ := M/𝔪M, and write m′_{i} for the image of m_{i} in M′. Prove that m′_{1}, …, m′_{n} ∈ M′ form a basis of the k-vector space M′ if and only if m_{1}, …, m_{n} form a minimal generating set of M (that is, no proper subset generates M), and prove that every minimal generating set of M has the same number of elements.

Exercise (10.14). — Let A be a local ring, k its residue field, M and N finitely generated modules. (1) Show that M = 0 if and only if M ⊗_{A} k = 0. (2) Show that M ⊗_{A} N ≠ 0 if M ≠ 0 and N ≠ 0.

Proposition (10.15). — Consider these conditions on an R-module P:

(1) P is free and of finite rank;

(2) P is projective and finitely generated;

(3) P is flat and finitely presented.

Then (1) implies (2), and (2) implies (3); all three are equivalent if R is local.

Proof: A free module is always projective by (5.21), and a projective module is always flat by (9.6). Further, each of the three conditions requires P to be finitely generated; so assume it is. Thus (1) implies (2).

Let p_{1}, …, p_{n} ∈ P generate, and let 0 → L → R^{n} → P → 0 be the short exact sequence defined by sending the ith standard basis element to p_{i}. Set F := R^{n}.

Assume P is projective. Then the sequence splits by (5.22). So (5.9) yields a surjection ρ: F → L. Hence L is finitely generated. Thus (2) implies (3).

Assume P is flat and R is local. Denote the residue field of R by k. Then, by (9.12)(1), the sequence 0 → L ⊗ k → F ⊗ k → P ⊗ k → 0 is exact. Now, F ⊗ k = (R ⊗ k)^{n} = k^{n} by (8.11) and the unitary law; so dim_{k} F ⊗ k = n. Finally, rechoose the p_{i} so that n is minimal. Then dim_{k} P ⊗ k = n, because the p_{i} ⊗ 1 form a basis by (10.13). Therefore, dim_{k} L ⊗ k = 0; so L ⊗ k = 0.

Assume P is finitely presented. Then L is finitely generated by (5.24). Hence L = 0 by (10.14)(1). So F = P. Thus (3) implies (1). □

Definition (10.16). — Let R be a ring, R′ an R-algebra. Then R′ is said to be module finite over R if R′ is a finitely generated R-module.

An element x ∈ R′ is said to be integral over R or integrally dependent on

September 3, 2012 11Nts.tex R if there exist a positive integer n and elements a_{i} ∈ R such that

x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0.  (10.16.1)

Such an equation is called an equation of integral dependence of degree n. If every x ∈ R′ is integral over R, then R′ is said to be integral over R.

Exercise (10.17). — Let G be a finite group acting on a domain R, and R′ the ring of invariants. Show every x ∈ R is integral over R′, in fact, over the subring R′′ generated by the elementary symmetric functions in the conjugates gx for g ∈ G.

Proposition (10.18). — Let R be a ring, R′ an R-algebra, n a positive integer, and x ∈ R′. Then the following conditions are equivalent:

(1) x satisfies an equation of integral dependence of degree n;

(2) R[x] is generated as an R-module by 1, x, …, x^{n−1};

(3) x lies in a subalgebra R′′ generated as an R-module by n elements;

(4) there is a faithful R[x]-module M generated over R by n elements.

Proof: Assume (1) holds. Say p(X) is a monic polynomial of degree n with p(x) = 0. For any m, let M_{m} ⊂ R[x] be the R-submodule generated by 1, …, x^{m}. For m ≥ n, clearly x^{m} −x^{m−n}p(x) is in M_{m−1}. But p(x) = 0. So also x^{m} ∈ M_{m−1}. So by induction, M_{m} = M_{n−1}. Hence M_{n−1} = R[x]. Thus (2) holds.

If (2) holds, then trivially (3) holds with R′′ := R[x].

If (3) holds, then (4) holds with M := R′′, as xM = 0 implies x = x · 1 = 0. Assume (4) holds. In (10.2), take φ := μ_{x}. We obtain a monic polynomial p of

degree n with p(x)M = 0. Since M is faithful, p(x) = 0. Thus (1) holds. □

Exercise (10.19). — Let k be a field, P := k[X] the polynomial ring in one variable, f ∈ P. Set R := k[X^{2}] ⊂ P. Using the free basis 1, X of P over R, find an explicit equation of integral dependence of degree 2 on R for f.

Corollary (10.20). — Let R be a ring, P the polynomial ring in one variable, and 𝔞 an ideal of P. Set R′ := P/𝔞, and let x be the image of X in R′. Let n be a positive integer. Then the following conditions are equivalent:

(1) 𝔞 = ⟨p⟩ where p is a monic polynomial of degree n;

(2) 1, x, …, x^{n−1} form a free basis of R′ over R;

(3) R′ is a free R-module of rank n.

Proof: Assume (1) holds. Then p(x) = 0 is an equation of integral dependence of degree n. So 1, x, …, x^{n−1} generate R′ by (1)⇒(2) of (10.18). Suppose

b_{1}x^{n−1} + ⋯ + b_{n} = 0

with the b_{i} ∈ R. Set q(X) := b_{1}X^{n−1} + ⋯ + b_{n}. Then q(x) = 0. So q ∈ 𝔞. Hence q = fp for some f ∈ P. But p is monic of degree n. Hence q = 0. Thus (2) holds.

Trivially, (2) implies (3).

Finally, assume (3) holds. Then (3)⇒(1) of (10.18) yields a monic polynomial p ∈ 𝔞 of degree n. Form the induced homomorphism ψ: P/⟨p⟩ → R′. It is obviously surjective. Since (1) implies (3), the quotient P/⟨p⟩ is free of rank n. So ψ is an isomorphism by (10.4). Hence ⟨p⟩ = 𝔞. Thus (1) holds. □

Lemma (10.21). — Let R be a ring, R′ a module-finite R-algebra, and M a finitely generated R′-module. Then M is a finitely generated R-module.

Proof: Say elements x_{i} generate R′ as a module over R, and say elements m_{j} generate M over R′. Then clearly the products x_{i}m_{j} generate M over R. □

Theorem (10.22) (Tower Law for Integrality). — Let R be a ring, R′ an algebra, and R′′ an R′-algebra. If x ∈ R′′ is integral over R′ and if R′ is integral over R, then x is integral over R.

Proof: Say x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0 with a_{i} ∈ R′. For m = 1, …, n, set R_{m} := R[a_{1}, …, a_{m}] ⊂ R′′. Then R_{m} is module finite over R_{m−1} by (1)⇒(2) of (10.18). So R_{m} is module finite over R by (10.21) and induction on m.

Moreover, x is integral over R_{n}. So R_{n}[x] is module finite over R_{n} by (1)⇒(2) of (10.18). Hence R_{n}[x] is module finite over R by (10.21). So x is integral over R by (3)⇒(1) of (10.18), as desired. □

Theorem (10.23). — Let R be a ring, R′ an R-algebra. Then the following conditions are equivalent:

(1) R′ is finitely generated as an R-algebra and is integral over R;

(2) R′ = R[x_{1}, …, x_{n}] with all x_{i} integral over R;

(3) R′ is module-finite over R.

Proof: Trivially, (1) implies (2).

Assume (2) holds. To prove (3), set R′′ := R[x_{1}] ⊂ R′. Then R′′ is module finite over R by (1)⇒(2) of (10.18). We may assume R′ is module finite over R′′ by induction on n. So (10.21) yields (3).

If (3) holds, then R′ is integral over R by (3)⇒(1) of (10.18); so (1) holds. □

Exercise (10.24). — Let R_{1}, …, R_{n} be R-algebras, integral over R. Show that their product ∏ R_{i} is a integral over R.

Definition (10.25). — Let R be a ring, R′ an algebra. The integral closure or normalization of R in R′ is the subset \overline{R} of elements that are integral over R. If R ⊂ R′ and R = \overline{R}, then R is said to be integrally closed in R′.

If R is a domain, then its integral closure \overline{R} in its fraction field Frac(R) is called simply its normalization, and R is said to be normal if R = \overline{R}.

Exercise (10.26). — For 1 ≤ i ≤ r, let R_{i} be a ring, R′_{i} an extension of R_{i}, and x_{i} ∈ R′_{i}. Set R := ∏ R_{i}, set R′ := ∏ R′_{i}, and set x := (x_{1}, …, x_{r}). Prove

(1) x is integral over R if and only if x_{i} is integral over R_{i} for each i;

(2) R is integrally closed in R′ if and only if each R_{i} is integrally closed in R′_{i}.

Corollary (10.27). — Let R be a ring, R′ an R-algebra, \overline{R} the integral closure of R in R′. Then \overline{R} is an R-algebra, and is integrally closed in R′.

Proof: Take a ∈ R and x, y ∈ \overline{R}. Then the ring R[x, y] is integral over R by (2)⇒(1) of (10.23). So ax and x + y and xy are integral over R. Thus \overline{R} is an R-algebra. Finally, \overline{R} is integrally closed in R′ owing to (10.22). □

Theorem (10.28) (Gauss). — A UFD is normal.

Proof: Let R be the UFD. Given x ∈ Frac(R), say x = r/s with r, s ∈ R relatively prime. Suppose x satisfies (10.16.1). Then

r^{n} = −(a_{1}r^{n−1} + ⋯ + a_{n}s^{n−1})s.

So any prime element dividing s also divides r. Hence s is a unit. Thus x ∈ R. □

Example (10.29). — (1) A polynomial ring in n variables over a field is a UFD, so normal by (10.28).

(2) The ring R := ℤ[\sqrt{5}] is not a UFD, since

(1 + \sqrt{5})(1 − \sqrt{5}) = −4 = −2 · 2,

and 1 + \sqrt{5}, and 1 − \sqrt{5} and 2 are irreducible, but not associates. However, set τ := (1 + \sqrt{5})/2, the “golden ratio.” The ring ℤ[τ] is known to be a PID; see [8, p. 292]. Hence, ℤ[τ] is a UFD, so normal by (10.28); hence, ℤ[τ] contains the normalization \overline{R} of R. On the other hand, τ^{2} − τ − 1 = 0; hence, ℤ[τ] ⊂ \overline{R}. Thus ℤ[τ] = \overline{R}.

(3) Let d ∈ ℤ be square-free. In the field K := ℚ(\sqrt{d}), form R := ℤ + ℤδ where

δ := \begin{cases}(1 + \sqrt{d})/2, & if d ≡ 1 (mod 4); \\ \sqrt{d}, & if not. \end{cases}

Then R is the normalization \overline{ℤ} of ℤ in K; see [1, pp. 412–3].

(4) Let k be a field, k[t] the polynomial ring in one variable. Set R := k[t^{2}, t^{3}]. Then Frac(R) = k(t). Further, t is integral over R as t satisfies X^{2} − t^{2} = 0; hence, k[t] ⊂ \overline{R}. However, k[t] is normal by (1); hence, k[t] ⊃ \overline{R}. Thus k[t] = \overline{R}.

Let k[X, Y] be the polynomial ring in two variables, and φ: k[X, Y] → R the k-algebra homomorphism defined by φ(X) := t^{2} and φ(Y) := t^{3}. Clearly φ is surjective. Set 𝔭 := Ker φ. Since R is a domain, but not a field, 𝔭 is prime by (2.9), but not maximal by (2.17). Clearly 𝔭 ⊃ ⟨Y^{2} − X^{3}⟩. Since Y^{2} − X^{3} is irreducible, (2.26) implies that 𝔭 = ⟨Y^{2} − X^{3}⟩. So k[X, Y]/⟨Y^{2} − X^{3}⟩ ⟶^{∼} R, which provides us with another description of R.

Exercise (10.30). — Let k be a field, X and Y variables. Set

R := k[X, Y]/⟨Y^{2} − X^{2} − X^{3}⟩,

and let x, y ∈ R be the residues of X, Y. Prove that R is a domain, but not a field. Set t := y/x ∈ Frac(R). Prove that k[t] is the integral closure of R in Frac(R).

11. Localization of Rings

Localization generalizes construction of the fraction field of a domain. We localize an arbitrary ring using as denominators the elements of any given multiplicative subset. The result is the universal example of an algebra where all these elements become units. When the multiplicative subset is the complement of a prime ideal, we obtain a local ring. We relate the ideals in the original ring to those in the localized ring. We end by localizing algebras and then varying the set of denominators.

(11.1) (Localization). — Let R be a ring, and S a multiplicative subset. Define a relation on R × S by (x, s) ∼ (y, t) if there is u ∈ S such that xtu = ysu.

This relation is an equivalence relation. Indeed, it is reflexive as 1 ∈ S and is trivially symmetric. As to transitivity, let (y, t) ∼ (z, r). Say yrv = ztv with v ∈ S. Then xturv = ysurv = ztvsu. Thus (x, s) ∼ (z, r).

Denote by S^{−1}R the set of equivalence classes, and by x/s the class of (x, s). Define x/s · y/t := xy/st. This product is well defined. Indeed, say y/t = z/r.

Then there is v ∈ S such that yrv = ztv. So xsyrv = xsztv. Thus xy/st = xz/sr. Define x/s + y/t := (tx + sy)/(st). Then, similarly, this sum is well defined. It is easy to check S^{−1}R is a ring, with 0/1 for 0 and 1/1 for 1. It is called the

ring of fractions with respect to S or the localization at S.

Let φ_{S}: R → S^{−1}R be the map given by φ_{S}(x) := x/1. Then φ_{S} is a ring map, and it carries elements of S to units in S^{−1}R as s/1 · 1/s = 1.

Exercise (11.2). — Let R be a ring, S a multiplicative subset. Prove S^{−1}R = 0 if and only if S contains a nilpotent element.

Exercise (11.3). — Let R be a ring, S a multiplicative subset, \overline{S} its saturation. Set T := (S^{−1}R)^{×}. Show T = \{ x/s | x ∈ \overline{S} and s ∈ S \}. Show φ_{S}^{−1}T = \overline{S}.

(11.4) (Total quotient ring). — Let R be a ring, S the set of all nonzerodivisors of R. Clearly S is a multiplicative subset. The map φ_{S}: R → S^{−1}R is injective, because if φ_{S}x = 0, then sx = 0 for some s ∈ S, and so x = 0. We call S^{−1}R the total quotient ring of R, and view R as a subring.

Let T ⊂ S be a multiplicative subset. Clearly, R ⊂ T^{−1}R ⊂ S^{−1}R.

Suppose R is a domain. Then S = R − {0}; so the total quotient ring is just the fraction field Frac(R), and φ_{S} is just the natural inclusion of R into Frac(R). Further, T^{−1}R is a domain by (2.3) as T^{−1}R ⊂ S^{−1}R = Frac(R).

Exercise (11.5). — Find all intermediate rings ℤ ⊂ R ⊂ ℚ, and describe each R as a localization of ℤ. As a starter, prove ℤ[2/3] = S^{−1}ℤ where S = {3^{i} | i ≥ 0}.

Theorem (11.6) (UMP). — Let R be a ring, S a multiplicative subset. Then S^{−1}R is the universal example of an R-algebra in which all the elements of S become units. In fact, given a ring map ψ: R → R′, then ψ(S) ⊂ R^{′×} if and only if there is a ring map ρ: S^{−1}R → R′ with ρφ_{S} = ψ; that is, this diagram commutes:

[figure: R; S^{−1}R; R′; φ_{S}; ψ; ρ]

Further, there is at most one ρ. Moreover, R′ may be noncommutative.

Proof: First, suppose that ρ exists. Let s ∈ S. Then ψ(s) = ρ(s/1). Hence ψ(s)ρ(1/s) = ρ(s/1 · 1/s) = 1. Thus ψ(S) ⊂ R^{′×}.

Next, note that ρ is determined by ψ as follows:

ρ(x/s) = ρ(x/1)ρ(1/s) = ψ(x)ψ(s)^{−1}.

Conversely, suppose ψ(S) ⊂ R^{′×}. Set ρ(x/s) := ψ(s)^{−1}ψ(x). Let’s check that ρ is well defined. Say x/s = y/t. Then there is u ∈ S such that xtu = ysu. Hence

ψ(x)ψ(t)ψ(u) = ψ(y)ψ(s)ψ(u).

Since ψ(u) is a unit, ψ(x)ψ(t) = ψ(y)ψ(s). Now, st = ts, so

ψ(t)^{−1}ψ(s)^{−1} = ψ(s)^{−1}ψ(t)^{−1}.

Hence ψ(x)ψ(s)^{−1} = ψ(y)ψ(t)^{−1}. Thus ρ is well defined. Clearly, ρ is a ring map. Clearly, ψ = ρφ_{S}. □

Corollary (11.7). — Let R be a ring, and S a multiplicative subset. Then the canonical map φ_{S}: R → S^{−1}R is an isomorphism if and only if S consists of units.

Proof: If φ_{S} is an isomorphism, then S consists of units, because φ_{S}(S) does so. Conversely, if S consists of units, then the identity map R → R has the UMP that characterizes φ_{S}; whence, φ_{S} is an isomorphism. □

Exercise (11.8). — Let R′ and R′′ be rings. Consider R := R′ × R′′ and set S := { (1, 1), (1, 0) }. Prove R′ = S^{−1}R.

Exercise (11.9). — Take R and S as in (11.8). On R × S, impose this relation:

(x, s) ∼ (y, t) if xt = ys.

Prove that it is not an equivalence relation.

Definition (11.10). — Let R be a ring, f ∈ R. Set S := {f^{n} | n ≥ 0}. We call the ring S^{−1}R the localization of R at f, and set R_{f} := S^{−1}R and φ_{f} := φ_{S}.

Proposition (11.11). — Let R be a ring, f ∈ R, and X a variable. Then

R_{f} = R[X]/⟨1 − fX⟩.

Proof: Set R′ := R[X]/⟨1 − fX⟩, and let φ: R → R′ be the canonical map. Let’s show that R′ has the UMP characterizing localization (11.6).

First, let x ∈ R′ be the residue of X. Then 1 − xφ(f) = 0. So φ(f) is a unit. So φ(f^{n}) is a unit for n ≥ 0.

Second, let ψ: R → R′′ be a homomorphism carrying f to a unit. Define θ: R[X] → R′′ by θ|R = ψ and θX = ψ(f)^{−1}. Then θ(1 − fX) = 0. So θ factors via a homomorphism ρ: R′ → R′′, and ψ = ρφ. Further, ρ is unique, since every element of R′ is a polynomial in x and since ρx = ψ(f)^{−1} as 1−(ρx)(ρφf) = 0. □

Proposition (11.12). — Let R be a ring, S a multiplicative subset, 𝔞 an ideal. (1) Then 𝔞S^{−1}R = {a/s ∈ S^{−1}R | a ∈ 𝔞 and s ∈ S}.

(2) Then 𝔞∩S ≠ ∅ if and only if 𝔞S^{−1}R = S^{−1}R if and only if φ_{S}^{−1}(𝔞S^{−1}R) = R.

Proof: Let a, b ∈ 𝔞 and x/s, y/t ∈ S^{−1}R. Then ax/s + by/t = (axt + bys)/st; further, axt + bys ∈ 𝔞 and st ∈ S. So 𝔞S^{−1}R ⊂ {a/s | a ∈ 𝔞 and s ∈ S}. But the opposite inclusion is trivial. Thus (1) holds.

As to (2), if 𝔞 ∩ S ∋ s, then 𝔞S^{−1}R ∋ s/s = 1, so 𝔞S^{−1}R = S^{−1}R; whence, φ_{S}^{−1}(𝔞S^{−1}R) = R. Conversely, suppose φ_{S}^{−1}(𝔞S^{−1}R) = R. Then 𝔞S^{−1}R ∋ 1. So (1) yields a ∈ 𝔞 and s ∈ S such that a/s = 1. So there exists a t ∈ S such that at = st. But at ∈ 𝔞 and st ∈ S. So 𝔞 ∩ S ≠ ∅. Thus (2) holds. □

Definition (11.13). — Let R be a ring, S a multiplicative subset, 𝔞 a subset of R. The saturation of 𝔞 with respect to S is the set denoted by 𝔞^{S} and defined by

𝔞^{S} := {a ∈ R | there is s ∈ S with as ∈ 𝔞}.

If 𝔞 = 𝔞^{S}, then we say 𝔞 is saturated.

Proposition (11.14). — Let R be a ring, S a multiplicative subset, 𝔞 an ideal. (1) Then Ker(φ_{S}) = ⟨0⟩^{S}. (2) Then 𝔞 ⊂ 𝔞^{S}. (3) Then 𝔞^{S} is an ideal.

Proof: Clearly, (1) holds, for a/1 = 0 if and only if there is s ∈ S with as = 0. Clearly, (2) holds as 1 ∈ S. Clearly, (3) holds, for if as, bt ∈ 𝔞, then (a + b)st ∈ 𝔞, and if x ∈ R, then xas ∈ 𝔞. □

Exercise (11.15). — Let R be a ring, S a multiplicative subset, 𝔞 and 𝔟 ideals. Show (1) if 𝔞 ⊂ 𝔟, then 𝔞^{S} ⊂ 𝔟^{S}; (2) (𝔞^{S})^{S} = 𝔞^{S}; and (3) (𝔞^{S}𝔟^{S})^{S} = (𝔞𝔟)^{S}.

Exercise (11.16). — Let R be a ring, S a multiplicative subset. Prove that

nil(R)(S^{−1}R) = nil(S^{−1}R).

Proposition (11.17). — Let R be a ring, S a multiplicative subset.

(1) Let 𝔟 be an ideal of S^{−1}R. Then

(a) φ_{S}^{−1}𝔟 = (φ_{S}^{−1}𝔟)^{S} and (b) 𝔟 = (φ_{S}^{−1}𝔟)(S^{−1}R).

(2) Let 𝔞 be an ideal of R. Then φ_{S}^{−1}(𝔞S^{−1}R) = 𝔞^{S}.

(3) Let 𝔭 be a prime ideal of R, and assume 𝔭 ∩ S = ∅. Then

(a) 𝔭 = 𝔭^{S} and (b) 𝔭S^{−1}R is prime.

Proof: To prove (1)(a), take a ∈ R and s ∈ S with as ∈ φ_{S}^{−1}𝔟. Then as/1 ∈ 𝔟; so a/1 ∈ 𝔟 because 1/s ∈ S^{−1}R. Hence a ∈ φ_{S}^{−1}𝔟. Therefore, (φ_{S}^{−1}𝔟)^{S} ⊂ φ_{S}^{−1}𝔟. The opposite inclusion holds as 1 ∈ S. Thus (1)(a) holds.

To prove (1)(b), take a/s ∈ 𝔟. Then a/1 ∈ 𝔟. So a ∈ φ_{S}^{−1}𝔟. Hence a/1 · 1/s is in (φ_{S}^{−1}𝔟)(S^{−1}R). Thus 𝔟 ⊂ (φ_{S}^{−1}𝔟)(S^{−1}R). Now, take a ∈ φ_{S}^{−1}𝔟. Then a/1 ∈ 𝔟. So 𝔟 ⊃ (φ_{S}^{−1}𝔟)(S^{−1}R). Thus (1)(b) holds too.

To prove (2), take a ∈ 𝔞^{S}. Then there is s ∈ S with as ∈ 𝔞. But a/1 = as/1·1/s. So a/1 ∈ 𝔞S^{−1}R. Thus φ_{S}^{−1}(𝔞S^{−1}R) ⊃ 𝔞^{S}. Now, take x ∈ φ_{S}^{−1}(𝔞S^{−1}R). Then x/1 = a/s with a ∈ 𝔞 and s ∈ S by (11.14)(1). Hence there is t ∈ S such that xst = at ∈ 𝔞. So x ∈ 𝔞^{S}. Thus φ_{S}^{−1}(𝔞S^{−1}R) ⊂ 𝔞^{S}. Thus (2) holds.

To prove (3), note 𝔭 ⊂ 𝔭^{S} as 1 ∈ S. Conversely, if sa ∈ 𝔭 with s ∈ S ⊂ R − 𝔭, then a ∈ 𝔭 as 𝔭 is prime. Thus (a) holds.

As for (b), say a/s · b/t ∈ 𝔭S^{−1}R. Then ab ∈ φ_{S}^{−1}(𝔭S^{−1}R), and the latter is equal to 𝔭^{S} by (2), so to 𝔭 by (a). Hence ab ∈ 𝔭, so either a ∈ 𝔭 or b ∈ 𝔭. So either a/s ∈ 𝔭S^{−1}R or b/t ∈ 𝔭S^{−1}R. Thus 𝔭S^{−1}R is prime. Thus (3) holds. □

Corollary (11.18). — Let R be a ring, S a multiplicative subset.

(1) Then 𝔞 ↦ 𝔞S^{−1}R is an inclusion-preserving bijection from the set of all ideals 𝔞 of R with 𝔞 = 𝔞^{S} to the set of all ideals 𝔟 of S^{−1}R. The inverse is 𝔟 ↦ φ_{S}^{−1}𝔟.

(2) Then 𝔭 ↦ 𝔭S^{−1}R is an inclusion-preserving bijection from the set of all primes of R with 𝔭 ∩ S = ∅ to the set of all primes 𝔮 of S^{−1}R. The inverse is 𝔮 ↦ φ_{S}^{−1}𝔮.

Proof: In (1), the maps are inverses by (11.17)(1), (2); clearly, they preserve inclusions. Further, (1) implies (2) by (11.17)(3), by (2.8), and by (11.12)(2). □

Definition (11.19). — Let R be a ring, 𝔭 a prime ideal. Set S := R − 𝔭. We call the ring S^{−1}R the localization of R at 𝔭, and set R_{𝔭} := S^{−1}R and φ_{𝔭} := φ_{S}.

Proposition (11.20). — Let R be a ring, 𝔭 a prime ideal. Then R_{𝔭} is local with maximal ideal 𝔭R_{𝔭}.

Proof: Let 𝔟 be a proper ideal of R_{𝔭}. Then φ_{𝔭}^{−1}𝔟 ⊂ 𝔭 owing to (11.12)(2). Hence (11.18)(1) yields 𝔟 ⊂ 𝔭R_{𝔭}. Thus 𝔭R_{𝔭} is a maximal ideal, and the only one.

Alternatively, let x/s ∈ R_{𝔭}. Suppose x/s is a unit. Then there is a y/t with xy/st = 1. So there is a u ∉ 𝔭 with xyu = stu. But stu ∉ 𝔭. Hence x ∉ 𝔭.

Conversely, let x ∉ 𝔭. Then s/x ∈ R_{𝔭}. So x/s is a unit in R_{𝔭} if and only if x ∉ 𝔭, so if and only if x/s ∉ 𝔭R_{𝔭}. Thus by (11.12)(1), the nonunits of R_{𝔭} form 𝔭R_{𝔭}, which is an ideal. Hence (3.4) yields the assertion. □

Proposition (11.21). — Let R be a domain, 𝔞 an ideal. Then 𝔞 = ⋂_{𝔪} 𝔞R_{𝔪} where 𝔪 runs through the maximal ideals and the intersection is taken in Frac(R).

Proof: Set I := ⋂_{𝔪} 𝔞R_{𝔪}. Clearly, 𝔞 ⊂ I. For the opposite inclusion, given x ∈ I, set 𝔟 := {y ∈ R | yx ∈ 𝔞}. Suppose x ∉ 𝔞. Then 1 ∉ 𝔟, so there is a maximal ideal 𝔫 ⊃ 𝔟 by (2.27). But x ∈ 𝔞R_{𝔫}. So there is s ∈ R − 𝔫 with sx ∈ 𝔞 by (11.12). Then s ∈ 𝔟 ⊂ 𝔫, a contradiction. Thus x ∈ 𝔞, as desired. □

(11.22) (Algebras). — Let R be a ring, S a multiplicative subset, R′ an R-algebra. It is easy to generalize (11.1) as follows. Define a relation on R′×S by (x, s) ∼ (y, t) if there is u ∈ S with xtu = ysu. It is easy to check, as in (11.1), that this relation is an equivalence relation.

Denote by S^{−1}R′ the set of equivalence classes, and by x/s the class of (x, s). Clearly, S^{−1}R′ is an S^{−1}R-algebra with addition and multiplication given by

x/s + y/t := (xt + ys)/(st) and x/s · y/t := xy/st.

We call S^{−1}R′ the localization of R′ with respect to S.

Let φ′_{S}: R′ → S^{−1}R′ be the map given by φ′_{S}(x) := x/1. Then φ′_{S} makes S^{−1}R′ into an R′-algebra, so also into an R-algebra, and φ′_{S} is an R-algebra map.

Note that elements of S become units in S^{−1}R′. Moreover, it is easy to check, as in (11.6), that S^{−1}R′ has the following UMP: φ′_{S} is an algebra map, and elements of S become units in S^{−1}R′; further, given an algebra map ψ: R′ → R′′ such that elements of S become units in R′′, there is a unique R-algebra map ρ: S^{−1}R′ → R′′ such that ρφ′_{S} = ψ; that is, the following diagram is commutative:

[figure: R′; S^{−1}R′; R′′; φ′_{S}; ψ; ρ]

In other words, S^{−1}R′ is the universal example of an R′-algebra in which the elements of S become units.

Let τ: R′ → R′′ be an R-algebra map. Then there is a commutative diagram of R-algebra maps

R′ ⟶^{τ} R′′

φ_{S} ↓  ↓^{φ′_{S}}

S^{−1}R′ ⟶^{S^{−1}τ} S^{−1}R′′

Further, S^{−1}τ is an S^{−1}R-algebra map.

Let T ⊂ R′ be the image of S ⊂ R. Then T is multiplicative. Further,

S^{−1}R′ = T^{−1}R′,

even though R′×S and R′×T are rarely equal, because the two UMPs are essentially the same; indeed, any ring map R′ → R′′ may be viewed as an R-algebra map, and trivially the elements of S become units in R′′ if and only if the elements of T do.

Exercise (11.23). — Let R′/R be a integral extension of rings, S a multiplicative subset of R. Show that S^{−1}R′ is integral over S^{−1}R.

Exercise (11.24). — Let R be a domain, K its fraction field, L a finite extension field, and \overline{R} the integral closure of R in L. Show L is the fraction field of \overline{R}. Show every element of L can, in fact, be expressed as a fraction b/a with b ∈ \overline{R} and a ∈ R.

Exercise (11.25). — Let R ⊂ R′ be domains, K and L their fraction fields. Assume that R′ is a finitely generated R-algebra, and that L is a finite dimensional K-vector space. Find an f ∈ R such that R′_{f} is module finite over R_{f}.

Proposition (11.26). — Let R be a ring, S a multiplicative subset. Let T′ be a multiplicative subset of S^{−1}R, and set T := φ_{S}^{−1}(T′). Assume S ⊂ T. Then

(T′)^{−1}(S^{−1}R) = T^{−1}R.

Proof: Let’s check (T′)^{−1}(S^{−1}R) has the UMP characterizing T^{−1}R. Clearly φ_{T′}φ_{S} carries T into ((T′)^{−1}(S^{−1}R))^{×}. Next, let ψ: R → R′ be a map carrying T into R^{′×}. We must show ψ factors uniquely through (T′)^{−1}(S^{−1}R).

First, ψ carries S into R^{′×} since S ⊂ T. So ψ factors through a unique map ρ: S^{−1}R → R′. Now, given r ∈ T′, write r = x/s. Then x/1 = s/1 · r ∈ T′ since S ⊂ T. So x ∈ T. Hence ρ(r) = ψ(x) · ρ(1/s) ∈ (R′)^{×}. So ρ factors through a unique map ρ′: (T′)^{−1}(S^{−1}R) → R′. Hence ψ = ρ′φ_{T′}φ_{S}, and ρ′ is clearly unique, as required. □

Corollary (11.27). — Let R be a ring, 𝔭 ⊂ 𝔮 prime ideals. Then R_{𝔭} is the localization of R_{𝔮} at the prime ideal 𝔭R_{𝔮}.

Proof: Set S := R − 𝔮 and T′ := R_{𝔮} − 𝔭R_{𝔮}. Set T := φ_{S}^{−1}(T′). Then T = R − 𝔭 by (11.18)(2). So S ⊂ T, and (11.26) yields the assertion. □

Exercise (11.28). — Let R be a ring, S and T multiplicative subsets. (1) Set T′ := φ_{S}(T) and assume S ⊂ T. Prove

T^{−1}R = T^{′−1}(S^{−1}R) = T^{−1}(S^{−1}R).

September 3, 2012 11Nts.tex (2) Set U := {st ∈ R | s ∈ S and t ∈ T }. Prove

T^{−1}(S^{−1}R) = S^{−1}(T^{−1}R) = U^{−1}R.

(3) Let S′ := {t′ ∈ R | t′t ∈ S for some t ∈ R}. Prove S^{′−1}R = S^{−1}R.

Proposition (11.29). — Let R be a ring, S a multiplicative subset, X a variable. Then (S^{−1}R)[X] = S^{−1}(R[X]).

Proof: Let’s check (S^{−1}R)[X] and S^{−1}(R[X]) have the same UMP: a ring map ψ: R → R′ factors uniquely through either one if ψ(S) ⊂ (R′)^{×} and if an x ∈ R′ is preassigned as the image of X. First, since ψ(S) ⊂ (R′)^{×}, there is a unique R-algebra map S^{−1}R → R′, so a unique (S^{−1}R)-algebra map (S^{−1}R)[X] → R′ sending X to x. Second, there is a unique R-algebra map R[X] → R′ sending X to x, so a unique R[X]-algebra map R[X] → R′ sending X to x, and so a unique R[X]-algebra map S^{−1}(R[X]) → R′ since ψ(S) ⊂ (R′)^{×}, as required. □

Corollary (11.30). — Let R be a ring, S a multiplicative subset, X a variable, 𝔭 an ideal of R[X]. Set R′ := S^{−1}R, and let φ: R[X] → R′[X] be the canonical map. Then 𝔭 is prime and 𝔭∩S = ∅ if and only if 𝔭R′[X] is prime and 𝔭 = φ^{−1}(𝔭R′[X]).

Proof: The assertion results directly from (11.30) and (11.18)(2). □

Exercise (11.31). — Let R be a domain, S a multiplicative subset with 0 ∉ S. Assume R is normal. Show that S^{−1}R is normal.

12. Localization of Modules

Formally, we localize a module just as we do a ring. However, the result is a module over the localized ring, and comes equipped with a linear map from the original module; in fact, the result is the universal module with these two properties. Further, as a functor, localization is the left adjoint of restriction of scalars. Hence, localization preserves direct limits, or equivalently, direct sums and cokernels. Therefore, by Watts’ Theorem, localization is naturally isomorphic to tensor product with the localized ring. Moreover, localization is exact; so the localized ring is flat. We end by discussing various compatibilities and examples.

Proposition (12.1). — Let R be a ring, S a multiplicative subset. Then a module M has a compatible S^{−1}R-module structure if and only if, for all s ∈ S, the multiplication map μ_{s}: M → M is bijective; if so, then the S^{−1}R-structure is unique.

Proof: Assume M has a compatible S^{−1}R-structure, and take s ∈ S. Then μ_{s} = μ_{s/1}. So μ_{s} · μ_{1/s} = μ_{(s/1)(1/s)} = 1. Similarly, μ_{1/s} · μ_{s} = 1. So μ_{s} is bijective.

Conversely, assume μ_{s} is bijective for all s ∈ S. Then μ_{R}: R → End_{ℤ}(M) sends S into the units of End_{ℤ}(M). Hence μ_{R} factors through a unique ring map μ_{S^{−1}R}: S^{−1}R → End_{ℤ}(M) by (11.6). Thus M has a unique compatible S^{−1}Rstructure by (4.5). □

(12.2) (Localization of modules). — Let R be a ring, S a multiplicative subset, M a module. Define a relation on M × S by (m, s) ∼ (n, t) if there is u ∈ S such that utm = usn. As in (11.1), this relation is an equivalence relation.

Denote by S^{−1}M the set of equivalence classes, and by m/s the class of (m, s). Then S^{−1}M is an S^{−1}R-module with addition given by m/s + n/t := (tm + sn)/st and scalar multiplication by a/s · m/t := am/st similar to (11.1). We call S^{−1}M the localization of M at S.

For example, let 𝔞 be an ideal. Then S^{−1}𝔞 = 𝔞S^{−1}R by (11.12)(1). Similarly, S^{−1}(𝔞M) = S^{−1}𝔞S^{−1}M = 𝔞S^{−1}M. Further, given an R-algebra R′, the S^{−1}Rmodule S^{−1}R′ constructed here underlies the S^{−1}R-algebra S^{−1}R′ of (11.22).

Define φ_{S}: M → S^{−1}M by φ_{S}(m) := m/1. Clearly, φ_{S} is R-linear.

Note that μ_{s}: S^{−1}M → S^{−1}M is bijective for all s ∈ S by (12.1).

If S = {f^{n} | n ≥ 0} for some f ∈ R, then we call S^{−1}M the localization of M at f, and set M_{f} := S^{−1}M and φ_{f} := φ_{S}.

Similarly, if S = R−𝔭 for some prime ideal 𝔭, then we call S^{−1}M the localization of M at 𝔭, and set M_{𝔭} := S^{−1}M and φ_{𝔭} := φ_{S}.

Theorem (12.3) (UMP). — Let R be a ring, S a multiplicative subset, and M a module. Then S^{−1}M is the universal example of an S^{−1}R-module equipped with an R-linear map from M.

Proof: The proof is like that of (11.6): given an R-linear map ψ: M → N where N is an S^{−1}R-module, it is easy to prove that ψ factors uniquely via the S^{−1}R-linear map ρ: S^{−1}M → N well defined by ρ(m/s) := 1/s · ψ(m). □

Exercise (12.4). — Let R be a ring, S a multiplicative subset, and M a module. Show that M = S^{−1}M if and only if M is an S^{−1}R-module.

Exercise (12.5). — Let R be a ring, S ⊂ T multiplicative subsets, M a module. Set T_{1} := φ_{S}(T) ⊂ S^{−1}R. Show T^{−1}M = T^{−1}(S^{−1}M) = T_{1}^{−1}(S^{−1}M).

Exercise (12.6). — Let R be a ring, S a multiplicative subset. Show that S becomes a filtered category when equipped as follows: given s, t ∈ S, set

Hom(s, t) := {x ∈ R | xs = t}.

Given a module M, define a functor S → ((R-mod)) as follows: for s ∈ S, set M_{s} := M; to each x ∈ Hom(s, t), associate μ_{x}: M_{s} → M_{t}. Define β_{s}: M_{s} → S^{−1}M by β_{s}(m) := m/s. Show the β_{s} induce an isomorphism \underrightarrow{lim} M_{s} ⟶^{∼} S^{−1}M.

Exercise (12.7). — Let R be a ring, S a multiplicative subset, M a module. Prove S^{−1}M = 0 if Ann(M) ∩ S ≠ ∅. Prove the converse if M is finitely generated.

(12.8) (Functoriality). — Let R be a ring, S a multiplicative subset, α: M → N an R-linear map. Then φ_{S}α carries M to the S^{−1}R-module S^{−1}N. So (12.3) yields a unique S^{−1}R-linear map S^{−1}α making the following diagram commutative:

M ⟶^{φ_{S}} S^{−1}M

↓^{α}  ↓_{S^{−1}α}

N ⟶^{φ_{S}} S^{−1}N

The construction in the proof of (12.3) yields

(S^{−1}α)(m/s) = α(m)/s.  (12.8.1)

Thus, canonically, we obtain the following map, and clearly, it is R-linear:

Hom_{R}(M, N) → Hom_{S^{−1}R}(S^{−1}M, S^{−1}N).  (12.8.2)

Any R-linear map β: N → P yields S^{−1}(βα) = (S^{−1}β)(S^{−1}α) owing to uniqueness or to (12.8.1). Thus S^{−1}(•) is a linear functor from ((R-mod)) to ((S^{−1}R-mod)).

Theorem (12.9). — Let R be a ring, S a multiplicative subset. Then the functor S^{−1}(•) is the left adjoint of the functor of restriction of scalars.

Proof: Let N be an S^{−1}R-module. Then N = S^{−1}N by (12.4), and the map (12.8.2) is bijective with inverse taking β: S^{−1}M → N to βφ_{S}: M → N. And (12.8.2) is natural in M and N by (6.3). Thus the assertion holds. □

Corollary (12.10). — Let R be a ring, S a multiplicative subset. Then the functor S^{−1}(•) preserves direct limits, or equivalently, direct sums and cokernels.

Proof: By (12.9), the functor is a left adjoint. Hence it preserves direct limits by (6.12); equivalently, it preserves direct sums and cokernels by (6.10). □

Exercise (12.11). — Let R be a ring, S a multiplicative subset, P a projective module. Then S^{−1}P is a projective S^{−1}R-module.

Corollary (12.12). — Let R be a ring, S a multiplicative subset. Then the functors S^{−1}(•) and S^{−1}R ⊗_{R} • are canonically isomorphic.

Proof: As S^{−1}(•) preserves direct sums and cokernels by (12.10), the assertion is an immediate consequence of Watts Theorem (8.15).

Alternatively, both functors are left adjoints of the same functor by (12.9) and by (8.10). So they are canonically isomorphic by (6.4). □

Exercise (12.13). — Let R be a ring, S a multiplicative subset, M and N modules. Show S^{−1}(M ⊗_{R}N) = S^{−1}M ⊗_{R}N = S^{−1}M ⊗_{S^{−1}R}S^{−1}N = S^{−1}M ⊗_{R}S^{−1}N.

Definition (12.14). — Let R be a ring, S a multiplicative subset, M a module. Given a submodule N, its saturation N^{S} is defined by

N^{S} := {m ∈ M | there is s ∈ S with sm ∈ N}.

If N = N^{S}, then we say N is saturated.

Proposition (12.15). — Let R be a ring, M a module, N and P submodules. Let S be a multiplicative subset, and K an S^{−1}R-submodule of S^{−1}M.

(1) Then (a) N^{S} is a submodule of M, and (b) S^{−1}N is a submodule of S^{−1}M.

(2) Then (a) φ_{S}^{−1}K = (φ_{S}^{−1}K)^{S} and (b) K = S^{−1}(φ_{S}^{−1}K).

(3) Then φ_{S}^{−1}(S^{−1}N) = N^{S}; in particular, Ker(φ_{S}) = 0^{S}.

(4) Then (a) (N ∩ P)^{S} = N^{S} ∩ P^{S} and (b) S^{−1}(N ∩ P) = S^{−1}N ∩ S^{−1}P.

(5) Then (a) (N + P)^{S} ⊃ N^{S} + P^{S} and (b) S^{−1}(N + P) = S^{−1}N + S^{−1}P.

Proof: Assertion (1)(b) holds because N×S is a subset of M×S and is equipped with the induced equivalence relation.

Assertions (1)(a), (2) and (3) can be proved as in (11.14)(3) and (11.17)(1), (2).

As to (4)(a), clearly (N ∩ P)^{S} ⊂ N^{S} ∩ P^{S}. Conversely, given m ∈ N^{S} ∩ P^{S}, there are s, t ∈ S with sm ∈ N and tm ∈ P. Then stm ∈ N ∩ P and st ∈ S. So m ∈ (N ∩ P)^{S}. Thus (a) holds. Alternatively, (4)(b) and (3) yield (4)(a).

As to (4)(b), since N ∩ P ⊂ N, P, using (1) yields S^{−1}(N ∩ P) ⊂ S^{−1}N ∩ S^{−1}P. But, given m/s = n/t ∈ S^{−1}N ∩ S^{−1}P, there is a u ∈ S with utm = usn ∈ N ∩ P. Hence utm/uts = usn/uts ∈ S^{−1}(N ∩ P). Thus (b) holds.

As to (5)(a), given n ∈ N^{S} and p ∈ P^{S}, there are s, t ∈ S with sn ∈ N and tp ∈ P. Then st ∈ S and st(n + p) ∈ N + P. Thus (5)(a) holds.

As to (5)(b), note N, P ⊂ N +P. So (1)(b) yields S^{−1}(N +P) ⊃ S^{−1}N +S^{−1}P. But the opposite inclusion holds as (n + p)/s = n/s + p/s. Thus (5)(b) holds. □

Theorem (12.16) (Exactness of Localization). — Let R be a ring, and S a multiplicative subset. Then the functor S^{−1}(•) is exact.

Proof: As S^{−1}(•) preserves injections by (12.15)(1) and cokernels by (12.10), it is exact by (9.3).

Alternatively, given an exact sequence M′ ⟶^{α} M ⟶^{β} M′′, for each s ∈ S, take a copy M′_{s} → M_{s} → M′′_{s}. Using (12.6), make S into a filtered category, and make these copies into a functor from S to the category of 3-term exact sequences; its limit is the following sequence, which is exact by (7.10), as desired:

S^{−1}M′ ⟶^{S^{−1}α} S^{−1}M ⟶^{S^{−1}β} S^{−1}M′′.

The latter argument can be made more direct as follows. Since βα = 0, we have (S^{−1}β)(S^{−1}α) = S^{−1}(βα) = 0. Hence Ker(S^{−1}β) ⊃ Im(S^{−1}α). Conversely, given m/s ∈ Ker(S^{−1}β), there is t ∈ S with tβ(m) = 0. So β(tm) = 0. So exactness yields m′ ∈ M′ with α(m′) = tm. So (S^{−1}α)(m′/ts) = m/s. Hence Ker(S^{−1}β) ⊂ Im(S^{−1}α). Thus Ker(S^{−1}β) = Im(S^{−1}α), as desired. □

Corollary (12.17). — Let R be a ring, S a multiplicative subset. Then S^{−1}R is flat over R.

Proof: The functor S^{−1}(•) is exact by (12.16), and is isomorphic to S^{−1}R⊗_{R}• by (12.12). Thus S^{−1}R is flat.

Alternatively, using (12.6), write S^{−1}R as a filtered direct limit of copies of R. But R is flat by (9.6). Thus S^{−1}R is flat by (9.13). □

Corollary (12.18). — Let R be a ring, S a multiplicative subset, 𝔞 an ideal, and M a module. Then S^{−1}(M/𝔞M) = S^{−1}M/S^{−1}(𝔞M) = S^{−1}M/𝔞S^{−1}M.

Proof: The assertion results from (12.16) and (12.2). □

Corollary (12.19). — Let R be a ring, 𝔭 a prime. Then Frac(R/𝔭) = R_{𝔭}/𝔭R_{𝔭}.

Proof: We have Frac(R/𝔭) = (R/𝔭)_{𝔭} = R_{𝔭}/𝔭R_{𝔭} by (11.22) and (12.18). □

Proposition (12.20). — Let R be a ring, M a module, S a multiplicative subset.

(1) Let m_{1}, …, m_{n} ∈ M. If M is finitely generated and if the m_{i}/1 ∈ S^{−1}M generate over S^{−1}R, then there’s f ∈ S so that the m_{i}/1 ∈ M_{f} generate over R_{f}.

(2) Assume M is finitely presented and S^{−1}M is a free S^{−1}R-module of rank n. Then there is h ∈ S such that M_{h} is a free R_{h}-module of rank n.

Proof: To prove (1), define α: R^{n} → M by α(e_{i}) := m_{i} with e_{i} the ith standard basis vector. Set C := Coker(α). Then S^{−1}C = Coker(S^{−1}α) by (12.10). Assume the m_{i}/1 ∈ S^{−1}M generate over S^{−1}R. Then S^{−1}α is surjective by (4.10)(1) as S^{−1}(R^{n}) = (S^{−1}R)^{n} by (12.10). Hence S^{−1}C = 0.

In addition, assume M is finitely generated. Then so is C. Hence, (12.7) yields f ∈ S such that C_{f} = 0. Hence α_{f} is surjective. So the m_{i}/1 generate M_{f} over R_{f} again by (4.10)(1). Thus (1) holds.

For (2), let m_{1}/s_{1}, …, m_{n}/s_{n} be a free basis of S^{−1}M over S^{−1}R. Then so is m_{1}/1, …, m_{n}/1 as the 1/s_{i} are units. Form α and C as above, and set K := Ker(α). Then (12.16) yields S^{−1}K = Ker(S^{−1}α) and S^{−1}C = Coker(S^{−1}α). But S^{−1}α is bijective. Hence S^{−1}K = 0 and S^{−1}C = 0.

Since M is finitely generated, C is too. Hence, as above, there is f ∈ S such that C_{f} = 0. Then 0 → K_{f} → R_{f}^{n} ⟶^{α_{f}} M_{f} → 0 is exact by (12.16). Take a finite presentation R^{p} → R^{q} → M → 0. By (12.16), it yields a finite presentation R_{f}^{p} → R_{f}^{q} → M_{f} → 0. Hence K_{f} is a finitely generated R_{f}-module by (5.24).

Let S_{1} ⊂ R_{f} be the image of S. Then (12.5) yields S_{1}^{−1}(K_{f}) = S^{−1}K. But S^{−1}K = 0. Hence there is g/1 ∈ S_{1} such that (K_{f})_{g/1} = 0. Set h := fg. Let’s show K_{h} = 0. Let x ∈ K. Then there is a such that (g^{a}x)/1 = 0 in K_{f}. Hence there is b such that f^{b}g^{a}x = 0 in K. Take c ≥ a, b. Then h^{c}x = 0. Thus K_{h} = 0. But C_{f} = 0 implies C_{h} = 0. Hence α_{h}: R_{h}^{n} → M_{h} is an isomorphism, as desired. □

Proposition (12.21). — Let R be a ring, S a multiplicative subset, and M and N modules. Then there is a canonical homomorphism

σ: S^{−1} Hom_{R}(M, N) → Hom_{S^{−1}R}(S^{−1}M, S^{−1}N).

Further, σ is injective if M is finitely generated; σ is an isomorphism if M is finitely presented.

Proof: The assertions result from (9.15) with R′ := S^{−1}R, since S^{−1}R is flat by (12.17) and since S^{−1}R ⊗ P = S^{−1}P for every R-module P by (12.12). □

Example (12.22). — Set R := ℤ and S := ℤ − ⟨0⟩ and M := ℚ/ℤ. Then M is faithful since z ∈ S implies z · (1/2z) = 1/2 ≠ 0; thus, μ_{R}: R → Hom_{R}(M, M) is injective. But S^{−1}R = ℚ. So (12.16) yields S^{−1} Hom_{R}(M, M) ≠ 0. On the other hand, S^{−1}M = 0 as s · r/s = 0 for any r/s ∈ M. So the map σ(M, M) of (12.21) is not injective. Thus (12.21)(2) can fail if M is not finitely generated.

Example (12.23). — Take R := ℤ and S := ℤ − 0 and M_{n} := ℤ/⟨n⟩ for n ≥ 2. Then S^{−1}M_{n} = 0 for all n as nm ≡ 0 (mod n) for all m. On the other hand, (1, 1, …)/1 is nonzero in S^{−1}(∏ M_{n}) as the kth component of m · (1, 1, …) is nonzero in ∏ M_{n} for k > m if m is nonzero. Thus S^{−1}(∏ M_{n}) ≠ ∏(S^{−1}M_{n}).

Also S^{−1}ℤ = ℚ. So (12.12) yields ℚ ⊗ (∏ M_{n}) ≠ ∏(ℚ ⊗ M_{n}), whereas (8.11) yields ℚ ⊗ (⨁ M_{n}) = ⨁(ℚ ⊗ M_{n}).

Exercise (12.24). — Set R := ℤ and S = ℤ − ⟨0⟩. Set M := ⨁_{n≥2} ℤ/⟨n⟩ and N := M. Show that the map σ of (12.21) is not injective.

13. Support

The spectrum of a ring is this topological space: its points are all the prime ideals; each closed set consists of those primes containing a given ideal. The support of a module is this subset: its points are the primes at which the localized module is nonzero. We relate the support to the closed set of the annihilator. We prove that a sequence is exact if and only if it is exact after localizing at each maximal ideal. Lastly, we prove that a module is finitely generated and projective if and only if it is locally free of finite rank.

(13.1) (Spectrum of a ring). — Let R be a ring. Its set of prime ideals is denoted Spec(R), and is called the (prime) spectrum of R.

Let 𝔞 be an ideal. Let V(𝔞) denote the subset of Spec(R) of those primes that contain 𝔞. We call V(𝔞) the variety of 𝔞.

Let 𝔟 be another ideal. Then by the Scheinnullstellensatz, V(𝔞) = V(𝔟) if and only if \sqrt{𝔞} = \sqrt{𝔟}. Further, (2.2) yields

V(𝔞) ∪ V(𝔟) = V(𝔞 ∩ 𝔟) = V(𝔞𝔟).

A prime ideal 𝔭 contains the ideals 𝔞_{λ} in an arbitrary collection if and only if 𝔭 contains their sum ∑ 𝔞_{λ}; hence,

⋂ V(𝔞_{λ}) = V(∑ 𝔞_{λ}).

Finally, V(R) = ∅, and V(⟨0⟩) = Spec(R). Thus the subsets V(𝔞) of Spec(R) are the closed sets of a topology; it is called the Zariski topology.

Given an element f ∈ R, we call the open set

D(f) := Spec(R) − V(⟨f⟩)

a principal open set. These sets form a basis for the topology of Spec(R); indeed, given any prime 𝔭 ⊅ 𝔞, there is an f ∈ 𝔞 − 𝔭, and so 𝔭 ∈ D(f) ⊂ Spec(R) − V(𝔞). Further, f, g ∉ 𝔭 if and only if fg ∉ 𝔭, for any f, g ∈ R and prime 𝔭; in other words,

D(f) ∩ D(g) = D(fg).  (13.1.1)

A ring map φ: R → R′ induces a set map

Spec(φ): Spec(R′) → Spec(R) by Spec(φ)(𝔭′) := φ^{−1}(𝔭′).

Clearly, Spec(φ)^{−1} V(𝔞) = V(𝔞R′); hence, Spec(φ) is continuous. Thus Spec(•) is a contravariant functor from ((Rings)) to ((Top spaces)).

For example, the quotient map R → R/𝔞 induces a closed embedding

Spec(R/𝔞) ↪ Spec(R),

whose image is V(𝔞), owing to (1.8) and (2.8). Furthermore, the localization map R → R_{f} induces an open embedding

Spec(R_{f}) ↪ Spec(R),

whose image is D(f), owing to (11.18).

Exercise (13.2). — Let R be a ring, 𝔭 ∈ Spec(R). Show that 𝔭 is a closed point — that is, {𝔭} is a closed set — if and only if 𝔭 is a maximal ideal.

Exercise (13.3). — Let R be a ring, R′ a flat algebra with structure map φ. Show that R′ is faithfully flat if and only if Spec(φ) is surjective.

Proposition (13.4). — Let R be a ring, X := Spec(R). Then X is quasicompact: if X = ⋃_{λ∈Λ} U_{λ} with U_{λ} open, then X = ⋃_{i=1}^{n} U_{λ_{i}} for some λ_{i} ∈ Λ.

Proof: Say U_{λ} = X − V(𝔞_{λ}). Then V(∑ 𝔞_{λ}) = ⋂ V(𝔞_{λ}) = ∅. So ∑ 𝔞_{λ} lies in no prime ideal. Hence there are λ_{1}, …, λ_{n} ∈ Λ and f_{λ_{i}} ∈ 𝔞_{λ_{i}} for all i with ∑ f_{λ_{i}} = 1. Hence ∑ 𝔞_{λ_{i}} = R; so V(∑ 𝔞_{λ_{i}}) = ⋂ V(𝔞_{λ_{i}}) = ∅; so X = ⋃ U_{λ_{i}}. □

Exercise (13.5). — Let R be a ring, X := Spec(R), and U an open subset. Show U is quasi-compact if and only if X − U = V (𝔞) where 𝔞 is finitely generated.

Exercise (13.6). — Let B be a Boolean ring, and set X := Spec(B). Show X is a compact Hausdorff space. (Following Bourbaki, “quasi-compact” is shortened to “compact” when the space is Hausdorff.) Further, show a subset U ⊂ X is both open and closed if and only if U = D(f) for some f ∈ B.

Exercise (13.7) (Stone’s Theorem). — Show every Boolean ring B is isomorphic to the ring of continuous functions from a compact Hausdorff space X to 𝔽_{2} with the discrete topology. Equivalently, show B is isomorphic to the ring R of open and closed subsets of X; in fact, X := Spec(B), and B ⟶^{∼} R is given by f ↦ D(f).

Definition (13.8). — Let R be a ring, M a module. Its support is the set

Supp(M) := { 𝔭 ∈ Spec(R) | M_{𝔭} ≠ 0 }.

Proposition (13.9). — Let R be a ring, M a module.

(1) Let 0 → L → M → N → 0 be exact. Then Supp(L) ∪ Supp(N) = Supp(M).

(2) Let M_{λ} be submodules with ∑ M_{λ} = M. Then ⋃ Supp(M_{λ}) = Supp(M).

(3) Then Supp(M) ⊂ V(Ann(M)), with equality if M is finitely generated.

Proof: Consider (1). For every prime 𝔭, the sequence 0 → L_{𝔭} → M_{𝔭} → N_{𝔭} → 0 is exact by (12.16). Hence M_{𝔭} ≠ 0 if and only if L_{𝔭} ≠ 0 or N_{𝔭} ≠ 0. Thus (1) holds.

In (2), M_{λ} ⊂ M. So (1) yields ⋃ Supp(M_{λ}) ⊂ Supp(M). To prove the opposite inclusion, take 𝔭 ∉ ⋃ Supp(M_{λ}). Then (M_{λ})_{𝔭} = 0 for all λ. By hypothesis, the natural map ⨁ M_{λ} → M is surjective. So ⨁(M_{λ})_{𝔭} → M_{𝔭} is surjective by (12.10). Hence M_{𝔭} = 0. Alternatively, given m/s ∈ M_{𝔭}, express m as a finite sum m = ∑ m_{λ} with m_{λ} ∈ M_{λ}. For each such λ, there is t_{λ} ∈ R − 𝔭 with t_{λ}m_{λ} = 0. Set t := ∏ t_{λ}. Then tm = 0 and t ∉ 𝔭. So m/s = 0 in M_{𝔭}. Hence again, M_{𝔭} = 0. Thus 𝔭 ∉ Supp(M), and so (2) holds.

Consider (3). Let 𝔭 be a prime. By (12.7), M_{𝔭} = 0 if Ann(M) ∩ (R − 𝔭) ≠ ∅, and the converse holds if M is finitely generated. But Ann(M) ∩ (R − 𝔭) ≠ ∅ if and only if Ann(M) ⊄ 𝔭. The assertion follows directly. □

Definition (13.10). — Let R be a ring, x ∈ R. We say x is nilpotent on a module M if there is n ≥ 1 with x^{n}m = 0 for all m ∈ M, that is, if x ∈ \sqrt{Ann(M)}. We denote the set of nilpotents on M by nil(M); that is, nil(M) := \sqrt{Ann(M)}.

Proposition (13.11). — Let R be a ring, M a finitely generated module. Then

nil(M) = ⋂_{𝔭∈Supp(M)} 𝔭.

Proof: First, nil(M) = ⋂_{𝔭⊃Ann(M)} 𝔭 by the Scheinnullstellensatz (3.22). But 𝔭 ⊃ Ann(M) if and only if 𝔭 ∈ Supp(M) by (13.9)(3). □

Proposition (13.12). — Let R be a ring, M and N modules. Then

Supp(M ⊗_{R} N) ⊂ Supp(M) ∩ Supp(N),  (13.12.1)

with equality if M and N are finitely generated.

Proof: First, (M ⊗_{R} N)_{𝔭} = M_{𝔭} ⊗_{R_{𝔭}} N_{𝔭} by (12.13); whence, (13.12.1) holds. The opposite inclusion follows from (10.14) if M and N are finitely generated. □

Corollary (13.13). — Let R be a ring, 𝔞 an ideal, M a module. Then

Supp(M/𝔞M) ⊂ Supp(M) ∩ V(𝔞).

with equality if M is finitely generated.

Proof: First, (8.13)(1) yields M/𝔞M = M ⊗ R/𝔞. But Ann(R/𝔞) = 𝔞; hence (13.9)(3) yields Supp(R/𝔞) = V(𝔞). Thus (13.12) yields the assertion. □

Exercise (13.14). — Let R be a ring, M a module, 𝔭 ∈ Supp(M). Prove

V(𝔭) ⊂ Supp(M).

Exercise (13.15). — Let ℤ be the integers, ℚ the rational numbers, and set M := ℚ/ℤ. Find Supp(M), and show that it is not Zariski closed.

Proposition (13.16). — Let R be a ring, M a module. These conditions are equivalent: (1) M = 0; (2) Supp(M) = ∅; (3) M_{𝔪} = 0 for every maximal ideal 𝔪.

Proof: Trivially, if (1) holds, then S^{−1}M = 0 for any multiplicative subset S. In particular, (2) holds. Trivially, (2) implies (3).

Finally, assume M ≠ 0, and take a nonzero m ∈ M, and set 𝔞 := Ann(m). Then 1 ∉ 𝔞, so 𝔞 lies in some maximal ideal 𝔪. Then, for all f ∈ R − 𝔪, we have fm ≠ 0. Hence m/1 ≠ 0 in M_{𝔪}. Thus (3) implies (1). □

Exercise (13.17). — Let R be a ring, P a module, and M, N submodules. Show M = N if M_{𝔪} = N_{𝔪} for every maximal ideal 𝔪. First assume M ⊂ N.

Exercise (13.18). — Prove these three conditions on a ring R are equivalent: (1) R is reduced.

(2) S^{−1}R is reduced for all multiplicatively closed sets S.

(3) R_{𝔪} is reduced for all maximal ideals 𝔪.

Exercise (13.19). — Let R be a ring, Σ the set of minimal primes. Prove this: (1) If R_{𝔭} is a domain for any prime 𝔭, then the 𝔭 ∈ Σ are pairwise comaximal.

(2) R = ∏_{i=1}^{n} R_{i} where R_{i} is a domain if and only if R_{𝔭} is a domain for any prime 𝔭 and Σ is finite. If so, then R_{i} = R/𝔭_{i} with {𝔭_{1}, …, 𝔭_{n}} = Σ.

Proposition (13.20). — A sequence of modules L ⟶^{α} M ⟶^{β} N is exact if and only if its localization L_{𝔪} ⟶^{α_{𝔪}} M_{𝔪} ⟶^{β_{𝔪}} N_{𝔪} is exact at each maximal ideal 𝔪.

Proof: If the sequence is exact, then so is its localization by (12.16).

Consider the converse. First Im(β_{𝔪}α_{𝔪}) = 0. But Im(β_{𝔪}α_{𝔪}) = (Im(βα))_{𝔪} by (12.16) and (9.3). Hence Im(βα) = 0 by (13.16). So βα = 0. Thus Im(α) ⊂ Ker(β).

Set H := Ker(β)/ Im(α). Then H_{𝔪} = Ker(β_{𝔪})/ Im(α_{𝔪}) by (12.16) and (9.3). So H_{𝔪} = 0 owing to the hypothesis. Hence H = 0 by (13.16), as required. □

September 3, 2012 11Nts.tex Exercise (13.21). — Let R be a ring, M a module. Prove elements m_{λ} ∈ M generate M if and only if, at every maximal ideal 𝔪, their images m_{λ} generate M_{𝔪}.

Proposition (13.22). — Let A be a semilocal ring, 𝔪_{1}, …, 𝔪_{n} its maximal ideals, M, N finitely presented modules. Assume M_{𝔪_{i}} ≅ N_{𝔪_{i}} for each i. Then M ≅ N.

Proof: For each i, take an isomorphism ψ_{i}: M_{𝔪_{i}} ⟶^{∼} N_{𝔪_{i}}. Then (12.21) yields s_{i} ∈ A − 𝔪_{i} and φ_{i}: M → N with (φ_{i})_{𝔪_{i}} = s_{i}ψ_{i}. However, (2.2) implies ⋂_{j≠i} 𝔪_{j} ⊄ 𝔪_{i}; so there’s x_{i} ∈ ⋂_{j≠i} 𝔪_{j} with x_{i} ∉ 𝔪_{i}. Set γ := ∑_{i} x_{i}φ_{i}.

For each i, set α_{i} := x_{i}s_{i}ψ_{i}. Then α_{i}: M_{𝔪_{i}} ⟶^{∼} N_{𝔪_{i}}. Set β_{i} := ∑_{j≠i} x_{j}s_{j}ψ_{j}.

Then β_{i}(M_{𝔪_{i}}) ⊂ 𝔪_{i}N_{𝔪_{i}} as x_{j} ∈ 𝔪_{i} for j ≠ i. Further, γ_{𝔪_{i}} = α_{i} + β_{i}. So γ_{𝔪_{i}} is an isomorphism by (10.12). Hence (13.20) implies γ: M ⟶^{∼} N. □

Proposition (13.23). — Let R be a ring, S a multiplicative subset, M a module. Then the following conditions are equivalent:

(1) M is flat over R.

(2) S^{−1}M is flat over S^{−1}R and over R.

(3) At every maximal ideal 𝔪, the localization M_{𝔪} is flat over R_{𝔪}.

Proof: Assume (1) holds. Let α: N′ → N be an injection of S^{−1}R-modules. Then M ⊗_{R} α: M ⊗_{R} N′ → M ⊗_{R} N is injective. Now, (12.4) yields α = S^{−1}α. So (12.13) yields M ⊗_{R}α = S^{−1}M ⊗_{S^{−1}R}α. Hence S^{−1}M ⊗_{S^{−1}R}α is injective; that is, S^{−1}M is flat over S^{−1}R. Therefore, S^{−1}M is flat over R by (12.17) and (9.7). Thus (2) holds. Trivially, (2) implies (3).

Assume (3) holds. Let α: N′ → N be an injection of R-modules. Then α_{𝔪} is injective by (13.20). So M_{𝔪} ⊗_{R_{𝔪}} α_{𝔪} is injective. Now, that map is equal to (M ⊗ α)_{𝔪} by (12.13), hence is injective. Therefore, M ⊗ α is injective by (13.20); that is, (1) holds. □

Exercise (13.24). — Let R be a ring, R′ a flat algebra, 𝔭′ a prine in R′, and 𝔭 its contraction in R. Prove that R′_{𝔭′} is a faithfully flat R_{𝔭}-algebra.

Definition (13.25). — Let R be a ring, M a module. We say M is locally finitely generated if each 𝔭 ∈ Spec(R) has a neighborhood on which M becomes finitely generated; more precisely, there exists f ∈ R − 𝔭 such that M_{f} is finitely generated over R_{f}. Similarly, we define the properties locally finitely presented, locally free of finite rank, and locally free of rank n.

Proposition (13.26). — Let R be a ring, M a module.

(1) If M is locally finitely generated, then it is finitely generated.

(2) If M is locally finitely presented, then it is finitely presented.

Proof: By (13.4), there are f_{1}, …, f_{n} ∈ R with ⋃ D(f_{i}) = Spec(R) and finitely many m_{i,j} ∈ M such that, for some n_{i,j} ≥ 0, the m_{i,j}/f_{i}^{n_{i,j}} generate M_{f_{i}}. Clearly, for each i, the m_{i,j}/1 also generate M_{f_{i}}.

Given any maximal ideal 𝔪, there is i such that f_{i} ∉ 𝔪. Let S_{1} be the image of R − 𝔪 in R_{f_{i}}. Then (12.5) yields M_{𝔪} = S_{1}^{−1}(M_{f_{i}}). Hence the m_{i,j}/1 generate M_{𝔪}. Thus (13.21) yields (1).

Assume M is locally finitely presented. Then M is finitely generated by (1). So there is a surjection R^{k} → → M. Let K be its kernel. Then K is locally finitely generated owing to (5.24). Hence K too is finitely generated by (1). So there is a surjection R^{ℓ} → → K. It yields the desired finite presentation R^{ℓ} → R^{k} → M → 0. Thus (2) holds. □

September 3, 2012 11Nts.tex Theorem (13.27). — These conditions on an R-module P are equivalent:

(1) P is finitely generated and projective.

(2) P is finitely presented and flat.

(3) P is finitely presented, and P_{𝔪} is free over R_{𝔪} at each maximal ideal 𝔪.

(4) P is locally free of finite rank.

(5) P is finitely generated, and for each 𝔭 ∈ Spec(R), there are f and n such that 𝔭 ∈ D(f) and P_{𝔮} is free of rank n over R_{𝔮} at each 𝔮 ∈ D(f).

Proof: Condition (1) implies (2) by (10.15).

Let 𝔪 be a maximal ideal. Then R_{𝔪} is local by (11.20). If P is finitely presented, then P_{𝔪} is finitely presented, because localization preserves direct sums and cokernels by (12.10).

Assume (2). Then P_{𝔪} is flat by (13.23), so free by (10.15). Thus (3) holds. Assume (3). Fix a surjective map α: M → N. Then α_{𝔪}: M_{𝔪} → N_{𝔪} is surjective.

So Hom(P_{𝔪}, α_{𝔪}): Hom(P_{𝔪}, M_{𝔪}) → Hom(P_{𝔪}, N_{𝔪}) is surjective by (5.22) and (5.21). But Hom(P_{𝔪}, α_{𝔪}) = Hom(P, α)_{𝔪} by (12.21) as P is finitely presented. Further, 𝔪 is arbitrary. Hence Hom(P, α) is surjective by (13.20). Therefore, P is projective by (5.22). Thus (1) holds.

Again assume (3). Given any prime 𝔭, take a maximal ideal 𝔪 containing it. By hypothesis, P_{𝔪} is free; its rank is finite as P_{𝔪} is finitely generated. By (12.20)(2), there is f ∈ R − 𝔪 such that P_{f} is free of finite rank over R_{f}. Thus (4) holds.

Assume (4). Then P is locally finitely presented. So P is finitely presented by (13.26)(2). Further, given 𝔭 ∈ Spec(R), there are f ∈ R − 𝔭 and n such that M_{f} is locally free of rank n over R_{f}. Given 𝔮 ∈ D(f), let S_{1} be the image of R − 𝔮 in R_{f}. Then (12.5) yields M_{𝔮} = S_{1}^{−1}(M_{f}). Hence M_{𝔮} is locally free of rank n over R_{𝔮}. Thus (5) holds. Further, (3) results from taking 𝔭 := 𝔪 and 𝔮 := 𝔪.

Finally, assume (5), and let’s prove (4). Given 𝔭 ∈ Spec(R), let f and n be provided by (5). Take a free basis p_{1}/f^{k_{1}}, …, p_{n}/f^{k_{n}} of P_{𝔭} over R_{𝔭}. The p_{i} define a map α: R^{n} → P, and α_{𝔭}: R_{𝔭}^{n} → P_{𝔭} is bijective, in particular, surjective.

As P is finitely generated, (12.20)(1) provides g ∈ R −𝔭 such that α_{g}: R_{g}^{n} → P_{g} is surjective. It follows that α_{𝔮}: R_{𝔮}^{n} → P_{𝔮} is surjective for every 𝔮 ∈ D(g). If also 𝔮 ∈ D(f), then by hypothesis P_{𝔮} ≃ R_{𝔮}^{n}. So α_{𝔮} is bijective by (10.4).

Set h := fg. Clearly, D(f) ∩ D(g) = D(h). By (13.1), D(h) = Spec(R_{h}). Clearly, α_{𝔮} = (α_{h})_{(𝔮R_{h})} for all 𝔮 ∈ D(h). Hence α_{h}: R_{h}^{n} → P_{h} is bijective owing to (13.20) with R_{h} for R. Thus (4) holds. □

Exercise (13.28). — Given n, prove an R-module P is locally free of rank n if and only if P is finitely generated and P_{𝔪} ≃ R_{𝔪}^{n} holds at each maximal ideal 𝔪.

Exercise (13.29). — Let A be a semilocal ring, P a locally free module of rank n. Show that P is free of rank n.

14. Krull–Cohen–Seidenberg Theory

Krull–Cohen–Seidenberg Theory relates the prime ideals in a ring to those in an integral extension. We prove each prime has at least one prime lying over it — that is, contracting to it. The overprime can be taken to contain any ideal that contracts to an ideal contained in the given prime; this stronger statement is known as the Going-up Theorem. Further, one prime is maximal if and only if the other is, and two overprimes cannot be nested. On the other hand, the Going-down Theorem asserts that, given nested primes in the subring and a prime lying over the larger, there is a subprime lying over the smaller, either if the subring is normal and the overring is a domain or if the extension is flat even if it’s not integral.

Lemma (14.1). — Let R ⊂ R′ be an integral extension of domains. Then R′ is a field if and only if R is.

Proof: First, suppose R′ is a field. Let x ∈ R be nonzero. Then 1/x ∈ R′, so satisfies an equation of integral dependence:

(1/x)^{n} + a_{1}(1/x)^{n−1} + ⋯ + a_{n} = 0

with n ≥ 1 and a_{i} ∈ R. Multiplying the equation by x^{n−1}, we obtain

1/x = −(a_{1} + a_{n−2}x + ⋯ + a_{n}x^{n−1}) ∈ R.

Conversely, suppose R is a field. Let y ∈ R′ be nonzero. Then y satisfies an equation of integral dependence

y^{n} + a_{1}y^{n−1} + ⋯ + a_{n−1}y + a_{n} = 0

with n ≥ 1 and a_{i} ∈ R. Rewriting the equation, we obtain

y(y^{n−1} + ⋯ + a_{n−1}) = −a_{n}.

Take n minimal. Then a_{n} ≠ 0 as R′ is a domain. So dividing by −a_{n}y, we obtain

1/y = (−1/a_{n})(y^{n−1} + ⋯ + a_{n−1}) ∈ R′.  □

Definition (14.2). — Let R be a ring, R′ an R-algebra, 𝔭 a prime of R, and 𝔭′ a prime of R′. We say 𝔭′ lies over 𝔭 if 𝔭′ contracts to 𝔭.

Theorem (14.3). — Let R ⊂ R′ be an integral extension of rings, and 𝔭 a prime of R. Let 𝔭′ ⊂ 𝔮′ be nested primes of R′, and 𝔞′ an arbitrary ideal of R′.

(1) (Maximality) Suppose 𝔭′ lies over 𝔭. Then 𝔭′ is maximal if and only if 𝔭 is.

(2) (Incomparability) Suppose both 𝔭′ and 𝔮′ lie over 𝔭. Then 𝔭′ = 𝔮′.

(3) (Lying over) Then there is a prime 𝔯′ of R′ lying over 𝔭.

(4) (Going up) Suppose 𝔞′ ∩ R ⊂ 𝔭. Then in (3) we can take 𝔯′ to contain 𝔞′.

Proof: Assertion (1) follows from (14.1) applied to the extension R/𝔭 ⊂ R′/𝔭′, which is integral as R ⊂ R′ is, since, if y ∈ R′ satisfies y^{n} + a_{1}y^{n−1} + ⋯ + a_{n} = 0, then reduction modulo 𝔭′ yields an equation of integral dependence over R/𝔭.

To prove (2), localize at R − 𝔭, and form this commutative diagram:

\begin{matrix}R′ ⟶ R′_{𝔭} \\ ↑ & ↑ \\ R ⟶ R_{𝔭} \end{matrix}

Here R_{𝔭} → R′_{𝔭} is injective by (12.15)(1), and the extension is integral by (11.23).

Here 𝔭′R′_{𝔭} and 𝔮′R′_{𝔭} are nested primes of R′_{𝔭} by (11.18)(2). By the same token, both lie over 𝔭R_{𝔭}, because both their contractions in R_{𝔭} contract to 𝔭 in R. Thus we may replace R by R_{𝔭} and R′ by R′_{𝔭}, and so assume R is local with 𝔭 as maximal ideal by (11.20). Then 𝔭′ is maximal by (1); whence, 𝔭′ = 𝔮′.

To prove (3), again we may replace R by R_{𝔭} and R′ by R′_{𝔭}: if 𝔯′′ is a prime ideal of R′_{𝔭} lying over 𝔭R_{𝔭}, then the contraction 𝔯′ of 𝔯′′ in R′ lies over 𝔭. So we may assume R is local with 𝔭 as unique maximal ideal. Now, R′ has a maximal ideal 𝔯′ by 2.28; further, 𝔯′ contracts to a maximal ideal 𝔯 of R by (1). Thus 𝔯 = 𝔭.

Finally, (4) follows from (3) applied to the extension R/(𝔞′ ∩ R) ⊂ R′/𝔞′. □

Exercise (14.4). — Let R ⊂ R′ be an integral extension of rings, and 𝔭 a prime of R. Suppose R′ has just one prime 𝔭′ over 𝔭. Show (a) that 𝔭′R′_{𝔭} is the only maximal ideal of R′_{𝔭}, (b) that R′_{𝔭′} = R′_{𝔭}, and (c) that R′_{𝔭′} is integral over R_{𝔭}.

Exercise (14.5). — Let R ⊂ R′ be an integral extension of domains, and 𝔭 a prime of R. Suppose R′ has at least two distinct primes 𝔭′ and 𝔮′ lying over 𝔭. Show that R′_{𝔭′} is not integral over R_{𝔭}. Show that, in fact, if y lies in 𝔮′, but not in 𝔭′, then 1/y ∈ R′_{𝔭′} is not integral over R_{𝔭}.

Exercise (14.6). — Let k be a field, and X an indeterminate. Set R′ := k[X], and Y := X^{2}, and R := k[Y]. Set 𝔭 := (Y − 1)R and 𝔭′ := (X − 1)R′. Is R′_{𝔭′} integral over R_{𝔭}? Explain.

Lemma (14.7). — Let R ⊂ R′ be a ring extension, X a variable, f ∈ R[X] a monic polynomial. Suppose f = gh with g, h ∈ R′[X] monic. Then the coefficients of g and h are integral over R.

Proof: Set R_{1} := R′[X]/⟨g⟩. Let x_{1} be the residue of X. Then 1, x_{1}, x_{1}^{2}, … form a free basis of R_{1} over R′ by (10.20) as g is monic; hence, R′ ⊂ R_{1}. Now, g(x_{1}) = 0; so g factors as (X − x_{1})g_{1} with g_{1} ∈ R_{1}[X] monic of degree 1 less than g. Repeat this process, extending R_{1}. Continuing, obtain g(X) = ∏(X − x_{i}) and h(X) = ∏(X − y_{j}) with all x_{i} and y_{j} in an extension of R′. The x_{i} and y_{j} are integral over R as they are roots of f. But the coefficients of g and h are polynomials in the x_{i} and y_{j}; so they too are integral over R. □

Proposition (14.8). — Let R be a normal domain, K := Frac(R), and L/K a field extension. Let y ∈ L be integral over R, and p ∈ K[X] its monic minimal polynomial. Then p ∈ R[X], and so p(y) = 0 is an equation of integral dependence.

Proof: Since y is integral, there is a monic polynomial f ∈ R[X] with f(y) = 0. Write f = pq with q ∈ K[X]. Then by (14.7) the coefficients of p are integral over R, so in R since R is normal. □

Theorem (14.9) (Going down for integral extensions). — Let R ⊂ R′ be an integral extension of domains, 𝔭 ⫋ 𝔮 nested primes of R, and 𝔮′ a prime of R′ lying over 𝔮. If R is normal, then there is a prime 𝔭′ lying over 𝔭 and contained in 𝔮′.

Proof: First, let us show 𝔭R′_{𝔮′} ∩ R = 𝔭. Take y ∈ 𝔭R′_{𝔮′} ∩ R. Say y = x/s with x ∈ 𝔭R′ and s ∈ R′ − 𝔮′. Say x = ∑_{i=1}^{m} y_{i}x_{i} with y_{i} ∈ 𝔭 and x_{i} ∈ R′, and set R′′ := R[x_{1}, …, x_{m}]. Then R′′ is a finite R-module by (10.23) and xR′′ ⊂ 𝔭R′′. Let f(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} be the characteristic polynomial of μ_{x}: R′′ →

September 3, 2012 11Nts.tex R′′. Then a_{i} ∈ 𝔭^{i} ⊂ 𝔭 by (10.1), and f(x) = 0 by the Determinant Trick (10.2).

Set K := Frac(R). Suppose f = gh with g, h ∈ K[X] monic. By (14.7) the coefficients of g, h lie in R as R is normal. Further, f ≡ X^{n} (mod 𝔭). So g ≡ X^{r} (mod 𝔭) and h ≡ X^{n−r} (mod 𝔭) for some r by unique factorization in Frac(R/𝔭)[X]. Hence g and h have all nonleading coefficients in 𝔭. Replace f by a monic factor of minimal degree. Then f is the minimal polynomial of x over K.

Recall s = x/y. So s satisfies the equation

s^{n} + b_{1}s^{n−1} + ⋯ + b_{n} = 0 with b_{i} := a_{i}/y^{i} ∈ K.

This equation is of minimal degree since y ∈ R ⊂ K and deg(f) is minimal for x. But s is integral over R. So all b_{i} are in R by (14.8).

Assume y ∉ 𝔭. Then b_{i} ∈ 𝔭 since a_{i} = b_{i}y^{i} ∈ 𝔭. So s^{n} ∈ 𝔭R′ ⊂ 𝔮R′ ⊂ 𝔮′. So s ∈ 𝔮′, a contradiction. Hence y ∈ 𝔭. Thus 𝔭R′_{𝔮′} ∩R ⊂ 𝔭. But the opposite inclusion holds trivially. Thus 𝔭R′_{𝔮′} ∩ R = 𝔭.

Hence, there is a prime 𝔭′′ of R′_{𝔮′} with 𝔭′′ ∩ R = 𝔭 by (3.10). Set 𝔭′ := 𝔭′′ ∩ R′. Then 𝔭′ ∩ R = 𝔭, and 𝔭′ ⊂ 𝔮′ by (11.18)(2), as desired. □

Lemma (14.10). — Always, a minimal prime consists entirely of zerodivisors.

Proof: Let R be the ring, 𝔭 the minimal prime. Then R_{𝔭} has only one prime 𝔭R_{𝔭} by (11.18)(2). So by the Scheinnullstellensatz, 𝔭R_{𝔭} consists entirely of nilpotents. Hence, given x ∈ 𝔭, there is s ∈ R − 𝔭 with sx^{n} = 0 for some n ≥ 1. Take n minimal. Then sx^{n−1} ≠ 0, but (sx^{n−1})x = 0. Thus x is a zerodivisor. □

Theorem (14.11) (Going down for Flat Algebras). — Let R be a ring, R′ a flat algebra. Let 𝔭 ⫋ 𝔮 be nested primes of R, and 𝔮′ a prime of R′ lying over 𝔮. Then there is a prime 𝔭′ of R′ that lies over 𝔭 and is contained in 𝔮′.

Proof: The canonical map R_{𝔮} → R′_{𝔮′} is faithfully flat by (13.24). Therefore, Spec(R′_{𝔮}) → Spec(R_{𝔮}) is surjective by (13.3). Thus (11.18) yields the desired 𝔭′.

Alternatively, R′ ⊗_{R} (R/𝔭) is flat over R/𝔭 by (9.8). Also, R′/𝔭R′ = R′ ⊗_{R} R/𝔭 by (8.13)(1). Hence, owing to (1.8), we may replace R by R/𝔭 and R′ by R′/𝔭R′, and thus assume R is a domain and 𝔭 = 0.

By (3.11), 𝔮′ contains a minimal prime 𝔭′ of R′. Let’s show that 𝔭′ lies over ⟨0⟩. Let x ∈ R be nonzero. Then the multiplication map μ_{x}: R → R is injective. Since R′ is flat, μ_{x}: R′ → R′ is also injective. Hence, (14.10) implies that x does not belong to the contraction of 𝔭′, as desired. □

Exercise (14.12). — Let R be a reduced ring, Σ the set of minimal primes. Prove that z.div(R) = ⋃_{𝔭∈Σ} 𝔭 and that R_{𝔭} = Frac(R/𝔭) for any 𝔭 ∈ Σ.

Exercise (14.13). — Let R be a ring, Σ the set of minimal primes, and K the total quotient ring. Assume Σ is finite. Prove these three conditions are equivalent:

(1) R is reduced.

(2) z.div(R) = ⋃_{𝔭∈Σ} 𝔭, and R_{𝔭} = Frac(R/𝔭) for each 𝔭 ∈ Σ.

(3) K/𝔭K = Frac(R/𝔭) for each 𝔭 ∈ Σ, and K = ∏_{𝔭∈Σ} K/𝔭K.

Exercise (14.14). — Let A be a reduced local ring with residue field k and finite set Σ of minimal primes. For each 𝔭 ∈ Σ, set K(𝔭) := Frac(A/𝔭). Let P be a finitely generated module. Show that P is free of rank r if and only if dim_{k}(P ⊗_{A} k) = r and dim_{K(𝔭)}(P ⊗_{A} K(𝔭)) = r for each 𝔭 ∈ Σ.

Exercise (14.15). — Let A be a reduced local ring with residue field k and a finite set of minimal primes. Let P be a finitely generated module, B an A-algebra with Spec(B) → Spec(A) surjective. Show that P is a free A-module of rank r if and only if P ⊗ B is a free B-module of rank r.

(14.16) (Arbitrary normal rings). — An arbitrary ring R is said to be normal if R_{𝔭} is a normal domain for every prime 𝔭. If R is a domain, then this definition recovers that in (10.24), owing to (11.31).

Exercise (14.17). — Let R be a ring, 𝔭_{1} …, 𝔭_{r} all its minimal primes, and K the total quotient ring. Prove that these three conditions are equivalent:

(1) R is normal.

(2) R is reduced and integrally closed in K.

(3) R is a finite product of normal domains R_{i}. Assume the conditions hold. Prove the R_{i} are equal to the R/𝔭_{j} in some order.

15. Noether Normalization

The Noether Normalization Lemma describes the basic structure of a finitely generated algebra over a field; namely, given a chain of ideals, there is a polynomial subring over which the algebra is module finite, and the ideals contract to ideals generated by initial segments of variables. After proving this lemma, we derive several versions of the Nullstellensatz. The most famous is Hilbert’s; namely, the radical of any ideal is the intersection of all the maximal ideals containing it.

Then we study the (Krull) dimension: the maximal length of any chain of primes. We prove our algebra is catenary; that is, if two chains have the same ends and maximal lengths, then the lengths are the same. Further, if the algebra is a domain, then its dimension is equal to the transcendence degree of its fraction field.

In an appendix, we give a simple direct proof of the Hilbert Nullstellensatz. At the same time, we prove it in significantly greater generality: for Jacobson rings.

Lemma (15.1) (Noether Normalization). — Let k be a field, R := k[x_{1}, …, x_{n}] a finitely generated k-algebra, and 𝔞_{1} ⊂ ⋯ ⊂ 𝔞_{r} a chain of proper ideals of R. Then there are algebraically independent elements t_{1}, …, t_{ν} ∈ R such that

(1) R is module finite over P := k[t_{1}, … t_{ν}] and

(2) for i = 1, ⋯, r, there is an h_{i} such that 𝔞_{i} ∩ P = ⟨t_{1}, …, t_{h_{i}}⟩.

If k is infinite, then we may choose the t_{i} to be k-linear combinations of the x_{i}.

Proof: Let R′ := k[X_{1}, …, X_{n}] be the polynomial ring, and φ: R′ → R the k-algebra map with φX_{i} := x_{i}. Set 𝔞′_{0} := Ker φ and 𝔞′_{i} := φ^{−1}𝔞_{i} for i = 1, ⋯, r. It suffices to prove the lemma for R′ and 𝔞′_{0} ⊂ ⋯ ⊂ 𝔞′_{r}: if t′_{i} ∈ R′ and h′_{i} work here, then t_{i} := φt′_{i+h′_{0}} and h_{i} := h′_{i} − h′_{0} work for R and the 𝔞_{i}, because the t_{i} are algebraically independent by (1.9), and clearly (1) and (2) hold. Thus we may assume the x_{i} are algebraically independent.

The proof proceeds by induction on r (and shows ν := n works now).

First, assume r = 1 and 𝔞_{1} = t_{1}R for some nonzero t_{1}. Then t_{1} ∉ k because 𝔞_{1} is proper. Suppose we have found t_{2}, …, t_{n} ∈ R so that x_{1} is integral over P := k[t_{1}, t_{2}, …, t_{n}] and so that P [x_{1}] = R. Then (10.23) yields (1).

Further, by the theory of transcendence bases [1, (8.3), p. 526], [6, Thm. 1.1, p. 356], the elements t_{1}, …, t_{n} are algebraically independent. Now, take x ∈ 𝔞_{1} ∩P. Then x = t_{1}x′ where x′ ∈ R ∩ Frac(P). Further, R ∩ Frac(P) = P because P is normal by (10.29) as P is a polynomial algebra. Hence 𝔞_{1} ∩ P = t_{1}P. Thus (2) holds too.

To find t_{2}, …, t_{n}, we are going to choose ℓ_{i} and set t_{i} := x_{i} − x_{1}^{ℓ_{i}}. Then clearly P [x_{1}] = R. Now, say t_{1} = ∑ a_{(j)}x_{1}^{j_{1}} ⋯ x_{n}^{j_{n}} with (j) := (j_{1}, …, j_{n}) and a_{(j)} ∈ k. Recall t_{1} ∉ k, and note that x_{1} satisfies this equation:

∑ a_{(j)}x_{1}^{j_{1}}(t_{2} + x_{1}^{ℓ_{2}})^{j_{2}} ⋯ (t_{n} + x_{1}^{ℓ_{n}})^{j_{n}} = t_{1}.

Set e(j) := j_{1} + ℓ_{2}j_{2} + ⋯ + ℓ_{n}j_{n}. Take ℓ > max{j_{i}} and ℓ_{i} := ℓ^{i}. Then the e(j) are distinct. Let e(j′) be largest among the e(j) with a_{(j)} ≠ 0. Then e(j′) > 0, and the above equation may be rewritten as follows:

a_{(j′)}x_{1}^{e(j′)} + ∑_{e<e(j′)} p_{e}x_{1}^{e} = 0 where p_{e} ∈ P. Thus x_{1} is integral over P, as desired.

Suppose k is infinite. We are going to reorder the x_{i}, choose a_{i} ∈ k, and set t_{i} := x_{i} − a_{i}x_{1}. Then clearly P [x_{1}] = R. Now, say t_{1} = H_{d} + ⋯ + H_{0} where H_{i} is homogeneous of degree i in x_{1}, …, x_{n} and where H_{d} ≠ 0. Then d > 0 as t_{1} ∉ k. Since k is infinite, we may reorder the x_{i} and take a_{i} ∈ k with H_{d}(1, a_{2}, …, a_{n}) ≠ 0. Then H_{d}(1, a_{2}, …, a_{n}) is the coefficient of x_{1}^{d} in H_{d}(x_{1}, t_{2} + a_{2}x_{1}, …, t_{n} + a_{n}x_{1}). So after we collect like powers of x_{1}, the equation

H_{d}(x_{1}, t_{2} + a_{2}x_{1}, …, t_{n} + a_{n}x_{1}) + ⋯ + H_{0}(x_{1}, t_{2} + a_{2}x_{1}, …, t_{n} + a_{n}x_{1}) + t_{1} = 0

becomes an equation of integral dependence for x_{1} over P, as desired.

Second, assume r = 1 and 𝔞_{1} is arbitrary. We may assume 𝔞_{1} ≠ 0. The proof proceeds by induction on n. The case n = 1 follows from the first case (but is simpler) because k[x_{1}] is a PID. Let t_{1} ∈ 𝔞_{1} be nonzero. By the first case, there exist elements u_{2}, …, u_{n} such that t_{1}, u_{2}, … u_{n} are algebraically independent and satisfy (1) and (2) with respect to R and t_{1}R. By induction, there are t_{2}, …, t_{n} satisfying (1) and (2) with respect to k[u_{2}, …, u_{n}] and 𝔞_{1} ∩ k[u_{2}, …, u_{n}].

Set P := k[t_{1}, …, t_{n}]. Since R is module finite over k[t_{1}, u_{2}, …, u_{n}] and the latter is module finite over P, the former is module finite over P by (10.22). Thus (1) holds, and so t_{1}, …, t_{n} are algebraically independent. Further, by assumption,

𝔞_{1} ∩ k[t_{2}, …, t_{n}] = ⟨t_{2}, …, t_{h}⟩

for some h. But t_{1} ∈ 𝔞_{1}. So 𝔞_{1} ∩ P ⊃ ⟨t_{1}, …, t_{h}⟩.

Conversely, given x ∈ 𝔞_{1} ∩ P, say x = ∑_{i=0}^{d} f_{i}t_{1}^{i} with f_{i} ∈ k[t_{2}, …, t_{n}]. Since t_{1} ∈ 𝔞_{1}, we have f_{0} ∈ 𝔞_{1} ∩ k[t_{2}, …, t_{n}]; so f_{0} ∈ ⟨t_{2}, …, t_{h}⟩. Hence x ∈ ⟨t_{1}, …, t_{h}⟩. Thus 𝔞_{1} ∩ P = ⟨t_{1}, …, t_{h}⟩. Thus (2) holds for r = 1.

Finally, assume the lemma holds for r − 1. Let u_{1}, …, u_{n} ∈ R be algebraically independent elements satisfying (1) and (2) for the sequence 𝔞_{1} ⊂ ⋯ ⊂ 𝔞_{r−1}, and set h := h_{r−1}. By the second case, there exist elements t_{h+1}, …, t_{n} satisfying (1) and (2) for k[u_{h+1}, …, u_{n}] and 𝔞_{r} ∩ k[u_{h+1}, …, u_{n}]. Then, for some h_{r},

𝔞_{r} ∩ k[t_{h+1}, …, t_{n}] = ⟨t_{h+1}, …, t_{h_{r}}⟩.

Set t_{i} := u_{i} for 1 ≤ i ≤ h. Set P := k[t_{1}, …, t_{n}]. Then, by assumption, R is module finite over k[u_{1}, …, u_{n}], and k[u_{1}, …, u_{n}] is module finite over P; hence, R is module finite over P by (10.22). Thus (1) holds, and t_{1}, …, t_{n} are algebraically independent over k.

Fix i with 1 ≤ i ≤ r. Set m := h_{i}. Then t_{1}, …, t_{m} ∈ 𝔞_{i}. Given x ∈ 𝔞_{i} ∩ P, say x = ∑ f_{(v)}t_{1}^{v_{1}} ⋯ t_{m}^{v_{m}} with (v) = (v_{1}, …, v_{m}) and f_{(v)} ∈ k[t_{m+1}, …, t_{n}]. Then f_{(0)} lies in 𝔞_{i} ∩ k[t_{m+1}, …, t_{n}]. We are going to see the latter intersection is equal to ⟨0⟩. It is so if i ≤ r − 1 because it lies in 𝔞_{i} ∩ k[u_{m+1}, …, u_{n}], which is equal to ⟨0⟩. Further, if i = r, then, by assumption, 𝔞_{i} ∩ k[t_{m+1}, …, t_{n}] = ⟨t_{m+1}, …, t_{m}⟩ = 0. Thus f_{(0)} = 0. Hence x ∈ ⟨t_{1}, …, t_{h_{i}}⟩. Thus 𝔞_{i} ∩ P ⊂ ⟨t_{1}, …, t_{h_{i}}⟩. So the two are equal. Thus (2) holds, and the proof is complete. □

Exercise (15.2). — Let k := 𝔽_{q} be the finite field with q elements, and k[X, Y] the polynomial ring. Set f := X^{q}Y − XY^{q} and R := k[X, Y]/⟨f⟩. Let x, y ∈ R be the residues of X, Y. For every a ∈ k, show that R is not module finite over P := k[y−ax]. (Thus, in (15.1), no k-linear combination works.) First, take a = 0.

Exercise (15.3). — Let k be a field, and X, Y, Z variables. Set

R := k[X, Y, Z]/⟨X^{2} − Y^{3} − 1, XZ − 1⟩,

and let x, y, z ∈ R be the residues of X, Y, Z. Fix a, b ∈ k, and set t := x + ay + bz and P := k[t]. Show that x and y are integral over P for any a, b and that z is integral over P if and only if b ≠ 0.

Theorem (15.4) (Weak Nullstellensatz). — Let k be a field, and R a finitely generated k-algebra. Suppose R is a field. Then R is a finite extension field of k.

Proof: By the Noether Normalization Lemma (15.1), R is module finite over a polynomial subring P := k[t_{1}, …, t_{ν}]. Then P ⊂ R is an integral extension by (10.18). Since R is a field, so is P by (14.1). Hence ν = 0. So P = k. Thus R is module finite over k, as asserted. □

Corollary (15.5). — Let k be a field, R := k[x_{1}, …, x_{n}] a finitely generated k-algebra, and 𝔪 a maximal ideal of R. Assume k is algebraically closed. Then there are a_{1}, …, a_{n} ∈ k such that 𝔪 = ⟨x_{1} − a_{1}, …, x_{n} − a_{n}⟩.

Proof: Set K := R/𝔪. Then K is a finite extension field of k by the Weak Nullstellensatz (15.4). But k is algebraically closed. Hence k = K. Let a_{i} ∈ k be the residue of x_{i}, and set 𝔫 := ⟨x_{1} − a_{1}, …, x_{n} − a_{n}⟩. Then 𝔫 ⊂ 𝔪.

Let R′ := k[X_{1}, …, X_{n}] be the polynomial ring, and φ: R′ → R the k-algebra map with φX_{i} := x_{i}. Set 𝔫′ := ⟨X_{1} − a_{1}, …, X_{n} − a_{n}⟩. Then φ(𝔫′) = 𝔫. But 𝔫′ is maximal by (2.19). So 𝔫 is maximal. Hence 𝔫 = 𝔪, as desired. □

Theorem (15.6) (Hilbert Nullstellensatz). — Let k be a field, and R a finitely generated k-algebra. Let 𝔞 be a proper ideal of R. Then

\sqrt{𝔞} = ⋂_{𝔪⊃𝔞} 𝔪

where 𝔪 runs through all maximal ideals containing 𝔞.

Proof: We may assume 𝔞 = 0 by replacing R by R/𝔞. Clearly \sqrt{0} ⊂ ⋂ 𝔪. Conversely, take f ∉ \sqrt{0}. Then R_{f} ≠ 0 by (11.2). So R_{f} has a maximal ideal 𝔫 by (2.28). Let 𝔪 be its contraction in R. Now, R is a finitely generated k-algebra by hypothesis; hence, R_{f} is one too owing to (11.11). Therefore, by the weak Nullstellensatz, R_{f}/𝔫 is a finite extension field of k.

Set K := R/𝔪. By construction, K is a k-subalgebra of R_{f}/𝔫. Therefore, K is a finite-dimensional k-vector space. So k ⊂ K is an integral extension by (10.18). Since k is a field, so is K by (14.1). Thus 𝔪 is maximal. But f/1 is a unit in R_{f}; so f/1 ∉ 𝔫. Hence f ∉ 𝔪. So f ∉ ⋂ 𝔪. Thus \sqrt{0} = ⋂ 𝔪. □

Exercise (15.7). — Let k be a field, K an algebraically closed extension field. (So K contains a copy of every finite extension field.) Let P := k[X_{1}, …, X_{n}] be the polynomial ring, and f, f_{1}, …, f_{r} ∈ P. Assume f vanishes at every zero in K^{n} of f_{1}, …, f_{r}; in other words, if (a) := (a_{1}, …, a_{n}) ∈ K^{n} and f_{1}(a) = 0, …, f_{r}(a) = 0, then f(a) = 0 too. Prove that there are polynomials g_{1}, …, g_{r} ∈ P and an integer N such that f^{N} = g_{1}f_{1} + ⋯ + g_{r}f_{r}.

Lemma (15.8). — Let k be a field, R a finitely generated k-algebra. Assume R is a domain. Let 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} be a chain of primes. Set K := Frac(R) and d := tr. deg_{k} K. Then r ≤ d, with equality if and only if the chain cannot be lengthened.

Proof: By the Noether Normalization Lemma (15.1), R is module finite over a polynomial subring P := k[t_{1}, …, t_{m}] such that 𝔭_{i}∩P = ⟨t_{1}, …, t_{h_{i}}⟩ for suitable h_{i}. Set M := Frac(P). Then m = tr. deg_{k} M. But P ⊂ R is an integral extension by (10.18). So M ⊂ K is algebraic. Hence m = d. Now, Incomparability (14.3)(2) yields h_{i} < h_{i+1} for all i. Hence r ≤ h_{r}. But h_{r} ≤ m and m = d. Thus r ≤ d.

If r = d, then r is maximal, as it was just proved that no chain can be longer. Conversely, assume r is maximal. Then 𝔭_{0} = 0 since R is a domain. So h_{0} = 0. Further, 𝔭_{r} is maximal since 𝔭_{r} is contained in some maximal ideal and it is prime. So 𝔭_{r} ∩ P is maximal by Maximality (14.3)(1). Hence h_{r} = m.

Suppose there is an i such that h_{i} + 1 < h_{i+1}. Then

(𝔭_{i} ∩ P) ⫋ ⟨t_{1}, …, t_{h_{i}+1}⟩ ⫋ (𝔭_{i+1} ∩ P).

Now, P/(𝔭_{i} ∩ P) is, by (1.9), equal to k[t_{h_{i}+1}, …, t_{m}]; the latter is a polynomial ring, so normal by (10.29)(1). Also, the extension P/(𝔭_{i} ∩ P) ⊂ R/𝔭_{i} is integral as P ⊂ R is. Hence, the Going-down Theorem (14.9) yields a prime 𝔭 with 𝔭_{i} ⊂ 𝔭 ⊂ 𝔭_{i+1} and 𝔭 ∩ P = ⟨t_{1}, …, t_{h_{i}+1}⟩. Then 𝔭_{i} ⫋ 𝔭 ⫋ 𝔭_{i+1}, contradicting the maximality of r. Thus h_{i} + 1 = h_{i+1} for all i. But h_{0} = 0. Hence r = h_{r}. But h_{r} = m and m = d. Thus r = d, as desired. □

Definition (15.9). — Given a ring R, its (Krull) dimension dim(R) is defined to be the supremum of the lengths r of all strictly ascending chains of primes:

dim(R) := sup{ r | there’s a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} in R }.

Exercise (15.10). — Let R be a domain of (finite) dimension r, and 𝔭 a nonzero prime. Prove that dim(R/𝔭) < r.

Exercise (15.11). — Let R′/R be an integral extension of rings. Prove that dim(R) = dim(R′).

Theorem (15.12). — Let k be a field, R a finitely generated k-algebra. If R is a domain, then dim(R) = tr. deg_{k}(Frac(R)).

Proof: The assertion is an immediate consequence of (15.8). □

Corollary (15.13). — Let k be a field, R a finitely generated k-algebra, and 𝔭 a prime of R. Suppose R is a domain. Then

dim(R_{𝔭}) + dim(R/𝔭) = dim(R).

If also 𝔭 is maximal, then dim(R_{𝔭}) = dim(R).

Proof: A chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭 ⫋ ⋯ ⫋ 𝔭_{r} in R gives rise to a pair of chains of primes, one in R_{𝔭} and one in R/𝔭,

𝔭_{0}R_{𝔭} ⫋ ⋯ ⫋ 𝔭R_{𝔭} and 0 = 𝔭/𝔭 ⫋ ⋯ ⫋ 𝔭_{r}/𝔭,

owing to (11.18) and to (1.8) and (2.7); conversely, every such pair of chains arises from a unique chain in R through 𝔭. But by (15.8), every strictly ascending chain through 𝔭 of maximal length is of length dim(R). The asserted equation follows.

If also 𝔭 is maximal, then clearly dim(R/𝔭) = 0, and so dim(R_{𝔭}) = dim(R). □

Definition (15.14). — We call a ring catenary if, given any two nested primes 𝔮 ⊂ 𝔭, there exists a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} of maximal length r with 𝔭_{0} = 𝔮 and 𝔭_{r} = 𝔭, and any two such chains have the same length r.

15. Noether Normalization 79

Theorem (15.15). — Over a field, a finitely generated algebra is catenary.

Proof: Let R be the algebra, and 𝔮 ⊂ 𝔭 two nested primes. Replacing R by R/𝔮, we may assume R is a domain. Then the proof of (15.13) shows that any chain of primes 0 ⫋ ⋯ ⫋ 𝔭 of maximal length is of length dim(R) − dim(R/𝔭). □ Exercise (15.16). — Let k be a field, R a finitely generated k-algebra, f ∈ R nonzero. Assume R is a domain. Prove that dim(R) = dim(R_{f}). Exercise (15.17). — Let k be a field, P := k[f] the polynomial ring in one variable f. Set 𝔭 := ⟨f⟩ and R := P_{𝔭}. Find dim(R) and dim(R_{f}). Exercise (15.18). — Let R be a ring, R[X] the polynomial ring. Prove

1 + dim(R) ≤ dim(R[X]) ≤ 1 + 2 dim(R).

15. Appendix: Jacobson Rings

(15.19) (Jacobson Rings). — We call a ring R Jacobson if, given any ideal 𝔞, its radical is equal to the intersection of all maximal ideals containing it; that is,

\sqrt{𝔞} = ⋂_{𝔪⊃𝔞} 𝔪.  (15.19.1)

Plainly, the nilradical of a Jacobson ring is equal to its Jacobson radical. Also, any quotient ring of a Jacobson ring is Jacobson too. In fact, a ring is Jacobson if and only if the the nilradical of every quotient ring is equal to its Jacobson radical.

In general, the right-hand side of (15.19.1) contains the left. So (15.19.1) holds if and only if every f outside \sqrt{𝔞} lies outside some maximal ideal 𝔪 containing 𝔞.

Recall the Scheinnullstellensatz, (3.22): it says \sqrt{𝔞} = ⋂_{𝔭⊃𝔞} 𝔭 with 𝔭 prime. Thus R is Jacobson if and only if (15.19.1) holds whenever 𝔞 is prime.

For example, a field k is Jacobson. More generally, a local ring A is Jacobson if and only if its maximal ideal is its only prime. Further, a Boolean ring B is Jacobson, as every prime is maximal by (2.16), and so trivially (15.19.1) holds whenever 𝔞 is prime. Moreover, the polynomial ring k[X_{1}, …, X_{n}] is Jacobson by (2.21). Finally, owing to the next lemma, both ℤ and k[X_{1}] are Jacobson.

Lemma (15.20). — Let R be a 1-dimensional domain. Assume every nonzero element lies in only finitely many maximal ideals. Then R is Jacobson if and only if the set {𝔪_{λ}}_{λ∈Λ} of maximal ideals is infinite.

Proof: If {𝔪_{λ}} is finite, take a nonzero x_{λ} ∈ 𝔪_{λ} for each λ. Set x := ∏ x_{λ}. Then x ≠ 0 and x ∈ ⋂ 𝔪_{λ}. But \sqrt{⟨0⟩} = ⟨0⟩ as R is a domain. So \sqrt{⟨0⟩} ≠ ⋂ 𝔪_{λ}. Thus R is not Jacobson.

If {𝔪_{λ}} is infinite, then ⋂ 𝔪_{λ} = ⟨0⟩ by hypothesis. But every nonzero prime is maximal as R is 1-dimensional. Thus (15.19.1) holds whenever 𝔞 is prime. □

Proposition (15.21). — A ring R is Jacobson if and only if, for any nonmaximal prime 𝔭 and any f ∉ 𝔭, the extension 𝔭R_{f} is not maximal.

Proof: Assume R is Jacobson. Take a nonmaximal prime 𝔭 and an f ∉ 𝔭. Then f ∉ 𝔪 for some maximal ideal 𝔪 containing 𝔭. So 𝔭R_{f} is not maximal by (11.18).

Conversely, let 𝔞 be an ideal, f ∉ \sqrt{𝔞}. Then (R/𝔞)_{f} ≠ 0. So there is a maximal ideal 𝔫 in (R/𝔞)_{f}. Let 𝔪 be its contraction in R. Then 𝔪 ⊃ 𝔞 and f ∉ 𝔪. Further, (4.8) and (12.18) yield R_{f}/𝔪R_{f} = (R/𝔞/𝔪/𝔞)_{f} = (R/𝔞)_{f}/𝔫. Since 𝔫 is maximal, R_{f}/𝔪R_{f} is a field. So 𝔪 is maximal by hypothesis. Thus R is Jacobson. □

Exercise (15.22). — Let X be a topological space. We say a subset Y is locally closed if Y is the intersection of an open set and a closed set; equivalently, Y is open in its closure \overline{Y}; equivalently, Y is closed in an open set containing it.

We say a subset X_{0} of X is very dense if X_{0} meets every nonempty locally closed subset Y. We say X is Jacobson if its set of closed points is very dense.

Show that the following conditions on a subset X_{0} of X are equivalent: (1) X_{0} is very dense.

(2) Every closed set F of X satisfies \overline{F ∩ X_{0}} = F.

(3) The map U ↦ U ∩ X_{0} from the open sets of X to those of X_{0} is bijective.

Exercise (15.23). — Let R be a ring, X := Spec(R), and X_{0} the set of closed points of X. Show that the following conditions are equivalent:

(1) R is a Jacobson ring.

(2) X is a Jacobson space.

(3) If y ∈ X is a point such that {y} is locally closed, then y ∈ X_{0}.

Theorem (15.24) (Generalized Hilbert Nullstellensatz). — Let R be a Jacobson ring, R′ a finitely generated algebra, 𝔪′ a maximal ideal of R′, and 𝔪 its contraction. Then (1) 𝔪 is maximal, and R′/𝔪′ is algebraic over R/𝔪, and (2) R′ is Jacobson.

Proof: To prove (1), replace R by R/𝔪 and R′ by R′/𝔪′. Then R is Jacobson, R′ is a field as well as a finitely generated algebra, and R ⊂ R′. We must show R is a field and R′/R is a finite field extension.

Write R′ = R[x_{1}, …, x_{n}] with x_{i} ≠ 0. Then R′ = R[x_{1}, …, x_{n−1}][x_{n}]. So the tower property for finite extensions (10.22) implies it suffices to prove (1) for n = 1.

Set x := x_{1} and Q := Frac(R). Then Q[x] = R′ as R ⊂ Q. But R′ is a field, and x ≠ 0; so 1/x ∈ Q[x]. Say 1/x = q_{0}x^{m} + ⋯ + q_{m} with q_{i} ∈ Q and q_{0} ≠ 0. Then

ax^{m+1} + a_{1}x^{m} + ⋯ + a_{m}x + a_{m+1} = 0 with a, a_{i} ∈ R and a ≠ 0.

So x is integral over R_{a}. Further R_{a}[x] = R′. Also R′ is a field. Hence R_{a} is a field by (14.1); so ⟨0⟩ ⊂ R_{a} is maximal. But R is a Jacobson domain. Hence, ⟨0⟩ ⊂ R is maximal by (15.21). So R is a field. So R = R_{a}. So R[x] = R′. Thus (1) holds.

To prove (2), let 𝔭′ ⊂ R′ be prime, and 𝔭 its contraction. Given a′ ∈ R′ − 𝔭′ with 𝔭′R′_{a′} maximal, apply (1) to R′_{a′}/R; thus 𝔭 is maximal and R′_{a′}/𝔭′R′_{a′} is integral (algebraic) over R/𝔭. But R′_{a′}/𝔭′R′_{a′} ⊃ R′/𝔭′. Hence R′/𝔭′ is integral over R/𝔭. So R′/𝔭′ is a field by (14.1). So 𝔭′ is maximal. Thus (15.21) yields (2). □

Exercise (15.25). — Let P := ℤ[X_{1}, …, X_{n}] be the polynomial ring. Assume f ∈ P vanishes at every zero in K^{n} of f_{1}, …, f_{r} ∈ P for every finite field K; that is, if (a) := (a_{1}, …, a_{n}) ∈ K^{n} and f_{1}(a) = 0, …, f_{r}(a) = 0 in K, then f(a) = 0 too. Prove there are g_{1}, …, g_{r} ∈ P and N ≥ 1 such that f^{N} = g_{1}f_{1} + ⋯ + g_{r}f_{r}.

Exercise (15.26). — Let R be a ring, R′ an algebra. Prove that if R′ is integral over R and R is Jacobson, then R′ is Jacobson.

Exercise (15.27). — Let R be a Jacobson ring, S a multiplicative subset, f ∈ R. True or false: prove or give a counterexample to each of the following statements:

(1) The localized ring R_{f} is Jacobson.

(2) The localized ring S^{−1}R is Jacobson.

(3) The filtered direct limit \underrightarrow{lim} R_{λ} of Jacobson rings R_{λ} is Jacobson.

Exercise (15.28). — Let R be a reduced Jacobson ring with a finite set Σ of minimal primes, and P a finitely generated module. Show that P is locally free of rank r if and only if dim_{R/𝔪}(P/𝔪P) = r for any maximal ideal 𝔪.

16. Chain Conditions

In a ring, often every ideal is finitely generated; if so, we call the ring Noetherian. Examples include the ring of integers and any field. We characterize Noetherian rings as those in which every ascending chain of ideals stabilizes, or equivalently, in which every set of ideals has one member maximal under inclusion. We prove the Hilbert Basis Theorem: if a ring is Noetherian, then so is any finitely generated algebra over it. We define and characterize Noetherian modules similarly, and we prove that, over a Noetherian ring, a module is Noetherian if and only if it is finitely generated. Lastly, we study Artinian rings and modules; in them, by definition, every descending chain of ideals, respectively of submodules, stabilizes.

(16.1) (Noetherian rings). — We call a ring Noetherian if every ideal is finitely generated.

A PID is, trivially, Noetherian. Examples include a field k, the polynomial ring k[X] in one variable, and the ring of integers ℤ.

Here are two standard examples of non-Noetherian rings. A third is given below in (16.6), and a fourth later in (18.26).

First, form the polynomial ring k[X_{1}, X_{2}, …] in infinitely many variables. It is non-Noetherian as ⟨X_{1}, X_{2}, … ⟩ is not finitely generated (but the ring is a UFD).

Second, in the polynomial ring k[X, Y], form this subring R and its ideal 𝔞:

R := {f := a + Xg | a ∈ k and g ∈ k[X, Y]} and
𝔞 := ⟨X, XY, XY^{2}, … ⟩.

Then 𝔞 is not generated by any f_{1}, …, f_{m} ∈ 𝔞. Indeed, let n be the highest power of Y occurring in any f_{i}. Then XY^{n+1} ∉ ⟨f_{1}, …, f_{m}⟩. Thus R is non-Noetherian.

Exercise (16.2). — Let 𝔞 be a finitely generated ideal in an arbitrary ring. Show every set that generates 𝔞 contains a finite subset that generates 𝔞.

Definition (16.3). — We say the ascending chain condition (acc) is satisfied if every ascending chain of ideals 𝔞_{0} ⊂ 𝔞_{1} ⊂ ⋯ stabilizes; that is, there is a j ≥ 0 such that 𝔞_{j} = 𝔞_{j+1} = · · ·.

We say the maximal condition (maxc) is satisfied if every nonempty set of ideals 𝒮 contains ones maximal for inclusion, that is, properly contained in no other in 𝒮.

Lemma (16.4). — Acc is satisfied if and only if maxc is.

Proof: Let 𝔞_{0} ⊂ 𝔞_{1} ⊂ ⋯ be a chain of ideals. If 𝔞_{j} is maximal, then trivially 𝔞_{j} = 𝔞_{j+1} = · · ·. Thus maxc implies acc.

Conversely, given a nonempty set of ideals 𝒮 with no maximal member, there’s 𝔞_{0} ∈ 𝒮; for each j ≥ 0, there’s 𝔞_{j+1} ∈ 𝒮 with 𝔞_{j} ⫋ 𝔞_{j+1}. So the Axiom of Countable Choice provides an infinite chain 𝔞_{0} ⫋ 𝔞_{1} ⫋ · · ·. Thus acc implies maxc. □

Proposition (16.5). — Given a ring R, the following conditions are equivalent:

(1) R is Noetherian; (2) acc is satisfied; (3) maxc is satisfied.

Proof: Assume (1) holds. Let 𝔞_{0} ⊂ 𝔞_{1} ⊂ ⋯ be a chain of ideals. Set 𝔞 := ⋃ 𝔞_{n}. Clearly, 𝔞 is an ideal. So by hypothesis, 𝔞 is finitely generated, say by x_{1}, …, x_{r}. For each i, there is an j_{i} such that x_{i} ∈ 𝔞_{j_{i}}. Set j := max{j_{i}}. Then x_{i} ∈ 𝔞_{j} for all i. So 𝔞 ⊂ 𝔞_{j} ⊂ 𝔞_{j+1} ⊂ ⋯ ⊂ 𝔞. So 𝔞_{j} = 𝔞_{j+1} = · · ·. Thus (2) holds.

Assume (2) holds. Then (3) holds by (16.4).

Assume (3) holds. Let 𝔞 be an ideal, a_{λ} for λ ∈ Λ generators, 𝒮 the set of ideals generated by finitely many a_{λ}. Let 𝔟 be a maximal element of 𝒮; say 𝔟 is generated by a_{λ_{1}}, …, a_{λ_{m}}. Then 𝔟 ⊂ 𝔟 + ⟨a_{λ}⟩ for any λ. So by maximality, 𝔟 = 𝔟 + ⟨a_{λ}⟩. Hence a_{λ} ∈ 𝔟. So 𝔟 = 𝔞; whence, 𝔞 is finitely generated. Thus (1) holds. □

Example (16.6). — In the field of rational functions k(X, Y), form this ring:

R := k[X, Y, X/Y, X/Y^{2}, X/Y^{3}, …].

Then R is non-Noetherian by (16.5). Indeed, X does not factor into irreducibles: X = (X/Y) · Y and X/Y = (X/Y^{2}) · Y and so on. Correspondingly, there is an ascending chain of ideals that does not stabilize:

⟨X⟩ ⫋ ⟨X/Y ⟩ ⫋ ⟨X/Y^{2}⟩ ⫋ · · · .

Proposition (16.7). — Let R be a Noetherian ring, S a multiplicative subset, 𝔞 an ideal. Then R/𝔞 and S^{−1}R are Noetherian.

Proof: If R satisfies the acc, so do R/𝔞 and S^{−1}R by (1.8) and by (11.18)(1). Alternatively, any ideal 𝔟/𝔞 of R/𝔞 is, clearly, generated by the images of gener-

ators of 𝔟. Similarly, any ideal 𝔟 of S^{−1}R is generated by the images of generators of φ_{S}^{−1}𝔟 by (11.17)(1)(b). □

Exercise (16.8). — Let R be a ring, X a variable, R[X] the polynomial ring. Prove this statement or find a counterexample: if R[X] is Noetherian, then so is R.

Theorem (16.9) (Cohen). — A ring is Noetherian if every prime ideal is finitely generated.

Proof: Let R be a ring. Suppose there are non-finitely-generated ideals. Given a nonempty set of them {𝔞_{λ}} that is linearly ordered by inclusion, set 𝔞 := ⋃ 𝔞_{λ}. If 𝔞 is finitely generated, then all the generators lie in some 𝔞_{λ}, so generate 𝔞_{λ}, a contradiction. Thus 𝔞 is non-finitely-generated. Hence, by Zorn’s Lemma, there is a maximal non-finitely-generated ideal 𝔭. In particular, 𝔭 ≠ R.

Assume every prime is finitely generated. Then there are a, b ∈ R−𝔭 with ab ∈ 𝔭. So 𝔭 + ⟨a⟩ is finitely generated, say by x_{1} + w_{1}a, …, x_{n} + w_{n}a with x_{i} ∈ 𝔭. Then {x_{1}, …, x_{n}, a} generate 𝔭 + ⟨a⟩.

Set 𝔟 = Ann((𝔭 + ⟨a⟩)/𝔭). Then 𝔟 ⊃ ⟨b⟩ + 𝔭 and b ∉ 𝔭. So 𝔟 is finitely generated, say by y_{1}, …, y_{m}. Take z ∈ 𝔭. Then z ∈ 𝔭 + ⟨a⟩, so write

z = a_{1}x_{1} + ⋯ + a_{n}x_{n} + ya

with a_{i}, y ∈ R. Then ya ∈ 𝔭. So y ∈ 𝔟. Hence y = b_{1}y_{1} + ⋯ + b_{m}y_{m} with b_{j} ∈ R. Thus 𝔭 is generated by {x_{1}, …, x_{n}, ay_{1}, …, ay_{m}}, a contradiction. Thus there are no non-finitely-generated ideals; in other words, R is Noetherian. □

Lemma (16.10). — If a ring R is Noetherian, then so is the polynomial ring R[X].

Proof: By way of contradiction, assume there is an ideal 𝔞 of R[X] that is not finitely generated. Set 𝔞_{0} := ⟨0⟩. For each i ≥ 1, choose inductively f_{i} ∈ 𝔞 − 𝔞_{i−1} of least degree d_{i}, and set 𝔞_{i} := ⟨f_{1}, …, f_{i}⟩. Let a_{i} be the leading coefficient of f_{i}, and 𝔟 the ideal generated by all the a_{i}. Since R is Noetherian, 𝔟 = ⟨a_{1}, …, a_{n}⟩ for some n by (16.2). Then a_{n+1} = r_{1}a_{1} + ⋯ + r_{n}a_{n} with r_{i} ∈ R.

By construction, d_{i} ≤ d_{i+1} for all i. Set

f := f_{n+1} − (r_{1}f_{1}X^{d_{n+1}−d_{1}} + ⋯ + r_{n}f_{n}X^{d_{n+1}−d_{n}}).

Then deg(f) < d_{n+1}, so f ∈ 𝔞_{n}. Therefore, f_{n+1} ∈ 𝔞_{n}, a contradiction. □

Theorem (16.11) (Hilbert Basis). — Let R be a Noetherian ring, R′ a finitely generated algebra. Then R′ is Noetherian.

Proof: Say x_{1}, …, x_{r} generate R′ over R, and let P := R[X_{1}, …, X_{r}] be the polynomial ring in r variables. Then P is Noetherian by (16.10) and induction on r. Assigning x_{i} to X_{i} defines an R-algebra map P → R′, and obviously, it is surjective. Hence R′ is Noetherian by (16.7). □

(16.12) (Noetherian modules). — We call a module M Noetherian if every submodule is finitely generated. In particular, a ring is Noetherian as a ring if and only if it is Noetherian as a module, because its submodules are just the ideals.

We say the ascending chain condition (acc) is satisfied in M if every ascending chain of submodules M_{0} ⊂ M_{1} ⊂ ⋯ stabilizes. We say the maximal condition (maxc) is satisfied in M if every nonempty set of submodules contains ones maximal under inclusion. It is simple to generalize (16.5): These conditions are equivalent:

(1) M is Noetherian; (2) acc is satisfied in M; (3) maxc is satisfied in M.

Lemma (16.13). — Let R be a ring, M a module. Nested submodules M_{1} ⊂ M_{2} of M are equal if both these equations hold:

M_{1} ∩ N = M_{2} ∩ N and (M_{1} + N)/N = (M_{2} + N)/N.

Proof: Given m_{2} ∈ M_{2}, there is m_{1} ∈ M_{1} with n := m_{2} − m_{1} ∈ N. Then n ∈ M_{2} ∩ N = M_{1} ∩ N. Hence m_{2} ∈ M_{1}. Thus M_{1} = M_{2}. □

Exercise (16.14). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence of R-modules, and M_{1}, M_{2} two submodules of M. Prove or give a counterexample to this statement: if β(M_{1}) = β(M_{2}) and α^{−1}(M_{1}) = α^{−1}(M_{2}), then M_{1} = M_{2}.

Proposition (16.15). — Let R be a ring, M a module, N a submodule. (1) Then M is finitely generated if N and M/N are finitely generated.

(2) Then M is Noetherian if and only if N and M/N are Noetherian.

Proof: Assertion (1) is equivalent to (5.6) owing to (5.2).

To prove (2), first assume M is Noetherian. A submodule N′ of N is also a submodule of M, so N′ is finitely generated; thus N is Noetherian. A submodule of M/N is finitely generated as its inverse image in M is so; thus M/N is Noetherian.

Conversely, assume N and M/N are Noetherian. Let P be a submodule of M. Then P ∩N and (P +N)/N are finitely generated. But P/(P ∩N) ⟶^{∼} (P +N)/N by (4.8.2). So (1) implies P is finitely generated. Thus M is Noetherian.

Here is a second proof of (2). First assume M is Noetherian. Then any ascending chain in N is also a chain in M, so it stabilizes. And any chain in M/N is the image of a chain in M, so it too stabilizes. Thus N and M/N are Noetherian.

September 3, 2012 11Nts.tex

Conversely, assume N and M/N are Noetherian. Given M_{1} ⊂ M_{2} ⊂ ⋯ ⊂ M, both (M_{1} ∩ N) ⊂ (M_{2} ∩ N) ⊂ ⋯ and (M_{1} + N)/N ⊂ (M_{2} + N)/N ⊂ ⋯ stabilize, say M_{j} ∩ N = M_{j+1} ∩ N = ⋯ and (M_{j} + N)/N = (M_{j+1} + N)/N = · · ·. Then M_{j} = M_{j+1} = ⋯ by (16.13). Thus M is Noetherian. □

Corollary (16.16). — Modules M_{1}, …, M_{r} are Noetherian if and only if their direct sum M_{1} ⊕ ⋯ ⊕ M_{r} is Noetherian.

Proof: The sequence 0 → M_{1} → M_{1} ⊕ (M_{2} ⊕ ⋯ ⊕ M_{r}) → M_{2} ⊕ ⋯ ⊕ M_{r} → 0 is exact. So the assertion results from (16.15)(2) by induction on r. □

Exercise (16.17). — Let R be a ring, 𝔞_{1}, …, 𝔞_{r} ideals such that each R/𝔞_{i} is a Noetherian ring. Prove (1) that ⨁ R/𝔞_{i} is a Noetherian R-module, and (2) that, if ⋂ 𝔞_{i} = 0, then R too is a Noetherian ring.

Theorem (16.18). — Let R be a Noetherian ring, and M a module. Then the following conditions on M are equivalent:

(1) M is Noetherian; (2) M is finitely generated; (3) M is finitely presented.

Proof: Assume (2). Then there is an exact sequence 0 → K → R^{n} → M → 0. Now, R^{n} is Noetherian by (16.16) and by (16.12). Hence K is finitely generated, so (3) holds; further, (1) holds by (16.15)(2). Trivially, (1) or (3) implies (2). □

Theorem (16.19) (E. Artin–Tate). — Let R ⊂ R′ ⊂ R′′ be rings. Assume R is Noetherian. Assume R′′ is module finite over R′, and R′′ is algebra finite over R. Then R′ is algebra finite over R.

Proof: Say x_{1}, …, x_{m} generate R′′ as an R-algebra, and y_{1}, …, y_{n} generate R′′ as an R′-module. Then there exist z_{ij} ∈ R′ and z_{ijk} ∈ R′ with

x_{i} = ∑ z_{ij}y_{j} and y_{i}y_{j} = ∑ z_{ijk}y_{k}.  (16.19.1)

Let R′_{0} be the R-algebra generated by the z_{ij} and the z_{ijk}. Since R is Noetherian, so is R′_{0} by the Hilbert Basis Theorem, (16.11).

Any x ∈ R′′ is a polynomial in the x_{i} with coefficients in R. Therefore, (16.19.1) implies that x is a linear combination of the y_{j} with coefficients in R′_{0}. But R′_{0} is a Noetherian ring, and R′ is an R′_{0}-submodule of R′′. Hence R′ is module finite over R′_{0} by (16.15). Since R′_{0} is algebra finite over R, it follows that R′ is too. □

Exercise (16.20). — Let G be a finite group acting on a domain R, and R′ the subring of invariants. Let k ⊂ R′ be a field. Using (10.17), prove this celebrated theorem of E. Noether (1926): if R is algebra finite over k, then so is R′.

Example (16.21). — Set δ := \sqrt{−5}, set R := ℤ[δ], and set 𝔭 := (2, 1 + δ). Let’s prove that 𝔭 is finitely presented and that 𝔭R_{𝔮} is free of rank 1 over R_{𝔮} for every maximal ideal 𝔮 of R, but that 𝔭 is not free. Thus the equivalent conditions of (13.27) do not imply that P is free.

Since ℤ is Noetherian and since R is generated over ℤ, the Hilbert Basis Theorem (16.11) yields that R is Noetherian. So since 𝔭 is generated by two elements, (16.18) yields that 𝔭 is finitely presented.

Recall from [1, pp. 417, 421, 425] that 𝔭 is maximal in R, but not principal. Now, 3 ∉ 𝔭; otherwise, 1 ∈ 𝔭 as 2 ∈ 𝔭, but 𝔭 ≠ R. So (1 − δ)/3 ∈ R_{𝔭}. Hence (1 + δ)R_{𝔭} contains (1 + δ)(1 − δ)/3, or 2. So (1 + δ)R_{𝔭} = 𝔭R_{𝔭}. Since R_{𝔭} is a domain, the map μ_{1+δ}: R_{𝔭} → 𝔭R_{𝔭} is injective, so bijective. Thus 𝔭R_{𝔭} is free of rank 1.

September 3, 2012 11Nts.tex

Let 𝔮 be a maximal ideal distinct from 𝔭. Then 𝔭 ∩ (R − 𝔮) ≠ ∅; so, 𝔭R_{𝔮} = R_{𝔮} by (11.12)(2). Thus 𝔭R_{𝔮} is free of rank 1.

Finally, suppose 𝔭 ≃ R^{n}. Set S := R − 0. Then S^{−1}R is the fraction field, K say, of R. So S^{−1}𝔭 ≃ K^{n}. But the inclusion 𝔭 ↪ R yields an injection S^{−1}𝔭 ↪ K. Hence S^{−1}𝔭 ⟶^{∼} K, since S^{−1}𝔭 is a nonzero K-vector space. Therefore, n = 1. So 𝔭 ≃ R. Hence 𝔭 is generated by one element. But 𝔭 is not principal. So there is a contradiction. Thus 𝔭 is not free.

Definition (16.22). — We say a module M is Artinian or the descending chain condition (dcc) is satisfied in M if every descending chain of submodules stabilizes.

We say the ring itself is Artinian if it is an Artinian module.

We say the minimal condition (minc) is satisfied in M if every nonempty set of submodules has a minimal member.

Proposition (16.23). — Let M_{1}, …, M_{r}, M be modules, N a submodule of M. (1) Then M is Artinian if and only if minc is satisfied in M.

(2) Then M is Artinian if and only if N and M/N are Artinian.

(3) Then M_{1}, …, M_{r} are Artinian if and only if M_{1} ⊕ ⋯ ⊕ M_{r} is Artinian.

Proof: It is easy to adapt the proof of (16.4), the second proof of (16.15)(2), and the proof of (16.16). □

Exercise (16.24). — Let k be a field, R an algebra. Assume that R is finite dimensional as a k-vector space. Prove that R is Noetherian and Artinian.

Exercise (16.25). — Let p be a prime number, and set M := ℤ[1/p]/ℤ ⊂ ℚ/ℤ. Prove that any ℤ-submodule N ⊂ M is either finite or all of M. Deduce that M is an Artinian ℤ-module, and that it is not Noetherian.

Exercise (16.26). — Let R be an Artinian ring. Prove that R is a field if it is a domain. Deduce that, in general, every prime ideal 𝔭 of R is maximal.

17. Associated Primes

Given a module, a prime is associated to it if the prime is equal to the annihilator of an element. Given a subset of the set of all associated primes, we prove there is a submodule whose own associated primes constitute that subset. If the ring is Noetherian, then the set of annihilators of elements has maximal members; we prove the latter are prime, so associated. Then the union of all the associated primes is the set of zerodivisors on the module. If also the module is finitely generated, then the intersection is the set of nilpotents. Lastly, we prove there is then a finite chain of submodules whose successive quotients are cyclic with prime annihilators; these primes include all associated primes, which are, therefore, finite in number.

Definition (17.1). — Let R be a ring, M a module. A prime ideal 𝔭 is said to be associated to M if there is a (nonzero) m ∈ M with 𝔭 = Ann(m). The set of associated primes is denoted by Ass(M) or Ass_{R}(M).

The primes that are minimal in Ass(M) are called the minimal primes of M; the others, the embedded primes.

Warning: following a old custom, we mean by the associated primes of an ideal 𝔞 not those of 𝔞 viewed as an abstract module, but rather those of R/𝔞.

Lemma (17.2). — Let R be a ring, M a module, and 𝔭 a prime ideal. Then 𝔭 ∈ Ass(M) if and only if there is an R-injection R/𝔭 ↪ M.

Proof: Assume 𝔭 = Ann(m) with m ∈ M. Define a map R → M by x ↦ xm. This map induces an R-injection R/𝔭 ↪ M.

Conversely, suppose there is an R-injection R/𝔭 ↪ M, and let m ∈ M be the image of 1. Then 𝔭 = Ann(m), so 𝔭 ∈ Ass(M). □

Proposition (17.3). — Let M be a module. Then Ass(M) ⊂ Supp(M).

Proof: Let 𝔭 ∈ Ass_{R}(M). Say 𝔭 = Ann(m). Then m/1 ∈ M_{𝔭} is nonzero as no x ∈ (R−𝔭) satisfies xm = 0. Alternatively, (17.2) yields an R-injection R/𝔭 ↪ M. It induces an injection (R/𝔭)_{𝔭} ↪ M_{𝔭} by (12.16). But (R/𝔭)_{𝔭} = Frac(R/𝔭) by (12.19). Thus M_{𝔭} ≠ 0 and so 𝔭 ∈ Supp(M). □

Lemma (17.4). — Let R be a ring, 𝔭 a prime ideal, m ∈ R/𝔭 a nonzero element. Then (1) Ann(m) = 𝔭 and (2) Ass(R/𝔭) = {𝔭}.

Proof: To prove (1), say m is the residue of y ∈ R. Let x ∈ R. Then xm = 0 if and only if xy ∈ 𝔭, so if and only if x ∈ 𝔭, as 𝔭 is prime and m ≠ 0. Thus (1) holds.

Trivially, (1) implies (2). □

Proposition (17.5). — Let M be a module, N a submodule. Then

Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(M/N).

Proof: Take m ∈ N. Then the annihilator of m is the same whether m is regarded as an element of N or of M. So Ass(N) ⊂ Ass(M).

Let 𝔭 ∈ Ass(M). Then (17.2) yields an R-injection R/𝔭 ↪ M. Denote its image by E. If E ∩ N = 0, then the composition R/𝔭 → M → M/N is injective; hence, 𝔭 ∈ Ass(M/N) by (17.2). Else, take a nonzero m ∈ E ∩ N. Then Ann(m) = 𝔭 by (17.4)(1). Thus 𝔭 ∈ Ass(N). □

September 3, 2012 11Nts.tex Exercise (17.6). — Given modules M_{1}, …, M_{r}, set M := M_{1} ⊕ ⋯ ⊕ M_{r}. Prove

Ass(M) = Ass(M_{1}) ∪ ⋯ ∪ Ass(M_{r}).

Exercise (17.7). — Take R := ℤ and M := ℤ/⟨2⟩ ⊕ ℤ. Find Ass(M) and find two submodules L, N ⊂ M with L + N = M but Ass(L) ∪ Ass(N) ⫋ Ass(M).

Proposition (17.8). — Let M be a module, and Ψ a subset of Ass(M). Then there is a submodule N of M with Ass(M/N) = Ψ and Ass(N) = Ass(M) − Ψ.

Proof: Given submodules N_{λ} of M totally ordered by inclusion, set N := ⋃ N_{λ}. Given 𝔭 ∈ Ass(N), say 𝔭 = Ann(m). Then m ∈ N_{λ} for some λ; so 𝔭 ∈ Ass(N_{λ}). Conversely, Ass(N_{λ}) ⊂ Ass(N) by (17.5). Thus Ass(N) = ⋃ Ass(N_{λ}).

So we may apply Zorn’s Lemma to obtain a submodule N of M that is maximal with Ass(N) ⊂ Ass(M) − Ψ. By (17.5), it suffices to show that Ass(M/N) ⊂ Ψ.

Take 𝔭 ∈ Ass(M/N). Then M/N has a submodule N′/N isomorphic to R/𝔭 by (17.2). So Ass(N′) ⊂ Ass(N) ∪ {𝔭} by (17.5) and (17.4)(2). Now, N′ ⫌ N and N is maximal with Ass(N) ⊂ Ass(M) − Ψ. Hence 𝔭 ∈ Ass(N′) ⊂ Ass(M), but 𝔭 ∉ Ass(M) − Ψ. Thus 𝔭 ∈ Ψ. □

Proposition (17.9). — Let R be a ring, S a multiplicative subset, M a module, and 𝔭 a prime ideal. If 𝔭 ∩ S = ∅ and 𝔭 ∈ Ass(M), then S^{−1}𝔭 ∈ Ass(S^{−1}M); the converse holds if 𝔭 is finitely generated.

Proof: Assume 𝔭 ∈ Ass(M). Then (17.2) yields an injection R/𝔭 ↪ M. It induces an injection S^{−1}(R/𝔭) ↪ S^{−1}M by (12.16). But S^{−1}(R/𝔭) = S^{−1}R/S^{−1}𝔭 by (12.18). Assume 𝔭 ∩ S = ∅ also. Then 𝔭S^{−1}R is prime by (11.17)(3)(b). But 𝔭S^{−1}R = S^{−1}𝔭 by (12.2). Thus S^{−1}𝔭 ∈ Ass(S^{−1}M).

Conversely, assume S^{−1}𝔭 ∈ Ass(S^{−1}M). Then there are m ∈ M and t ∈ S with S^{−1}𝔭 = Ann(m/t). Say 𝔭 = ⟨x_{1}, …, x_{n}⟩. Fix i. Then x_{i}m/t = 0. So there is s_{i} ∈ S with s_{i}x_{i}m = 0. Set s := ∏ s_{i}. Then x_{i} ∈ Ann(sm). Thus 𝔭 ⊂ Ann(sm).

Take b ∈ Ann(sm). Then bsm/st = 0. So b/1 ∈ S^{−1}𝔭. So b ∈ 𝔭 by (11.17)(1)(a) and (11.17)(3)(a). Thus 𝔭 ⊃ Ann(sm). So 𝔭 = Ann(sm). Thus 𝔭 ∈ Ass(M).

Finally, 𝔭 ∩ S = ∅ by (11.18)(2), as S^{−1}𝔭 is prime. □

Exercise (17.10). — Let R be a ring, and suppose R_{𝔭} is a domain for every prime 𝔭. Prove every associated prime of R is minimal.

Lemma (17.11). — Let R be a ring, M a module, and 𝔞 an ideal. Suppose 𝔞 is maximal in the set of annihilators of nonzero elements m of M. Then 𝔞 ∈ Ass(M).

Proof: Say 𝔞 := Ann(m) with m ≠ 0. Then 1 ∉ 𝔞 as m ≠ 0. Now, take b, c ∈ R with bc ∈ 𝔞, but c ∉ 𝔞. Then bcm = 0, but cm ≠ 0. Plainly, 𝔞 ⊂ Ann(cm). So 𝔞 = Ann(cm) by maximality. But b ∈ Ann(cm), so b ∈ 𝔞. Thus 𝔞 is prime. □

Proposition (17.12). — Let R be a Noetherian ring, M a module. Then M = 0 if and only if Ass(M) = ∅.

Proof: Obviously, if M = 0, then Ass(M) = ∅. Conversely, suppose M ≠ 0. Let 𝒮 be the set of annihilators of nonzero elements of M. Then 𝒮 has a maximal element 𝔞 by (16.5). By (17.11), 𝔞 ∈ Ass(M). Thus Ass(M) ≠ ∅. □

Definition (17.13). — Let R be a ring, M a module, x ∈ R. We say x is a zerodivisor on M if there is a nonzero m ∈ M with xm = 0; otherwise, we say x is a nonzerodivisor. We denote the set of zerodivisors by z.div(M).

Proposition (17.14). — Let R be a Noetherian ring, M a module. Then

z.div(M) = ⋃_{𝔭∈Ass(M)} 𝔭.

Proof: Given x ∈ z.div(M), say xm = 0 where m ∈ M and m ≠ 0. Then x ∈ Ann(m). But Ann(m) is contained in an ideal 𝔭 that is maximal among annihilators of nonzero elements because of (16.5); hence, 𝔭 ∈ Ass(M) by (17.11). Thus z.div(M) ⊂ ⋃ 𝔭. The opposite inclusion results from the definitions. □

Exercise (17.15). — Let R be a Noetherian ring, M a module, N a submodule, x ∈ R. Show that, if x ∉ 𝔭 for any 𝔭 ∈ Ass(M/N), then xM ∩ N = xN.

Lemma (17.16). — Let R be a Noetherian ring, M a module. Then

Supp(M) = ⋃_{𝔮∈Ass(M)} V(𝔮) ⊃ Ass(M).

Proof: Let 𝔭 be a prime. Then R_{𝔭} is Noetherian by (16.7) as R is. So M_{𝔭} ≠ 0 if and only if Ass_{R_{𝔭}}(M_{𝔭}) ≠ ∅ by (17.12). But R is Noetherian; so Ass_{R_{𝔭}}(M_{𝔭}) ≠ ∅ if and only if there is 𝔮 ∈ Ass(M) with 𝔮∩(R−𝔭) = ∅, or 𝔮 ⊂ 𝔭, owing to (11.18)(2) and (17.9). Thus 𝔭 ∈ Supp(M) if and only if 𝔭 ∈ V(𝔮) for some 𝔮 ∈ Ass(M). □

Theorem (17.17). — Let R be a Noetherian ring, M a module, 𝔭 ∈ Supp(M). Then 𝔭 contains some 𝔮 ∈ Ass(M); if 𝔭 is minimal in Supp(M), then 𝔭 ∈ Ass(M).

Proof: By (17.16), 𝔮 exists. Also, 𝔮 ∈ Supp(M); so 𝔮 = 𝔭 if 𝔭 is minimal. □

Theorem (17.18). — Let R be a Noetherian ring, and M a finitely generated module. Then

nil(M) = ⋂_{𝔭∈Ass(M)} 𝔭.

Proof: Since M is finitely generated, nil(M) = ⋂_{𝔭∈Supp(M)} 𝔭 by (13.11). Since R is Noetherian, given 𝔭 ∈ Supp(M), there is 𝔮 ∈ Ass(M) with 𝔮 ⊂ 𝔭 by (17.16). The assertion follows. □

Lemma (17.19). — Let R be a Noetherian ring, M a finitely generated module. Then there exists a chain of submodules

0 = M_{0} ⊂ M_{1} ⊂ ⋯ M_{n−1} ⊂ M_{n} = M

with M_{i}/M_{i−1} ≃ R/𝔭_{i} for some prime 𝔭_{i} for i = 1, …, n. For any such chain,

Ass(M) ⊂ {𝔭_{1}, …, 𝔭_{n}} ⊂ Supp(M).  (17.19.1)

Proof: Among all submodules of M having such a chain, there is a maximal submodule N by (16.18) and (16.12). Suppose M/N ≠ 0. Then by (17.12), the quotient M/N contains a submodule N′/N isomorphic to R/𝔭 for some prime 𝔭. Then N ⫋ N′, contradicting maximality. Hence N = M. Thus a chain exists.

The first inclusion of (17.19.1) follows by induction from (17.5) and (17.4)(2). Now, 𝔭_{i} ∈ Supp(R/𝔭_{i}) owing to (12.19). Thus (17.19.1) follows from (13.9)(1).

□

Theorem (17.20). — Let R be a Noetherian ring, and M a finitely generated module. Then the set Ass(M) is finite.

Proof: The assertion follows directly from (17.19). □

Exercise (17.21). — Let R be a Noetherian ring, 𝔞 an ideal. Prove the primes minimal containing 𝔞 are associated to 𝔞. Prove such primes are finite in number.

Exercise (17.22). — Take R := ℤ and M := ℤ in (17.19). Determine when a chain 0 ⊂ M_{1} ⫋ M is acceptable, and show that then 𝔭_{2} ∉ Ass(M).

Exercise (17.23). — Take R := ℤ and M := ℤ/⟨12⟩ in (17.19). Find all three acceptable chains, and show that, in each case, {𝔭_{i}} = Ass(M).

Proposition (17.24). — Let R be a Noetherian ring, and M and N finitely generated modules. Then

Ass(Hom(M, N)) = Supp(M) ⋂ Ass(N).

Proof: Take 𝔭 ∈ Ass(Hom(M, N)). Then (17.2) yields an injective R-map R/𝔭 ↪ Hom(M, N). Set k(𝔭) := Frac(R/𝔭). Then k(𝔭) = R_{𝔭}/𝔭R_{𝔭} by (12.19). Now, M is finitely presented by (16.18) as R is Noetherian; hence,

Hom(M, N)_{𝔭} = Hom_{R_{𝔭}}(M_{𝔭}, N_{𝔭})  (17.24.1)

by (12.21)(2). Therefore, by exactness, localizing yields an injection

φ: k(𝔭) ↪ Hom_{R_{𝔭}}(M_{𝔭}, N_{𝔭}).

Hence M_{𝔭} ≠ 0; so 𝔭 ∈ Supp(M). For any m ∈ M_{𝔭} with φ(1)(m) ≠ 0, the map k(𝔭) → N_{𝔭} given by x ↦ φ(x)(m) is nonzero, so an injection. Hence by (17.2), we have 𝔭R_{𝔭} ∈ Ass(N_{𝔭}). Therefore, also 𝔭 ∈ Ass(N) by (17.9).

Conversely, take 𝔭 ∈ Supp(M) ∩ Ass(N). Then M_{𝔭} ≠ 0. So by Nakayama’s Lemma, M_{𝔭}/𝔭M_{𝔭} is a nonzero vector space over k(𝔭). Take any nonzero R-map M_{𝔭}/𝔭M_{𝔭} → k(𝔭), precede it by the canonical map M_{𝔭} → M_{𝔭}/𝔭M_{𝔭}, and follow it by an R-injection k(𝔭) ↪ N_{𝔭}, which exists by (17.2) and (17.9). We obtain a nonzero element of Hom_{R_{𝔭}}(M_{𝔭}, N_{𝔭}), annihilated by 𝔭R_{𝔭}. But 𝔭R_{𝔭} is maximal, so is the entire annihilator. So 𝔭R_{𝔭} ∈ Ass(Hom_{R_{𝔭}}(M_{𝔭}, N_{𝔭})). Hence 𝔭 ∈ Ass(Hom(M, N)) by (17.24.1) and (17.9). □

Proposition (17.25). — Let R be a Noetherian ring, 𝔭 a prime, M a finitely generated module, and x, y ∈ 𝔭 nonzerodivisors on M. Then 𝔭 ∈ Ass(M/xM) if and only if 𝔭 ∈ Ass(M/yM).

Proof: Form the sequence 0 → K → M/xM ⟶^{μ_{y}} M/xM with K := Ker(μ_{y}). Apply the functor Hom(R/𝔭, •) to that sequence, and get this one:

0 → Hom(R/𝔭, K) → Hom(R/𝔭, M/xM) ⟶^{μ_{y}} Hom(R/𝔭, M/xM).

It is exact by (5.17). But y ∈ 𝔭; so the right-hand map vanishes. Thus

Hom(R/𝔭, K) ⟶^{∼} Hom(R/𝔭, M/xM).

Form the following commutative diagram with exact rows:

0 ⟶ M ⟶^{μ_{x}} M ⟶ M/xM ⟶ 0

μ_{y} ↓ ^{μ_{y}} ↓  ^{μ_{y}} ↓

0 ⟶ M ⟶^{μ_{x}} M ⟶ M/xM ⟶ 0

The Snake Lemma (5.12) yields an exact sequence 0 → K → M/yM ⟶^{μ_{x}} M/yM. Hence, similarly, Hom(R/𝔭, K) ⟶^{∼} Hom(R/𝔭, M/yM). Therefore,

Hom(R/𝔭, M/yM) = Hom(R/𝔭, M/xM).  (17.25.1)

Finally, 𝔭 ∈ Supp(R/𝔭) by (13.9)(3). Hence (17.24) yields the assertion. □

18. Primary Decomposition

Primary decomposition of a submodule generalizes factorization of an integer into powers of primes. A submodule is called primary if the quotient module has only one associated prime. We characterize these submodules in various ways over a Noetherian ring, emphasizing the case of ideals. A primary decomposition is a representation of a submodule as a finite intersection of primary submodules. The decomposition is called irredundant, or minimal, if cannot be reduced. We consider several illustrative examples in a polynomial ring.

Then we prove existence and uniqueness theorems for a proper submodule of a finitely generated module over a Noetherian ring. The celebrated Lasker–Noether Theorem asserts the existence of an irredundant primary decomposition. The First Uniqueness Theorem asserts the uniqueness of the primes that arise; they are just the associated primes of the quotient. The Second Uniqueness Theorem asserts the uniqueness of the primary components whose primes are minimal among these associated primes; the other primary components may vary.

Definition (18.1). — Let R be a ring, M a module, Q a submodule. If Ass(M/Q) consists of a single prime 𝔭, we say Q is primary or 𝔭-primary in M.

Example (18.2). — A prime 𝔭 is 𝔭-primary, as Ass(R/𝔭) = {𝔭} by (17.4)(2).

Proposition (18.3). — Let R be a Noetherian ring, M a finitely generated module, Q a submodule. If Q is 𝔭-primary, then 𝔭 = nil(M/Q).

Proof: The assertion holds as nil(M/Q) = ⋂_{𝔮∈Ass(M/Q)} 𝔮 by (17.18). □

Theorem (18.4). — Let R be a Noetherian ring, M a nonzero finitely generated module, Q a submodule. Set 𝔭 := nil(M/Q). Then these conditions are equivalent:

(1) 𝔭 is prime and Q is 𝔭-primary. (2) 𝔭 = z.div(M/Q).

(3) Given x ∈ R and m ∈ M with xm ∈ Q but m ∉ Q, necessarily x ∈ 𝔭.

Proof: Recall 𝔭 = ⋂_{𝔮∈Ass(M/Q)} 𝔮 by (17.18), and z.div(M/Q) = ⋃_{𝔮∈Ass(M/Q)} 𝔮 by (17.14). Thus 𝔭 ⊂ z.div(M/Q).

Further, (2) holds if Ass(M/Q) = {𝔭}, that is, if (1) holds.

Conversely, if x ∈ 𝔮 ∈ Ass(M/Q), but x ∉ 𝔮′ ∈ Ass(M/Q), then x ∉ 𝔭, but x ∈ z.div(M/Q); hence, (2) implies (1). Thus (1) and (2) are equivalent.

Clearly, (3) means every zerodivisor on M/Q is nilpotent, or 𝔭 ⊃ z.div(M/Q). But the opposite inclusion always holds. Thus (2) and (3) are equivalent. □

Corollary (18.5). — Let R be a Noetherian ring, and 𝔮 a proper ideal. Set 𝔭 := \sqrt{𝔮}. Then 𝔮 is primary (in R) if and only if, given x, y ∈ R with xy ∈ 𝔮 but x ∉ 𝔮, necessarily y ∈ 𝔭; if so, then 𝔭 is prime and 𝔮 is 𝔭-primary.

Proof: Clearly 𝔮 = Ann(R/𝔮). So 𝔭 = nil(R/𝔮). So the assertions result directly from (18.4) and (18.3). □

Exercise (18.6). — Let R be a ring, and 𝔭 = ⟨p⟩ a principal prime generated by a nonzerodivisor p. Show every positive power 𝔭^{n} is 𝔭-primary. Show conversely, if R is Noetherian, then every 𝔭-primary ideal 𝔮 is equal to some power 𝔭^{n}.

Exercise (18.7). — Let k be a field, and k[X, Y] the polynomial ring. Let 𝔞 be the ideal ⟨X^{2}, XY ⟩. Show 𝔞 is not primary, but \sqrt{𝔞} is prime. Show 𝔞 satisfies this condition: ab ∈ 𝔞 implies a^{2} ∈ 𝔞 or b^{2} ∈ 𝔞.

Exercise (18.8). — Let φ: R → R′ be a homomorphism of Noetherian rings, and 𝔮 ⊂ R′ a 𝔭-primary ideal. Show that φ^{−1}𝔮 ⊂ R is φ^{−1}𝔭-primary. Show that the converse holds if φ is surjective.

Proposition (18.9). — Let R be a Noetherian ring, M a finitely generated module, Q a submodule. Set 𝔭 := nil(M/Q). If 𝔭 is maximal, then Q is 𝔭-primary.

Proof: Since 𝔭 = ⋂_{𝔮∈Ass(M/Q)} 𝔮 by (17.18), if 𝔭 is maximal, then 𝔭 = 𝔮 for any 𝔮 ∈ Ass(M/Q), or {𝔭} = Ass(M/Q), as desired. □

Corollary (18.10). — Let R be a Noetherian ring, 𝔮 an ideal. Set 𝔭 := \sqrt{𝔮}. If 𝔭 is maximal, then 𝔮 is 𝔭-primary.

Proof: Since 𝔭 = nil(R/𝔮), the assertion is a special case of (18.9). □

Corollary (18.11). — Let R be a Noetherian ring, 𝔪 a maximal ideal. An ideal 𝔮 is 𝔪-primary if and only if there exists n ≥ 1 such that 𝔪^{n} ⊂ 𝔮 ⊂ 𝔪.

Proof: The condition 𝔪^{n} ⊂ 𝔮 ⊂ 𝔪 just means that 𝔪 := \sqrt{𝔮} by (3.26). So the assertion results from (18.5) and (18.10). □

Lemma (18.12). — Let R be a Noetherian ring, 𝔭 a prime ideal, M a module. Let Q_{1} and Q_{2} be 𝔭-primary submodules; set Q := Q_{1} ∩ Q_{2}. Then Q is 𝔭-primary.

Proof: Form the canonical map M → M/Q_{1} ⊕ M/Q_{2}. Its kernel is Q, so it induces an injection M/Q ↪ M/Q_{1} ⊕ M/Q_{2}. Hence (17.12) and (17.5) yield

∅ ≠ Ass(M/Q) ⊂ Ass(M/Q_{1}) ∪ Ass(M/Q_{2}).

Since the latter two sets are each equal to {𝔭}, so is Ass(M/Q), as desired. □

(18.13) (Primary decomposition). — Let R be a ring, M a module, and N a submodule. A primary decomposition of N is a decomposition

N = Q_{1} ∩ ⋯ ∩ Q_{r} with the Q_{i} primary.

We call the decomposition irredundant or minimal if these conditions are satisfied:

(1) N ≠ ⋂_{j≠i} Q_{j}, or equivalently, ⋂_{j≠i} Q_{j} ⊄ Q_{i} for i = 1, …, r.
(2) Say Q_{i} is 𝔭_{i}-primary for i = 1, …, r. Then 𝔭_{1}, …, 𝔭_{r} are distinct.

If R is Noetherian, then owing to (18.12), any primary decomposition can be made irredundant by intersecting all the primary submodules with the same prime and then discarding those of them that are not needed.

Example (18.14). — Let k be a field, R := k[X, Y] the polynomial ring, and 𝔞 := ⟨X^{2}, XY ⟩. Below, it is proved that, for any n ≥ 1,

𝔞 = ⟨X⟩ ∩ ⟨X^{2}, XY, Y^{n}⟩ = ⟨X⟩ ∩ ⟨X^{2}, Y ⟩.  (18.14.1)

Here ⟨X^{2}, XY, Y^{n}⟩ and ⟨X^{2}, Y ⟩ contain ⟨X, Y ⟩^{n}, so are ⟨X, Y ⟩-primary by (18.11). Thus (18.14.1) shows infinitely many distinct primary decompositions of 𝔞. They are clearly irredundant. Note: the ⟨X, Y ⟩-primary component is not unique!

Plainly, 𝔞 ⊂ ⟨X⟩ and 𝔞 ⊂⟨X^{2}, XY, Y^{n}⟩ ⊂ ⟨X^{2}, Y ⟩. To see 𝔞 ⊃ ⟨X⟩ ∩ ⟨X^{2}, Y ⟩,

September 3, 2012 11Nts.tex take F ∈ ⟨X⟩ ∩ ⟨X^{2}, Y ⟩. Then F = GX = AX^{2} + BY where A, B, G ∈ R. Then X(G − AX) = BY. So X | B. Say B = B′X. Then F = AX^{2} + B′XY ∈ 𝔞.

Example (18.15). — Let k be a field, P := k[X, Y, Z] the polynomial ring. Set R := P/⟨XZ − Y^{2}⟩. Let x, y, z be the residues of X, Y, Z in R. Set 𝔭 := ⟨x, y⟩. Clearly 𝔭^{2} = ⟨x^{2}, xy, y^{2}⟩ = x⟨x, y, z⟩. Let’s show that 𝔭^{2} = ⟨x⟩ ∩ ⟨x^{2}, y, z⟩ is an irredundant primary decomposition.

First note the inclusions x⟨x, y, z⟩ ⊂ ⟨x⟩ ∩ ⟨x, y, z⟩^{2} ⊂ ⟨x⟩ ∩ ⟨x^{2}, y, z⟩.

Conversely, given f ∈ ⟨x⟩ ∩ ⟨x^{2}, y, z⟩, represent f by GX with G ∈ P. Then

GX = AX^{2} + BY + CZ + D(XZ − Y^{2}) with A, B, C, D ∈ P.

So (G − AX)X = B′Y + C′Z with B′, C′ ∈ P. Say G − AX = A′′ + B′′Y + C′′Z with A′′ ∈ k[X] and B′′, C′′ ∈ P. Then

A′′X = −B′′XY − C′′XZ + B′Y + C′Z = (B′ − B′′X)Y + (C′ − C′′X)Z;

whence, A′′ = 0. Therefore, GX ∈ X⟨X, Y, Z⟩. Thus 𝔭^{2} =⟨x⟩ ∩ ⟨x^{2}, y, z⟩.

The ideal ⟨x⟩ is ⟨x, y⟩-primary in R by (18.8). Indeed, the preimage in P of ⟨x⟩ is ⟨X, Y^{2}⟩ and of ⟨x, y⟩ is ⟨X, Y ⟩. Further, ⟨X, Y^{2}⟩ is ⟨X, Y ⟩-primary, as under the map φ: P → k[Y, Z] with φ(X) = 0, clearly ⟨X, Y^{2}⟩ = φ^{−1}⟨Y^{2}⟩ and ⟨X, Y ⟩ = φ^{−1}⟨Y ⟩; moreover, ⟨Y^{2}⟩ is ⟨Y ⟩-primary by (18.5), or by (18.6).

Finally ⟨x, y, z⟩^{2} ⊂ ⟨x^{2}, y, z⟩ ⊂ ⟨x, y, z⟩ and ⟨x, y, z⟩ is maximal. So ⟨x^{2}, y, z⟩ is ⟨x, y, z⟩-primary by (18.11).

Thus 𝔭^{2} = ⟨x⟩ ∩ ⟨x^{2}, y, z⟩ is a primary decomposition. It is clearly irredundant.

Exercise (18.16). — Let k be a field, R := k[X, Y, Z] be the polynomial ring. Set 𝔞 := ⟨XY, X − Y Z⟩, set 𝔮_{1} := ⟨X, Z⟩ and set 𝔮_{2} := ⟨Y^{2}, X − Y Z⟩. Show that 𝔞 = 𝔮_{1} ∩ 𝔮_{2} and that this expression is an irredundant primary decomposition.

Exercise (18.17). — Let R := R′ × R′′ be a product of two domains. Find an irredundant primary decomposition of ⟨0⟩.

Lemma (18.18). — Let R be a ring, M a module, N = Q_{1} ∩ ⋯ ∩ Q_{r} a primary decomposition in M. Say Q_{i} is 𝔭_{i}-primary for i = 1, …, r. Then

Ass(M/N) ⊆ {𝔭_{1}, …, 𝔭_{r}}.  (18.18.1)

If equality holds and if 𝔭_{1}, …, 𝔭_{r} are distinct, then the decomposition is irredundant; the converse holds if R is Noetherian.

Proof: Since N = ⋂ Q_{i}, the canonical map is injective: M/N ↪ ⨁ M/Q_{i}. So (17.5) and (17.6) yield Ass(M/N) ⊆ ⋃ Ass(M/Q_{i}). Thus (18.18.1) holds.

If N = Q_{2} ∩ ⋯ ∩ Q_{r}, then Ass(M/N) ⊆ {𝔭_{2}, …, 𝔭_{r}} too. Thus if equality holds in (18.18.1) and if 𝔭_{1}, …, 𝔭_{r} are distinct, then N = Q_{1} ∩ ⋯ ∩ Q_{r} is irredundant.

Conversely, assume N = Q_{1} ∩⋯∩Q_{r} is irredundant. Given i, set P_{i} := ⋂_{j≠i} Q_{j}. Then P_{i} ∩ Q_{i} = N and P_{i}/N ≠ 0. Consider these two canonical injections:

P_{i}/N ↪ M/Q_{i} and P_{i}/N ↪ M/N.

Assume R is Noetherian. Then Ass(P_{i}/N) ≠ ∅ by (17.12). So the first injection yields Ass(P_{i}/N) = {𝔭_{i}} by (17.5); then the second yields 𝔭_{i} ∈ Ass(M/N). Thus Ass(M/N) ⊇ {𝔭_{1}, …, 𝔭_{r}}, and (18.18.1) yields equality, as desired. □

Theorem (18.19) (First Uniqueness). — Let R be a Noetherian ring, and M a module. Let N = Q_{1} ∩ ⋯ ∩ Q_{r} be an irredundant primary decomposition in M; say Q_{i} is 𝔭_{i}-primary for i = 1, …, r. Then 𝔭_{1}, …, 𝔭_{r} are uniquely determined; in fact, they are just the distinct associated primes of M/N.

Proof: The assertion is just part of (18.18). □

Theorem (18.20) (Lasker–Noether). — Over a Noetherian ring, each proper submodule of a finitely generated module has an irredundant primary decomposition.

Proof: Let M be the module, N the submodule. By (17.20), M/N has finitely many distinct associated primes, say 𝔭_{1}, …, 𝔭_{r}. Owing to (17.8), for each i, there is a 𝔭_{i}-primary submodule Q_{i} of M with Ass(Q_{i}/N) = Ass(M/N) − {𝔭_{i}}. Set P := ⋂ Q_{i}. Fix i. Then P/N ⊂ Q_{i}/N. So Ass(P/N) ⊂ Ass(Q_{i}/N) by (17.5). But i is arbitrary. Hence Ass(P/N) = ∅. Therefore, P/N = 0 by (17.12). Finally, the decomposition N = ⋂ Q_{i} is irredundant by (18.18). □

Exercise (18.21). — Let R be a Noetherian ring, 𝔞 an ideal, and M a finitely generated module. Consider the following submodule of M:

Γ_{𝔞}(M) := ⋃_{n≥1}{m ∈ M | 𝔞^{n}m = 0 for some n ≥ 1}.

(1) For any decomposition 0 = ⋂ Q_{i} with Q_{i} 𝔭_{i}-primary, show Γ_{𝔞}(M) = ⋂_{𝔞⊄𝔭_{i}} Q_{i}.

(2) Show Γ_{𝔞}(M) is the set of all m ∈ M such that m/1 ∈ M_{𝔭} vanishes for every prime 𝔭 with 𝔞 ⊄ 𝔭. (Thus Γ_{𝔞}(M) is the set of all m whose support lies in V(𝔞).)

Lemma (18.22). — Let R be a Noetherian ring, S a multiplicative subset, 𝔭 a prime ideal, M a module, and Q a 𝔭-primary submodule. If S ∩ 𝔭 ≠ ∅, then S^{−1}Q = S^{−1}M and Q^{S} = M. If S ∩ 𝔭 = ∅, then S^{−1}Q is S^{−1}𝔭-primary and Q^{S} = φ_{S}^{−1}(S^{−1}Q) = Q.

Proof: Every prime of S^{−1}R is of the form S^{−1}𝔮 where 𝔮 is a prime of R with S ∩ 𝔮 = ∅ by (11.18)(2) and (12.2). And S^{−1}𝔮 ∈ Ass(S^{−1}(M/Q)) if and only if 𝔮 ∈ Ass(M/Q), that is, 𝔮 = 𝔭, by (17.9).

However, S^{−1}(M/Q) = S^{−1}M/S^{−1}Q by (12.16). Therefore, if S ∩ 𝔭 ≠ ∅, then Ass(S^{−1}M/S^{−1}Q) = ∅; whence, (17.12) yields S^{−1}M/S^{−1}Q = 0. Otherwise, if S ∩ 𝔭 = ∅, then Ass(S^{−1}M/S^{−1}Q) = {S^{−1}𝔭}; whence, S^{−1}Q is S^{−1}𝔭-primary.

Finally, Q^{S} = φ_{S}^{−1}(S^{−1}Q) by (12.15)(3). So if S^{−1}Q = S^{−1}M, then Q^{S} = M. Now, suppose S ∩ 𝔭 = ∅. Given m ∈ Q^{S}, there is s ∈ S with sm ∈ Q. But s ∉ 𝔭. Further, 𝔭 = z.div(M/Q) owing to (17.14). Therefore, m ∈ Q. Thus Q^{S} ⊂ Q. But Q^{S} ⊃ Q as 1 ∈ S. Thus Q^{S} = Q. □

Proposition (18.23). — Let R be a Noetherian ring, S a multiplicative subset, M a finitely generated module. Let N = Q_{1} ∩ ⋯ ∩ Q_{r} ⊂ M be an irredundant primary decomposition. Say Q_{i} is 𝔭_{i}-primary for all i, and S ∩ 𝔭_{i} = ∅ just for i ≤ h. Then

S^{−1}N = S^{−1}Q_{1} ∩ ⋯ ∩ S^{−1}Q_{h} ⊂ S^{−1}M and N^{S} = Q_{1} ∩ ⋯ ∩ Q_{h} ⊂ M

are irredundant primary decompositions.

Proof: By (12.15)(4)(b), S^{−1}N = S^{−1}Q_{1} ∩ ⋯ ∩ S^{−1}Q_{r}. Further, by (18.22), S^{−1}Q_{i} is S^{−1}𝔭_{i}-primary for i ≤ h, and S^{−1}Q_{i} = S^{−1}M for i > h. Therefore, S^{−1}N = S^{−1}Q_{1} ∩ ⋯ ∩ S^{−1}Q_{h} is a primary decomposition.

It is irredundant by (18.18). Indeed, Ass(S^{−1}M/S^{−1}N) = {S^{−1}𝔭_{1}, …, S^{−1}𝔭_{h}}

September 3, 2012 11Nts.tex by an argument like that in the first part of (18.22). Further, S^{−1}𝔭_{1}, …, S^{−1}𝔭_{h} are distinct by (11.18)(2) as the 𝔭_{i} are distinct.

Apply φ_{S}^{−1} to S^{−1}N = S^{−1}Q_{1} ∩ ⋯ ∩ S^{−1}Q_{h}. Owing to (12.15)(3), we get N^{S} = Q_{1}^{S} ∩⋯∩Q_{h}^{S}. But Q_{i}^{S} = Q_{i} by (18.22). So N^{S} = Q_{1} ∩⋯∩Q_{h} is a primary decomposition. It is irredundant as, clearly, (18.13)(1) and (2) hold for it, since they hold for N = Q_{1} ∩ ⋯ ∩ Q_{r}. □

Theorem (18.24) (Second Uniqueness). — Let R be a ring, M a module, N a submodule. Assume R is Noetherian and M is finitely generated. Let 𝔭 be a minimal prime of M/N. Then, in any irredundant primary decomposition of N in M, the 𝔭-primary component Q is uniquely determined; in fact, Q = N^{S} where S := R−𝔭.

Proof: In (18.23), take S := R − 𝔭. Then h = 1 as 𝔭 is minimal. □

Exercise (18.25). — Let R be a Noetherian ring, M a finitely generated module, N a submodule. Prove N = ⋂_{𝔭∈Ass(M/N)} φ_{𝔭}^{−1}(N_{𝔭}).

Exercise (18.26). — Let R be a Noetherian ring, 𝔭 a prime. Its nth symbolic power 𝔭^{(n)} is defined as the saturation (𝔭^{n})^{S} where S := R − 𝔭.

(1) Show 𝔭^{(n)} is the 𝔭-primary component of 𝔭^{n}.

(2) Show 𝔭^{(m+n)} is the 𝔭-primary component of 𝔭^{(n)}𝔭^{(m)}.

(3) Show 𝔭^{(n)} = 𝔭^{n} if and only if 𝔭^{n} is 𝔭-primary.

(4) Given a 𝔭-primary ideal 𝔮, show 𝔮 ⊃ 𝔭^{(n)} for all large n.

Exercise (18.27). — Let R be a Noetherian ring, ⟨0⟩ = 𝔮_{1}∩⋯∩𝔮_{n} an irredundant primary decomposition. Set 𝔭_{i} := \sqrt{𝔮_{i}} for i = 1, …, n.

(1) Suppose 𝔭_{i} is minimal for some i. Show 𝔮_{i} = 𝔭_{i}^{(r)} for all large r.

(2) Suppose 𝔭_{i} is not minimal for some i. Show that replacing 𝔮_{i} by 𝔭_{i}^{(r)} for large r gives infinitely many distinct irredundant primary decompositions of ⟨0⟩.

Theorem (18.28) (Krull Intersection). — Let R be a Noetherian ring, 𝔞 an ideal, and M a finitely generated module. Set N := ⋂_{n≥0} 𝔞^{n}M. Then there exists x ∈ 𝔞 such that (1 + x)N = 0.

Proof: By (16.18), N is finitely generated. So the desired x ∈ 𝔞 exists by (10.3) provided N = 𝔞N. Clearly N ⊃ 𝔞N. To prove N ⊂ 𝔞N, use (18.20): take a primary decomposition 𝔞N = ⋂ Q_{i} with Q_{i} 𝔭_{i}-primary. Fix i. If there’s a ∈ 𝔞 − 𝔭_{i}, then aN ⊂ Q_{i}, and so (18.4) yields N ⊂ Q_{i}. If 𝔞 ⊂ 𝔭_{i}, then there’s n_{i} with 𝔞^{n_{i}}M ⊂ Q_{i} by (18.3) and (3.25), and so again N ⊂ Q_{i}. Thus N ⊂ ⋂ Q_{i} = 𝔞N, as desired. □

Exercise (18.29). — Let R be a Noetherian ring, 𝔪 ⊂ rad(R) an ideal, M a finitely generated module, and M′ a submodule. Considering M/M′, show that

M′ = ⋂_{n≥0}(𝔪^{n}M + M′).

Example (18.30) (Another non-Noetherian ring). — Let R denote the ring of C^{∞} functions on the real line, 𝔪 the ideal of all f ∈ R that vanish at the origin. Note that 𝔪 is maximal, as f ↦ f(0) defines an isomorphism R/𝔪 ⟶^{∼} ℝ.

Let f ∈ R and n ≥ 1. Then, Taylor’s Theorem yields

f(x) = f(0) + f′(0)x + ⋯ + \frac{f^{(n−1)}(0)}{(n−1)!}x^{n−1} + x^{n}f_{n}(x)

where f_{n}(x) := ∫_{0}^{1} \frac{(1−t)^{n−1}}{(n−1)!} f^{(n)}(xt) dt.

Here f_{n} is C^{∞} too, since we can differentiate under the integral sign by [5, (7.1), p. 276]. So, if f ∈ 𝔪, then f(x) = xf_{1}(x). Thus 𝔪 ⊂ ⟨x⟩. But, obviously, 𝔪 ⊃ ⟨x⟩. Hence 𝔪 = ⟨x⟩. Therefore, 𝔪^{n} = ⟨x^{n}⟩.

If the first n − 1 derivatives of f vanish at 0, then Taylor’s Theorem yields f ∈ ⟨x^{n}⟩. Conversely, assume f(x) = x^{n}g(x) for some g ∈ R. By Leibniz’s Rule,

f^{(k)}(x) = ∑_{j=0}^{k} \binom{k}{j} \frac{n!}{(n−j+1)!}x^{n−j+1}g^{(k−j)}(x).

Hence f^{(k)} vanishes at 0 if n > k. Thus ⟨x^{n}⟩ consists of the f ∈ R whose first n − 1 derivatives vanish at 0. But ⟨x^{n}⟩ = 𝔪^{n}. Thus ⋂_{n≥0} 𝔪^{n} consists of those f ∈ R all of whose derivatives vanish at 0.

There is a well-known nonzero C^{∞}-function all of whose derivatives vanish at 0:

h(x) := \begin{cases}e^{−1/x^{2}} & if x ≠ 0, \\ 0 & if x = 0; \end{cases}

see [5, Ex. 7, p. 82]. Thus ⋂_{n≥0} 𝔪^{n} ≠ 0.

Given g ∈ 𝔪, let’s show (1 + g)h ≠ 0. Since g(0) = 0 and g is continuous, there is δ > 0 such that |g(x)| < 1/2 if |x| < δ. Hence 1 + g(x) ≥ 1/2 if |x| < δ. Hence (1 + g(x))h(x) > (1/2)h(x) > 0 if 0 < |x| < δ. Thus (1 + g)(⋂ 𝔪^{n}) ≠ 0. Thus the Krull Intersection Theorem (18.28) fails for R, and so R is non-Noetherian.

19. Length

The length of a module is a generalization of the dimension of a vector space. The length is the number of links in a composition series, which is a finite chain of submodules whose successive quotients are simple — that is, their only proper submodules are zero. Our main result is the Jordan–Hölder Theorem: any two composition series do have the same length and even the same successive quotients; further, their annihilators are just the primes in the support of the module, and the module is equal to the product of its localizations at these primes. Consequently, the length is finite if and only if the module is both Artinian and Noetherian. We also prove Akizuki’s Theorem: a ring is Artinian if and only if it is Noetherian and every prime is maximal. Consequently, a ring is Artinian if and only if its length is finite; if so, then it is the product of Artinian local rings.

(19.1) (Length). — Let R be a ring, and M a module. We call M simple if it is nonzero and its only proper submodule is 0. We call a chain of submodules,

M = M_{0} ⊃ M_{1} ⊃ ⋯ ⊃ M_{m} = 0  (19.1.1)

a composition series of length m if each successive quotient M_{i−1}/M_{i} is simple. Finally, we define the length ℓ(M) to be the infimum of all those lengths:

ℓ(M) := inf{ m | M has a composition series of length m }.  (19.1.2)

By convention, if M has no composition series, then ℓ(M) := ∞; further, ℓ(0) := 0.

For example, if R is a field, then M is a vector space and ℓ(M) = dim_{R}(M). Further, the chains in (17.23) are composition series, but those in (17.22) are not.

Exercise (19.2). — Let R be a ring, M a module. Prove these statements: (1) If M is simple, then any nonzero element m ∈ M generates M.

(2) M is simple if and only if M ≃ R/𝔪 for some maximal ideal 𝔪, and if so, then 𝔪 = Ann(M).

(3) If M has finite length, then M is finitely generated.

Theorem (19.3) (Jordan–Hölder). — Let R be a ring, and M a module with a composition series (19.1.1). Then any chain of submodules can be refined to a composition series, and every composition series is of the same length ℓ(M). Also,

Supp(M) = { 𝔪 ∈ Spec(R) | 𝔪 = Ann(M_{i−1}/M_{i}) for some i };

the 𝔪 ∈ Supp(M) are maximal; there is a canonical isomorphism

M ⟶^{∼} ∏_{𝔪∈Supp(M)} M_{𝔪};

and ℓ(M_{𝔪}) is equal to the number of i with 𝔪 = Ann(M_{i−1}/M_{i}).

Proof: First, let M′ be a proper submodule of M. Let’s show that

ℓ(M′) < ℓ(M).  (19.3.1)

To do so, set M′_{i} := M_{i} ∩ M′. Then M′_{i−1} ∩ M_{i} = M′_{i}. So

M′_{i−1}/M′_{i} = (M′_{i−1} + M_{i})/M_{i} ⊂ M_{i−1}/M_{i}.

September 3, 2012 11Nts.tex Since M_{i−1}/M_{i} is simple, either M′_{i−1}/M′_{i} = 0, or M′_{i−1}/M′_{i} = M_{i−1}/M_{i} and so

M′_{i−1} + M_{i} = M_{i−1}.  (19.3.2)

If (19.3.2) holds and if M_{i} ⊂ M′, then M_{i−1} ⊂ M′. Hence, if (19.3.2) holds for all i, then M ⊂ M′, a contradiction. Therefore, there is an i with M′_{i−1}/M′_{i} = 0. Now, M′ = M′_{0} ⊃ ⋯ ⊃ M′_{m} = 0. Omit M′_{i} if M′_{i−1}/M′_{i} = 0. Thus M′ has a composition series of length strictly less than m. Therefore, ℓ(M′) < m for any choice of (19.1.1). Thus (19.3.1) holds.

Next, given a chain N_{0} ⫌ ⋯ ⫌ N_{n} = 0, let’s prove n ≤ ℓ(M) by induction on ℓ(M). If ℓ(M) = 0, then M = 0; so also n = 0. Assume ℓ(M) ≥ 1. If n = 0, then we’re done. If n ≥ 1, then ℓ(N_{1}) < ℓ(M) by (19.3.1); so n − 1 ≤ ℓ(N_{1}) by induction. Thus n ≤ ℓ(M).

If N_{i−1}/N_{i} is not simple, then there is N′ with N_{i−1} ⫌ N′ ⫌ N_{i}. The new chain can have length at most ℓ(M) by the previous paragraph. Repeating, we can refine the given chain into a composition series in at most ℓ(M) − n steps.

Suppose the given chain is a composition series. Then ℓ(M) ≤ n by (19.1.2). But we proved n ≤ ℓ(M) above. Thus n = ℓ(M), and the first assertion is proved.

To proceed, fix a prime 𝔭. Exactness of Localization, (12.16), yields this chain:

M_{𝔭} = (M_{0})_{𝔭} ⊃ (M_{1})_{𝔭} ⊃ ⋯ ⊃ (M_{m})_{𝔭} = 0.  (19.3.3)

Now, consider a maximal ideal 𝔪. If 𝔭 = 𝔪, then (R/𝔪)_{𝔭} ≃ R/𝔪 by (12.4). If 𝔭 ≠ 𝔪, then there is s ∈ 𝔪 − 𝔭; so (R/𝔪)_{𝔭} = 0.

Set 𝔪_{i} := Ann(M_{i−1}/M_{i}). So M_{i−1}/M_{i} ≃ R/𝔪_{i} and 𝔪_{i} is maximal by (19.2)(2). Then Exactness of Localization yields (M_{i−1}/M_{i})_{𝔭} = (M_{i−1})_{𝔭}/(M_{i})_{𝔭}. Hence

(M_{i−1})_{𝔭}/(M_{i})_{𝔭} = \begin{cases}0, & if 𝔭 ≠ 𝔪_{i}; \\ M_{i−1}/M_{i} ≃ R/𝔪_{i}, & if 𝔭 = 𝔪_{i}. \end{cases}

Thus Supp(M) = {𝔪_{1}, …, 𝔪_{m}}.

If we omit the duplicates from the chain (19.3.3), then we get a composition series from the (M_{i})_{𝔭} with M_{i−1}/M_{i} ≃ R/𝔭. Thus the number of such i is ℓ(M_{𝔭}).

Finally, consider the canonical map φ: M → ∏_{𝔪∈Supp(M)} M_{𝔪}. To prove φ is an isomorphism, it suffices, by (13.20), to prove φ_{𝔭} is for each maximal ideal 𝔭. Now, localization commutes with finite product by (12.10). Therefore,

φ_{𝔭}: M_{𝔭} ⟶ (∏_{𝔪} M_{𝔪})_{𝔭} = ∏_{𝔪}(M_{𝔪})_{𝔭} = M_{𝔭}

as (M_{𝔪})_{𝔭} = 0 if 𝔪 ≠ 𝔭 and (M_{𝔪})_{𝔭} = M_{𝔭} if 𝔪 = 𝔭 by the above. Thus φ_{𝔭} = 1. □

Exercise (19.4). — Let R be a Noetherian ring, M a finitely generated module. Prove the equivalence of the following three conditions:

(1) that M has finite length;

(2) that Supp(M) consists entirely of maximal ideals;

(3) that Ass(M) consists entirely of maximal ideals. Prove that, if the conditions hold, then Ass(M) and Supp(M) are equal and finite.

Exercise (19.5). — Let R be a Noetherian ring, 𝔮 a 𝔭-primary ideal. Consider chains of primary ideals from 𝔮 to 𝔭. Show (1) all such chains have length at most ℓ(A) where A := (R/𝔮)_{𝔭} and (2) all maximal chains have length exactly ℓ(A).

Corollary (19.6). — A module M is both Artinian and Noetherian if and only if M is of finite length.

Proof: Any chain M ⊃ N_{0} ⫌ ⋯ ⫌ N_{n} = 0 has n < ℓ(M) by the Jordan–Hölder Theorem, (19.3). So if ℓ(M) < ∞, then M satisfies both the dcc and the acc.

Conversely, assume M is both Artinian and Noetherian. Form a chain as follows. Set M_{0} := M. For i ≥ 1, if M_{i−1} ≠ 0, take a maximal M_{i} ⫋ M_{i−1} by the maxc. By the dcc, this recursion terminates. Then the chain is a composition series. □

Example (19.7). — Any simple ℤ-module is finite owing to (19.2)(2). Hence, a ℤ-module is of finite length if and only if it is finite. In particular, ℓ(ℤ) = ∞.

Of course, ℤ is Noetherian, but not Artinian.

Let p ∈ ℤ be a prime, and set M := ℤ[1/p]/ℤ. Then M is an Artinian ℤ-module, but not Noetherian by (16.25). Since M is infinite, ℓ(M) = ∞.

Exercise (19.8). — Let k be a field, and R a finitely generated k-algebra. Prove that R is Artinian if and only if R is a finite-dimensional k-vector space.

Theorem (19.9) (Additivity of Length). — Let M be a module, and M′ a submodule. Then ℓ(M) = ℓ(M′) + ℓ(M/M′).

Proof: If M has a composition series, then the Jordan–Hölder Theorem yields another one of the form M = M_{0} ⊃ ⋯ ⊃ M′ ⊃ ⋯ ⊃ M_{m} = 0. The latter yields a pair of composition series: M/M′ = M_{0}/M′ ⊃ ⋯ ⊃ M′/M′ = 0 and M′ ⊃ ⋯ ⊃ M_{m} = 0. Conversely, every such pair arises from a unique composition series in M through M′. Therefore, ℓ(M) < ∞ if and only if ℓ(M/M′) < ∞ and ℓ(M′) < ∞; furthermore, if so, then ℓ(M) = ℓ(M′) + ℓ(M/M′), as desired. □

Exercise (19.10). — Let k be a field, A a local k-algebra. Assume the map from k to the residue field is bijective. Given an A-module M, prove ℓ(M) = dim_{k}(M).

Theorem (19.11) (Akizuki). — A ring R is Artinian if and only if R is Noetherian and dim(R) = 0. If so, then R has only finitely many primes.

Proof: If dim(R) = 0, then every prime is maximal. If also R is Noetherian, then R has finite length by (19.4). Thus R is Artinian by (19.6).

Conversely, suppose R is Artinian. Let 𝔪 be a minimal product of maximal ideals of R. Then 𝔪^{2} = 𝔪. Let 𝒮 be the set of ideals 𝔞 contained in 𝔪 such that 𝔞𝔪 ≠ 0. If 𝒮 ≠ ∅, take 𝔞 ∈ 𝒮 minimal. Then 𝔞𝔪^{2} = 𝔞𝔪 ≠ 0; hence, 𝔞𝔪 = 𝔞 by minimality of 𝔞. For any x ∈ 𝔞, if x𝔪 ≠ 0, then 𝔞 = ⟨x⟩ by minimality of 𝔞.

Let 𝔫 be any maximal ideal. Then 𝔫𝔪 = 𝔪 by minimality of 𝔪. But 𝔫𝔪 ⊂ 𝔫. Thus 𝔪 ⊂ rad(R). But 𝔞 = ⟨x⟩. So Nakayama’s Lemma yields 𝔞 = 0, a contradiction. So x𝔪 = 0 for any x ∈ 𝔞. Thus 𝔞𝔪 = 0, a contradiction. Hence 𝒮 = ∅. Therefore, 𝔪^{2} = 0. But 𝔪^{2} = 𝔪. Thus 𝔪 = 0. Say 𝔪 = 𝔪_{1} ⋯ 𝔪_{r} with 𝔪_{i} maximal.

Set 𝔞_{i} := 𝔪_{1} ⋯ 𝔪_{i} for 1 ≤ i ≤ r. Consider the chain

R =: 𝔞_{0} ⊃ 𝔞_{1} ⊃ ⋯ ⊃ 𝔞_{r} = 0.

Fix i. Set V_{i} := 𝔞_{i−1}/𝔞_{i}. Then V_{i} is a vector space over R/𝔪_{i}. Given linearly independent elements x_{1}, x_{2}, … ∈ V_{i}, let W_{j} ⊂ V_{i} be the subspace spanned by x_{j}, x_{j+1}, . . .. The W_{j} form a descending chain. It must stabilize as R is Artinian. Thus dim(V_{i}) < ∞. Hence ℓ(R) < ∞ by (19.9). So R is Noetherian by (19.6). So, by (19.4), every prime is maximal, and there are only finitely many primes. □

Exercise (19.12). — Prove these conditions on a Noetherian ring R equivalent: (1) that R is Artinian;

September 3, 2012 11Nts.tex (2) that Spec(R) is discrete and finite;

(3) that Spec(R) is discrete.

Exercise (19.13). — Let R be an Artinian ring. Show that rad(R) is nilpotent.

Corollary (19.14). — Let R be an Artinian ring, and M a finitely generated module. Then M has finite length, and Ass(M) and Supp(M) are equal and finite.

Proof: By (19.11) every prime is maximal, so Supp(M) consists of maximal ideals. Also R is Noetherian by (19.11). Hence (19.4) yields the assertions. □

Corollary (19.15). — A ring R is Artinian if and only if ℓ(R) < ∞.

Proof: Simply take M := R in (19.14) and (19.6). □

Exercise (19.16). — Let R be a ring, 𝔭 a prime ideal, and R′ a module-finite R-algebra. Show that R′ has only finitely many primes 𝔭′ over 𝔭, as follows: reduce to the case that R is a field by localizing at 𝔭 and passing to the residue rings.

Corollary (19.17). — A ring R is Artinian if and only if R is a finite product of Artinian local rings; if so, then R = ∏_{𝔪∈Spec(R)} R_{𝔪}.

Proof: A finite product of rings is Artinian if and only if each factor is Artinian by (16.23)(3). If R is Artinian, then ℓ(R) < ∞ by (19.15); whence, R = ∏ R_{𝔪} by the Jordan–Hölder Theorem. Thus the assertion holds. □

Exercise (19.18). — Let R be a Noetherian ring, and M a finitely generated module. Prove the following four conditions are equivalent:

(1) that M has finite length;

(2) that M is annihilated by some finite product of maximal ideals ∏ 𝔪_{i};

(3) that every prime 𝔭 containing Ann(M) is maximal;

(4) that R/Ann(M) is Artinian.

20. Hilbert Functions

The Hilbert Function of a graded module lists the lengths of its components. The corresponding generating function is called the Hilbert Series. This series is, under suitable hypotheses, a rational function, according to the Hilbert–Serre Theorem, which we prove. Passing to an arbitrary module, we study its Hilbert– Samuel Series, namely, the generating function of the colengths of the submodules in a filtration. We prove Samuel’s Theorem: if the ring is Noetherian, if the module is finitely generated, and if the filtration is stable, then the Hilbert–Samuel Series is a rational function with poles just at 0 and 1. In the same setup, we prove the Artin–Rees Lemma: given any submodule, its induced filtration is stable.

In a brief appendix, we study further one notion that arose: homogeneity.

(20.1) (Graded rings and modules). — We call a ring R graded if there are additive subgroups R_{n} for n ≥ 0 with R = ⨁ R_{n} and R_{m}R_{n} ⊂ R_{m+n} for all m, n.

For example, a polynomial ring R with coefficient ring R_{0} is graded if R_{n} is the R_{0}-submodule generated by the monomials of (total) degree n.

In general, R_{0} is a subring. Obviously, R_{0} is closed under addition and under multiplication, but we must check 1 ∈ R_{0}. So say 1 = ∑ x_{m} with x_{m} ∈ R_{m}. Given z ∈ R, say z = ∑ z_{n} with z_{n} ∈ R_{n}. Fix n. Then z_{n} = 1 · z_{n} = ∑ x_{m}z_{n} with x_{m}z_{n} ∈ R_{m+n}. So ∑_{m>0} x_{m}z_{n} = z_{n} − x_{0}z_{n} ∈ R_{n}. Hence x_{m}z_{n} = 0 for m > 0. But n is arbitrary. So x_{m}z = 0 for m > 0. But z is arbitrary. Taking z := 1 yields x_{m} = x_{m} · 1 = 0 for m > 0. Thus 1 = x_{0} ∈ R_{0}.

We call an R-module M (compatibly) graded if there are additive subgroups M_{n} for n ∈ ℤ with M = ⨁ M_{n} and R_{m}M_{n} ⊂ M_{m+n} for all m, n. We call M_{n} the nth homogeneous component; we say its elements are homogeneous. Obviously, M_{n} is an R_{0}-module.

Given m ∈ ℤ, set M(m) := ⨁ M_{m+n}. Then M(m) is another graded module; its nth graded component M(m)_{n} is M_{m+n}. Thus M(m) is obtained from M by shifting m places to the left.

Lemma (20.2). — Let R = ⨁ R_{n} be a graded ring, and M = ⨁ M_{n} a graded R-module. If R is a finitely generated R_{0}-algebra and if M is a finitely generated R-module, then each M_{n} is a finitely generated R_{0}-module.

Proof: Say R = R_{0}[x_{1}, …, x_{r}]. If x_{i} = ∑_{j} x_{ij} with x_{ij} ∈ R_{j}, then replace the x_{i} by the nonzero x_{ij}. Similarly, say M is generated over R by m_{1}, …, m_{s} with m_{i} ∈ M_{l_{i}}. Then any m ∈ M_{n} is a sum m = ∑ f_{i}m_{i} where f_{i} ∈ R. Say f_{i} = ∑ f_{ij} with f_{ij} ∈ R_{j}, and replace f_{i} by f_{ik} with k := n − l_{i} or by 0 if n < l_{i}. Then f_{i} is an R_{0}-linear combination of monomials x_{1}^{i_{1}} ⋯ x_{r}^{i_{r}} ∈ R_{k}; hence, m is one of the products x_{1}^{i_{1}} ⋯ x_{r}^{i_{r}}m_{i} ∈ M_{n}. Thus M_{n} is a finitely generated R_{0}-module. □

(20.3) (Hilbert functions). — Let R = ⨁ R_{n} be a graded ring, and M = ⨁ M_{n} a graded R-module. Assume R_{0} is Artinian, R is a finitely generated R_{0}-algebra, and M is a finitely generated R-module. Then each M_{n} is a finitely generated R_{0}-module by (20.2), so is of finite length ℓ(M_{n}) by (19.14). We call n ↦ ℓ(M_{n}) the Hilbert Function of M and its generating function

H(M, t) := ∑_{n∈ℤ} ℓ(M_{n})t^{n}

September 3, 2012 11Nts.tex the Hilbert Series of M. This series is a rational function by (20.7) below.

If R = R_{0}[x_{1}, …, x_{r}] with x_{i} ∈ R_{1}, then by (20.8) below, the Hilbert Function is, for n ≫ 0, a polynomial h(M, n), which we call the Hilbert Polynomial of M.

Example (20.4). — Let R := R_{0}[X_{1}, …, X_{r}] be the polynomial ring, graded by degree. Then R_{n} is free over R_{0} on the monomials of degree n, so of rank \binom{r−1+n}{r−1}.

Assume R_{0} is Artinian. Then ℓ(R_{n}) = ℓ(R_{0})\binom{r−1+n}{r−1} by Additivity of Length, (19.9). Thus the Hilbert Function is, for n ≥ 0, a polynomial of degree r − 1.

Formal manipulation yields \binom{r−1+n}{r−1} = (−1)^{n}\binom{−r}{n}. Therefore, Newton’s binomial theorem for negative exponents yields this computation for the Hilbert Series:

H(R, t) = ∑_{n≥0} ℓ(R_{0})\binom{r−1+n}{r−1}t^{n} = ∑_{n≥0} ℓ(R_{0})\binom{−r}{n}(−t)^{n} = ℓ(R_{0})/(1 − t)^{r}.

Exercise (20.5). — Let k be a field, k[X, Y] the polynomial ring. Show ⟨X, Y^{2}⟩ and ⟨X^{2}, Y^{2}⟩ have different Hilbert Series, but the same Hilbert Polynomial.

Exercise (20.6). — Let R = ⨁ R_{n} be a graded ring, M = ⨁ M_{n} a graded Rmodule. Let N = ⨁ N_{n} be a homogeneous submodule; that is, N_{n} = N ∩ M_{n}. Assume R_{0} is Artinian, R is a finitely generated R_{0}-algebra, and M is a finitely generated R-module. Set

N′ := { m ∈ M | there is k_{0} such that R_{k}m ∈ N for all k ≥ k_{0} }.

(1) Prove that N′ is a homogeneous submodule of M with the same Hilbert Polynomial as N, and that N′ is the largest such submodule.

(2) Let N = ⋂ Q_{i} be a decomposition with Q_{i} 𝔭_{i}-primary. Set R_{+} := ⨁_{n>0} R_{n}. Prove that N′ = ⋂_{𝔭_{i}⊅R_{+}} Q_{i}.

Theorem (20.7) (Hilbert–Serre). — Let R = ⨁ R_{n} be a graded ring, and let M = ⨁ M_{n} be a graded R-module. Assume R_{0} is Artinian, R is a finitely generated R_{0}-algebra, and M is a finitely generated R-module. Then

H(M, t) = e(t)/t^{l}(1 − t^{k_{1}}) ⋯ (1 − t^{k_{r}})

with e(t) ∈ ℤ[t], with l ≥ 0, and with k_{1}, …, k_{r} ≥ 1.

Proof: Say R = R_{0}[x_{1}, …, x_{r}] with x_{i} ∈ R_{k_{i}}. First, assume r = 0. Say M is generated over R by m_{1}, …, m_{s} with m_{i} ∈ M_{l_{i}}. Then R = R_{0}. So M_{n} = 0 for n < l_{0} := min{l_{i}} and for n > max{l_{i}}. Hence t^{−l_{0}}H(M, t) is a polynomial.

Next, assume r ≥ 1 and form the exact sequence

0 → K → M(−k_{1}) ⟶^{μ_{x_{1}}} M → L → 0

where μ_{x_{1}} is the map of multiplication by x_{1}. Since x_{1} ∈ R_{k_{1}}, the grading on M induces a grading on K and on L. Further, μ_{x_{1}} acts as 0 on both K and L.

Since R_{0} is Artinian, R_{0} is Noetherian by Akizuki’s Theorem, (19.11). So, since R is a finitely generated R_{0}-algebra, R is Noetherian by (16.11). Since M is a finitely generated R-module, obviously so is M(−k_{1}). Hence, so are both K and L by (16.15)(2). Set R′ := R_{0}[x_{2}, …, x_{r}]. Since x_{1} acts as 0 on K and L, they are finitely generated R′-modules. Therefore, H(K, t) and H(L, t) are defined, and they may be written in the desired form by induction on r.

By definition, M(−k_{1})_{n} := M_{n−k_{1}}; hence, H(M(−k_{1}), t) = t^{k_{1}}H(M, t). Therefore, Additivity of Length, (19.9), and the previous paragraph yield

(1 − t^{k_{1}})H(M, t) = H(L, t) − H(K, t) = e(t)/t^{l}(1 − t^{k_{2}}) ⋯ (1 − t^{k_{r}}).

September 3, 2012 11Nts.tex Thus the assertion holds. □

Corollary (20.8). — Under the conditions of (20.7), assume that M ≠ 0 and R = R_{0}[x_{1}, …, x_{r}] with x_{i} ∈ R_{1}. Then H(M, t) can be written uniquely in the form e(t)/t^{l}(1 − t)^{d} where e(t) ∈ ℤ[t] with e(0) ≠ 0 and e(1) ≠ 0 and where l ∈ ℤ and r ≥ d ≥ 0; further, there is a polynomial h(M, n) ∈ ℚ[n] with degree d − 1 and leading coefficient e(1)/(d − 1) ! such that ℓ(M_{n}) = h(M, n) for n ≥ deg(e(t)) − l.

Proof: We may take k_{i} = 1 for all i in the proof of (20.7). Hence H(M, t) is of the form e(t)(1−t)^{s}/t^{l}(1−t)^{r} with e(0) ≠ 0 and e(1) ≠ 0 and l ∈ ℤ. Set d := r −s. Then d ≥ 0 since H(M, 1) > 0 as M ≠ 0. Thus H(M, t) is of the asserted form. This form is unique owing to the uniqueness of factorization of polynomials.

Say e(t) = ∑_{i=0}^{N} e_{i}t^{i}. Now, (1 − t)^{−d} = ∑ \binom{−d}{n}(−t)^{n} = ∑ \binom{d−1+n}{d−1}t^{n}. Hence ℓ(M_{n}) = ∑_{i=0}^{N} e_{i}\binom{d−1+n+l−i}{d−1} for n + l ≥ N. But \binom{d−1+n−i}{d−1} = n^{d−1}/(d − 1) ! + · · ·. Therefore, ℓ(M_{n}) = e(1)n^{d−1}/(d − 1) ! + ⋯, as asserted. □

Exercise (20.9). — Let k be a field, P := k[X, Y, Z] the polynomial ring in three variables, f ∈ P a homogeneous polynomial of degree d ≥ 1. Set R := P/⟨f⟩. Find the coefficients of the Hilbert Polynomial h(R, n) explicitly in terms of d.

Exercise (20.10). — Under the conditions of (20.8), assume there is a homogeneous nonzerodivisor f ∈ R with M_{f} = 0. Prove deg(h(R, n)) > deg(h(M, n)); start with the case M := R/⟨f^{k}⟩.

(20.11) (Filtrations). — Let R be an arbitrary ring, 𝔮 an ideal, and M a module. A filtration of M is an infinite descending chain of submodules:

M = M_{0} ⊃ M_{1} ⊃ M_{2} ⊃ · · · .  (20.11.1)

We call it a 𝔮-filtration if 𝔮M_{n} ⊂ M_{n+1} for all n, and a stable 𝔮-filtration if also 𝔮M_{n} = M_{n+1} for n ≫ 0, or equivalently, if also there is an m with 𝔮^{n}M_{m} = M_{n+m} for n ≥ 0. For example, setting M_{n} := 𝔮^{n}M, we get a stable 𝔮-filtration; we call it the 𝔮-adic filtration.

The 𝔮-adic filtration of R yields a graded ring G_{𝔮}(R) or G(R), defined by

G(R) := ⨁_{n≥0} G(R)_{n} where G(R)_{n} := 𝔮^{n}/𝔮^{n+1}.

We obtain the product of an element in 𝔮^{i}/𝔮^{i+1} and one in 𝔮^{j}/𝔮^{j+1} by choosing a representative of each, forming their product, and taking its residue in 𝔮^{i+j}/𝔮^{i+j+1}. We call G(R) the associated graded ring.

Similarly, if (20.11.1) is a 𝔮-filtration, then we obtain a graded G(R)-module

G_{𝔮}(M) := G(M) := ⨁_{n≥0} G(M)_{n} where G(M)_{n} := M_{n}/M_{n+1}.

If all the quotients M/M_{n} of the filtration (20.11.1) are of finite length, then we call n ↦ ℓ(M/M_{n}) the Hilbert–Samuel Function, and the generating function

P (M_{•}, t) := ∑_{n≥0} ℓ(M/M_{n})t^{n}

the Hilbert–Samuel Series. If the function n ↦ ℓ(M/M_{n}) is, for n ≫ 0, a polynomial p(M_{•}, n), then we call it the Hilbert–Samuel Polynomial. If the filtration is the 𝔮-adic filtration, we also denote P (M_{•}, t) and p(M_{•}, n) by P_{𝔮}(M, t) and p_{𝔮}(M, n).

Lemma (20.12). — Let R be a Noetherian ring, 𝔮 an ideal, M a finitely generated module with a stable 𝔮-filtration. Then G(R) is generated as an R/𝔮-algebra by finitely many elements of 𝔮/𝔮^{2}, and G(M) is a finitely generated G(R)-module.

Proof: Since R is Noetherian, 𝔮 is a finitely generated ideal, say by x_{1}, …, x_{r}. Then, clearly, the residues of the x_{i} in 𝔮/𝔮^{2} generate G(R) as a R/𝔮-algebra.

Say the filtration is M = M_{0} ⊃ M_{1} ⊃ · · ·. Since it is stable, there is an m with 𝔮^{n}M_{m} = M_{n+m} for n ≥ 0. Hence G(M) is generated by M_{0}/M_{1}, …, M_{m}/M_{m+1} over G(R). But R is Noetherian and M is finitely generated over R; hence, every M_{i} is finitely generated over R. Therefore, every M_{n}/M_{n+1} is finitely generated over R/𝔮. Thus G(M) is a finitely generated G(R)-module. □

Theorem (20.13) (Samuel). — Let R be a Noetherian ring, 𝔮 an ideal, M a finitely generated module with a stable 𝔮-filtration M = M_{0} ⊃ M_{1} ⊃ · · ·. Assume ℓ(M/𝔮M) < ∞. Then ℓ(M_{n}/M_{n+1}) < ∞ and ℓ(M/M_{n}) < ∞ for every n; further,

P (M_{•}, t) = H(G(M), t) t/(1 − t).

Proof: Set 𝔞 := Ann(M). Set R′ := R/𝔞 and 𝔮′ := (𝔞 + 𝔮)/𝔞. Then R′/𝔮′ is Noetherian as R is. Further, M can be viewed as a finitely generated R′-module, and M = M_{0} ⊃ M_{1} ⊃ ⋯ as a stable 𝔮′-filtration. So G(R′) is generated as a R′/𝔮′- algebra by finitely many elements of degree 1, and G(M) is a finitely generated G(R′)-module by (20.12). Therefore, each M_{n}/M_{n+1} is a finitely generated R′/𝔮′- module by (20.2) or by the proof of (20.12).

On the other hand, (13.1) and (13.9)(3) and (13.13) yield, respectively,

V(𝔞 + 𝔮) = V(𝔞) ⋂ V(𝔮) = Supp(M) ⋂ V(𝔮) = Supp(M/𝔮M).

Hence V(𝔞 + 𝔮) consists entirely of maximal ideals, because Supp(M/𝔮M) does by (19.4) as ℓ(M/𝔮M) < ∞. Thus dim(R′/𝔮′) = 0. But R′/𝔮′ is Noetherian. Therefore, R′/𝔮′ is Artinian by Akizuki’s Theorem, (19.11).

Therefore, ℓ(M_{n}/M_{n+1}) < ∞ for every n by (19.14). Form the exact sequence

0 → M_{n}/M_{n+1} → M/M_{n+1} → M/M_{n} → 0.

Then Additivity of Length, (19.9), yields

ℓ(M_{n}/M_{n+1}) = ℓ(M/M_{n+1}) − ℓ(M/M_{n}).

So induction on n yields ℓ(M/M_{n+1}) < ∞ for every n. Further, multiplying that equation by t^{n} and summing over n yields the desired expression in another form:

H(G(M), t) = (t^{−1} − 1)P (M_{•}, t) = P (M_{•}, t) (1 − t)/t.  □

Corollary (20.14). — Under the conditions of (20.13), assume 𝔮 is generated by r elements and M ≠ 0. Then P (M_{•}, t) can be written uniquely in the form e(t)/t^{l−1}(1 − t)^{d+1} where e(t) ∈ ℤ[t] with e(0) ≠ 0 and e(1) ≠ 0 and where l ∈ ℤ and r ≥ d ≥ 0; further, there is a polynomial p(M_{•}, n) ∈ ℚ[n] with degree d and leading coefficient e(1)/d! such that ℓ(M/M_{n}) = p(M_{•}, n) for n ≥ deg(e(t)) − l.

Finally, p_{𝔮}(M, n)−p(M_{•}, n) is a polynomial with degree at most d−1 and nonnegative leading coefficient; further, d and e(1) are the same for every stable 𝔮-filtration.

Proof: The proof of (20.13) shows that G(R′) and G(M) satisfy the hypotheses of (20.8). So (20.8) yields a certain form for H(G(M), t). Then (20.13) yields the asserted form for P (M_{•}, t). In turn, that form yields the asserted polynomial p(M_{•}, n) by the argument in the second paragraph of the proof of (20.8).

September 3, 2012 11Nts.tex Finally, as M = M_{0} ⊃ M_{1} ⊃ ⋯ is a stable 𝔮-filtration, there’s an m such that

M_{n} ⊃ 𝔮^{n}M ⊃ 𝔮^{n}M_{m} = M_{n+m}

for all n ≥ 0. Dividing into M and extracting lengths yields

ℓ(M/M_{n}) ≤ ℓ(M/𝔮^{n}M) ≤ ℓ(M/M_{n+m}).

Therefore, for large n, we get

p(M_{•}, n) ≤ p_{𝔮}(M, n) ≤ p(M_{•}, n + m).

The two extremes are polynomials in n with the same degree and leading coefficient, say d and c. Dividing by n^{d} and letting n → ∞, we conclude that the polynomial p_{𝔮}(M, n) also has degree d and leading coefficient c.

Thus the degree and leading coefficient are the same for every stable 𝔮-filtration. Further p_{𝔮}(M, n)−p(M_{•}, n) has degree at most d−1 and positive leading coefficient, owing to cancellation of the two leading terms and to the first inequality. □

Exercise (20.15). — Let R be a Noetherian ring, 𝔮 an ideal, and M a finitely generated module. Assume ℓ(M/𝔮M) < ∞. Set 𝔪 := \sqrt{𝔮}. Show

deg p_{𝔪}(M, n) = deg p_{𝔮}(M, n).

(20.16) (Rees Algebras). — Let R be an arbitrary ring, 𝔮 an ideal. The sum

ℛ(𝔮) := ⨁_{n≥0} 𝔮^{n}

is, canonically, a graded ring, with R as zeroth graded component and 𝔮 as first. We call ℛ(𝔮) the Rees Algebra of 𝔮.

Let M be a module with a 𝔮-filtration M = M_{0} ⊃ M_{1} ⊃ · · ·. Then the sum

ℛ(M_{•}) := ⨁_{n≥0} M_{n}

is canonically a module over the Rees Algebra ℛ(𝔮).

Lemma (20.17). — Let R be a Noetherian ring, 𝔮 an ideal, and M a finitely generated module with a 𝔮-filtration. Then ℛ(𝔮) is generated as an R-algebra by finitely many elements of 𝔮, and ℛ(M_{•}) is a finitely generated ℛ(𝔮)-module if and only if the filtration is stable.

Proof: Say the filtration is M = M_{0} ⊃ M_{1} ⊃ · · ·. Suppose ℛ(M_{•}) is generated over ℛ(𝔮) by m_{1}, …, m_{s}. Say m_{i} = ∑_{j=0}^{μ} m_{ij} with m_{ij} ∈ M_{j}. Then for any n ≥ 0, any m ∈ M_{n+μ} is a sum m = ∑ f_{ij}m_{ij} where f_{ij} ∈ 𝔮^{n+μ−j}. But 𝔮^{n+μ−j} = 𝔮^{n}𝔮^{μ−j}. Thus M_{n+μ} = 𝔮^{n}M_{μ}; that is, the filtration is stable.

The rest of the proof is similar to that of (20.12), but simpler. □

Lemma (20.18) (Artin–Rees). — Let R be a Noetherian ring, M a finitely generated module, N a submodule, 𝔮 an ideal, M = M_{0} ⊃ M_{1} ⊃ ⋯ a stable 𝔮-filtration. For n ≥ 0, set N_{n} := N ∩ M_{n}. Then N = N_{0} ⊃ N_{1} ⊃ ⋯ is a stable 𝔮-filtration.

Proof: By (20.17), the Rees Algebra ℛ(𝔮) is finitely generated over R, so Noetherian by (16.11). By (20.17), the module ℛ(M_{•}) is finitely generated over ℛ(𝔮), so Noetherian by (16.18). Clearly, N = N_{0} ⊃ N_{1} ⊃ ⋯ is a 𝔮-filtration; hence, ℛ(N_{•}) is a submodule of ℛ(M_{•}), so Noetherian by (16.15)(2), so finitely generated by (16.18). Hence, N = N_{0} ⊃ N_{1} ⊃ ⋯ is stable by (20.17), as desired. □

Exercise (20.19). — Derive the Krull Intersection Theorem, (18.28), from the Artin–Rees Lemma, (20.18).

Proposition (20.20). — Let R be a Noetherian ring, 𝔮 an ideal, and

0 → M′ → M → M′′ → 0

an exact sequence of finitely generated modules. Then M/𝔮M has finite length if and only if M′/𝔮M′ and M′′/𝔮M′′ do. If so, then the polynomial

p_{𝔮}(M′, n) − p_{𝔮}(M, n) + p_{𝔮}(M′′, n)

has degree at most deg(p_{𝔮}(M′, n))−1 and has positive leading coefficient; also then

deg p_{𝔮}(M, n) = max{ deg p_{𝔮}(M′, n), deg p_{𝔮}(M′′, n) }.

Proof: First off, (13.13) and (13.9)(1) and (13.13) again yield

Supp(M/𝔮M) = Supp(M) ⋂ V(𝔮) = (Supp(M′) ⋃ Supp(M′′)) ⋂ V(𝔮)
= (Supp(M′) ⋂ V(𝔮)) ⋃(Supp(M′′) ⋂ V(𝔮))
= Supp(M′/𝔮M′) ⋃ Supp(M′′/𝔮M′′).

Hence M/𝔮M has finite length if and only if M′/𝔮M′ and M′′/𝔮M′′ do by (19.4).

For n ≥ 0, set M′_{n} := M′ ⋂ 𝔮^{n}M. Then M′ = M′_{0} ⊃ M′_{1} ⊃ ⋯ is a stable 𝔮-filtration by the Artin–Rees Lemma. Form this canonical commutative diagram:

0 ⟶ M′_{n} ⟶ 𝔮^{n}M ⟶ 𝔮^{n}M′′ ⟶ 0

↓  ↓  ↓
0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

Its rows are exact. So the Nine Lemma yields this exact sequence:

0 → M′/M′_{n} → M/𝔮^{n}M → M′′/𝔮^{n}M′′ → 0.

Assume M/𝔮M has finite length. Then Additivity of Length and (20.14) yield

p(M′_{•}, n) − p_{𝔮}(M, n) + p_{𝔮}(M′′, n) = 0.  (20.20.1)

Hence p_{𝔮}(M′, n) − p_{𝔮}(M, n) + p_{𝔮}(M′′, n) is equal to p_{𝔮}(M′, n) − p(M′_{•}, n). But by (20.14) again, the latter is a polynomial with degree at most deg(p_{𝔮}(M′, n)) − 1 and positive leading coefficient.

Finally, deg p_{𝔮}(M, n) = max{deg p(M′_{•}, n), deg p_{𝔮}(M′′, n)} owing to (20.20.1), as the leading coefficients of p(M′_{•}, n) and p_{𝔮}(M′′, n) are both positive, so cannot cancel. But deg p(M′_{•}, n) = deg p_{𝔮}(M′, n) by (20.14), completing the proof. □

20. Appendix: Homogeneity

(20.21) (Homogeneity). — Let R be a graded ring, and M = ⨁ M_{n} a graded module. We call the M_{n} the homogeneous components of M.

Given m ∈ M, write m = ∑ m_{n} with m_{n} ∈ M_{n}. Call the finitely many nonzero m_{n} the homogeneous components of m. Say that a component m_{n} is homogeneous of degree n. If n is lowest, call m_{n} the initial component of m.

Call a submodule N ⊂ M homogeneous if, whenever m ∈ N, also m_{n} ∈ N, or equivalently, N = ⨁(M_{n} ∩ N).

Call a map α: M′ → M of graded modules with components M′_{n} and M_{n} homogeneous of degree r if α(M′_{n}) ⊂ M_{n+r} for all n. If so, then clearly Ker(α) is a homogeneous submodule of M. Further, Coker(α) is canonically graded, and the quotient map M → Coker(α) is homogeneous of degree 0.

Exercise (20.22). — Let R = ⨁ R_{n} be a graded ring, M = ⨁_{n≥n_{0}} M_{n} a graded module, 𝔞 ⊂ ⨁_{n>0} R_{n} a homogeneous ideal. Assume M = 𝔞M. Show M = 0.

Exercise (20.23). — Let R = ⨁ R_{n} be a Noetherian graded ring, M = ⨁ M_{n} a finitely generated graded R-module, N = ⨁ N_{n} a homogeneous submodule. Set

N′ := { m ∈ M | R_{n}m ∈ N for all n ≫ 0 }.

Show that N′ is the largest homogeneous submodule of M containing N and having, for all n ≫ 0, its degree-n homogeneous component N′_{n} equal to N_{n}.

Proposition (20.24). — Let R be a Noetherian graded ring, M a nonzero finitely generated graded module, Q a homogeneous submodule. Suppose Q possesses this property: given any homogeneous x ∈ R and homogeneous m ∈ M with xm ∈ Q but m ∉ Q, necessarily x ∈ 𝔭 := nil(M/Q). Then 𝔭 is prime, and Q is 𝔭-primary.

Proof: Given x ∈ R and m ∈ M, decompose them into their homogeneous components: x = ∑_{i≥r} x_{i} and m = ∑_{j≥s} m_{j}. Suppose xm ∈ Q, but m ∉ Q. Then m_{t} ∉ Q for some t; take t minimal. Set m′ := ∑_{j<t} m_{j}. Then m′ ∈ Q. Set m′′ := m − m′. Then xm′′ ∈ Q.

Either x_{s}m_{t} vanishes or it’s the initial component of xm′′. But Q is homogeneous. So x_{s}m_{t} ∈ Q. But m_{t} ∉ Q. Hence x_{s} ∈ 𝔭 by the hypothesis. Say x_{s}, …, x_{u} ∈ 𝔭 with u maximal. Set x′ := ∑_{i=s}^{u} x_{i}. Then x′ ∈ 𝔭. So x^{′k} ∈ Ann(M/Q) for some k ≥ 1. So x^{′k}m′′ ∈ Q. Set x′′ := x − x′. Since xm′′ ∈ Q, also x^{′′k}m′′ ∈ Q.

Suppose x ∉ 𝔭. Then x′′ ≠ 0. And its initial component is x_{v} with v > u. Either x′′_{v}m′′_{t} vanishes or it is the initial component of xm. But Q is homogeneous. So x_{v}m_{t} ∈ Q. But m_{t} ∉ Q. Hence x_{v} ∈ 𝔭 by the hypothesis, contradicting v > u. Thus x ∈ 𝔭. Thus Q is 𝔭-primary by (18.4). □

Exercise (20.25). — Let R be a graded ring, 𝔞 a homogeneous ideal, and M a graded module. Prove that \sqrt{𝔞} and Ann(M) and nil(M) are homogeneous.

Exercise (20.26). — Let R be a graded ring, M a graded module, and Q a primary submodule. Let Q^{∗} ⊂ Q be the submodule generated by the homogeneous elements of Q. Then Q^{∗} is primary.

Theorem (20.27). — Let R be a Noetherian graded ring, M a finitely generated graded module, N a homogeneous submodule. Then all the associated primes of M/N are homogeneous, and N admits an irredundant primary decomposition in which all the primary submodules are homogeneous.

Proof: Let N = ⋂ Q_{j} be any primary decomposition; one exists by (18.20). Let Q_{j}^{∗} ⊂ Q_{j} be the submodule generated by the homogeneous elements of Q_{j}. Trivially, ⋂ Q_{j}^{∗} ⊂ ⋂ Q_{j} = N ⊂ ⋂ Q_{j}^{∗}. Further, each Q_{j}^{∗} is clearly homogeneous, and is primary by (20.26). Thus N = ⋂ Q_{j}^{∗} is a primary decomposition into homogeneous primary submodules. And, owing to (18.18), it is irredundant if N = ⋂ Q_{j} is, as both decompositions have minimal length. Finally, M/Q_{j}^{∗} is graded by (20.21); so each associated prime is homogeneous by (18.19) and (20.25). □

(20.28) (Graded Domains). — Let R = ⨁_{n≥0} R_{n} be a graded domain, and set K := Frac(R). We call z ∈ K homogeneous of degree n ∈ ℤ if z = x/y with x ∈ R_{m} and y ∈ R_{m−n}. Clearly, n is well defined.

Let K_{n} be the set of all such z, plus 0. Then K_{m}K_{n} ⊂ K_{m+n}. Clearly, the canonical map ⨁_{n∈ℤ} K_{n} → K is injective. Thus ⨁_{n≥0} K_{n} is a graded subring of K. Further, K_{0} is a field.

The n with K_{n} ≠ 0 form a subgroup of ℤ. So by renumbering, we may assume K_{1} ≠ 0. Fix any nonzero x ∈ K_{1}. Clearly, x is transcendental over K_{0}. If z ∈ K_{n}, then z/x^{n} ∈ K_{0}. Hence R ⊂ K_{0}[x]. So (2.3) yields K = K_{0}(x).

Any w ∈ ⨁ K_{n} can be written w = a/b with a, b ∈ R and b homogeneous: say w = ∑(a_{n}/b_{n}) with a_{n}, b_{n} ∈ R homogeneous; set b := ∏ b_{n} and a := ∑(a_{n}b/b_{n}).

Theorem (20.29). — Let R be a Noetherian graded domain, K := Frac(R), and \overline{R} the integral closure of R in K. Then \overline{R} is a graded subring of K.

Proof: Use the setup of (20.28). Since K_{0}[x] is a polynomial ring over a field, it is normal by (10.29). Hence \overline{R} ⊂ K_{0}[x]. So every y ∈ R can be written as y = ∑_{i=r}^{r+n} y_{i}, with y_{i} homogeneous and nonzero. Let’s show y_{i} ∈ \overline{R} for all i.

Since y is integral over R, the R-algebra R[y] is module finite by (10.18). So (20.28) yields a homogeneous b ∈ R with bR[y] ⊂ R. Hence by^{j} ∈ R for all j ≥ 0. But R is graded. Hence by_{r}^{j} ∈ R. Set z := 1/b. Then y_{r}^{j} ∈ Rz. Since R is Noetherian, the R-algebra R[y_{r}] is module finite. Hence y_{r} ∈ \overline{R}. Then y − y_{r} ∈ \overline{R}. Thus y_{i} ∈ \overline{R} for all i by induction on n. Thus \overline{R} is graded. □

Exercise (20.30). — Under the conditions of (20.8), assume that R is a domain and that its integral closure \overline{R} in Frac(R) is a finitely generated R-module.

(1) Prove that there is a homogeneous f ∈ R with R_{f} = \overline{R}_{f}.

(2) Prove that the Hilbert Polynomials of R and \overline{R} have the same degree and same leading coefficient.

21. Dimension

The dimension of a module is defined as the sup of the lengths of the chains of primes in its support. The Dimension Theorem, which we prove, characterizes the dimension of a nonzero finitely generated semilocal module over a Noetherian ring in two ways. First, the dimension is the degree of the Hilbert–Samuel Polynomial formed with the radical of the ring. Second, the dimension is the smallest number of elements in the radical that span a submodule of finite colength.

Next, in an arbitrary Noetherian ring, we study the height of a prime: the length of the longest chain of subprimes. We bound the height by the minimal number of generators of an ideal over which the prime is minimal. In particular, when this number is 1, we obtain Krull’s Principal Ideal Theorem. Finally, we study regular local rings: Noetherian local rings whose maximal ideal has the minimum number of generators, namely, the dimension.

(21.1) (Dimension of a module). — Let R be a ring, and M a nonzero module. The dimension of M, denoted dim(M), is defined by this formula:

dim(M) := sup{ r | there’s a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} in Supp(M) }.

Assume R is Noetherian, and M is finitely generated. Then M has finitely many minimal (associated) primes by (17.19). They are also the minimal primes 𝔭_{0} ∈ Supp(M) by (17.16). Thus (1.8) yields

dim(M) = max{ dim(R/𝔭_{0}) | 𝔭_{0} ∈ Supp(M) is minimal }.  (21.1.1)

(21.2) (Parameters). — Let R be a ring, M a nonzero module. Denote the intersection of the maximal ideals in Supp(M) by rad(M), and call it the radical of M. If there are only finitely many such maximal ideals, call M semilocal.Call an ideal 𝔮 a parameter ideal of M if 𝔮 ⊂ rad(M) and M/𝔮M is Artinian.

Assume M is finitely generated. Then Supp(M) = V(Ann(M)) by (13.9)(3). Hence M is semilocal if and only if R/ Ann(M) is a semilocal ring.

Assume, in addition, R is Noetherian; so M is Noetherian by (16.18). Fix an ideal 𝔮. Then (19.6) yields that M/𝔮M is Artinian if and only if ℓ(M/𝔮M) < ∞.

However, ℓ(M/𝔮M) < ∞ if and only if Supp(M/𝔮M) consists of finitely many maximal ideals by (19.4) and (17.20). Further, (13.13), (13.9)(3), and (13.1) yield

Supp(M/𝔮M) = Supp(M) ⋂ V(𝔮) = V(Ann(M)) ⋂ V(𝔮) = V(Ann(M) + 𝔮).

Set 𝔮′ := Ann(M) + 𝔮. Thus M/𝔮M is Artinian if and only if V(𝔮′) consists of finitely many maximal ideals; so by (19.11), if and only if R/𝔮′ is Artinian. But (19.18) implies that R/𝔮′ is Artinian if and only if 𝔮′ contains a product of maximal ideals each of which contains 𝔮′. Then each lies in Supp(M), so contains rad(M).

Set 𝔪 := rad(M). Thus if R/𝔮′ is Artinian, then 𝔮′ ⊃ 𝔪^{n} for some n > 0.

Assume, in addition, M is semilocal, so that Supp(M) contains only finitely many maximal ideals. Then their product is contained in 𝔪. Thus, conversely, if 𝔮′ ⊃ 𝔪^{n} for some n > 0, then R/𝔮′ is Artinian. Thus 𝔮 is a parameter ideal if and only if

𝔪 ⊃ 𝔮′ ⊃ 𝔪^{n} for some n,  (21.2.1)

or by (3.26) if and only if 𝔪 = \sqrt{𝔮′}, or by (13.1) if and only if V(𝔪) = V(𝔮′). In particular, 𝔪^{n} is a parameter ideal for any n.

Assume 𝔮 is a parameter ideal. Then the Hilbert–Samuel polynomial p_{𝔮}(M, n) exists by (20.14). Similarly, p_{𝔪}(M, n) exists, and the two polynomials have the same degree by (20.15) since 𝔪 = \sqrt{𝔮′} and p_{𝔮′}(M, n) = p_{𝔮}(M, n). Thus the degree is the same for every parameter ideal. Denote this common degree by d(M).

Alternatively, d(M) can be viewed as the order of pole at 1 of the Hilbert series H(G_{𝔮}(M), t). Indeed, that order is 1 less than the order of pole at 1 of the Hilbert– Samuel series P_{𝔮}(M, t) by (20.13). In turn, the latter order is d(M)+1 by (20.14).

Denote by s(M) the smallest s such that there are x_{1}, …, x_{s} ∈ 𝔪 with

ℓ(M/⟨x_{1}, …, x_{s}⟩M) < ∞.  (21.2.2)

By convention, if ℓ(M) < ∞, then s(M) = 0. We say that x_{1}, …, x_{s} ∈ 𝔪 form a system of parameters (sop) for M if s = s(M) and (21.2.2) holds. Note that a sop generates a parameter ideal.

Lemma (21.3). — Let R be a Noetherian ring, M a nonzero Noetherian semilocal module, 𝔮 a parameter ideal of M, and x ∈ rad(M). Set K := Ker(M ⟶^{μ_{x}} M).

(1) Then s(M) ≤ s(M/xM) + 1.

(2) Then dim(M/xM) ≤ dim(M) − 1 if x ∉ 𝔭 for any 𝔭 ∈ Supp(M) with dim(R/𝔭) = dim(M).

(3) Then deg(p_{𝔮}(K, n) − p_{𝔮}(M/xM, n)) ≤ d(M) − 1.

Proof: For (1), set s := s(M/xM). There are x_{1}, …, x_{s} ∈ rad(M/xM) with

ℓ(M/⟨x, x_{1}, …, x_{s}⟩M) < ∞.

Now, Supp(M/xM) = Supp(M) ∩ V(⟨x⟩) by (13.13). However, x ∈ rad(M). Hence, Supp(M/xM) and Supp(M) have the same maximal ideals. Therefore, rad(M/xM) = rad(M). Hence s(M) ≤ s + 1. Thus (1) holds.

To prove (2), take a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} in Supp(M/xM). Now, Supp(M/xM) = Supp(M) ∩ V(⟨x⟩) by (13.13). So x ∈ 𝔭_{0} ∈ Supp(M). So, by hypothesis, dim(R/𝔭_{0}) < dim(M). Hence r ≤ dim(M) − 1. Thus (2) holds.

To prove (3), set xM := Im(μ_{x}), and form these two exact sequences:

0 → K → M → xM → 0, and 0 → xM → M → M/xM → 0.

Then (20.20) yields d(K) ≤ d(M) and d(xM) ≤ d(M). So by (20.20) again, both p_{𝔮}(K, n) + p_{𝔮}(xM, n) − p_{𝔮}(M, n) and p_{𝔮}(xM, n) + p_{𝔮}(M/xM, n) − p_{𝔮}(M, n) are of degree at most d(M) − 1. So their difference is too. Thus (3) holds. □

Theorem (21.4) (Dimension). — Let R be a Noetherian ring, M a nonzero finitely generated semilocal module. Then

dim(M) = d(M) = s(M) < ∞.

Proof: Let’s prove a cycle of inequalities. Set 𝔪 := rad(M). First, let’s prove dim(M) ≤ d(M). We proceed by induction on d(M). Suppose d(M) = 0. Then ℓ(M/𝔪^{n}M) stabilizes. So 𝔪^{n}M = 𝔪^{n+1}M for some n. Hence 𝔪^{n}M = 0 by Nakayama’s Lemma applied over the semilocal ring R/ Ann(M). Hence ℓ(M) < ∞. So dim(M) = 0 by (19.4).

Suppose d(M) ≥ 1. Take 𝔭_{0} ∈ Ass(M) with dim(R/𝔭_{0}) = dim(M). Then M has a submodule N isomorphic to R/𝔭_{0} by (17.2). Further, d(N) ≤ d(M) by (20.20).

Take a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} in Supp(N). If r = 0, then r ≤ d(M).

September 3, 2012 11Nts.tex Suppose r ≥ 1. Then there’s an x_{1} ∈ 𝔭_{1} − 𝔭_{0}. Further, since 𝔭_{0} is not maximal, for each maximal ideal 𝔫 in Supp(M), there is an x_{𝔫} ∈ 𝔫 − 𝔭_{0}. Set x := x_{1} ∏ x_{𝔫}. Then x ∈ (𝔭_{1} ∩ 𝔪) − 𝔭_{0}. Then 𝔭_{1} ⫋ ⋯ ⫋ 𝔭_{r} lies in Supp(N) ⋂ V(⟨x⟩). But the latter is equal to Supp(N/xN) by (13.13). So r − 1 ≤ dim(N/xN).

However, μ_{x} is injective on N as N ≃ R/𝔭_{0} and x ∉ 𝔭_{0}. So (21.3)(3) yields d(N/xN) ≤ d(N) − 1. But d(N) ≤ d(M). So dim(N/xN) ≤ d(N/xN) by the induction hypothesis. Therefore, r ≤ d(M). Thus dim(M) ≤ d(M).

Second, let’s prove d(M) ≤ s(M). Let 𝔮 be a parameter ideal of M with s(M) generators. Then d(M) := deg p_{𝔮}(M, n). But deg p_{𝔮}(M, n) ≤ s(M) owing to (20.14). Thus d(M) ≤ s(M).

Finally, let’s prove s(M) ≤ dim(M). Set r := dim(M), which is finite since r ≤ d(M) by the first step. The proof proceeds by induction on r. If r = 0, then M has finite length by (19.4); so s(M) = 0.

Suppose r ≥ 1. Let 𝔭_{1}, …, 𝔭_{k} be the primes of Supp(M) with dim(R/𝔭_{i}) = r. No 𝔭_{i} is maximal as r ≥ 1. So 𝔪 lies in no 𝔭_{i}. Hence, by Prime Avoidance (3.15), there is an x ∈ 𝔪 such that x ∉ 𝔭_{i} for all i. So (21.3)(1), (2) yield s(M) ≤ s(M/xM) + 1 and dim(M/xM) + 1 ≤ r. By the induction hypothesis, s(M/xM) ≤ dim(M/xM). Hence s(M) ≤ r, as desired. □

Corollary (21.5). — Let R be a Noetherian ring, M a nonzero Noetherian semilocal module, x ∈ rad(M). Then dim(M/xM) ≥ dim(M) − 1, with equality if x ∉ 𝔭 for 𝔭 ∈ Supp(M) with dim(R/𝔭) = dim(M); equality holds if x ∉ z.div(M).

Proof: By (21.3)(1), we have s(M/xM) ≥ s(M)−1. So the asserted inequality holds by (21.4). If x ∉ 𝔭 ∈ Supp(M) when dim(R/𝔭) = dim(M), then (21.3)(2) yields the opposite inequality, so equality. Finally, if x ∉ z.div(M), then x ∉ 𝔭 for any 𝔭 ∈ Supp(M) with dim(R/𝔭) = dim(M) owing to (17.17) and (17.14). □

(21.6) (Height). — Let R be a ring, and 𝔭 a prime. The height of 𝔭, denoted ht(𝔭), is defined by this formula:

ht(𝔭) := sup{ r | there’s a chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} = 𝔭 }.

The bijective correspondence 𝔭 ↦ 𝔭R_{𝔭} of (11.18)(2) yields this formula:

ht(𝔭) = dim(R_{𝔭}).  (21.6.1)

Corollary (21.7). — Let R be a Noetherian ring, 𝔭 a prime. Then ht(𝔭) ≤ r if and only if 𝔭 is minimal containing an ideal generated by r elements.

Proof: Assume 𝔭 is minimal containing an ideal 𝔞 generated by r elements. Now, any prime of R_{𝔭} containing 𝔞R_{𝔭} is of the form 𝔮R_{𝔭} where 𝔮 is a prime of R with 𝔞 ⊂ 𝔮 ⊂ 𝔭 by (11.18). So 𝔮 = 𝔭. Hence 𝔭R_{𝔭} = \sqrt{𝔞R_{𝔭}} by the Scheinnullstellensatz. Hence r ≥ s(R_{𝔭}) by (21.2). But s(R_{𝔭}) = dim(R_{𝔭}) by (21.4), and dim(R_{𝔭}) = ht(𝔭) by (21.6.1). Thus ht(𝔭) ≤ r.

Conversely, assume ht(𝔭) ≤ r. Then R_{𝔭} has a parameter ideal 𝔟 generated by r elements, say y_{1}, …, y_{r} by (21.6.1) and (21.4). Say y_{i} = x_{i}/s_{i} with s_{i} ∉ 𝔭. Set 𝔞 := ⟨x_{1}, …, x_{r}⟩. Then 𝔞R_{𝔭} = 𝔟.

Suppose there is a prime 𝔮 with 𝔞 ⊂ 𝔮 ⊂ 𝔭. Then 𝔟 = 𝔞R_{𝔭} ⊂ 𝔮R_{𝔭} ⊂ 𝔭R_{𝔭}, and 𝔮R_{𝔭} is prime by (11.18)(2). But \sqrt{𝔟} = 𝔭R_{𝔭}. So 𝔮R_{𝔭} = 𝔭R_{𝔭}. Hence 𝔮 = 𝔭 by (11.18)(2). Thus 𝔭 is minimal containing 𝔞, which is generated by r elements. □

Exercise (21.8). — Let R be a Noetherian ring, and 𝔭 be a prime minimal containing x_{1}, …, x_{r}. Given r′ with 1 ≤ r′ ≤ r, set R′ := R/⟨x_{1}, …, x_{r′}⟩ and 𝔭′ := 𝔭/⟨x_{1}, …, x_{r′}⟩. Assume ht(𝔭) = r. Prove ht(𝔭′) = r − r′.

Theorem (21.9) (Krull Principal Ideal). — Let R be a Noetherian ring, x ∈ R, and 𝔭 a prime minimal containing x. If x ∉ z.div(R), then ht(𝔭) = 1.

Proof: We have ht(𝔭) ≤ 1 by (21.7). But if ht(𝔭) = 0, then 𝔭 ∈ Ass(R) by (17.17), and so x ∈ z.div(R) by (17.14). □

Exercise (21.10). — Let R be a Noetherian ring, 𝔭 a prime ideal with ht(𝔭) ≥ 2. Prove 𝔭 is the union of infinitely many distinct prime ideals 𝔮 with ht(𝔮) = 1.

Exercise (21.11). — Let R be a Noetherian ring with only finitely many prime ideals. Show dim(R) ≤ 1.

Exercise (21.12). — Let R be a domain. Prove that, if R is a UFD, then every height-1 prime is principal, and that the converse holds if R is Noetherian.

Exercise (21.13). — (1) Let A be a Noetherian local ring, and 𝔭 a principal prime of height at least 1. Prove that A is a domain.

(2) Let k be a field, P := k[[X]] the formal power series ring in one variable. Set R := P × P. Prove that P is Noetherian and semilocal, and that P contains a principal prime 𝔭 of height 1, but that P is not a domain.

Exercise (21.14). — Let R be a finitely generated algebra over a field. Assume R is a domain of dimension r. Let x ∈ R be neither 0 nor a unit. Set R′ := R/⟨x⟩. Prove that r − 1 is the length of any chain of primes in R′ of maximal length.

Corollary (21.15). — Let A and B be Noetherian local rings, 𝔪 and 𝔫 their maximal ideals. Let φ: A → B be a local homomorphism. Then

dim(B) ≤ dim(A) + dim(B/𝔪B),

with equality if B is flat over A.

Proof: Set s := dim(A). By (21.4), there is a parameter ideal 𝔮 generated by s elements. Then 𝔪/𝔮 is nilpotent by (21.2.1). Hence 𝔪B/𝔮B is nilpotent. It follows that dim(B/𝔪B) = dim(B/𝔮B). But (21.5) yields dim(B/𝔮B) ≥ dim(B)−s. Thus the inequality holds.

Assume B is flat over A. Let 𝔭 ⊃ 𝔪B be a prime with dim(B/𝔭) = dim(B/𝔪B). Then dim(B) ≥ dim(B/𝔭) + ht(𝔭) because the concatenation of a chain of primes containing 𝔭 of length dim(B/𝔭) with a chain of primes contained in 𝔭 of length ht(𝔭) is a chain of primes of B of length ht(𝔭)+dim(B/𝔭). Hence it suffices to show that ht(𝔭) ≥ dim(A).

As 𝔫 ⊃ 𝔭 ⊃ 𝔪B and as φ is local, φ^{−1}(𝔭) = 𝔪. Since B is flat over A, (14.11) and induction yield a chain of primes of B descending from 𝔭 and lying over any given chain in A. Thus ht(𝔭) ≥ dim(A), as desired. □

Exercise (21.16). — Let R be a Noetherian ring. Prove that

dim(R[X]) = dim(R) + 1.

Exercise (21.17). — Let A be a Noetherian local ring of dimension r. Let 𝔪 be the maximal ideal, and k := A/𝔪 the residue class field. Prove that

r ≤ dim_{k}(𝔪/𝔪^{2}),

with equality if and only if 𝔪 is generated by r elements.

(21.18) (Regular local rings). — Let A be a Noetherian local ring of dimension r. We say A is regular if its maximal ideal is generated by r elements. Then any r generators are said to form a regular system of parameters.

By (21.17), A is regular if and only if r = dim_{k}(𝔪/𝔪^{2}).

For example, a field is a regular local ring of dimension 0, and is the only one.

Lemma (21.19). — Let A be a Noetherian semilocal ring of dimension r, and 𝔮 a parameter ideal. Then deg h(G_{𝔮}(R), n) = r − 1.

Proof: By (20.8), deg h(G_{𝔮}(A), r) is equal to 1 less than the order of pole at 1 of the Hilbert series H(G_{𝔮}(A), t). But that order is equal to d(A) by (21.2). Further, d(A) = r by the Dimension Theorem, (21.4). Thus the assertion holds.

□

Proposition (21.20). — Let A be a Noetherian local ring of dimension r, and 𝔪 its maximal ideal. Then A is regular if and only if its associated graded ring G_{𝔪}(A) is a polynomial ring; if so, then the number of variables is r.

Proof: Assume G(A) is a polynomial ring in s variables. Then dim(𝔪/𝔪^{2}) = s. By (20.4), deg h(G_{𝔪}(A), n) = s − 1. So s = r by (21.19). So A is regular by (21.18).

Conversely, assume A is regular. Let x_{1}, …, x_{r} be a regular sop, and x′_{i} ∈ 𝔪/𝔪^{2} the residue of x_{i}. Set k := A/𝔪, and let P := k[X_{1}, …, X_{r}] be the polynomial ring. Form the k-algebra homomorphism φ: P → G(A) with φ(X_{i}) = x′_{i}.

Then φ is surjective as the x_{i} generate G(A). Set 𝔞 := Ker φ. Let P = ⨁ P_{n} be the grading by total degree. Then φ preserves the gradings of P and G(A). So 𝔞 inherits a grading: 𝔞 = ⨁ 𝔞_{n}. So for n ≥ 0, there’s this canonical exact sequence:

0 → 𝔞_{n} → P_{n} → 𝔪^{n}/𝔪^{n+1} → 0.  (21.20.1)

Suppose 𝔞 ≠ 0. Then there’s a nonzero f ∈ 𝔞_{m} for some m. Take n ≥ m. Then P_{n−m}f ⊂ 𝔞_{n}. Since P is a domain, P_{n−m} ⟶^{∼} P_{n−m}f. Therefore, (21.20.1) yields

dim_{k}(𝔪^{n}/𝔪^{n+1}) = dim_{k}(P_{n}) − dim_{k}(𝔞_{n})
≤ dim_{k}(P_{n}) − dim_{k}(P_{n−m}) = \binom{r−1+n}{r−1} − \binom{r−1+n−m}{r−1}.

The expression on the right is a polynomial in n of degree r − 2.

On the other hand, dim_{k}(𝔪^{n}/𝔪^{n+1}) = h(G(A), n) for n ≫ 0 by (20.8). Further, deg h(G(A), n) = r − 1 by (21.19). However, it follows from the conclusion of the preceding paragraph that deg h(G(A), n) ≤ r − 2. We have a contradiction! Hence 𝔞 = 0. Thus φ is injective, so bijective, as desired. □

Exercise (21.21). — Let A be a Noetherian local ring of dimension r, and let x_{1}, …, x_{s} ∈ A with s ≤ r. Set 𝔞 := ⟨x_{1}, …, x_{s}⟩ and B := A/𝔞. Prove equivalent:

(1) A is regular, and there are x_{s+1}, …, x_{r} ∈ A with x_{1}, …, x_{r} a regular sop.

(2) B is regular of dimension r − s.

Theorem (21.22). — A regular local ring A is a domain.

Proof: Use induction on r := dim(A). If r = 0, then A is a field, so a domain. Assume r ≥ 1. Let x be a member of a regular sop. Then A/⟨x⟩ is regular of

dimension r − 1 by (21.21). By induction, A/⟨x⟩ is a domain. So ⟨x⟩ is prime. Thus A is a domain by (21.13). □

Lemma (21.23). — Let A be a local ring, 𝔪 its maximal ideal, 𝔞 a proper ideal. Set 𝔫 := 𝔪/𝔞 and k := A/𝔪. Then this sequence of k-vector spaces is exact:

0 → (𝔪^{2} + 𝔞)/𝔪^{2} → 𝔪/𝔪^{2} → 𝔫/𝔫^{2} → 0.

Proof: The assertion is very easy to check. □

Proposition (21.24). — Let A be a regular local ring of dimension r, and 𝔞 an ideal. Set B := A/𝔞, and assume B is regular of dimension r − s. Then 𝔞 is generated by s elements, and any such s elements form part of a regular sop.

Proof: In its notation, (21.23) yields dim((𝔪^{2} + 𝔞)/𝔪^{2}) = s. Hence, any set of generators of 𝔞 includes s members of a regular sop of A. Let 𝔟 be the ideal the s generate. Then A/𝔟 is regular of dimension r − s by (21.21). By (21.22), both A/𝔟 and B are domains of dimension r − s; whence, (15.10) implies 𝔞 = 𝔟. □

22. Completion

Completion is used to simplify a ring and its modules beyond localization. First, we discuss the topology of a filtration, and use Cauchy sequences to construct the completion. Then we discuss the inverse limit, the dual notion of the direct limit; thus we obtain an alternative construction. We conclude that, if we use the adic filtration of an ideal, then the functor of completion is exact on finitely generated modules over a Noetherian ring. Further, then the completion of a Noetherian ring is Noetherian; if the ideal is maximal, then the completion is local. We end with a useful version of the Cohen Structure Theorem for complete Noetherian local rings.

(22.1) (Topology and completion). — Let R be a ring, M a module equipped with a filtration M = M_{0} ⊃ M_{1} ⊃ · · ·. Then M has a topology: the open sets are the arbitrary unions of the sets m + M_{n}. Indeed, the intersection of two open sets is open, because the intersection of two unions is the union of the pairwise intersections; further, if the intersection U of m + M_{n} and m′ + M_{n′} is nonempty and if n ≥ n′, then U = m + M_{n}, because, if say m′′ ∈ U, then

m + M_{n} = m′′ + M_{n} ⊂ m′′ + M_{n′} = m′ + M_{n′}.  (22.1.1)

The addition map M × M → M, given by (m, m′) ↦ m + m′, is continuous, as

(m + M_{n}) + (m′ + M_{n}) ⊂ (m + m′) + M_{n}.

So, with m′ fixed, the translation m ↦ m + m′ is a homeomorphism M → M. (Similarly, inversion m ↦ −m is a homeomorphism; so M is a topological group.)

Let 𝔞 be an ideal, and give R the 𝔞-adic filtration. If the filtration on M is an 𝔞-filtration, then scalar multiplication (x, m) ↦ xm too is continuous, because

(x + 𝔞^{n})(m + M_{n}) ⊂ xm + M_{n}.

Further, if the filtration is 𝔞-stable, then it yields the same topology as the 𝔞-adic filtration, because

M_{n} ⊃ 𝔞^{n}M ⊃ 𝔞^{n}M_{n′} = M_{n+n′}.

Thus any two stable 𝔞-filtrations give the same topology, called the 𝔞-adic topology.

When 𝔞 is given, it is conventional to use the 𝔞-adic filtration and 𝔞-adic topology unless there’s explicit mention to the contrary. Further, if R is semi-local, then it is conventional to take 𝔞 := rad(R).

Let N be a submodule of M. Then the closure \overline{N} of N is equal to ⋂_{n≥0}(N +M_{n}), because m ∉ \overline{N} means there’s n ≥ 0 with (m + M_{n}) ∩ N = ∅, or m ∉ (N + M_{n}). In particular, each M_{n} is closed, and {0} is closed if and only if ⋂ M_{n} = {0}.

Further, M is separated — that is, Hausdorff — if and only if {0} is closed. For, if {0} is closed, then so is each {m}. Hence, given m′ ≠ m, there’s n′ so that m ∉ (m′ +M_{n′}). Take n ≥ n′. Then (m+M_{n})∩(m′ +M_{n′}) = ∅ owing to (22.1.1).

Finally, M is discrete — that is, every {m} is both open and closed — if and only if {0} is open.

A sequence (m_{n})_{n≥0} in M is called Cauchy if, given n_{0}, there’s n_{1} with

m_{n} − m_{n′} ∈ M_{n_{0}}, or simply m_{n} − m_{n+1} ∈ M_{n_{0}}, for all n, n′ ≥ n_{1};

September 3, 2012 11Nts.tex the two conditions are equivalent because M_{n_{0}} is a subgroup and

m_{n} − m_{n′} = (m_{n} − m_{n+1}) + (m_{n+1} − m_{n+2}) + ⋯ + (m_{n′−1} − m_{n′}).

An m ∈ M is called a limit of (m_{n}) if, given n_{0}, there’s n_{1} with m − m_{n} ∈ M_{n_{0}} for all n ≥ n_{1}. If every Cauchy sequence has a limit, then M is called complete.

The Cauchy sequences form a module under termwise addition and termwise scalar multiplication. The sequences with 0 as a limit form a submodule. The quotient module is denoted M̂ and is called the completion. There is a canonical homomorphism, which carries m ∈ M to the class of the constant sequence (m):

κ: M → M̂ by κ(m) := (m).

It is straightforward to check that the notions of Cauchy sequence and limit depend only on the topology. Similarly, M̂ is separated and complete with respect to the filtration M̂ = M̂_{0} ⊃ M̂_{1} ⊃ ⋯, and κ is the universal example of a continuous homomorphism from M into a separated and complete, filtered module.

Again, let 𝔞 be an ideal. Under termwise multiplication of Cauchy sequences, R̂ is a ring, κ: R → R̂ is a ring homomorphism, and M̂ is an R̂-module. Further, M ↦ M̂ is a linear functor from ((R-mod)) to ((R̂-mod)).

For example, let k be a ring, and R := k[X_{1}, …, X_{r}] the polynomial ring in r variables. Set 𝔞 := ⟨X_{1}, …, X_{r}⟩. Then a sequence (m_{n})_{n≥0} of polynomials is Cauchy if and only if, given n_{0}, there’s n_{1} such that, for all n ≥ n_{1}, the m_{n} agree in degree less than n_{0}. Thus R̂ is just the power series ring k[[X_{1}, …, X_{r}]].

For another example, take a prime integer p, and set 𝔞 := ⟨p⟩. Then a sequence (m_{n})_{n≥0} of integers is Cauchy if and only if, given n_{0}, there’s n_{1} such that, for all n, n′ ≥ n_{1}, the difference m_{n} − m_{n′} is a multiple of p^{n_{0}}. The completion of ℤ is called the p-adic integers, and consists of the sums ∑_{i=0}^{∞} z_{i}p^{i} with 0 ≤ z_{i} < p.

Proposition (22.2). — Let R be a ring, and 𝔞 an ideal. Then 𝔞̂ ⊂ rad(R̂).

Proof: Recall from (22.1) that R̂ is complete in the 𝔞̂-adic topology. Hence for x ∈ 𝔞̂, we have 1/(1 − x) = 1 + x + x^{2} + ⋯ in R̂. Thus 𝔞̂ ⊂ rad(R̂) by (3.2). □

Exercise (22.3). — In the 2-adic integers, evaluate the sum 1 + 2 + 4 + 8 + · · ·.

Exercise (22.4). — Let R be a ring, 𝔞 an ideal, and M a module. Prove the following three conditions are equivalent:

(1) κ: M → M̂ is injective;  (2) ⋂ 𝔞^{n}M = ⟨0⟩;  (3) M is separated.

Corollary (22.5). — Let R be a Noetherian ring, 𝔞 ⊂ rad(R) an ideal, and M a finitely generated module. Then M ⊂ M̂.

Proof: The assertion results from (22.4), (18.28) or (20.19), and (3.2). □

(22.6) (Inverse limits). — Let R be a ring. Given modules Q_{n} equipped with homomorphisms α_{n}^{n+1} : Q_{n+1} → Q_{n} for n ≥ 0, their inverse limit \underleftarrow{lim} Q_{n} is the submodule of ∏ Q_{n} of all vectors (q_{n}) with α_{n}^{n+1}(q_{n+1}) = q_{n} for all n. Note that

\underleftarrow{lim} Q_{n} = Ker(θ)  (22.6.1)

where θ: ∏ Q_{n} → ∏ Q_{n} is the map defined by θ(q_{n}) := (q_{n} − α_{n}^{n+1}q_{n+1}).

Clearly, \underleftarrow{lim} Q_{n} has this UMP: given maps β_{n}: P → Q_{n} with α_{n}^{n+1}β_{n+1} = β_{n}, there’s a unique map β: P → \underleftarrow{lim} Q_{n} with π_{n}β = β_{n} for all n.

Further, the UMP yields these two natural module isomorphisms:

\underleftarrow{lim} Hom(P, Q_{n}) = Hom(P, \underleftarrow{lim} Q_{n}),
\underleftarrow{lim} Hom(Q_{n}, N) = Hom(\underrightarrow{lim} Q_{n}, N).

(The notion of inverse limit is formally dual to that of direct limit.)

For example, let k be a ring, and R := k[X_{1}, …, X_{r}] the polynomial ring in r variables. Set 𝔪 := ⟨X_{1}, …, X_{r}⟩ and R_{n} := R/𝔪^{n+1}. Then R_{n} is just the R-algebra of polynomials of degree at most n, and the canonical map α_{n}^{n+1} : R_{n+1} → R_{n} is just truncation. Thus \underleftarrow{lim} R_{n} is equal to the power series ring k[[X_{1}, …, X_{r}]].

For another example, take a prime integer p, and set ℤ_{n} := ℤ/⟨p^{n+1}⟩. Then ℤ_{n} is just the ring of sums ∑_{i=0}^{n} z_{i}p^{i} with 0 ≤ z_{i} < p, and the canonical map α_{n}^{n+1} : ℤ_{n+1} → ℤ_{n} is just truncation. Thus \underleftarrow{lim} ℤ_{n} is just the ring of p-adic integers.

In general, consider exact sequences of modules

0 → Q′_{n} ⟶^{β_{n}} Q_{n} ⟶^{γ_{n}} Q′′_{n} → 0

and commutative diagrams

0 ⟶ Q′_{n+1} ⟶^{β_{n+1}} Q_{n+1} ⟶^{γ_{n+1}} Q′′_{n+1} ⟶ 0

α_{n}^{′n+1}↓  ^{α_{n}^{n+1}}↓  ^{α_{n}^{′′n+1}}↓

0 ⟶ Q′_{n} ⟶^{β_{n}} Q_{n} ⟶^{γ_{n}} Q′′_{n} ⟶ 0

Then the induced sequence

0 → \underleftarrow{lim} Q′_{n} \underrightarrow{β̂} \underleftarrow{lim} Q_{n} \underrightarrow{γ̂} \underleftarrow{lim} Q′′_{n}  (22.6.2)

is exact; further, γ̂ is surjective if all the α_{n}^{′n+1} are surjective.

Indeed, the above commutative diagrams yield the following one:

0 ⟶ ∏ Q′_{n} \underrightarrow{∏ β_{n}} ∏ Q_{n} \underrightarrow{∏ γ_{n}} ∏ Q′′_{n} ⟶ 0
θ′↓  ^{θ}↓  _{θ′′}↓
0 ⟶ ∏ Q′_{n} \underrightarrow{∏ β_{n}} ∏ Q_{n} \underrightarrow{∏ γ_{n}} ∏ Q′′_{n} ⟶ 0

Owing to (22.6.1), the Snake Lemma (5.12) yields the exact sequence (22.6.2) and an injection Coker(γ̂) ↪ Coker(θ′). Also, Coker(θ′) = 0 if the α_{n}^{′n+1} are surjective, because given (q′_{n}) ∈ ∏ Q′_{n} , we can solve the equations p′_{n} −α_{n}^{′n+1}(p′_{n+1}) = q′_{n} recursively to get (p′_{n}) ∈ ∏ Q′_{n} with θ′(p′_{n}) = (q′_{n}). Thus γ̂ is surjective.

Proposition (22.7). — Let R be a ring, M a module, M = M_{0} ⊃ M_{1} ⊃ ⋯ a filtration. Then M̂ ⟶^{∼} \underleftarrow{lim}(M/M_{n}).

Proof: First, let’s define a map α: M̂ → \underleftarrow{lim}(M/M_{n}). Given a Cauchy sequence (m_{ν}), let q_{n} be the image of m_{ν} in M/M_{n} for ν ≫ 0. Then q_{n} is independent of ν, because the sequence is Cauchy. Clearly, q_{n} is the residue of q_{n+1} in M/M_{n}. Also, (m_{ν}) has 0 as a limit if and only if q_{n} = 0 for all n. Define α by α(m_{ν}) := (q_{n}). Clearly, α is well defined, linear, and injective.

As to surjectivity, given (q_{n}) ∈ \underleftarrow{lim}(M/M_{n}), let m_{ν} ∈ M represent q_{ν} ∈ M/M_{ν} for each ν. Then m_{μ} − m_{ν} ∈ M_{ν} for μ ≥ ν because the residue of q_{n} in M/M_{ν} is q_{ν}. Hence (m_{ν}) is Cauchy. Thus α is surjective, so an isomorphism. □

Exercise (22.8). — Let A be a Noetherian semilocal ring, and 𝔪_{1}, …, 𝔪_{m} all its maximal ideals. Prove that Â = ∏ Â_{𝔪_{i}}.

Exercise (22.9). — Let R be a ring, M a module, M = M_{0} ⊃ M_{1} ⊃ ⋯ a filtration, and N ⊂ M a submodule. Filter N by N_{n} := N ∩ M_{n}. Assume N ⊃ M_{n} for n ≥ n_{0} for some n_{0}. Prove N̂ ⊂ M̂ and M̂/N̂ = M/N and G(M̂) = G(M).

Exercise (22.10). — (1) Let R be a ring, 𝔞 an ideal. If G_{𝔞}(R) is a domain, show R̂ is an domain. If also ⋂_{n≥0} 𝔞^{n} = 0, show R is a domain.

(2) Use (1) to give an alternative proof that a regular local ring is a domain.

Proposition (22.11). — Let A be a ring, 𝔪 a maximal ideal. Then Â is a local ring with maximal ideal 𝔪̂.

Proof: First, Â/𝔪̂ = A/𝔪 by (22.9); so 𝔪̂ is maximal. Next, rad(Â) ⊃ 𝔪̂ by (22.2). Finally, let 𝔪′ be any maximal ideal of Â. Then 𝔪′ ⊃ rad(Â). Hence 𝔪′ = 𝔪̂. Thus 𝔪̂ is the only maximal ideal. □

Exercise (22.12). — Let A be a semilocal ring, 𝔪_{1}, …, 𝔪_{m} all its maximal ideals, and set 𝔪 := rad(A). Prove that Â is a semilocal ring, that 𝔪̂_{1}, …, 𝔪̂_{m} are all its maximal ideals, and that 𝔪̂ = rad(Â).

(22.13) (Completion, units, and localization). — Let R be a ring, 𝔞 an ideal, and κ: R → R̂ the canonical map. Given t ∈ R, for each n denote by t_{n} ∈ R/𝔞^{n} the residue of t. Let’s show that κ(t) is a unit if and only if each t_{n} is.

Indeed, by (22.7), we may regard R̂ as a submodule of ∏ R/𝔞^{n}. Then each t_{n} is equal to the projection of κ(t). Hence t_{n} is a unit if κ(t) is. Conversely, assume t_{n} is a unit for each n. Then there are u_{n} ∈ R with u_{n}t ≡ 1 (mod 𝔞^{n}). By the uniqueness of inverses, u_{n+1} ≡ u_{n} in R/𝔞^{n} for each n. Set u := (u_{n}) ∈ ∏ R/𝔞^{n}. Then u ∈ R̂, and uκ(t) = 1. Thus κ(t) is a unit.

Set T := κ^{−1}(R̂^{×}). Then by the above, T consists of the t ∈ R whose residue t_{n} ∈ R/𝔞^{n} is a unit for each n. So (2.29) and (1.8) yield

T = { t ∈ R | t lies in no maximal ideal containing 𝔞 }.  (22.13.1)

Set S := 1 + 𝔞. Then S ⊂ T owing to (22.13.1) as no maximal ideal can contain both x and 1 + x. Hence the UMP of localization (11.6) yields this diagram:

[figure: R; S^{−1}R; T^{−1}R; R̂; κ; φ_{S}; φ_{T}; σ; τ]

Further, S and T map into (R/𝔞^{n})^{×}; hence, (11.7), (11.22), and (12.18) yield:

R/𝔞^{n} = S^{−1}R/𝔞^{n}S^{−1}R = T^{−1}R/𝔞^{n}T^{−1}R.

Therefore, R̂ is, by (22.7), equal to the completion of each of S^{−1}R and T^{−1}R in their 𝔞S^{−1}R-adic and 𝔞T^{−1}R-adic topologies.

For example, take 𝔞 to be a maximal ideal 𝔪. Then T = R − 𝔪 by (22.13.1). Thus R̂ is equal to the completion of the localization R_{𝔪}.

Finally, assume R is Noetherian. Let’s prove that σ and τ are injective. Indeed, say τσ(x/s) = 0. Then κ(x) = 0 as κ(s) is a unit. So x ∈ ⋂ 𝔞^{n}. Hence the Krull Intersection Theorem, (18.28) or (20.19), yields an s′ ∈ S with s′x = 0. So x/s = 0 in S^{−1}R. Thus σ is injective. Similarly, τ is injective.

September 3, 2012 11Nts.tex Theorem (22.14) (Exactness of completion). — Let R be a Noetherian ring, 𝔞 an ideal. Then on the finitely generated modules M, the functor M ↦ M̂ is exact.

Proof: Let 0 → M′ → M → M′′ → 0 be an exact sequence of modules. Set M′_{n} := M′ ∩ 𝔞^{n}M. Then we obtain these exact sequences:

0 → M′/M′_{n} → M/𝔞^{n}M → M′′/𝔞^{n}M′′ → 0.

The maps M′/M′_{n+1} → M′/M′_{n} are surjective. So (22.6) yields this exact sequence:

0 → \underleftarrow{lim} M′/M′_{n} → \underleftarrow{lim} M/𝔞^{n}M → \underleftarrow{lim} M′′/𝔞^{n}M′′ → 0.

Assume R is Noetherian and M is finitely generated. Then M′ = M′_{0} ⊃ M′_{1} ⊃ ⋯ is an 𝔞-stable filtration by the Artin–Rees Lemma (20.18). Hence, (22.1) and (22.7) yield the desired exactness of the sequencce 0 → M̂′ → M̂ → M̂′′ → 0. □

Exercise (22.15). — Let A be a Noetherian semilocal ring. Prove that an element x ∈ A is a nonzerodivisor if and only if its image x̂ ∈ Â is also.

Exercise (22.16). — Let p ∈ ℤ be prime. For n > 0, define a ℤ-linear map

α_{n}: ℤ/⟨p⟩ → ℤ/⟨p^{n}⟩ by α_{n}(1) = p^{n−1}.

Set A := ⨁_{n≥1} ℤ/⟨p⟩ and B := ⨁_{n≥1} ℤ/⟨p^{n}⟩. Set α := ⨁ α_{n}; so α: A → B. (1) Show that the p-adic completion Â is just A.

(2) Show that, in the topology on A induced by the p-adic topology on B, the completion \overline{A} is equal to ∏_{n=1}^{∞} ℤ/⟨p⟩.

(3) Show that the natural sequence of p-adic completions

0 → Â \underrightarrow{α̂} B̂ \underrightarrow{κ̂} (B/A)

is not exact at B̂. (Thus p-adic completion is not right-exact on ((ℤ-mod)).)

Corollary (22.17). — Let R be a Noetherian ring, 𝔞 an ideal, and M a finitely generated module. Then the natural map is an isomorphism:

R̂ ⊗ M ⟶^{∼} M̂.

Proof: By (22.14), the functor M ↦ M̂ is exact on the category of finitely generated modules, and so (8.16) yields the conclusion. □

Exercise (22.18). — Let R be a ring, 𝔞 an ideal. Show that M ↦ M̂ preserves surjections, and that R̂ ⊗ M → M̂ is surjective if M is finitely generated.

Corollary (22.19). — Let R be a Noetherian ring, 𝔞 and 𝔟 ideals, M a finitely generated module. Then, using the 𝔞-adic topology, we have

(1) (𝔟M) = 𝔟M̂ = 𝔟̂ M̂ and (2) (𝔟^{n}) = 𝔟^{n}R̂ = (𝔟R̂)^{n} = (𝔟̂)^{n} for any n ≥ 0.

Proof: In general, the inclusion 𝔟M → M induces a commutative square

R̂ ⊗ (𝔟M) ⟶ R̂ ⊗ M
↓  ↓
(𝔟M) ⟶ M̂

It is not hard to see that top map’s image is 𝔟(R̂ ⊗ M).

In the present case, the two vertical maps are isomorphisms by (22.17), and the

September 3, 2012 11Nts.tex bottom map is injective by (22.14). Thus (𝔟M) = 𝔟M̂.

Taking R for M yields 𝔟̂ = 𝔟R̂. Hence 𝔟M̂ = 𝔟R̂ M̂ = 𝔟̂ M̂. Thus (1) holds.

In (1), taking 𝔟^{n} for 𝔟 and R for M yields (𝔟^{n}) = 𝔟^{n}R̂. In particular, 𝔟̂ = 𝔟R̂; so (𝔟R̂)^{n} = (𝔟̂)^{n}. But 𝔟^{n}R′ = (𝔟R′)^{n} for any R-algebra R′. Thus (2) holds. □

Corollary (22.20). — Let R be a Noetherian ring, 𝔞 an ideal. Then R̂ is flat.

Proof: Let 𝔟 be any ideal. Then R̂ ⊗ 𝔟 = 𝔟̂ by (22.17), and 𝔟̂ = 𝔟R̂ by (22.19)(2). Thus R̂ is flat by the Ideal Criterion (9.20). □

Exercise (22.21). — Let R be a Noetherian ring, and 𝔞 and 𝔟 ideals. Assume 𝔞 ⊂ rad(R), and use the 𝔞-adic topology. Prove 𝔟 is principal if 𝔟R̂ is.

Lemma (22.22). — Let R be a ring, β: M → N a map of filtered modules (so β preserves the filtration). If G(β) is injective or surjective, then so is β̂.

Proof: Consider the following commutative diagram of exact sequences: 0 ⟶ M_{n}/M_{n+1} ⟶ M/M_{n+1} ⟶ M/M_{n} ⟶ 0

G_{n}(β)↓  ^{β_{n+1}}↓  ^{β_{n}}↓
0 ⟶ N_{n}/N_{n+1} ⟶ N/N_{n+1} ⟶ N/N_{n} ⟶ 0

Apply the Snake Lemma (5.12). It yields the following exact sequence:

Ker G_{n}(β) → Ker β_{n+1} → Ker β_{n} → Coker G_{n}(β) → Coker β_{n+1} → Coker β_{n}.

Assume G(β) is injective. Then Ker G_{n}(β) = 0. Hence induction on n yields Ker β_{n} = 0 for all n. Thus β̂ is injective by (22.6).

Assume G(β) is surjective. Then Coker G_{n}(β) = 0. So Ker β_{n+1} → Ker β_{n} is surjective for all n. Also, induction on n yields Coker β_{n} = 0 for all n; that is,

0 ⟶ Ker β_{n} ⟶ M/M_{n} ⟶^{β_{n}} N/N_{n} ⟶ 0

is exact. Thus β̂ is surjective by (22.6). □

Lemma (22.23). — Let R be a complete ring, M a separated module. If G(M) is finitely generated over G(R), then M is finitely generated over R and complete.

Proof: Take finitely many homogeneous generators of G(M). Lift them to M. The lifts define a map α: R^{n} → M, and G(α) is surjective. So α̂ is surjective by (22.22). Now, form this canonical commutative diagram:

R^{n} ⟶^{κ_{R^{n}}} R̂^{n}

α↓  α̂↓

M ⟶^{κ_{M}} M̂

Since R is complete, κ_{R^{n}} is surjective by (22.1). Since M is separated, κ_{M} is injective by (22.4). Hence κ_{M} is an isomorphism and α is surjective, as desired. □

Exercise (22.24) (Nakayama’s Lemma for a complete ring). — Let R be a ring, 𝔞 an ideal, and M a module. Assume R is complete, and M separated. Show m_{1}, …, m_{n} ∈ M generate if their images in M/𝔞M generate.

Proposition (22.25). — Let R be a complete ring, M a separated module. If G(M) is a Noetherian G(R)-module, then M is a complete Noetherian R-module.

September 3, 2012 11Nts.tex

Proof: Given a submodule N ⊂ M, set N_{n} := M_{n} ∩ N. Then G(N) ⊂ G(M). As G(M) is Noetherian, G(N) is finitely generated. Hence N is finitely generated and complete by (22.23). Thus M is Noetherian and complete. □

Theorem (22.26). — Let R be a ring, 𝔞 an ideal. If R is Noetherian, so is R̂.

Proof: Assume R is Noetherian. Then G(R) is finitely generated as an (R/𝔞)- algebra by (20.12). So G(R) is Noetherian by the Hilbert Basis Theorem, (16.11). But G(R) = G(R̂) by (22.9). Hence R̂ is Noetherian by (22.25) with R̂ for R and for M. □

Example (22.27). — Let k be a Noetherian ring, P := k[X_{1}, …, X_{r}] the polynomial ring, and A := k[[X_{1}, …, X_{r}]] the formal power series ring. Then A is the completion of P in the ⟨X_{1}, …, X_{r}⟩-adic topology by (22.1). Further, P is Noetherian by the Hilbert Basis Theorem, (20.12). Thus A is Noetherian by (22.26).

Assume k is a domain. Then A is a domain. Indeed, A is one if r = 1, because

(a_{m}X_{1}^{m} + ⋯)(b_{n}X_{1}^{n} + ⋯) = a_{m}b_{n}X_{1}^{m+n} + · · · .

If r > 1, then A = k[[X_{1}, …, X_{i}]] [[X_{i+1}, …, X_{r}]]; so A is a domain by induction.

Set 𝔭_{i} := ⟨X_{i+1}, …, X_{r}⟩. Then A/𝔭_{i} = k[[X_{1}, …, X_{i}]] by (3.7). Hence 𝔭_{i} is prime. So 0 = 𝔭_{r} ⫋ ⋯ ⫋ 𝔭_{0} is a chain of primes of length r. Thus dim(A) ≥ r.

Assume k is a field. Then A is local with maximal ideal ⟨X_{1}, …, X_{r}⟩ and with residue field k by the above and either by (22.11) or again by (3.7). Therefore, dim(A) ≤ r by (21.17). Thus A is regular of dimension r.

Exercise (22.28). — Let A be a Noetherian local ring, 𝔪 the maximal ideal. Prove (1) that Â is a Noetherian local ring with 𝔪̂ as maximal ideal, (2) that dim(A) = dim(Â), and (3) that A is regular if and only if Â is regular.

Theorem (22.29) (UMP of Formal Power Series). — Let R be a ring, R′ an R-algebra, 𝔟 an ideal of R′, and x_{1}, …, x_{n} ∈ 𝔟. Let P := R[[X_{1}, …, X_{n}]] be the formal power series ring. If R′ is separated and complete, then there is a unique R-algebra map π: P → R′ with π(X_{i}) = x_{i} for 1 ≤ i ≤ n.

Proof: For each m, there’s a unique R-algebra map R[X_{1}, …, X_{n}] → R′/𝔟^{m} sending X_{i} to the residue of x_{i}. This map induces a map

P/⟨X_{1}, … X_{n}⟩^{m} = R[X_{1}, …, X_{n}]/⟨X_{1}, … X_{n}⟩^{m} ⟶ R′/𝔟^{m}.

Taking inverse limits yields π owing to (22.6) and (22.7). □

Theorem (22.30) (Cohen Structure). — Let A be a complete Noetherian local ring with maximal ideal 𝔪. Assume that A contains a coefficient field k; that is, k ⟶^{∼} A/𝔪. Then A ≃ k[[X_{1}, …, X_{n}]]/𝔞 for some variables X_{i} and ideal 𝔞. Further, if A is regular of dimension r, then A ≃ k[[X_{1}, …, X_{r}]].

Proof: Take generators x_{1}, …, x_{n} ∈ 𝔪. Let π: k[[X_{1}, …, X_{n}]] → A be the map with π(X_{i}) = x_{i} of (22.29). Then G(π) is surjective. Hence, π is surjective by (22.22). Set 𝔞 := Ker(π). Then k[[X_{1}, …, X_{n}]]/𝔞 ⟶^{∼} A.

Assume A is regular of dimension r. Take n = r. Then G(A) is a polynomial ring in r variables over k by (21.20). And G(k[[X_{1}, …, X_{r}]]) is too by (22.6). Since G(π) is surjective, its kernel is a minimal prime, so equal to ⟨0⟩. Hence G(π) is bijective. So π is bijective by (22.22). Thus k[[X_{1}, …, X_{r}]] ⟶^{∼} A. □

23. Discrete Valuation Rings

A discrete valuation is a homomorphism from the multiplicative group of a field to the additive group integers such that the value of a sum is at least the minimum value of the summands. The corresponding discrete valuation ring consists of the elements whose values are nonnegative, plus 0. We characterize these rings in various ways; notably, we prove they are the normal Noetherian local domains of dimension 1. Then we prove that any normal Noetherian domain is the intersection of all the discrete valuation rings obtained by localizing at its height-1 primes. Finally, we prove Serre’s Criterion for normality of Noetherian domains.

(23.1) (Discrete Valuations). — Let K be a field. We define a discrete valuation of K to be a surjective function v: K^{×} → ℤ such that, for every x, y ∈ K^{×},

(1) v(x · y) = v(x) + v(y), (2) v(x + y) ≥ min{v(x), v(y)} if x ≠ −y. (23.1.1)

Condition (1) just means v is a group homomorphism. Hence, for any x ∈ K^{×},

(1) v(1) = 0 and (2) v(x^{−1}) = −v(x).  (23.1.2)

As a convention, we define v(0) := ∞. Consider the sets

A := {x ∈ K | v(x) ≥ 0} and 𝔪 := {x ∈ K | v(x) > 0}.

Clearly, A is a subring, so a domain, and 𝔪 is an ideal. Further, 𝔪 is nonzero as v is surjective. We call A the discrete valuation ring (DVR) of v.

Notice that, if x ∈ K, but x ∉ A, then x^{−1} ∈ 𝔪; indeed, v(x) < 0, and so v(x^{−1}) = −v(x) > 0. Hence, Frac(A) = K. Further,

A^{×} = {x ∈ K | v(x) = 0} = A − 𝔪.

Indeed, if x ∈ A^{×}, then v(x) ≥ 0 and −v(x) = v(x^{−1}) ≥ 0; so v(x) = 0. Conversely, if v(x) = 0, then v(x^{−1}) = −v(x) = 0; so x^{−1} ∈ A, and so x ∈ A^{×}. Therefore, by the nonunit criterion, A is a local domain, not a field, and 𝔪 is its maximal ideal.

An element t ∈ 𝔪 with v(t) = 1 is called a (local) uniformizing parameter. Such a t is irreducible, as t = ab with v(a) ≥ 0 and v(b) ≥ 0 implies v(a) = 0 or v(b) = 0 since 1 = v(a) + v(b). Further, any x ∈ K^{×} has the unique factorization x = ut^{n} where u ∈ A^{×} and n := v(x); indeed, v(u) = 0 as u = xt^{−n}. In particular, t_{1} is uniformizing parameter if and only if t_{1} = ut with u ∈ A^{×}; also, A is a UFD.

Moreover, A is a PID; in fact, any nonzero ideal 𝔞 of A has the form

𝔞 = ⟨t^{m}⟩ where m := min{ v(x) | x ∈ 𝔞 }.  (23.1.3)

Indeed, given a nonzero x ∈ 𝔞, say x = ut^{n} where u ∈ A^{×}. Then t^{n} ∈ 𝔞. So n ≥ m. Set y := ut^{n−m}. Then y ∈ A and x = yt^{m}, as desired.

In particular, 𝔪 = ⟨t⟩ and dim(A) = 1. Thus A is regular local of dimension 1.

Example (23.2). — The prototype is this example. Let k be a field, t a variable, and K := k((t)) the field of formal Laurent series x := ∑_{i≥n} a_{i}t^{i} with n ∈ ℤ and with a_{i} ∈ k and a_{n} ≠ 0. Set v(x) := n, the “order of vanishing” of x. Clearly, v is a discrete valuation, the formal power series ring k[[t]] is its DVR, and 𝔪 := ⟨t⟩ is its maximal ideal.

The preceding example can be extended to cover any DVR A that contains a field k with k ⟶^{∼} A/⟨t⟩ where t is a uniformizing power. Indeed, A is a subring

September 3, 2012 11Nts.tex of its completion Â by (22.5), and Â = k[[t]] by the proof of the Cohen Structure Theorem (22.30). Further, clearly, the valuation on Â restricts to that on A.

A second old example is this. Let p ∈ ℤ be prime. Given x ∈ ℚ, write x = ap^{n}/b with a, b ∈ ℤ relatively prime and prime to p. Set v(x) := n. Clearly, v is a discrete valuation, the localization ℤ_{⟨p⟩} is its DVR, and pℤ_{⟨p⟩} is its maximal ideal. We call v the p-adic valuation of ℚ.

Lemma (23.3). — Let A be a local domain, 𝔪 its maximal ideal. Assume that 𝔪 is nonzero and principal and that ⋂_{n≥0} 𝔪^{n} = 0. Then A is a DVR.

Proof: Given a nonzero x ∈ A, there is an n ≥ 0 such that x ∈ 𝔪^{n} − 𝔪^{n+1}. Say 𝔪 = ⟨t⟩. Then x = ut^{n}, and u ∉ 𝔪, so u ∈ A^{×}. Set K := Frac(A). Given x ∈ K^{×}, write x = y/z where y = bt^{m} and z = ct^{k} with b, c ∈ A^{×}. Then x = ut^{n} with u := b/c ∈ A^{×} and n := m − k ∈ ℤ. Define v: K^{×} → ℤ by v(x) := n. If ut^{n} = wt^{h} with n ≥ h, then (u/w)t^{n−h} = 1, and so n = h. Thus v is well defined.

Since v(t) = 1, clearly v is surjective. To verify (23.1.1), take x = ut^{n} and y = wt^{h} with u, w ∈ A^{×}. Then xy = (uw)t^{n+h}. Thus (1) holds. To verify (2), we may assume n ≥ h. Then x + y = t^{h}(ut^{n−h} + w). Hence

v(x + y) ≥ h = min{n, h} = min{v(x), v(y)}.

Thus (2) holds. So v: K^{×} → ℤ is a valuation. Clearly, A is the DVR of v. □

(23.4) (Depth). — Let R be a ring, M a nonzero module, and x_{1}, …, x_{n} ∈ R. Set M_{i} := M/⟨x_{1}, …, x_{i}⟩. We say the sequence x_{1}, …, x_{n} is regular on M, or is an M-sequence, and call n its length if M_{n} ≠ 0 and x_{i} ∉ z.div(M_{i−1}) for all i.

We call the supremum of the lengths n of the M-sequences found in an ideal 𝔞 the depth of 𝔞 on M, and denote it by depth(𝔞, M). By convention, depth(𝔞, M) = 0 means 𝔞 contains no nonzerodivisor on M.

When M is semilocal, we call the depth of rad(M) on M simply the depth of M and denote it by depth(M). If depth(M) = dim(M), we call M Cohen– Macaulay.

Lemma (23.5). — Let A be a Noetherian local ring, 𝔪 its maximal ideal, and M a nonzero finitely generated module.

(1) Then depth(M) = 0 if and only if 𝔪 ∈ Ass(M).

(2) Then depth(M) = 1 if and only if there is an x ∈ 𝔪 with x ∉ z.div(M) and 𝔪 ∈ Ass(M/xM).

(3) Then depth(M) ≤ dim(M).

Proof: Consider (1). If 𝔪 ∈ Ass(M), then it is immediate from the definitions that 𝔪 ⊂ z.div(M) and so depth(M) = 0.

‘1Conversely, assume depth(M) = 0. Then 𝔪 ⊂ z.div(M). Since A is Noetherian, z.div(M) = ⋃_{𝔭∈Ass(M)} 𝔭 by (17.14). Since M is also finitely generated, Ass(M) is finite by (17.20). Hence 𝔪 = 𝔭 for some 𝔭 ∈ Ass(M) by Prime Avoidance, (3.15).

Consider (2). Assume depth(M) = 1. Then there is an M-sequence of length 1, but none longer. So there is an x ∈ 𝔪 with x ∉ z.div(M) and depth(M/xM) = 0. Then 𝔪 ∈ Ass(M/xM) by (1).

Conversely, assume there is x ∈ 𝔪 with x ∉ z.div(M). Then depth(M) ≥ 1 by definition. Assume 𝔪 ∈ Ass(M/xM). Then given any y ∈ 𝔪 with y ∉ z.div(M), also 𝔪 ∈ Ass(M/yM) by (17.25). So depth(M/yM) = 0 by (1). So there is no z ∈ 𝔪 such that y, z is an M-sequence. Thus depth(M) ≤ 1. Thus depth(M) = 1.

September 3, 2012 11Nts.tex

Consider (3). Given any M-sequence x_{1}, …, x_{n}, set M_{i} := M/⟨x_{1}, …, x_{i}⟩. Then dim(M_{i+1}) = dim(M_{i}) − 1 by (21.5). Hence dim(M) − n = dim(M_{n}) ≥ 0. But depth(M) := sup{n}. Thus (3) holds. □

Exercise (23.6). — Let R be a ring, M a module, and x, y ∈ R.

(1) Prove that, if x, y form an M-sequence, then, given any m, n ∈ M such that xm = yn, there exists p ∈ M such that m = yp and n = xp.

(2) Prove the converse of (1) if R is local, and x, y lie in its maximal ideal 𝔪, and M is Noetherian.

Exercise (23.7). — Let R be a local ring, 𝔪 its maximal ideal, M a Noetherian module, x_{1}, …, x_{n} ∈ 𝔪, and σ a permutation of 1, …, n. Assume x_{1}, …, x_{n} form an M-sequence, and prove x_{σ1}, …, x_{σn} do too; first, say σ transposes i and i + 1.

Exercise (23.8). — Prove that a Noetherian local ring A of dimension r ≥ 1 is regular if and only if its maximal ideal 𝔪 is generated by an A-sequence.

Theorem (23.9) (Characterization of DVRs). — Let A be a local ring, 𝔪 its maximal ideal. Assume A is Noetherian. Then these five conditions are equivalent:

(1) A is a DVR.

(2) A is a normal domain of dimension 1.

(3) A is a normal domain of depth 1.

(4) A is a regular local ring of dimension 1.

(5) 𝔪 is principal and of height at least 1.

Proof: Assume (1). Then A is UFD by (23.1); so A is normal by (10.28). Further, A has just two primes, ⟨0⟩ and 𝔪; so dim(A) = 1. Thus (2) holds. Further,

(4) holds by (23.1). Clearly, (4) implies (5).

Assume (2). Take a nonzero x ∈ 𝔪. Then A/⟨x⟩ ≠ 0, so Ass(A/⟨x⟩) ≠ ∅ by (17.12). Now, A is a local domain of dimension 1, so A has just two primes, ⟨0⟩ and 𝔪. Clearly, ⟨0⟩ ∉ Ass(A/⟨x⟩). Hence, 𝔪 ∈ Ass(A/⟨x⟩). Thus (3) holds.

Assume (3). By (23.5)(2), there are x, y ∈ 𝔪 such that x is nonzero and y has residue \overline{y} ∈ A/⟨x⟩ with 𝔪 = Ann(\overline{y}). So y𝔪 ⊂ ⟨x⟩. Set z := y/x ∈ Frac(A). Then z𝔪 = (y𝔪)/x ⊂ A. Suppose z𝔪 ⊂ 𝔪. Then z is integral over A by (10.18). But A is normal, so z ∈ A. So y = zx ∈ ⟨x⟩, a contradiction. Hence, 1 ∈ z𝔪; so there is t ∈ 𝔪 with zt = 1. Given w ∈ 𝔪, therefore w = (wz)t with wz ∈ A. Thus 𝔪 is principal. Finally, ht(𝔪) ≥ 1 because x ∈ 𝔪 and x ≠ 0. Thus (5) holds.

Assume (5). The Krull Intersection Theorem (18.28) yields an x ∈ 𝔪 with (1 + x) ⋂ 𝔪^{n} = 0. Then 1 + x ∈ A^{×}. So ⋂ 𝔪^{n} = 0. Further, A is a domain by (21.13)(1). Hence (1) holds by (23.3). □

Exercise (23.10). — Let A be a DVR with fraction field K, and f ∈ A a nonzero nonunit. Prove A is a maximal proper subring of K. Prove dim(A) ≠ dim(A_{f}).

Exercise (23.11). — Let k be a field, P := k[X, Y] the polynomial ring in two variables, f ∈ P an irreducible polynomial. Say f = ℓ(X, Y) + g(X, Y) with ℓ(X, Y) = aX + bY for a, b ∈ k and with g ∈ ⟨X, Y ⟩^{2}. Set R := P/⟨f⟩ and 𝔭 := ⟨X, Y ⟩/⟨f⟩. Prove that R_{𝔭} is a DVR if and only if ℓ ≠ 0. (Thus R_{𝔭} is a DVR if and only if the plane curve C : f = 0 ⊂ k^{2} is nonsingular at (0, 0).)

Exercise (23.12). — Let k be a field, A a ring intermediate between the polynomial ring and the formal power series ring in one variable: k[X] ⊂ A ⊂ k[[X]]. Suppose that A is local with maximal ideal ⟨X⟩. Prove that A is a DVR. (Such local rings arise as rings of power series with curious convergence conditions.)

Exercise (23.13). — Let L/K be an algebraic extension of fields, X_{1}, …, X_{n} variables, P and Q the polynomial rings over K and L in X_{1}, …, X_{n}.

(1) Let 𝔮 be a prime of Q, and 𝔭 its contraction in P. Prove ht(𝔭) = ht(𝔮).

(2) Let f, g ∈ P be two polynomials with no common prime factor in P. Prove that f and g have no common prime factor q ∈ Q.

(23.14) (Serre’s Conditions). — Let R be a Noetherian ring. We say Serre’s Condition (R_{n}) holds if, for any prime 𝔭 of height m ≤ n, the localization R_{𝔭} is regular of dimension m. We say Serre’s Condition (S_{n}) holds if, for any prime 𝔭 of any height m, the depth of 𝔭 on R_{𝔭} is at least min{m, n}, or equivalently, if

depth(R_{𝔭}) ≥ min{dim(R_{𝔭}), n}

as x_{1}, …, x_{r} ∈ 𝔭 is an R_{𝔭}-sequence if and only if x_{1}/t_{i}, …, x_{r}/t_{r} is for any t_{i} ∉ 𝔭.

For example, (R_{0}) holds if and only if R_{𝔭} is a field for any minimal prime 𝔭. Also, (R_{1}) holds if and only if (R_{0}) does and R_{𝔭} is a DVR for any 𝔭 of height-1.

Note depth(R_{𝔭}) ≤ dim(R_{𝔭}) by (23.5)(3). Hence (S_{n}) holds if and only if R_{𝔭} is Cohen–Macaulay when depth(R_{𝔭}) < n. In particular, (S_{1}) holds if and only if 𝔭 is minimal when 𝔭 ∈ Ass(R) by (17.14); that is, there are no embedded primes.

Exercise (23.15). — Let R be a Noetherian ring. Show that R is reduced if and only if (R_{0}) and (S_{1}) hold.

Lemma (23.16). — Let R be a Noetherian domain. Set

Φ := { 𝔭 prime | ht(𝔭) = 1 } and Σ := { 𝔭 prime | depth(R_{𝔭}) = 1 }.

Then Φ ⊂ Σ, and Φ = Σ if and only if (S_{2}) holds. Further, R = ⋂_{𝔭∈Σ} R_{𝔭}.

Proof: Given 𝔭 ∈ Φ, set 𝔮 := 𝔭R_{𝔭}. Take 0 ≠ x ∈ 𝔮. Then 𝔮 is minimal over ⟨x⟩. So 𝔮 ∈ Ass(R_{𝔭}/⟨x⟩) by (17.17). Hence depth(R_{𝔭}) = 1 by (23.5)(2). Thus Φ ⊂ Σ.

However, (S_{1}) holds by (23.15). Hence (S_{2}) holds if and only if Φ ⊃ Σ. Thus Φ = Σ if and only if R satisfies (S_{2}).

Further, R ⊂ R_{𝔭} for any prime 𝔭 by (11.4); so R ⊂ ⋂_{𝔭∈Σ} R_{𝔭}. As to the opposite inclusion, take an x ∈ ⋂_{𝔭∈Σ} R_{𝔭}. Say x = a/b with a, b ∈ R and b ≠ 0. Then a ∈ bR_{𝔭} for all 𝔭 ∈ Σ. But 𝔭 ∈ Σ if 𝔭 ∈ Ass(R_{𝔭}/bR_{𝔭}) by (23.5)(2). So a ∈ bR by (18.25). Thus x ∈ R, as desired. □

Theorem (23.17). — Let R be a normal Noetherian domain. Then

R = ⋂_{𝔭∈Φ} R_{𝔭} where Φ := { 𝔭 prime | ht(𝔭) = 1 }.

Proof: As R is normal, so is R_{𝔭} for any prime 𝔭 by (11.31). So depth(R_{𝔭}) = 1 if and only if dim(R_{𝔭}) = 1 by (23.9). Thus (23.16) yields the assertion. □

Theorem (23.18) (Serre’s Criterion). — Let R be a Noetherian domain. Then R is normal if and only if (R_{1}) and (S_{2}) hold.

Proof: As R is a domain, (R_{0}) and (S_{1}) hold by (23.15). If R is normal, then so is R_{𝔭} for any prime 𝔭 by (11.31); whence, (R_{1}) and (S_{2}) hold by (23.9).

Conversely, assume R satisfies (R_{1}) and (S_{2}). Let x be integral over R. Then x is integral over R_{𝔭} for any prime 𝔭. Now, R_{𝔭} is a DVR for all 𝔭 of height 1 as R satisfies (R_{1}). Hence, x ∈ R_{𝔭} for all 𝔭 of height 1, so for all 𝔭 of depth 1 as R satisfies (S_{2}). So x ∈ R owing to (23.16). Thus R is normal. □

Example (23.19). — Let k be an algebraically closed field, P := k[X, Y] the polynomial ring in two variables, f ∈ P irreducible. Then dim(P) = 2 by (15.12). Set R := P/⟨f⟩. Then R is a domain.

Let 𝔭 ⊂ R be a nonzero prime. Say 𝔭 = 𝔪/⟨f⟩. Then 0 ⫋ ⟨f⟩ ⫋ 𝔪 is a chain of primes of length 2, the maximum. Thus 𝔪 is maximal, and dim(R) = 1.

Hence 𝔪 = ⟨X − a, Y − b⟩ for some a, b ∈ k by (15.5). Write

f(X, Y) = ∂f/∂X(a, b)(X − a) + ∂f/∂Y (a, b)(Y − b) + g

where g ∈ 𝔪^{2}. Then R_{𝔭} is a DVR if and only if ∂f/∂X(a, b) and ∂f/∂Y (a, b) are not both equal to zero owing to (23.11) applied after making the change of variables X′ := X − a and Y′ := Y − b.

Clearly, R satisfies (S_{2}). Further, R satisfies (R_{1}) if and only if R_{𝔭} is a DVR for every nonzero prime 𝔭. Hence, by Serre’s Criterion, R is normal if and only if ∂f/∂X and ∂f/∂Y do not both belong to any maximal ideal 𝔪 of P containing f. (Put geometrically, R is normal if and only if the plane curve C : f = 0 ⊂ k^{2} is nonsingular everywhere.) Thus R is normal if and only if ⟨f, ∂f/∂X, ∂f/∂Y ⟩ = 1.

Exercise (23.20). — Prove that a Noetherian domain R is normal if and only if, given any prime 𝔭 associated to a principal ideal, 𝔭R_{𝔭} is principal.

Exercise (23.21). — Let R be a Noetherian ring, K its total quotient ring,

Φ := { 𝔭 prime | ht(𝔭) = 1 } and Σ := { 𝔭 prime | depth(R_{𝔭}) = 1 }.

Assuming (S_{1}) holds in R, prove Φ ⊂ Σ, and prove Φ = Σ if and only if (S_{2}) holds. Further, without assuming (S_{1}) holds, prove this canonical sequence is exact:

R → K → ∏_{𝔭∈Σ} K_{𝔭}/R_{𝔭}.  (23.21.1)

Exercise (23.22). — Let R be a Noetherian ring, and K its total quotient ring. Set Φ := { 𝔭 prime | ht(𝔭) = 1 }. Prove these three conditions are equivalent:

(1) R is normal.

(2) (R_{1}) and (S_{2}) hold.

(3) (R_{1}) and (S_{1}) hold, and R → K → ∏_{𝔭∈Φ} K_{𝔭}/R_{𝔭} is exact.

24. Dedekind Domains

Dedekind domains are defined as the normal Noetherian domains of dimension 1. We prove they are the Noetherian domains whose localizations at nonzero primes are discrete valuation rings. Next we prove the Main Theorem of Classical Ideal Theory: in a Dedekind domain, every nonzero ideal factors uniquely into primes. Then we prove that a normal domain has a module-finite integral closure in any finite separable extension of its fraction field by means of Artin’s Character Theorem and the trace pairing of a separable extension. We conclude that a ring of algebraic integers is a Dedekind domain and that, if a domain is a finitely generated algebra over a field of characteristic 0, then in any algebraic extension of its fraction field — in particular, in the fraction field itself — the integral closure is a finitely generated module over the domain and is a finitely generated algebra over the field.

Definition (24.1). — A domain R is said to be Dedekind if it is Noetherian, normal, and of dimension 1.

Example (24.2). — Examples of Dedekind domains include the integers ℤ, the Gaussian integers ℤ[ \sqrt{−1} ], the polynomial ring k[X] in one variable over a field, and any DVR. Indeed, those rings are PIDs, and every PID R is a Dedekind domain: R is Noetherian by definition; R is a UFD, so normal by Gauss’s Theorem, (10.28); and R is of dimension 1 since every nonzero prime is maximal by (2.23).

On the other hand, any local Dedekind domain is a DVR by (23.9).

Example (24.3). — Let d ∈ ℤ be a square-free integer. Set R := ℤ + ℤη where

η := \begin{cases}(1 + \sqrt{d})/2 & if d ≡ 1 (mod 4); \\ \sqrt{d} & if not. \end{cases}

Then R is the integral closure of ℤ in ℚ(\sqrt{d}) by [1, Prp. (6.14), p. 412]; so R is normal. Also, dim(R) = dim(ℤ) by (15.11); so dim(R) = 1. Finally, R is Noetherian by (16.11) as ℤ is so and as R := ℤ + ℤη. Thus R is Dedekind.

Example (24.4). — Let k be an algebraically closed field, P := k[X, Y] the polynomial ring in two variables, f ∈ P irreducible. By (23.19), R is a Noetherian domain of dimension 1, and R is Dedekind if and only if ⟨f, ∂f/∂X, ∂f/∂Y ⟩ = 1.

Exercise (24.5). — Let R be a domain, S a multiplicative subset.

(1) Assume dim(R) = 1. Prove dim(S^{−1}R) = 1 if and only if there is a nonzero prime 𝔭 with 𝔭 ∩ S = ∅.

(2) Assume dim(R) ≥ 1. Prove dim(R) = 1 if and only if dim(R_{𝔭}) = 1 for every nonzero prime 𝔭.

Exercise (24.6). — Let R be a Dedekind domain, S a multiplicative subset. Prove S^{−1}R is a Dedekind domain if and only if there’s a nonzero prime 𝔭 with 𝔭 ∩ S = ∅.

Proposition (24.7). — Let R be a Noetherian domain, not a field. Then R is a Dedekind domain if and only if R_{𝔭} is a DVR for every nonzero prime 𝔭.

Proof: If R is Dedekind, then R_{𝔭} is too by (24.6); so R_{𝔭} is a DVR by (23.9). Conversely, suppose R_{𝔭} is a DVR for every nonzero prime 𝔭. Then, trivially, R

satisfies (R_{1}) and (S_{2}); so R is normal by Serre’s Criterion. Since R is not a field, dim(R) ≥ 1; whence, dim(R) = 1 by (24.5)(2). Thus R is Dedekind. □

Exercise (24.8). — Let R be a Dedekind domain, and 𝔞, 𝔟, 𝔠 ideals. By first reducing to the case that R is local, prove that

𝔞 ∩ (𝔟 + 𝔠) = (𝔞 ∩ 𝔟) + (𝔞 ∩ 𝔠),
𝔞 + (𝔟 ∩ 𝔠) = (𝔞 + 𝔟) ∩ (𝔞 + 𝔠).

Proposition (24.9). — In a Noetherian domain R of dimension 1, every ideal 𝔞 ≠ 0 has a unique factorization 𝔞 = 𝔮_{1} ⋯ 𝔮_{r} with the 𝔮_{i} primary and their primes 𝔭_{i} distinct; further, {𝔭_{1}, …, 𝔭_{r}} = Ass(R/𝔞) and 𝔮_{i} = 𝔞R_{𝔭_{i}} ∩ R for each i.

Proof: The Lasker–Noether Theorem, (18.20), yields an irredundant primary decomposition 𝔞 = ⋂ 𝔮_{i}. Say 𝔮_{i} is 𝔭_{i}-primary. Then by (18.18) the 𝔭_{i} are distinct and {𝔭_{i}} = Ass(R/𝔞).

The 𝔮_{i} are pairwise comaximal for the following reason. Suppose 𝔮_{i} + 𝔮_{j} lies in a maximal ideal 𝔪. Now, 𝔭_{i} := \sqrt{𝔮_{i}} by (18.5); so 𝔭_{i}^{n_{i}} ⊂ 𝔮_{i} for some n_{i} by (3.25). Hence 𝔭_{i}^{n_{i}} ⊂ 𝔪. So 𝔭_{i} ⊂ 𝔪 by (2.2). But 0 ≠ 𝔞 ⊂ 𝔭_{i}; hence, 𝔭_{i} is maximal since dim(R) = 1. Therefore, 𝔭_{i} = 𝔪. Similarly, 𝔭_{j} = 𝔪. Hence i = j. Thus the 𝔮_{i} are pairwise comaximal. So the Chinese Remainder Theorem, (1.13), yields 𝔞 = ∏_{i} 𝔮_{i}.

As to uniqueness, let 𝔞 = ∏ 𝔮_{i} be any factorization with the 𝔮_{i} primary and their primes 𝔭_{i} distinct. The 𝔭_{i} are minimal containing 𝔞 as dim(R) = 1; so the 𝔭_{i} are associated primes by (17.17). By the above reasoning, the 𝔮_{i} are pairwise comaximal and so ∏ 𝔮_{i} = ⋂ 𝔮_{i}. Hence 𝔞 = ⋂ 𝔮_{i} is an irredundant primary decomposition by (18.18). So the 𝔭_{i} are unique by the First Uniqueness Theorem, (18.19), and 𝔮_{i} = 𝔞R_{𝔭_{i}} ∩R by the Second Uniqueness Theorem, (18.24), and by (12.15)(3). □

Theorem (24.10) (Main Theorem of Classical Ideal Theory). — Let R be a domain. Assume R is Dedekind. Then every nonzero ideal 𝔞 has a unique factorization into primes 𝔭. In fact, if v_{𝔭} denotes the valuation of R_{𝔭}, then

𝔞 = ∏ 𝔭^{v_{𝔭}(𝔞)} where v_{𝔭}(𝔞) := min{ v_{𝔭}(a) | a ∈ 𝔞 }.

Proof: Using (24.9), write 𝔞 = ∏ 𝔮_{i} with the 𝔮_{i} primary, their primes 𝔭_{i} distinct and unique, and 𝔮_{i} = 𝔞R_{𝔭_{i}} ∩ R. Then R_{𝔭_{i}} is a DVR by (24.7). So (23.1.3) yields 𝔞R_{𝔭_{i}} = 𝔭_{i}^{m_{i}}R_{𝔭_{i}} with m_{i} := min{ v_{𝔭_{i}}(a/s) | a ∈ 𝔞 and s ∈ R − 𝔭_{i} }. But v_{𝔭_{i}}(1/s) = 0. So v_{𝔭_{i}}(a/s) = v_{𝔭_{i}}(a). Hence m_{i} := min{ v_{𝔭_{i}}(a) | a ∈ 𝔞 }. Now, 𝔭_{i}^{m_{i}} is primary by (18.10) as 𝔭_{i} is maximal; so 𝔭_{i}^{m_{i}}R_{𝔭_{i}} ∩ R = 𝔭_{i}^{m_{i}} by (18.22). Thus 𝔮_{i} = 𝔭_{i}^{m_{i}}. □

Corollary (24.11). — A Noetherian domain R of dimension 1 is Dedekind if and only if every primary ideal is a power of its radical.

Proof: If R is Dedekind, every primary ideal is a power of its radical by (24.10). Conversely, given a nonzero prime 𝔭, set 𝔪 := 𝔭R_{𝔭}. Then 𝔪 ≠ 0. So 𝔪 ≠ 𝔪^{2} by

Nakayama’s Lemma. Take t ∈ 𝔪 − 𝔪^{2}. Then 𝔪 is the only prime containing t, as dim(R_{𝔭}) = 1 by (24.5)(2). So tR_{𝔭} is 𝔪-primary by (18.10). Set 𝔮 := tR_{𝔭} ∩ R. Then 𝔮 is 𝔭-primary by (18.8). So 𝔮 = 𝔭^{n} for some n by hypothesis. But 𝔮R_{𝔭} = tR_{𝔭} by (11.17)(3)(b). So tR_{𝔭} = 𝔪^{n}. But t ∉ 𝔪^{2}. So n = 1. So R_{𝔭} is a DVR by (23.9). Thus R is Dedekind by (24.7). □

September 3, 2012 11Nts.tex Exercise (24.12). — Prove that a semilocal Dedekind domain A is a PID. Begin by proving that each maximal ideal is principal.

Exercise (24.13). — Let R be a Dedekind domain, 𝔞 and 𝔟 two nonzero ideals. Prove (1) every ideal in R/𝔞 is principal, and (2) 𝔟 is generated by two elements.

Lemma (24.14) (E. Artin). — Let L be a field, G a group, σ_{i}: G → L^{×} distinct homomorphisms. Then the σ_{i} are linearly independent over L in the vector space of set maps σ: G → L under valuewise addition and scalar multiplication.

Proof: Suppose there’s an equation ∑_{i=1}^{m} a_{i}σ_{i} = 0 with nonzero a_{i} ∈ L. Take m ≥ 1 minimal. Now, σ_{i} ≠ 0 as σ_{i}: G → L^{×}; so m ≥ 2. Since σ_{1} ≠ σ_{2}, there’s an x ∈ G with σ_{1}(x) ≠ σ_{2}(x). Then ∑_{i=1}^{m} a_{i}σ_{i}(x)σ_{i}(y) = ∑_{i=1}^{m} a_{i}σ_{i}(xy) = 0 for every y ∈ G since σ_{i} is a homomorphism.

Set τ_{i}(x) := 1 − σ_{i}(x)/σ_{1}(x). Then

∑_{i=1}^{m} a_{i}τ_{i}(x)σ_{i} = ∑_{i=1}^{m} a_{i}σ_{i} − \frac{1}{σ_{1}(x)} ∑_{i=1}^{m} a_{i}σ_{i}(x)σ_{i} = 0.

But τ_{1}(x) = 0 and τ_{2}(x) ≠ 0, contradicting the minimality of m. □

(24.15) (Trace). — Let L/K be a finite Galois field extension. Its trace is this:

tr: L → K by tr(x) := ∑_{σ∈Gal(L/K)} σ(x).

Clearly, tr is K-linear. It is nonzero by (24.14) applied with G := L^{×}. Consider the symmetric K-bilinear Trace Pairing:

L × L → K by (x, y) ↦ tr(xy).  (24.15.1)

It is nondegenerate for this reason. Since tr is nonzero, there is a z ∈ L with tr(z) ≠ 0. Now, given x ∈ L^{×}, set y := z/x. Then tr(xy) ≠ 0, as desired.

Lemma (24.16). — Let R be a normal domain, K its fraction field, L/K a finite Galois field extension, and x ∈ L integral over R. Then tr(x) ∈ R.

Proof: Let x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0 be an equation of integral dependence for x over R. Let σ ∈ Gal(L/K). Then

(σx)^{n} + a_{1}(σx)^{n−1} + ⋯ + a_{n} = 0;

so σx is integral over R. Hence tr(x) is integral over R, and lies in K. Thus tr(x) ∈ R since R is normal. □

Theorem (24.17) (Finiteness of integral closure). — Let R be a normal Noetherian domain, K its fraction field, L/K a finite separable field extension, and R′ the integral closure of R in L. Then R′ is module finite over R.

Proof: Let L_{1} be the Galois closure of L/K, and R′_{1} the integral closure of R in L_{1}. Let z_{1}, …, z_{n} ∈ L_{1} form a K-basis. Using (11.24), write z_{i} = y_{i}/a_{i} with y_{i} ∈ R′_{1} and a_{i} ∈ R. Clearly, y_{1}, …, y_{n} form a basis of L_{1}/K contained in R′_{1}.

Let x_{1}, …, x_{n} form the dual basis with respect to the Trace Paring, (24.15.1), so that tr(x_{i}y_{j}) = δ_{ij}. Given b ∈ R′, write b = ∑ c_{i}x_{i} with c_{i} ∈ K. Fix j. Then

tr(by_{j}) = tr(∑ c_{i}x_{i}y_{j}) = ∑ c_{i} tr(x_{i}y_{j}) = c_{j} for each j.

But by_{j} ∈ R′_{1}. So c_{j} ∈ R by (24.16). Thus R′ ⊂ ∑ Rx_{i}. Since R is Noetherian, R′ is a finitely generated R-module, as desired. □

Corollary (24.18). — Let R be a Dedekind domain, K its fraction field, L/K a finite separable field extension. Then the integral closure R′ of R in L is Dedekind.

Proof: First, R′ is module finite over R by (24.17); so R′ is Noetherian by (16.18). Second, R′ is normal by (10.27). Finally, dim(R′) = dim(R) by (15.11), and dim(R) = 1 as R is Dedekind. Thus R is Dedekind. □

Theorem (24.19). — A ring of algebraic integers is a Dedekind domain.

Proof: By (24.2), ℤ is a Dedekind domain; whence, so is its integral closure in any field that is a finite extension of ℚ by (24.18). □

Theorem (24.20) (Noether). — Let k be a field of characteristic 0, and R a domain that is a finitely generated k-algebra. Set K := Frac(R). Let L/K be a finite field extension (possibly L = K), and let R′ be the integral closure of R in L. Then R′ is a finitely generated R-module and a finitely generated k-algebra.

Proof: By the Noether Normalization Lemma, (15.1), R is a module-finite k-algebra over a polynomial subring P. Then P is normal by Gauss’s Theorem, (10.28), and Noetherian by the Hilbert Basis Theorem, (16.11); also, L/ Frac(P) is a finite field extension, which is separable as k is of characteristic 0. Hence, R′ is module finite over P by (24.17). The assertion follows. □

(24.21) (Other cases). — In (24.18), even if L/K is inseparable, the integral closure R′ of R in L is still Dedekind, as is proved below in Lecture 26.

However, Akizuki constructed an example of a DVR R and a finite inseparable extension L/ Frac(R) such that the integral closure of R is a DVR, but is not module finite over R. The construction is nicely explained in [7, Secs. 9.4(1) and 9.5]. Thus separability is a necessary hypothesis in (24.17).

Noether’s Theorem, (24.20), remains valid in positive characteristic, but the proof is more involved. See [3, (13.13), p. 297].

25. Fractional Ideals

A fractional ideal is defined to be a submodule of the fraction field of a domain. A fractional ideal is called invertible if its product with another fractional ideal is equal to the given domain. We characterize the invertible fractional ideals as those that are nonzero, finitely generated, and principal locally at every maximal ideal. We prove that, in a Dedekind domain, any two nonzero integral (that is, ordinary) ideals have an invertible fractional ideal as their quotient. We characterize Dedekind domains as those domains whose integral ideals are, equivalently, all invertible, all projective, or all finitely generated and flat. Further, we prove a Noetherian domain is Dedekind if and only if every torsion-free module is flat. Finally, we prove the ideal class group is equal to the Picard group; the former is the group of invertible fractional ideals modulo those that are principal, and the latter is the group, under tensor product, of isomorphism classes of modules local free of rank 1.

Definition (25.1). — Let R be a domain, and set K := Frac(R). We call an R-submodule M of K a fractional ideal. We call M integral if M ⊂ R. We call M principal if there is an x ∈ K with M = Rx.

Given another fractional ideal N, form these two new fractional ideals:

MN := { ∑ x_{i}y_{i} | x_{i} ∈ M and y_{i} ∈ N } and (M : N) := { z ∈ K | zN ⊂ M }.

We call them the product of M and N and the quotient of M by N.

Exercise (25.2). — Let R be a domain, M and N nonzero fractional ideals. Prove that M is principal if and only if there exists some isomorphism M ≃ R. Construct the following canonical surjection and canonical isomorphism:

π: M ⊗ N → → MN and φ: (M : N) ⟶^{∼} Hom(N, M).

Proposition (25.3). — Let R be a domain, and K := Frac(R). Consider these finiteness conditions on a fractional ideal M:

(1) There exist integral ideals 𝔞 and 𝔟 with 𝔟 ≠ 0 and (𝔞 : 𝔟) = M.

(2) There exists an x ∈ K^{×} with xM ⊂ R.

(3) There exists a nonzero x ∈ R with xM ⊂ R.

(4) M is finitely generated. Then (1), (2), and (3) are equivalent, and they are implied by (4). Further, all four conditions are equivalent for every M if and only if R is Noetherian.

Proof: Assume (1) holds. Take any nonzero x ∈ 𝔟. Given m ∈ M, clearly xm ∈ 𝔞 ⊂ R; so xM ⊂ R. Thus (2) holds.

Assume (2) holds. Write x = a/b with a, b ∈ R and a, b ≠ 0. Then aM ⊂ bR ⊂ R. Thus (3) holds.

If (3) holds, then xM and xR are integral, and M = (xM : xR); thus (1) holds. Assume (4) holds. Say y_{1}/x_{1}, …, y_{n}/x_{n} ∈ K^{×} generate M with x_{i}, y_{i} ∈ R. Set

x := ∏ x_{i}. Then x ≠ 0 and xM ⊂ R. Thus (3) holds.

Assume (3) holds and R is Noetherian. Then xM ⊂ R. So xM is finitely generated, say by y_{1}, …, y_{n}. Then y_{1}/x, …, y_{n}/x generate M. Thus (4) holds.

Finally, assume all four conditions are equivalent for every M. If M is integral, then (3) holds with x := 1, and so (4) holds. Thus R is Noetherian. □

September 3, 2012 11Nts.tex Lemma (25.4). — Let R be a domain, M and N fractional ideals. Let S be a multiplicative subset. Then

S^{−1}(MN) = (S^{−1}M)(S^{−1}N) and S^{−1}(M : N) ⊂ (S^{−1}M : S^{−1}N),

with equality if N is finitely generated.

Proof: Given x ∈ S^{−1}(MN), write x = (∑ m_{i}n_{i})/s with m_{i} ∈ M, with n_{i} ∈ N, and with s ∈ S. Then x = ∑(m_{i}/s)(n_{i}/1), and so x ∈ (S^{−1}M)(S^{−1}N). Thus S^{−1}(MN) ⊂ (S^{−1}M)(S^{−1}N).

Conversely, given x ∈ (S^{−1}M)(S^{−1}N), say x = ∑(m_{i}/s_{i})(n_{i}/t_{i}) with m_{i} ∈ M and n_{i} ∈ N and s_{i}, t_{i} ∈ S. Set s := ∏ s_{i} and t := ∏ t_{i}. Then

x = ∑(m_{i}n_{i}/s_{i}t_{i}) = ∑ m′_{i}n′_{i}/st ∈ S^{−1}(MN)

with m′_{i} ∈ M and n′_{i} ∈ N. Thus S^{−1}(MN) ⊃ (S^{−1}M)(S^{−1}N), so equality holds.

Given z ∈ S^{−1}(M : N), write z = x/s with x ∈ (M : N) and s ∈ S. Given y ∈ S^{−1}N, write y = n/t with n ∈ N and t ∈ S. Then z · n/t = xn/st and xn ∈ M and st ∈ S. So z ∈ (S^{−1}M : S^{−1}N). Thus S^{−1}(M : N) ⊂ (S^{−1}M : S^{−1}N).

Conversely, say N is generated by n_{1}, …, n_{r}. Given z ∈ (S^{−1}M : S^{−1}N), write zn_{i}/1 = m_{i}/s_{i} with m_{i} ∈ M and s_{i} ∈ S. Set s := ∏ s_{i}. Then sz · n_{i} ∈ M. So sz ∈ (M : N). Hence z ∈ S^{−1}(M : N), as desired. □

Definition (25.5). — Let R be a domain. We call a fractional ideal M locally principal if, for every maximal ideal 𝔪, the localization M_{𝔪} is principal over R_{𝔪}.

Exercise (25.6). — Let R be a domain, M and N fractional ideals. Prove that the map π: M ⊗ N → MN is an isomorphism if M is locally principal.

(25.7) (Invertible fractional ideals). — Let R be a domain. A fractional ideal M is said to be invertible if there is some fractional ideal M^{−1} with MM^{−1} = R.

For example, a nonzero principal ideal Rx is invertible, as (Rx)(R · 1/x) = R.

Proposition (25.8). — Let R be a domain, M an invertible fractional ideal. Then M^{−1} is unique; in fact, M^{−1} = (R : M).

Proof: Clearly M^{−1} ⊂ (R : M) as MM^{−1} = R. But, if x ∈ (R : M), then x · 1 ∈ (R : M)MM^{−1} ⊂ M^{−1}, so x ∈ M^{−1}. Thus (R : M) ⊂ M^{−1}, as desired. □

Lemma (25.9). — An invertible ideal is finitely generated and nonzero.

Proof: Let R be the domain, M the ideal. Say 1 = ∑ m_{i}n_{i} with m_{i} ∈ M and n_{i} ∈ M^{−1}. Let m ∈ M. Then m = ∑ m_{i}mn_{i}. But mn_{i} ∈ R as m ∈ M and n_{i} ∈ M^{−1}. So the m_{i} generate M. Trivially, M ≠ 0. □

Lemma (25.10). — Let A be a local domain. Then a fractional ideal M is invertible if and only if M is principal and nonzero.

Proof: Assume M is invertible. Say 1 = ∑ m_{i}n_{i} with m_{i} ∈ M and n_{i} ∈ M^{−1}. As A is local, A − A^{×} is an ideal. So there’s a j with m_{j}n_{j} ∈ A^{×}. Let m ∈ M. Then mn_{j} ∈ A. Set a := (mn_{j})(m_{j}n_{j})^{−1} ∈ A. Then m = am_{j}. Thus M = Am_{j}.

Conversely, if M is principal and nonzero, then it’s always invertible by (25.7). □

Exercise (25.11). — Let R be a UFD. Show that a fractional ideal M is invertible if and only if M is principal and nonzero.

Theorem (25.12). — Let R be a domain, M a fractional ideal. Then M is invertible if and only if M is finitely generated and locally principal.

Proof: Say MN = R. Then M is finitely generated and nonzero by (25.9). Let S be a multiplicative subset. Then (S^{−1}M)(S^{−1}N) = S^{−1}R by (25.4). Let 𝔪 be a maximal ideal. Then, therefore, M_{𝔪} is an invertible fractional ideal over R_{𝔪}. Thus M_{𝔪} is principal by (25.10), as desired.

Conversely, set 𝔞 := M(R : M) ⊂ R. Assume M is finitely generated. Then (25.4) yields 𝔞_{𝔪} = M_{𝔪}(R_{𝔪} : M_{𝔪}). In addition, assume M_{𝔪} is principal and nonzero. Then (25.7) and (25.8) yield 𝔞_{𝔪} = R_{𝔪}. Hence (13.16) yields 𝔞 = R, as desired. □

Theorem (25.13). — Let R be a Dedekind domain, 𝔞, 𝔟 nonzero integral ideals. Set M := (𝔞 : 𝔟). Then M is invertible, and has a unique factorization into powers of primes 𝔭. In fact, if v_{𝔭} denotes the valuation of R_{𝔭}, then

M = ∏ 𝔭^{v_{𝔭}(M)} where v_{𝔭}(M) := min{ v_{𝔭}(x) | x ∈ M }.

Finally, v_{𝔭}(M) = min{v_{𝔭}(x_{i})} if the x_{i} generate M.

Proof: First, R is Noetherian. So (25.2) yields that M is finitely generated and that there is a nonzero x ∈ R with xM ⊂ R. Hence, each localization xM_{𝔭} is principal by (23.1.3). Thus M is invertible by (25.12).

Next, the Main Theorem of Classical Ideal Theory, (24.10), yields ⟨x⟩ = ∏ 𝔭^{v_{𝔭}(x)} and xM = ∏ 𝔭^{v_{𝔭}(xM)}. Since v_{𝔭}(xM) = v_{𝔭}(x) + v_{𝔭}(M), we can cancel the v_{𝔭}(x) to conclude M = ∏ 𝔭^{v_{𝔭}(M)}.

Finally, given x ∈ M, say x = ∑_{i=1}^{n} a_{i}x_{i} with a_{i} ∈ R. Then (23.1.1) yields

v_{𝔭}(x) ≥ min{v_{𝔭}(a_{i}x_{i})} ≥ min{v_{𝔭}(x_{i})}

by induction on n. Thus v_{𝔭}(M) = min{v_{𝔭}(x_{i})}. □

Exercise (25.14). — Show that a ring is a PID if and only if it’s a Dedekind domain and a UFD.

(25.15) (Invertible modules). — Let R be an arbitrary ring. We call a module M invertible if there is another module N with M ⊗ N ≃ R.

For example, suppose R is a domain. Let M be an invertible fractional ideal; say N is a fractional ideal with MN = R. Then M is locally principal by (25.12). So M ⊗ N = MN by (25.6). Thus M is an invertible abstract module.

Exercise (25.16). — Let R be an ring, M an invertible module. Prove that M is finitely generated, and that, if R is local, then M is free of rank 1.

Exercise (25.17). — Show these conditions on an R-module M are equivalent: (1) M is invertible.

(2) M is finitely generated, and M_{𝔪} ≃ R_{𝔪} at each maximal ideal 𝔪.

(3) M is locally free of rank 1. Assuming these conditions hold, show that M ⊗ Hom(M, R) = R.

Lemma (25.18). — Let R be a domain, M a fractional ideal. Then M is an invertible fractional ideal if and only if M is a projective abstract module.

Proof: Assume M is an invertible fractional ideal. Then M is an invertible abstract module by (25.15). Hence M is locally free of rank 1 by (25.17). So M is projective by (13.27).

Conversely, assume M is projective. Then by (5.22), there exists a module M′ with M ⊕ M′ ≃ R^{⊕Λ}. Let ρ: R^{⊕Λ} → M be the projection, and set x_{λ} := ρ(e_{λ}). Define φ_{λ}: M ↪ R^{⊕Λ} → R as the composition of the injection with the projection φ_{λ} on the λth factor. Then for all x ∈ M, we have x = ∑_{λ∈Λ} φ_{λ}(x)x_{λ} and φ_{λ}(x) = 0 for almost all λ.

Fix a nonzero y ∈ M. For λ ∈ Λ, set q_{λ} := \frac{1}{y} φ_{λ}(y) ∈ Frac(R). Set N := ∑ Rq_{λ}. Then for any nonzero x ∈ M, let’s check that xq_{λ} = φ_{λ}(x). Write x = a/b and y = c/d with a, b, c, d ∈ R. Then a, c ∈ M; whence, adφ(y) = φ(ac) = bcφ(x). Thus xq_{λ} = φ_{λ}(x) ∈ R. Hence M · N ⊂ R. But y = ∑ φ_{λ}(y)y_{λ}, so 1 = y_{λ}q_{λ}. Thus M · N = R. □

Theorem (25.19). — Let R be a domain. Then the following are equivalent: (1) R is Dedekind;

(2) every integral ideal is invertible;

(3) every integral ideal is projective;

(4) every integral ideal is finitely generated and flat.

Proof: Let 𝔞 be an integral ideal. Assume (1). Since 𝔞 = (𝔞 : R), it is invertible by (25.13). Thus (2) holds.

Conversely, assume (2). Then 𝔞 is finitely generated by (25.9). Thus R is Noetherian. Let 𝔭 be any nonzero prime of R. Then by hypothesis, 𝔭 is invertible. So by (25.12), it is locally principal. So R_{𝔭} is a DVR by (23.9). Hence R is Dedekind by (24.7). Thus (1) holds. Thus (1) and (2) are equivalent.

Recall that (2) and (3) are equivalent by (25.18). But (2) implies that R is Noetherian by (25.9). Thus (3) and (4) are equivalent by (16.18) and (13.27). □

Theorem (25.20). — A Noetherian domain R is Dedekind if and only if every torsion-free module is flat.

Proof: (Of course, as R is a domain, every flat module is torsion free by (9.22).) Assume R is Dedekind. Let M be a torsion-free module, 𝔪 a maximal ideal.

Let’s see that M_{𝔪} is torsion free over R_{𝔪}. Let z ∈ R_{𝔪} be nonzero, and say z = x/s with x, s ∈ R and s ∉ 𝔪. Then μ_{x}: M → M is injective as M is torsion free. So μ_{x}: M_{𝔪} → M_{𝔪} is injective by the Exactness of Localization. But μ_{x/s} = μ_{x}μ_{1/s} and μ_{1/s} is invertible. So μ_{x/s} is injective. Thus M_{𝔪} is torsion free.

Since R is Dedekind, R_{𝔪} is a DVR by (24.7), so a PID by (24.1). Hence M_{𝔪} is flat over R_{𝔪} by (9.22). But 𝔪 is arbitrary. Therefore, M is flat over R by (13.23).

Conversely, assume every torsion-free module is flat. Then, in particular, every integral ideal is flat. But R is Noetherian. Thus R is Dedekind by (25.19). □

(25.21) (The Picard Group). — Let R be a ring. We denote the collection of isomorphism classes of invertible modules by Pic(R). By (25.16), every invertible module is finitely generated, so isomorphic to a quotient of R^{n} for some integer n. Hence, Pic(R) is a set. Further, Pic(R) is, clearly, a group under tensor product with the class of R as identity. We call Pic(R) the Picard Group of R.

Assume R is a domain, and set K := Frac(R). Given an invertible module M, we can embed M into K as follows. Set S := R − 0, and form the canonical map M → S^{−1}M. It is injective owing to (12.15) if the multiplication map μ_{x}: M → M

September 3, 2012 11Nts.tex is injective for any x ∈ S. Fix x, and let’s prove μ_{x} is injective.

Let 𝔪 be a maximal ideal. Clearly, M_{𝔪} is an invertible R_{𝔪}-module. So M_{𝔪} ≃ R_{𝔪} by (25.16). Hence μ_{x}: M_{𝔪} → M_{𝔪} is injective. Therefore, μ_{x}: M → M is injective by (13.20). Thus M embeds canonically into S^{−1}M. Now, S^{−1}M is a localization of M_{𝔪}, so is a 1-dimensional K-vector space, again as M_{𝔪} ≃ R_{𝔪}. Choose an isomorphism S^{−1}M ≃ K. It yields the desired embedding of M into K.

Since M is invertible, M is finitely generated by (25.16). Further, as noted, M_{𝔪} ≃ R_{𝔪} at each maximal ideal 𝔪. Say x ∈ M_{𝔪} corresponds to 1 ∈ R_{𝔪}. Then yx ∈ M_{𝔪} corresponds to y ∈ R_{𝔪}. Thus M is locally principal. So, by (25.12), M is also invertible as a fractional ideal.

The invertible fractional ideals M, clearly, form a group ℱ(R). Sending an M to its isomorphism class yields a map κ: ℱ(R) → Pic(R) by (25.15). By the above, κ is surjective. Further, κ is a group homomorphism by (25.6). It’s not hard to check that its kernel is the group 𝒫(R) of principal ideals and that 𝒫(R) = K^{×}/R^{×}. We call ℱ(R)/𝒫(R) the Ideal Class Group of R. Thus ℱ(R)/𝒫(R) = Pic(R); in other words, the Ideal Class Group is canonically isomorphic to the Picard Group.

Every invertible fractional ideal is, by (25.12), finitely generated and nonzero, so of the form (𝔞 : 𝔟) where 𝔞 and 𝔟 are integral and nonzero by (25.3). Conversely, by (25.13) and (25.19), every fractional ideal of this form is invertible if and only if R is Dedekind. In fact, then ℱ(R) is the free abelian group on the prime ideals. Further, then Pic(R) = 0 if and only if R is UFD, or equivalently by (25.14), a PID. See [1, Ch. 11, Sects. 10–11, pp. 424–437] for a discussion of the case in which R is a ring of quadratic integers, including many examples where Pic(R) ≠ 0.

26. Arbitrary Valuation Rings

A valuation ring is, by definition, a subring of a field whose elements either lie in the subring or their reciprocals do. Valuation rings are normal local domains. They are maximal under domination, that is, inclusion of both the local rings and their maximal ideals. Given any subring, its normalization is equal to the intersection of all the valuation rings containing it. We end with the Krull–Akizuki Theorem: given a 1-dimensional Noetherian domain, a finite extension of its fraction field, and a proper subring of the extension containing the domain, that subring too is 1dimensional and Noetherian. We conclude that, if we normalize a Dedekind domain in any finite extension of its fraction field, we obtain another Dededind domain.

Definition (26.1). — A subring V of a field K is said to be a valuation ring of K if, whenever z ∈ K − V, then 1/z ∈ V.

Proposition (26.2). — Let V be a valuation ring of a field K, and set

𝔪 := {1/z | z ∈ K − V } ∪ {0}.

Then V is local, 𝔪 is its maximal ideal, and K is its fraction field.

Proof: Clearly 𝔪 = V −V^{×}. Let’s show 𝔪 is an ideal. Take a nonzero a ∈ V and nonzero x, y ∈ 𝔪. Suppose ax ∉ 𝔪. Then ax ∈ V^{×}. So a(1/ax) ∈ V. So 1/x ∈ V. So x ∈ V^{×}, a contradiction. Thus ax ∈ 𝔪. Now, by hypothesis, either x/y ∈ V or y/x ∈ V. Say y/x ∈ V. Then 1 + (y/x) ∈ V. So x + y = (1 + (y/x))x ∈ 𝔪. Thus 𝔪 is an ideal. Hence V is local and 𝔪 is its maximal ideal by (3.4). Finally, K is its fraction field, because whenever z ∈ K − V, then 1/z ∈ V. □

Exercise (26.3). — Let V be a domain. Show that V is a valuation ring if and only if, given any two ideals 𝔞 and 𝔟, either 𝔞 lies in 𝔟 or 𝔟 lies in 𝔞.

Exercise (26.4). — Let V be a valuation ring, 𝔪 its maximal ideal, and 𝔭 ⊂ 𝔪 another prime ideal. Prove that V_{𝔭} is a valuation ring, that its maximal ideal 𝔭V_{𝔭} is equal to 𝔭, and that V/𝔭 is a valuation ring of the field V_{𝔭}/𝔭.

Exercise (26.5). — Prove that a valuation ring V is normal.

Lemma (26.6). — Let R be a domain, 𝔞 an ideal, K := Frac(R), and x ∈ K^{×}. Then either 1 ∉ 𝔞R[x] or 1 ∉ 𝔞R[1/x].

Proof: Assume 1 ∈ 𝔞R[x] and 1 ∈ 𝔞R[1/x]. Then there are equations

1 = a_{0} + ⋯ + a_{n}x^{n} and 1 = b_{0} + ⋯ + b_{m}/x^{m} with all a_{i}, b_{j} ∈ 𝔞.

Assume n, m minimal and m ≤ n. Multiply through by 1 − b_{0} and a_{n}x^{n}, getting

1 − b_{0} = (1 − b_{0})a_{0} + ⋯ + (1 − b_{0})a_{n}x^{n} and
(1 − b_{0})a_{n}x^{n} = a_{n}b_{1}x^{n−1} + ⋯ + a_{n}b_{m}x^{n−m}.

Combine the latter equations, getting

1 − b_{0} = (1 − b_{0})a_{0} + ⋯ + (1 − b_{0})a_{n−1}x^{n−1} + a_{n}b_{1}x^{n−1} + ⋯ + a_{n}b_{m}x^{n−m}.

Simplify, getting an equation of the form 1 = c_{0} +⋯+c_{n−1}x^{n−1} with c_{i} ∈ 𝔞, which contradicts the minimality of n. □

Lemma (26.7). — Let A, B be local rings, and 𝔪, 𝔫 their maximal ideals. If B ⊃ A, then these conditions are equivalent: (1) 𝔫∩A = 𝔪; (2) 1 ∉ 𝔪B; (3) 𝔪 ⊂ 𝔫.

Proof: Assume B ⊃ A. If (1) holds, then 𝔪B ⊂ 𝔫, so (2) holds. If (2) holds, then 𝔪B ⊂ 𝔫, so (3) holds. If (3) holds, then 𝔪 ⊂ 𝔫 ∩ A ⊊ A, so (1) holds. □

(26.8) (Domination). — Let A, B be local rings, and 𝔪, 𝔫 their maximal ideals. We say B dominates A if B ⊃ A and 𝔫 ∩ A = 𝔪.

Proposition (26.9). — Let K be a field, A a local subring. Then A is dominated by a valuation ring V of K.

Proof: Let 𝔪 be the maximal ideal of A. Let 𝒮 be the set of subrings R of K with R ⊃ A and 1 ∉ 𝔪R. Then A ∈ 𝒮. Order 𝒮 by inclusion. Let {R_{λ}} be a totally ordered subset. Set R := ⋃ R_{λ}. If 1 ∈ 𝔪R, then

1 = a_{1}x_{1} + ⋯ + a_{n}x_{n} with a_{i} ∈ 𝔪 and x_{i} ∈ R.

But then there is λ such that x_{i} ∈ R_{λ} for all i; so 1 ∈ 𝔪R_{λ}, a contradiction. Thus R ∈ 𝒮. Hence, by Zorn’s Lemma, 𝒮 has a maximal element V.

For any nonzero x ∈ K, set V′ := V [x] and V′′ := V [1/x]. By (26.6), either 1 ∉ 𝔪V′ or 1 ∉ 𝔪V′′. Hence by maximality, either V = V′ or V = V′′. So either x ∈ V or 1/x ∈ V. Thus V is a valuation ring. So V is local by (26.2), and dominates A by (26.8) as 1 ∉ 𝔪V. □

Exercise (26.10). — Let K be a field, 𝒮 the set of local subrings with fraction field K, ordered by domination. Show its maximal elements are the valuation rings.

Theorem (26.11). — Let K be a field, and R a subring of K. Then the integral closure R′ of R in K is the intersection of all valuation rings V of K containing R. Further, if R is local, then the V dominating R suffice.

Proof: Every valuation ring V is normal by (26.5). So if V ⊃ R, then V ⊃ R′. Thus ⋂_{V ⊃R} V ⊃ R′.

To prove the opposite inclusion, take any x ∈ K − R′. To find a valuation ring V with V ⊃ R and x ∉ V, set y := 1/x. If 1/y ∈ R[y], then for some n,

1/y = a_{0}y^{n} + a_{1}y^{n−1} + ⋯ + a_{n} with a_{λ} ∈ R.

Multiplying by x^{n} yields x^{n+1} − a_{n}x^{n} − ⋯ − a_{0} = 0. So x ∈ R′, a contradiction.

Thus 1 ∉ yR[y]. So there is a maximal ideal 𝔪 of R[y] containing y. Then the composition R → R[y] → R[y]/𝔪 is surjective as y ∈ 𝔪. So 𝔪 ∩ R is a maximal ideal of R. By (26.9), there is a valuation ring V that dominates R[y]_{𝔪}; whence, if R is local, then V also dominates R. But y ∈ 𝔪; so x = 1/y ∉ V, as desired. □

(26.12) (Valuations). — We call an additive abelian group Γ totally ordered if Γ has a subset Γ_{+} that is closed under addition and satisfies −Γ_{+} ⊔ {0} ⊔ Γ_{+} = Γ.

Given x, y ∈ Γ, write x > y if x − y ∈ Γ_{+}. Note that either x > y or x = y or y > x. Note that, if x > y, then x + z > y + z for any z ∈ Γ.

Let V be a domain, and set K := Frac(V) and Γ := K^{×}/V^{×}. Write the group Γ additively, and let v: K^{×} → Γ be the quotient map. It is a homomorphism:

v(xy) = v(x) + v(y).  (26.12.1)

Set Γ_{+} := v(V −0)−0. Then Γ_{+} is closed under addition. Clearly, V is a valuation ring if and only if −Γ_{+} ⊔ {0} ⊔ Γ_{+} = Γ, so if and only if Γ is totally ordered.

September 3, 2012 11Nts.tex Assume V is a valuation ring. Let’s prove that, for all x, y ∈ K^{×},

v(x + y) ≥ min{v(x), v(y)} if x ≠ −y.  (26.12.2)

Indeed, say v(x) ≥ v(y). Then z := x/y ∈ V. So v(z + 1) ≥ 0. Hence

v(x + y) = v(z + 1) + v(y) ≥ v(y) = min{v(x), v(y)},

Note that (26.12.1) and (26.12.2) are the same as (1) and (2) of (23.1).

Conversely, start with a field K, with a totally ordered additive abelian group Γ, and with a surjective homomorphism v: K^{×} → Γ satisfying (26.12.2). Set

V := {x ∈ K^{×} | v(x) ≥ 0} ∪ {0}.

Then V is a valuation ring, and Γ = K^{×}/V^{×}. We call such a v a valuation of K, and Γ the value group of v or of V.

For example, a DVR V of K is just a valuation ring with value group ℤ, since any x ∈ K^{×} has the form x = ut^{n} with u ∈ V^{×} and n ∈ ℤ.

Example (26.13). — Fix totally ordered additive abelian group Γ, and a field k. Form the k-vector space R with basis the symbols X^{a} for a ≥ 0 in Γ. Define X^{a}X^{b} := X^{a+b}, and extend this product to R by linearity. Then R is a k-algebra with X_{0} = 1. We call R the group algebra of Γ.Define v: (R − 0) → Γ by

v(∑ r_{a}X^{a}) := min{a | r_{a} ≠ 0}.

Then for x, y ∈ (R − 0), clearly v(xy) = v(x) + v(y) because k is a domain and Γ is ordered. Hence R is a domain. Moreover, if v(x + y) = a, then either v(x) ≤ a or v(y) ≤ a. Thus v(x + y) ≥ min{v(x), v(y)}.

Set K := Frac(R), and extend v to a map v: K^{×} → Γ by v(x/y) := v(x) − v(y) if y ≠ 0. Clearly v is well defined, surjective, and a homomorphism. Further, for x, y ∈ K^{×}, clearly v(x + y) ≥ min{v(x), v(y)}. Thus v is a valuation with group Γ.

Set R′ := {x ∈ R | v(x) ≥ 0} and 𝔭 := {x ∈ R | v(x) > 0}. Clearly, R′ is a ring, and 𝔭 is a prime of R′. Further, R′_{𝔭} is the valuation ring of v.

There are many choices for Γ other than ℤ. Examples include the additive rationals, the additive reals, its subgroup generated by two incommensurate reals, and the lexicographically ordered product of any two totally ordered abelian groups.

Proposition (26.14). — Let v be a valuation of a field K, and x_{1}, …, x_{n} ∈ K^{×} with n ≥ 2. Set m := min{v(x_{i})}.

(1) If n = 2 and if v(x_{1}) ≠ v(x_{2}), then v(x_{1} + x_{2}) = m.

(2) If x_{1} + ⋯ + x_{n} = 0, then m = v(x_{i}) = v(x_{j}) for some i ≠ j.

Proof: For (1), say v(x_{1}) > v(x_{2}); so v(x_{2}) = m. Set z := x_{1}/x_{2}. Then v(z) > 0. Also v(−z) = v(z) + v(−1) > 0. Now,

0 = v(1) = v(z + 1 − z) ≥ min{v(z + 1), v(−z)} ≥ 0.

Hence v(z + 1) = 0. Now, x_{1} + x_{2} = (z + 1)x_{2}. Therefore, v(x_{1} + x_{2}) = v(x_{2}) = m. Thus (1) holds.

For (2), reorder the x_{i} so v(x_{i}) = m for i ≤ k and v(x_{i}) > m for i > k. By induction, (26.12.2) yields v(x_{k+1} + ⋯ + x_{n}) ≥ min_{i>k}{v(x_{i})}. Therefore, v(x_{k+1} +⋯+x_{n}) > m. If k = 1, then (1) yields v(0) = v(x_{1} +(x_{2} +⋯+x_{n})) = m, a contradiction. So k > 1, and v(x_{1}) = v(x_{2}) = m, as desired. □

Exercise (26.15). — Let V be a valuation ring, such as a DVR, whose value group Γ is Archimedean; that is, given any nonzero α, β ∈ Γ, there’s n ∈ ℤ such that nα > β. Show that V is a maximal proper subring of its fraction field K.

Exercise (26.16). — Let V be a valuation ring. Show that (1) every finitely generated ideal 𝔞 is principal, and

(2) V is Noetherian if and only if V is a DVR.

Lemma (26.17). — Let R be a 1-dimensional Noetherian domain, K its fraction field, M a torsion-free module, and x ∈ R nonzero. Then ℓ(R/xR) < ∞. Further,

ℓ(M/xM) ≤ dim_{K}(M ⊗_{R} K) ℓ(R/xR),  (26.17.1)

with equality if M is finitely generated.

Proof: Set r := dim_{K}(M ⊗_{R} K). If r = ∞, then (26.17.1) is trivial; so we may assume r < ∞.

Set S := R−{0}. Given any module N, set N_{K} := S^{−1}N. Recall N_{K} = N ⊗_{R} K. First, assume M is finitely generated. Choose any K-basis m_{1}/s_{1}, …, m_{r}/s_{r} of

M_{K} with m_{i} ∈ M and s_{i} ∈ S. Then m_{1}/1, …, m_{r}/1 is also a basis. Define an R-map α: R^{r} → M by sending the standard basis elements to the m_{i}. Then its localization α_{K} is an K-isomorphism. But Ker(α) is a submodule of R^{r}, so torsion free. Further, S^{−1} Ker(α) = Ker(α_{K}) = 0. Hence Ker(α) = 0. Thus α is injective.

Set N := Coker(α). Then N_{K} = 0, and N is finitely generated. Hence, Supp(N) is a proper closed subset of Spec(R). But dim(R) = 1 by hypothesis. Hence, Supp(N) consists entirely of maximal ideals. So ℓ(N) < ∞ by (19.4).

Similarly, Supp(R/xR) is closed and proper in Spec(R). So ℓ(R/xR) < ∞.

Consider the standard exact sequence:

0 → N′ → N → N → N/xN → 0 where N′ := Ker(μ_{x}).

Apply Additivity of Length, (19.9); it yields ℓ(N′) = ℓ(N/xN).

Since M is torsion free, μ_{x}: M → M is injective. Consider this commutative diagram with exact rows:

0 ⟶ R^{r} ⟶^{α} M ⟶ N ⟶ 0

μ_{x}↓ ^{μ_{x}}↓ ^{μ_{x}}↓

0 ⟶ R^{r} ⟶^{α} M ⟶ N ⟶ 0

Apply the snake lemma (5.12). It yields this exact sequence:

0 → N′ → (R/xR)^{r} → M/xM → N/xN → 0.

Hence ℓ(M/xM) = ℓ((R/xR)^{r}) by additivity. But ℓ((R/xR)^{r}) = r ℓ(R/xR) also by additivity. Thus equality holds in (26.17.1) when M is finitely generated.

Second, assume M is arbitrary, but (26.17.1) fails. Then M possesses a finitely generated submodule M′ whose image H in M/xM satisfies ℓ(H) > rℓ(R/xR). Now, M_{K} ⊃ M′_{K}; so r ≥ dim_{K}(M′_{K}). Therefore,

ℓ(M′/xM′) ≥ ℓ(H) > r ℓ(R/xR) ≥ dim_{K}(M′_{K}) ℓ(R/xR).

However, together these inequalities contradict the first case with M′ for M. □

Theorem (26.18) (Krull–Akizuki). — Let R be a 1-dimensional Noetherian domain, K its fraction field, K′ a finite extension field, and R′ a proper subring of K′ containing R. Then R′ is, like R, a 1-dimensional Noetherian domain.

Proof: Given a nonzero ideal 𝔞′ of R′, take any nonzero x ∈ 𝔞′. Since K′/K is finite, there is an equation a_{n}x^{n} + ⋯ + a_{0} = 0 with a_{i} ∈ R and a_{0} ≠ 0. Then a_{0} ∈ 𝔞′ ∩ R. Further, (26.17) yields ℓ(R/a_{0}R) < ∞.

Clearly, R′ is a domain, so a torsion free R-module. Further, R′ ⊗_{R} K ⊂ K′; hence, dim_{K}(R′ ⊗_{R} K) < ∞. Therefore, (26.17) yields ℓ_{R}(R′/a_{0}R′) < ∞.

But 𝔞′/a_{0}R′ ⊂ R′/a_{0}R′. So ℓ_{R}(𝔞′/a_{0}R′) < ∞. So 𝔞′/a_{0}R′ is finitely generated over R by (19.2)(3). Hence 𝔞′ is finitely generated over R′. Thus R′ is Noetherian.

Set R′′ := R′/a_{0}R′. Clearly, ℓ_{R′′}R′′ ≤ ℓ_{R}R′′. So ℓ_{R′′}R′′ < ∞. So, in R′′, every prime is maximal by (19.4). So if 𝔞′ is prime, then 𝔞′/a_{0}R′ is maximal, whence 𝔞′ maximal. So in R, every nonzero prime is maximal. Thus R′ is 1-dimensional. □

Corollary (26.19). — Let R be a 1-dimensional Noetherian domain, such as a Dedekind domain. Let K be its fraction field, K′ a finite extension field, and R′ the normalization of R in K′. Then R′ is Dedekind.

Proof: Since R is 1-dimensional, it’s not a field. But R′ is the normalization of R. So R′ is not a field by (14.1). Hence, R′ is Noetherian and 1-dimensional by Theorem (26.18). Thus R′ is Dedekind by Definition (24.1). □

Corollary (26.20). — Let K′/K be a field extension, and V′ a valuation ring of K′ not containing K. Set V := V′ ∩ K. Then V is a DVR if and only if V′ is.

Proof: It follows easily from Definition (26.1) that V is a valuation ring, and from Subsection (26.12) that its value group is a subgroup of that of V′. Now, a nonzero subgroup of ℤ is a copy of ℤ. Thus V is a DVR if V′ is.

Conversely, assume V is a DVR, so Noetherian and 1-dimensional. Now, V′ does not contain K, so is proper in K′.- Hence, V′ is Noetherian by Theorem (26.18), so a DVR by Exercise (26.16)(2). □

Solutions

1. Rings and Ideals

Exercise (1.6). — Let R be a ring, 𝔞 an ideal, and P := R[X_{1}, …, X_{n}] the polynomial ring. Construct an isomorphism ψ from P/𝔞P onto (R/𝔞)[X_{1}, …, X_{n}].

Solution: Let κ: R → R/𝔞 be the quotient map. Form the homomorphism φ: P → (R/𝔞)[X_{1}, …, X_{n}] such that φ|R = κ and φ(X_{i}) = X_{i}. Then

φ(∑ a_{(i_{1},...,i_{n})}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}}) = ∑ κ(a_{(i_{1},...,i_{n})})X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}}.

Since κ is surjective, so is φ. Since Ker(κ) = 𝔞, it follows that

Ker(φ) = ∑ 𝔞X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}} = 𝔞P.

Therefore, φ induces the desired isomorphism ψ by (1.5.1). □

Exercise (1.9). — Let R be ring, and P := R[X_{1}, …, X_{n}] the polynomial ring. Let m ≤ n and a_{1}, …, a_{m} ∈ R. Set 𝔭 := ⟨X_{1} − a_{1}, …, X_{m} − a_{m}⟩. Prove that P/𝔭 = R[X_{m+1}, …, X_{n}].

Solution: First, assume m = n. Set P′ := R[X_{1}, …, X_{n−1}] and

𝔭′ := ⟨X_{1} − a_{1}, …, X_{n−1} − a_{n−1}⟩ ⊂ P′.

By induction on n, we may assume P′/𝔭′ = R. However, P = P′[X_{n}]. Hence P/𝔭′P = (P′/𝔭′)[X_{n}] by (1.6). Thus P/𝔭′P = R[X_{n}].

We have P/𝔭 = (P/𝔭′P)/𝔭(P/𝔭′P) by (1.8). But 𝔭 = 𝔭′P + ⟨X_{n} − a_{n}⟩P. Hence 𝔭(P/𝔭′P) = ⟨X_{n} − a_{n}⟩(P/𝔭′P). So P/𝔭 = R[X_{n}]/⟨X_{n} − a_{n}⟩. So P/𝔭 = R by (1.7).

In general, P = (R[X_{1}, …, X_{m}])[X_{m+1}, …, X_{n}]. Thus P/𝔭 = R[X_{m+1}, …, X_{n}] by (1.6). □

Exercise (1.13) (Chinese Remainder Theorem). — Let R be a ring. (1) Let 𝔞 and 𝔟 be comaximal ideals; that is, 𝔞 + 𝔟 = R. Prove

(a) 𝔞𝔟 = 𝔞 ∩ 𝔟 and (b) R/𝔞𝔟 = (R/𝔞) × (R/𝔟).

(2) Let 𝔞 be comaximal to both 𝔟 and 𝔟′. Prove 𝔞 is also comaximal to 𝔟𝔟′.

(3) Let 𝔞, 𝔟 be comaximal, and m, n ≥ 1. Prove 𝔞^{m} and 𝔟^{n} are comaximal.

(4) Let 𝔞_{1}, …, 𝔞_{n} be pairwise comaximal. Prove (a) 𝔞_{1} and 𝔞_{2} ⋯ 𝔞_{n} are comaximal;

(b) 𝔞_{1} ∩ ⋯ ∩ 𝔞_{n} = 𝔞_{1} ⋯ 𝔞_{n};

(c) R/(𝔞_{1} ⋯ 𝔞_{n}) ⟶^{∼} ∏(R/𝔞_{i}).

Solution: To prove (1)(a), note that always 𝔞𝔟 ⊆ 𝔞 ∩ 𝔟. Conversely, 𝔞 + 𝔟 = R implies x+y = 1 with x ∈ 𝔞 and y ∈ 𝔟. So given z ∈ 𝔞∩𝔟, we have z = xz+yz ∈ 𝔞𝔟.

To prove (1)(b), form the map R → R/𝔞 × R/𝔟 that carries an element to its pair of residues. The kernel is 𝔞 ∩ 𝔟, which is 𝔞𝔟 by (1). So we have an injection

φ: R/𝔞𝔟 ↪ R/𝔞 × R/𝔟.

To show that φ is surjective, take any element (x̄, ȳ) in R/𝔞 × R/𝔟. Say x̄ and ȳ are the residues of x and y. Since 𝔞 + 𝔟 = R, we can find a ∈ 𝔞 and b ∈ 𝔟 such that a + b = y − x. Then φ(x + a) = (x̄, ȳ), as desired. Thus (1) holds.

To prove (2), note that

R = (𝔞 + 𝔟)(𝔞 + 𝔟′) = (𝔞^{2} + 𝔟𝔞 + 𝔞𝔟′) + 𝔟𝔟′ ⊆ 𝔞 + 𝔟𝔟′ ⊆ R.

To prove (3), note that (2) implies 𝔞 and 𝔟^{n} are comaximal for any n ≥ 1 by induction on n. Hence, 𝔟^{n} and 𝔞^{m} are comaximal for any m ≥ 1.

To prove (4)(a), assume 𝔞_{1} and 𝔞_{2} ⋯ 𝔞_{n−1} are comaximal by induction on n. By hypothesis, 𝔞_{1} and 𝔞_{n} are comaximal. Thus (2) yields (a).

To prove (4)(b) and (4)(c), again proceed by induction on n. Thus (1) yields

𝔞_{1} ∩ (𝔞_{2} ∩ ⋯ ∩ 𝔞_{n}) = 𝔞_{1} ∩ (𝔞_{2} ⋯ 𝔞_{n}) = 𝔞_{1}𝔞_{2} ⋯ 𝔞_{n};
R/(𝔞_{1} ⋯ 𝔞_{n}) ⟶^{∼} R/𝔞_{1} × R/(𝔞_{2} ⋯ 𝔞_{n}) ⟶^{∼} ∏(R/𝔞_{i}).  □

Exercise (1.14). — First, given a prime number p and a k ≥ 1, find the idempotents in ℤ/⟨p^{k}⟩. Second, find the idempotents in ℤ/⟨12⟩. Third, find the number of idempotents in ℤ/⟨n⟩ where n = ∏_{i=1}^{N} p_{i}^{n_{i}} with p_{i} distinct prime numbers.

Solution: First, let m ∈ ℤ be idempotent modulo p^{k}. Then m(m−1) is divisible by p^{k}. So either m or m − 1 is divisible by p^{k}, as m and m − 1 have no common prime divisor. Hence 0 and 1 are the only idempotents in ℤ/⟨p^{k}⟩.

Second, since −3 + 4 = 1, the Chinese Remainder Theorem (1.13) yields

ℤ/⟨12⟩ = ℤ/⟨3⟩ × ℤ/⟨4⟩.

Hence m is idempotent modulo 12 if and only if m is idempotent modulo 3 and modulo 4. By the previous case, we have the following possibilities:

m ≡ 0 (mod 3) and m ≡ 0 (mod 4);

m ≡ 1 (mod 3) and m ≡ 1 (mod 4);

m ≡ 1 (mod 3) and m ≡ 0 (mod 4);

m ≡ 0 (mod 3) and m ≡ 1 (mod 4).

Therefore, m ≡ 0, 1, 4, 9 (mod 12).

Third, for each i, the two numbers p_{1}^{n_{1}} ⋯ p_{i−1}^{n_{i−1}} and p_{i}^{n_{i}} have no common prime divisor. Hence some linear combination is equal to 1 by the Euclidean Algorithm. So the principal ideals they generate are comaximal. Hence by induction on N, the Chinese Remainder Theorem yields

ℤ/⟨n⟩ = ∏_{i=1}^{N} ℤ/⟨p_{i}^{n_{i}}⟩.

So m is idempotent modulo n if and only if m is idempotent modulo p^{n_{i}} for all i; hence, if and only if m is 0 or 1 modulo p^{n_{i}} for all i by the first case. Thus there are 2^{N} idempotents in ℤ/⟨n⟩. □

Exercise (1.15). — Let R := R′ × R′′ be a product of rings, 𝔞 ⊂ R an ideal. Show 𝔞 = 𝔞′ × 𝔞′′ with 𝔞′ ⊂ R′ and 𝔞′′ ⊂ R′′ ideals. Show R/𝔞 = (R′/𝔞′) × (R′′/𝔞′′).

Solution: Set 𝔞′ := {x′ | (x′, 0) ∈ 𝔞} and 𝔞′′ := {x′′ | (0, x′′) ∈ 𝔞}. Clearly 𝔞′ ⊂ R′ and 𝔞′′ ⊂ R′′ are ideals. Clearly,

𝔞 ⊃ 𝔞′ × 0 + 0 × 𝔞′′ = 𝔞′ × 𝔞′′.

The opposite inclusion holds, because if 𝔞 ∋ (x′, x′′), then

𝔞 ∋ (x′, x′′) · (1, 0) = (x′, 0) and 𝔞 ∋ (x′, x′′) · (0, 1) = (0, x′′).

Finally, the equation R/𝔞 = (R/𝔞′) × (R/𝔞′′) is now clear from the construction of the residue class ring. □

Exercise (1.16). — Let R be a ring, and e, e′ idempotents. (See (10.6) also.) (1) Set 𝔞 := ⟨e⟩. Show 𝔞 is idempotent; that is, 𝔞^{2} = 𝔞.

(2) Let 𝔞 be a principal idempotent ideal. Show 𝔞 = ⟨f⟩ with f idempotent.

(3) Assume ⟨e⟩ = ⟨e′⟩. Show e = e′.

(4) Set e′′ := e + e′ − ee′. Show ⟨e, e′⟩ = ⟨e′′⟩ and e′′ is idempotent.

(5) Let e_{1}, …, e_{r} be idempotents. Show ⟨e_{1}, …, e_{r}⟩ = ⟨f⟩ with f idempotent.

(6) Assume R is Boolean. Show every finitely generated ideal is principal.

Solution: For (1), note 𝔞^{2} ⊂ 𝔞 always. Conversely, xe = xe^{2} for any x ∈ R; so 𝔞 ⊂ 𝔞^{2}. Thus (1) holds.

For (2), say 𝔞 = ⟨g⟩. Then 𝔞^{2} = ⟨g^{2}⟩. But 𝔞^{2} = 𝔞. So g = xg^{2} for some x. Set f := xg. Then f ∈ 𝔞; so ⟨f⟩ ⊂ 𝔞. And g = fg. So 𝔞 ⊂ ⟨f⟩. Thus (2) holds.

For (3), say e′ = xe. So e′ = xe^{2} = e′e. By symmetry, e = ee′. Thus (3) holds. For (4), note ⟨e′′⟩ ⊂ ⟨e, e′⟩. Conversely, ee′′ = e^{2} + ee′ − e^{2}e′ = e + ee′ − ee′ = e.

By symmetry, e′e′′ = e′. So ⟨e, e′⟩ ⊂ ⟨e′′⟩ and e^{′′2} = ee′′ + e′e′′ − ee′e′′ = e′′. Thus (4) holds.

For (5), induct on r. Thus (4) yields (5).

For (6), recall that every element of R is idempotent. Thus (5) yields (6). □

2. Prime Ideals

Exercise (2.2). — Let 𝔞 and 𝔟 be ideals, and 𝔭 a prime ideal. Prove that these conditions are equivalent: (1) 𝔞 ⊂ 𝔭 or 𝔟 ⊂ 𝔭; and (2) 𝔞 ∩ 𝔟 ⊂ 𝔭; and (3) 𝔞𝔟 ⊂ 𝔭.

Solution: Trivially, (1) implies (2). If (2) holds, then (3) follows as 𝔞𝔟 ⊂ 𝔞 ∩ 𝔟. Finally, assume 𝔞 ⊄ 𝔭 and 𝔟 ⊄ 𝔭. Then there are x ∈ 𝔞 and y ∈ 𝔟 with x, y ∉ 𝔭. Hence, since 𝔭 is prime, xy ∉ 𝔭. However, xy ∈ 𝔞𝔟. Thus (3) implies (1). □

Exercise (2.4). — Given a prime number p and an integer n ≥ 2, prove that the residue ring ℤ/⟨p^{n}⟩ does not contain a domain.

Solution: Any subring of ℤ/⟨p^{n}⟩ must contain 1, and 1 generates ℤ/⟨p^{n}⟩ as an abelian group. So ℤ/⟨p^{n}⟩ contains no proper subrings. However, ℤ/⟨p^{n}⟩ is not a domain, because in it, p · p^{n−1} = 0 but neither p nor p^{n−1} is 0. □

Exercise (2.5). — Let R := R′ × R′′ be a product of two rings. Show that R is a domain if and only if either R′ or R′′ is a domain and the other is 0.

Solution: Assume R is a domain. As (1, 0) · (0, 1) = (0, 0), either (1, 0) = (0, 0) or (0, 1) = (0, 0). Correspondingly, either R′ = 0 and R = R′′, or R′′ = 0 and R = R′′. The assertion is now obvious. □

Exercise (2.10). — Let R be a ring, 𝔭 a prime ideal, R[X] the polynomial ring. Show that 𝔭R[X] and 𝔭R[X] + ⟨X⟩ are prime ideals of R[X].

Solution: Note R[X]/𝔭R[X] = (R/𝔭)[X] by (1.6). But R/𝔭 is a domain by (2.9). So R[X]/𝔭R[X] is a domain by (2.3). Thus 𝔭R[X] is prime by (2.9).

Note (𝔭R[X] + ⟨X⟩)/𝔭R[X] is equal to ⟨X⟩ ⊂ (R/𝔭)[X]. But (R/𝔭)[X]/⟨X⟩ is equal to R/𝔭 by (1.7). So R[X]/(𝔭R[X] + ⟨X⟩) is equal to R/𝔭 by (1.8). But R/𝔭 is a domain by (2.9). Thus 𝔭R[X] + ⟨X⟩ is prime again by (2.9). □

Exercise (2.11). — Let R be a domain, and R[X_{1}, …, X_{n}] the polynomial ring in n variables. Let m ≤ n, and set 𝔭 := ⟨X_{1}, …, X_{m}⟩. Prove 𝔭 is a prime ideal.

Solution: Simply combine (2.9), (2.3), and (1.9) □

Exercise (2.12). — Let R := R′ × R′′ be a product of rings. Show every prime ideal of R has the form 𝔭′ × R′′ with 𝔭′ ⊂ R′ prime or R′ × 𝔭′′ with 𝔭′′ ⊂ R′′ prime.

Solution: Simply combine (1.15), (2.9), and (2.5). □

Exercise (2.16). — Let k be a field, R a nonzero ring, φ: k → R a ring map. Prove φ is injective.

Solution: By (1.1), 1 ≠ 0 in R. So Ker(φ) ≠ k. So Ker(φ) = 0 by (2.15). Thus φ is injective. □

Exercise (2.18). — Let B be a Boolean ring. Show that every prime 𝔭 is maximal, and B/𝔭 = 𝔽_{2}.

Solution: Take any z ∈ B/𝔭. Then z(z − 1) = 0. But B/𝔭 is a domain. So z = 0 or z = 1. Thus B/𝔭 = 𝔽_{2}. Clearly, 𝔽_{2} is a field. Thus (2.17) yields (1). □

Exercise (2.20). — Prove the following statements or give a counterexample. (1) The complement of a multiplicative subset is a prime ideal.

(2) Given two prime ideals, their intersection is prime.

(3) Given two prime ideals, their sum is prime.

(4) Given a ring map φ: R → R′, the operation φ^{−1} carries maximal ideals of R′ to maximal ideals of R.

(5) In (1.8), κ^{−1} takes maximal ideals of R/𝔞 to maximal ideals of R.

Solution: (1) False. In the ring ℤ, consider the set S of powers of 2. The complement T of S contains 3 and 5, but not 8; so T is not an ideal.

(2) False. In the ring ℤ, consider the prime ideals ⟨2⟩ and ⟨3⟩; their intersection ⟨2⟩ ∩ ⟨3⟩ is equal to ⟨6⟩, which is not prime.

(3) False. Since 2 · 3 − 5 = 1, we have ⟨3⟩ + ⟨5⟩ = ℤ.

(4) False. Let φ: ℤ → ℚ be the inclusion map. Then φ^{−1}⟨0⟩ = ⟨0⟩.

(5) True. The assertion is immediate from (1.8). □

Exercise (2.21). — Let k be a field, P := k[X_{1}, …, X_{n}] the polynomial ring, f ∈ P nonzero. Let d be the highest power of any variable appearing in f.

(1) Let S ⊂ k have at least dn + 1 elements. Proceeding by induction on n, find a_{1}, …, a_{n} ∈ S with f(a_{1}, …, a_{n}) ≠ 0.

(2) Using the algebraic closure K of k, find a maximal ideal 𝔪 of P with f ∉ 𝔪.

Solution: Consider (1). Assume n = 1. Then f has at most d roots by [Artin, (1.8), p. 392]. So f(a_{1}) ≠ 0 for some a_{1} ∈ S.

Assume n > 1. Say f = ∑_{j} g_{j}X_{1}^{j} with g_{j} ∈ k[X_{2}, …, X_{n}]. But f ≠ 0. So g_{i} ≠ 0 for some i. By induction, g_{i}(a_{2}, …, a_{n}) ≠ 0 for some a_{2}, …, a_{n} ∈ S. By n = 1, find a_{1} ∈ S such that f(a_{1}, …, a_{n}) = ∑_{j} g_{j}(a_{2}, …, a_{n})a_{1}^{j} ≠ 0. Thus (1) holds.

Consider (2). As K is infinite, (1) yields a_{1}, …, a_{n} ∈ K with f_{i}(a_{1}, …, a_{n}) ≠ 0. Define φ: P → K by φ(X_{i}) = a_{i}. Then Im(φ) ⊂ K is the k-subalgebra generated by the a_{i}. It is a field by [Artin, (2.6), p. 495]. Set 𝔪 := Ker(φ). Hence 𝔪 is a maximal ideal, and f_{i} ∉ 𝔪 as φ(f_{i}) = f_{i}(a_{1}, …, a_{n}) ≠ 0. Thus (2) holds. □

Exercise (2.24). — Prove that, in a PID, elements x and y are relatively prime (share no prime factor) if and only if the ideals ⟨x⟩ and ⟨y⟩ are comaximal.

Solution: Say ⟨x⟩ + ⟨y⟩ = ⟨d⟩. Then d = gcd(x, y), as is easy to check. The assertion is now obvious. □

Exercise (2.27). — Preserve the setup of (2.26). Let f := a_{0}X^{n} + ⋯ + a_{n} be a polynomial of positive degree n. Assume that R has infinitely many prime elements p, or simply that there is a p such that p ∤ a_{0}. Show that ⟨f⟩ is not maximal.

Solution: Set 𝔞 := ⟨p, f⟩. Then 𝔞 ⫌ ⟨f⟩, because p is not a multiple of f. Set k := R/⟨p⟩. Since p is irreducible, k is a domain by (2.6) and (2.8). Let f′ ∈ k[X] denote the image of f. By hypothesis, deg(f′) = n ≥ 1. Hence f′ is not a unit by (2.3) since k is a domain. Therefore, ⟨f′⟩ is proper. But P/𝔞 ⟶^{∼} k[X]/⟨f′⟩ by (1.6) and (1.8). So 𝔞 is proper. Thus ⟨f⟩ is not maximal. □

3. Radicals

Exercise (3.6). — Let A be a ring, 𝔪 a maximal ideal such that 1 + m is a unit for every m ∈ 𝔪. Prove A is local. Is this assertion still true if 𝔪 is not maximal?

Solution: Take y ∈ A. Let’s prove that, if y ∉ 𝔪, then y is a unit. Since 𝔪 is maximal, ⟨y⟩ + 𝔪 = A. Hence there exist x ∈ R and m ∈ 𝔪 such that xy + m = 1, or in other words, xy = 1 − m. So xy is a unit by hypothesis; whence, y is a unit. Thus A is local by (3.4).

The assertion is not true if 𝔪 is not maximal. Indeed, take any ring that is not local, for example ℤ, and take 𝔪 := ⟨0⟩. □

Exercise (3.10). — Let φ: R → R′ be a map of rings, 𝔭 an ideal of R. Prove (1) there is an ideal 𝔮 of R′ with φ^{−1}(𝔮) = 𝔭 if and only if φ^{−1}(𝔭R′) = 𝔭;

(2) if 𝔭 is prime with φ^{−1}(𝔭R′) = 𝔭, then there’s a prime 𝔮 of R′ with φ^{−1}(𝔮) = 𝔭.

Solution: In (1), given 𝔮, note φ(𝔭) ⊂ 𝔮, as always φ(φ^{−1}(𝔮)) ⊂ 𝔮. So 𝔭R′ ⊂ 𝔮. Hence φ^{−1}(𝔭R′) ⊂ φ^{−1}(𝔮) = 𝔭. But, always 𝔭 ⊂ φ^{−1}(𝔭R′). Thus φ^{−1}(𝔭R′) = 𝔭. The converse is trivial: take 𝔮 := 𝔭R′.

In (2), set S := φ(R − 𝔭). Then S ∩ 𝔭R′ = ∅, as φ(x) ∈ 𝔭R′ implies x ∈ φ^{−1}(𝔭R′) and φ^{−1}(𝔭R′) = 𝔭. So there’s a prime 𝔮 of R′ containing 𝔭R′ and disjoint from S by (3.9). So φ^{−1}(𝔮) ⊃ φ^{−1}(𝔭R′) = 𝔭 and φ^{−1}(𝔮) ∩ (R − 𝔭) = ∅. Thus φ^{−1}(𝔮) = 𝔭. □

Exercise (3.11). — Use Zorn’s lemma to prove that any prime ideal 𝔭 contains a prime ideal 𝔮 that is minimal containing any given subset 𝔰 ⊂ 𝔭.

Solution: Let 𝒮 be the set of all prime ideals 𝔮 such that 𝔰 ⊂ 𝔮 ⊂ 𝔭. Then 𝔭 ∈ 𝒮, so 𝒮 ≠ ∅. Order 𝒮 by reverse inclusion. To apply Zorn’s Lemma, we must show that, for any decreasing chain {𝔮_{λ}} of prime ideals, the intersection 𝔮 := ⋂ 𝔮_{λ} is a prime ideal. Plainly 𝔮 is always an ideal. So take x, y ∉ 𝔮. Then there exists λ such that x, y ∉ 𝔮_{λ}. Since 𝔮_{λ} is prime, xy ∉ 𝔮_{λ}. So xy ∉ 𝔮. Thus 𝔮 is prime. □

Exercise (3.13). — Let R be a ring, S a subset. Show that S is saturated multiplicative if and only if R − S is a union of primes.

Solution: First, take x ∈ R − S. Assume S is multiplicative. Then xy ∉ S for all y ∈ R. So ⟨x⟩ ∩ S = ∅. Assume S is saturated too. Then (3.9) gives a prime 𝔭 ⊃ ⟨x⟩ with 𝔭 ∩ S = ∅. Thus R − S is a union of primes.

Conversely, assume R − S is a union of primes 𝔭. Then 1 ∈ S as 1 lies in no 𝔭. Take x, y ∈ R. Then x, y ∈ S if and only if x, y lie in no 𝔭; if and only if xy lies in no 𝔭, as every 𝔭 is prime; if and only if xy ∈ S. Thus S is saturated multiplicative. □

Exercise (3.14). — Let R be a ring, and S a multiplicative subset. Define its saturation to be the subset

\overline{S} := \{ x ∈ R | there is y ∈ R with xy ∈ S \}.

(1) Show (a) that \overline{S} ⊃ S, and (b) that \overline{S} is saturated multiplicative, and (c) that any saturated multiplicative subset T containing S also contains \overline{S}.

(2) Show that R − \overline{S} is the union U of all the primes 𝔭 with 𝔭 ∩ S = ∅.

(3) Let 𝔞 be an ideal; assume S = 1 + 𝔞; set W := ⋃_{𝔭∈V(𝔞)} 𝔭. Show R − \overline{S} = W.

Solution: Consider (1). Trivially, if x ∈ S, then x · 1 ∈ S. Thus (a) holds. Hence 1 ∈ \overline{S} as 1 ∈ S. Now, take x, x′ ∈ \overline{S}. Then there are y, y′ ∈ R with

xy, x′y′ ∈ S. But S is multiplicative. So (xx′′)(yy′) ∈ S. Hence xx′ ∈ \overline{S}. Thus \overline{S} is multiplicative. Conversely, take x, x′ ∈ R with xx′ ∈ \overline{S}. Then there is y ∈ R with xx′y ∈ S. So x, x′ ∈ \overline{S}. Thus S is saturated. Thus (b) holds

Finally, given x ∈ \overline{S}, there is y ∈ R with xy ∈ S. So xy ∈ T. But T is saturated multiplicative. So x ∈ T. Thus T ⊃ \overline{S}. Thus (c) holds.

Consider (2). Plainly, R−U contains S. Further, R−U is saturated multiplicative by (3.13). So R − U ⊃ \overline{S} by (1)(c). Thus U ⊂ R − \overline{S}. Conversely, R − \overline{S} is a union of primes 𝔭 by (3.13). Plainly, 𝔭 ∩ S = ∅ for all 𝔭. So U ⊃ R − \overline{S}. Thus (2) holds.

For (3), first take a prime 𝔭 with 𝔭 ∩ S = ∅. Then 1 ∉ 𝔭 + 𝔞. So 𝔭 + 𝔞 lies in a maximal ideal 𝔪 by (3.9). Then 𝔞 ⊂ 𝔪; so 𝔪 ∈ V(𝔞). Also, 𝔭 ⊂ 𝔪. Thus U ⊂ W.

Conversely, take 𝔭 ∈ V(𝔞). Then S ⊂ 1 + 𝔭. But 𝔭 ∩ (1 + 𝔭) = ∅. So 𝔭 ∩ S = ∅. Thus U ⊃ W. Thus U = W. Thus (2) yields (3). □

Exercise (3.16). — Let k be an infinite field.

(1) Let V be a vector space, W_{1}, …, W_{r} proper subspaces. Show ⋃ W_{i} ≠ V.

(2) In (1), let W ⊂ ⋃ W_{i} be a subspace. Show W ⊂ W_{i} for some i.

(3) Let R a k-algebra, 𝔞, 𝔞_{1}, …, 𝔞_{r} ideals. If 𝔞 ⊂ ⋃ 𝔞_{i}, show 𝔞 ⊂ 𝔞_{i} for some i.

Solution: For (1), for all i, take v_{i} ∈ V − W_{i}. Form their span V′ ⊂ V. Set n := dim V′ and W′_{i} := W_{i} ∩ V′. Then n < ∞, and it suffices to show ⋃ W′_{i} ≠ V′.

Identify V′ with k^{n}. Form the polynomial ring P := k[X_{1}, …, X_{n}]. For each i, take a linear form f_{i} ∈ P that vanishes on W′_{i}. Set f := f_{1} ⋯ f_{r}. Then (2.21)(1) yields a_{1}, …, a_{n} ∈ k with f(a_{1}, …, a_{n}) ≠ 0. Then (a_{1}, …, a_{n}) ∈ V′ − ⋃ W′_{i}.

For (2), for all i, set U_{i} := W ∩ W_{i}. Then ⋃ U_{i} = W. So (1) implies U_{i} = W for some i. Thus W ⊂ W_{i}.

Finally, as every ideal is a k-vector space, (3) is a special case of (2). □

Exercise (3.17). — Let k be a field, R := k[X, Y] the polynomial ring in two variables, 𝔪 := ⟨X, Y ⟩. Show 𝔪 is a union of smaller primes.

Solution: Since R is a UFD, and 𝔪 is maximal, so prime, any nonzero f ∈ 𝔪 has a prime factor p ∈ 𝔪. Thus 𝔪 = ⋃_{p}⟨p⟩, but 𝔪 ≠ ⟨p⟩ as 𝔪 is not principal. □

Exercise (3.19). — Find the nilpotents in ℤ/⟨n⟩. In particular, take n = 12.

Solution: An integer m is nilpotent modulo n if and only if some power m^{k} is divisible by n. The latter holds if and only if every prime factor of n occurs in m. In particular, in ℤ/⟨12⟩, the nilpotents are 0 and 6. □

Exercise (3.20). — Let φ: R → R′ be a ring map, 𝔟 ⊂ R′ a subset. Prove

φ^{−1}\sqrt{𝔟} = \sqrt{φ^{−1}𝔟}.

Solution: Below, (1) is clearly equivalent to (2); and (2), to (3); and so forth:

(1) x ∈ φ^{−1}\sqrt{𝔟};  (4) φ(x^{n}) ∈ 𝔟 for some n;
(2) φx ∈ \sqrt{𝔟};  (5) x^{n} ∈ φ^{−1}𝔟 for some n;
(3) (φx)^{n} ∈ 𝔟 for some n; (6) x ∈ \sqrt{φ^{−1}𝔟}. □

Exercise (3.21). — Let R be a ring, 𝔞 ⊂ \sqrt{⟨0⟩} an ideal, and P := R[Y] the polynomial ring in one variable. Let u ∈ R be a unit, and x ∈ R a nilpotent.

(1) Prove (a) that u + x is a unit in R and (b) that u + xY is a unit in P.

(2) Suppose w ∈ R maps to a unit of R/𝔞. Prove that w is a unit in R.

Solution: In (1), say x^{n} = 0. Set y := −xu^{−1}. Then (a) holds as

(u + x) · u^{−1}(1 + y + y^{2} + ⋯ + y^{n−1}) = 1.

Now, u is also a unit in P, and (xY)^{n} = 0; hence, (a) implies (b).

In (2), say wy ∈ R maps to 1 ∈ R/𝔞. Set z := wy − 1. Then z ∈ 𝔞, so z is nilpotent. Hence, 1+z is a unit by (1)(a). So wy is a unit. Then w·y(wy)^{−1} = 1. □

Exercise (3.23). — Let B be a Boolean ring. Show that rad(B) = nil(B) = ⟨0⟩.

Solution: By (3.22), nil(B) = ⋂ 𝔭 where 𝔭 runs through all the primes of B. Every 𝔭 is maximal by (2.18); the converse holds by (2.22). Thus rad(B) = nil(B).

Let f ∈ nil(B). Then f^{n} = 0 for some n ≥ 1 by (3.18). But f^{2} = f by (1.2). So f = 0. Thus nil(B) = ⟨0⟩. □

Exercise (3.25). — Let R be a ring, and 𝔞 an ideal. Assume \sqrt{𝔞} is finitely generated. Show (\sqrt{𝔞})^{n} ⊂ 𝔞 for all large n.

Solution: Let x_{1}, …, x_{m} be generators of \sqrt{𝔞}. For each i, there is n_{i} such that x_{i}^{n_{i}} ∈ 𝔞. Let n > ∑(n_{i} − 1). Given a ∈ \sqrt{𝔞}, write a = ∑_{i=1}^{m} y_{i}x_{i} with y_{i} ∈ R. Then a^{n} is a linear combination of terms of the form x_{1}^{j_{1}} ⋯ x_{m}^{j_{m}} with ∑_{i=1}^{m} j_{i} = n. Hence j_{i} ≥ n_{i} for some i, because if j_{i} ≤ n_{i} − 1 for all i, then ∑ j_{i} ≤ ∑(n_{i} − 1). Thus a^{n} ∈ 𝔞, as desired. □

Exercise (3.26). — Let R be a ring, 𝔮 an ideal, 𝔭 a finitely generated prime. Prove that 𝔭 = \sqrt{𝔮} if and only if there is n ≥ 1 such that 𝔭 ⊃ 𝔮 ⊃ 𝔭^{n}.

Solution: If 𝔭 = \sqrt{𝔮}, then 𝔭 ⊃ 𝔮 ⊃ 𝔭^{n} by (3.25). Conversely, if 𝔮 ⊃ 𝔭^{n}, then clearly \sqrt{𝔮} ⊃ 𝔭 . Further, since 𝔭 is prime, if 𝔭 ⊃ 𝔮, then 𝔭 ⊃ \sqrt{𝔮}. □

Exercise (3.28). — Let R be a ring. Assume R is reduced and has finitely many minimal prime ideals 𝔭_{1}, …, 𝔭_{n}. Prove that φ: R → ∏(R/𝔭_{i}) is injective, and for each i, there is some (x_{1}, …, x_{n}) ∈ Im(φ) with x_{i} ≠ 0 but x_{j} = 0 for j ≠ i.

Solution: Clearly Ker(φ) = ⋂ 𝔭_{i}. Now, R is reduced and the 𝔭_{i} are its minimal primes; hence, (3.22) and (3.11) yield

⟨0⟩ = \sqrt{⟨0⟩} = ⋂ 𝔭_{i}.

Thus Ker(φ) = ⟨0⟩, and so φ is injective.

Finally, fix i. Since 𝔭_{i} is minimal, 𝔭_{i} ⊅ 𝔭_{j} for j ≠ i; say a_{j} ∈ 𝔭_{j} − 𝔭_{i}. Set a := ∏_{j≠i} a_{j}. Then a ∈ 𝔭_{j} − 𝔭_{i} for all j ≠ i. Thus Im(φ) meets R/𝔭_{i}. □

4. Modules

Exercise (4.3). — Let R be a ring, M a module. Consider the set map

θ: Hom(R, M) → M defined by θ(ρ) := ρ(1).

Show that θ is an isomorphism, and describe its inverse.

Solution: First off, θ is R-linear, because

θ(xρ + x′ρ′) = (xρ + x′ρ′)(1) = xρ(1) + x′ρ′(1) = xθ(ρ) + x′θ(ρ′).

Set H := Hom(R, M). Define η: M → H by η(m)(x) := xm. It is easy to check that ηθ = 1_{H} and θη = 1_{M}. Thus θ and η are inverse isomorphisms by (4.2). □

Exercise (4.12). — Let R be a domain, and x ∈ R nonzero. Let M be the submodule of Frac(R) generated by 1, x^{−1}, x^{−2}, . . .. Suppose that M is finitely generated. Prove that x^{−1} ∈ R, and conclude that M = R.

Solution: Suppose M is generated by m_{1}, …, m_{k}. Say m_{i} = ∑_{j=0}^{n_{i}} a_{ij}x^{−j} for some n_{i} and a_{ij} ∈ R. Set n := max{n_{i}}. Then 1, x^{−1}, …, x^{−n} generate M. So

x^{−(n+1)} = a_{n}x^{−n} + ⋯ + a_{1}x^{−1} + a_{0}

for some a_{i} ∈ R. Thus

x^{−1} = a_{n} + ⋯ + a_{1}x^{n−1} + a_{0}x^{n} ∈ R.

Finally, as x^{−1} ∈ R and R is a ring, also 1, x^{−1}, x^{−2}, … ∈ R; so M ⊂ R. Conversely, M ⊃ R as 1 ∈ M. Thus M = R. □

Exercise (4.14). — Let Λ be an infinite set, R_{λ} a ring for λ ∈ Λ. Endow ∏ R_{λ} and ⨁ R_{λ} with componentwise addition and multiplication. Show that ∏ R_{λ} has a multiplicative identity (so is a ring), but that ⨁ R_{λ} does not (so is not a ring).

Solution: Consider the vector (1) whose every component is 1. Obviously, (1) is a multiplicative identity of ∏ R_{λ}. On the other hand, no restricted vector (e_{λ}) can be a multiplicative identity in ⨁ R_{λ}; indeed, because Λ is infinite, e_{μ} must be zero for some μ. So (e_{λ}) · (x_{λ}) ≠ (x_{λ}) if x_{μ} ≠ 0. □

Exercise (4.15). — Let L, M, and N be modules. Consider a diagram

L ⟶^{α} ⟵_{ρ} M ⟶^{β} ⟵_{σ} N

where α, β, ρ, and σ are homomorphisms. Prove that

M = L ⊕ N and α = ι_{L}, β = π_{N}, σ = ι_{N}, ρ = π_{L}

if and only if the following relations hold:

βα = 0, βσ = 1, ρσ = 0, ρα = 1, and αρ + σβ = 1.

Solution: If M = L ⊕ N and α = ι_{L}, β = π_{N}, σ = ι_{N}, ρ = π_{L}, then the definitions immediately yield αρ + σβ = 1 and βα = 0, βσ = 1, ρσ = 0, ρα = 1.

Conversely, assume αρ + σβ = 1 and βα = 0, βσ = 1, ρσ = 0, ρα = 1. Consider the maps φ: M → L ⊕ N and θ: L ⊕ N → M given by φm := (ρm, βm) and θ(l, n) := αl + σn. They are inverse isomorphisms, because

φθ(l, n) = (ραl + ρσn, βαl + βσn) = (l, n) and θφm = αρm + σβm = m.

Lastly, β = π_{N}φ and ρ = π_{L}φ by definition of φ, and α = θι_{L} and σ = θι_{N} by definition of θ. □

Exercise (4.16). — Let N be a module, Λ a nonempty set, M_{λ} a module for λ ∈ Λ. Prove that the injections ι_{κ}: M_{κ} → ⨁ M_{λ} induce an injection

⨁ Hom(N, M_{λ}) ↪ Hom(N, ⨁ M_{λ}),

and that it is an isomorphism if N is finitely generated.

Solution: For λ ∈ Λ, let α_{λ}: N → M_{λ} be maps, almost all 0. Then

(∑ ι_{λ}α_{λ})(n) = (α_{λ}(n)) ∈ ⨁ M_{λ}.

So if ∑ ι_{λ}α_{λ} = 0, then α_{λ} = 0 for all λ. Thus the ι_{κ} induce an injection.

Assume N is finitely generated, say by n_{1}, …, n_{k}. Let α: N → ⨁ M_{λ} be a map. Then each α(n_{i}) lies in a finite direct subsum of ⨁ M_{λ}. So α(N) lies in one too. Set α_{κ} := π_{κ}α for all κ ∈ Λ. Then almost all α_{κ} vanish. So (α_{κ}) lies in ⨁ Hom(N, M_{λ}), and ∑ ι_{κ}α_{κ} = α. Thus the ι_{κ} induce a surjection, so an isomorphism. □

Exercise (4.17). — Let N be a module, Λ a nonempty set, M_{λ} a module for λ ∈ Λ. Prove that the injections ι_{κ}: M_{κ} → ⨁ M_{λ} induce an injection

⨁ Hom(N, M_{λ}) ↪ Hom(N, ⨁ M_{λ}),

and that it is an isomorphism if N is finitely generated.

Solution: First, 𝔞(⨁ M_{λ}) ⊂ ⨁ 𝔞M_{λ} because a · (m_{λ}) = (am_{λ}). Conversely, 𝔞(⨁ M_{λ}) ⊃ ⨁ 𝔞M_{λ} because (a_{λ}m_{λ}) = ∑ a_{λ}ι_{λ}m_{λ} since the sum is finite.

Second, 𝔞(∏ M_{λ}) ⊂ ∏ 𝔞M_{λ} as a(m_{λ}) = (am_{λ}). Conversely, say 𝔞 is generated by f_{1}, …, f_{n}. Then 𝔞(∏ M_{λ}) ⊃ ∏ 𝔞M_{λ}. Indeed, take (m′_{λ}) ∈ ∏ 𝔞M_{λ}. Then for each λ, there is n_{λ} such that m′_{λ} = ∑_{j=1}^{n_{λ}} a_{λj}m_{λj} with a_{λj} ∈ 𝔞 and m_{λj} ∈ M_{λ}. Write a_{λj} = ∑_{i=1}^{n} x_{λji}f_{i} with the x_{λji} scalars. Then

(m′_{λ}) = (∑_{j=1}^{n_{λ}} ∑_{i=1}^{n} f_{i}x_{λji}m_{λj}) = ∑_{i=1}^{n} f_{i}(∑_{j=1}^{n_{λ}} x_{λji}m_{λj}) ∈ 𝔞(∏ M_{λ}).  □

5. Exact Sequences Exercise (5.5). — Let M′ and M′′ be modules, N ⊂ M′ a submodule. Set M := M′ ⊕ M′′. Using (5.2)(1) and (5.3) and (5.4), prove M/N = M′/N ⊕ M′′.

Solution: By (5.2)(1) and (5.3), the two sequences 0 → M′′ → M′′ → 0 and 0 → N → M′ → M′/N → 0 are exact. So by (5.4), the sequence

0 → N → M′ ⊕ M′′ → (M′/N) ⊕ M′′ → 0

is exact. Thus (5.3) yields the assertion. □

Exercise (5.6). — Let 0 → M′ → M → M′′ → 0 be a short exact sequence. Prove that, if M′ and M′′ are finitely generated, then so is M.

Solution: Let m′′_{1}, …, m′′_{n} ∈ M map to elements generating M′′. Let m ∈ M, and write its image in M′′ as a linear combination of the images of the m′′_{i}. Let m′′ ∈ M be the same combination of the m′′_{i}. Set m′ := m − m′′. Then m′ maps to 0 in M′′; so m′ is the image of an element of M′.

Let m′_{1}, …, m′_{l} ∈ M be the images of elements generating M′. Then m′ is a linear combination of the m′_{j}. So m is a linear combination of the m′′_{i} and m′_{j}. Thus the m′_{i} and m′′_{j} together generate M. □

Exercise (5.10). — Let M′, M′′ be modules, and set M := M′ ⊕ M′′. Let N be a submodule of M containing M′, and set N′′ := N ∩ M′′. Prove N = M′ ⊕ N′′.

Solution: Form the sequence 0 → M′ → N → π_{M′′}N → 0. It splits by (5.9) as (π_{M′}|N) ∘ ι_{M′} = 1_{M′}. Finally, if (m′, m′′) ∈ N, then (0, m′′) ∈ N as M′ ⊂ N; hence, π_{M′′}N = N′′. □

Exercise (5.11). — Criticize the following misstatement of (5.9): given a short exact sequence 0 → M′ ⟶^{α} M ⟶^{β} M′′ → 0, there is an isomorphism M ≃ M′ ⊕ M′′ if and only if there is a section σ: M′′ → M of β.

Solution: We have α: M′ → M, and ι_{M′}: M′ → M′ ⊕ M′′, but (5.9) requires that they be compatible with the isomorphism M ≃ M′ ⊕ M′′, and similarly for β: M → M′′ and π_{M′′}: M′ ⊕ M′′ → M′′.

Let’s construct a counterexample (due to B. Noohi). For each integer n ≥ 2, let M_{n} be the direct sum of countably many copies of ℤ/⟨n⟩. Set M := ⨁ M_{n}.

First, let us check these two statements:

(1) For any finite abelian group G, we have G ⊕ M ≃ M.

(2) For any finite subgroup G ⊂ M, we have M/G ≃ M. Statement (1) holds since G is isomorphic to a direct sum of copies of ℤ/⟨n⟩ for

various n by the structure theorem for finite abelian groups [1, (6.4), p. 472], [4, Thm. 13.3, p. 200].

To prove (2), write M = B ⨁ M′, where B contains G and involves only finitely many components of M. Then M′ ≃ M. Therefore, (5.10) and (1) yield

M/G ≃ (B/G) ⊕ M′ ≃ M.

To construct the counterexample, let p be a prime number. Take one of the ℤ/⟨p^{2}⟩ components of M, and let M′ ⊂ ℤ/⟨p^{2}⟩ be the cyclic subgroup of order p. There is no retraction ℤ/⟨p^{2}⟩ → M′, so there is no retraction M → M′ either, since the latter would induce the former. Finally, take M′′ := M/M′. Then (1) and (2) yield M ≃ M′ ⊕ M′′. □

Exercise (5.13). — Referring to (4.8), give an alternative proof that β is an isomorphism by applying the Snake Lemma to the diagram

0 ⟶ M ⟶ N ⟶ N/M ⟶ 0

↓  ^{κ}↓  ^{β}↓

0 ⟶ M/L ⟶ N/L ⟶^{λ} (N/L)/(M/L) ⟶ 0

Solution: The Snake Lemma yields an exact sequence,

L ⟶^{1} L → Ker(β) → 0;

hence, Ker(β) = 0. Moreover, β is surjective because κ and λ are. □

Exercise (5.14) (Five Lemma). — Consider this commutative diagram:

M_{4} ⟶^{α_{4}} M_{3} ⟶^{α_{3}} M_{2} ⟶^{α_{2}} M_{1} ⟶^{α_{1}} M_{0}

γ_{4}↓  ^{γ_{3}}↓  ^{γ_{2}}↓  ^{γ_{1}}↓  ^{γ_{0}}↓

N_{4} ⟶^{β_{4}} N_{3} ⟶^{β_{3}} N_{2} ⟶^{β_{2}} N_{1} ⟶^{β_{1}} N_{0}

Assume it has exact rows. Via a chase, prove these two statements:

(1) If γ_{3} and γ_{1} are surjective and if γ_{0} is injective, then γ_{2} is surjective.

(2) If γ_{3} and γ_{1} are injective and if γ_{4} is surjective, then γ_{2} is injective.

Solution: Let’s prove (1). Take n_{2} ∈ N_{2}. Since γ_{1} is surjective, there is m_{1} ∈ M_{1} such that γ_{1}(m_{1}) = β_{2}(n_{2}). Then γ_{0}α_{1}(m_{1}) = β_{1}γ_{1}(m_{1}) = β_{1}β_{2}(n_{2}) = 0 by commutativity and exactness. Since γ_{0} is injective, α_{1}(m_{1}) = 0. Hence exactness yields m_{2} ∈ M_{2} with α_{2}(m_{2}) = m_{1}. So β_{2}(γ_{2}(m_{2}) − n_{2}) = γ_{1}α_{2}(m_{2}) − β_{2}(n_{2}) = 0.

Hence exactness yields n_{3} ∈ N_{3} with β_{3}(n_{3}) = γ_{2}(m_{2})−n_{2}. Since γ_{3} is surjective, there is m_{3} ∈ M_{3} with γ_{3}(m_{3}) = n_{3}. Then γ_{2}α_{3}(m_{3}) = β_{3}γ_{3}(m_{3}) = γ_{2}(m_{2}) − n_{2}. Hence γ_{2}(m_{2} − α_{3}(m_{3})) = n_{2}. Thus γ_{2} is surjective.

The proof of (2) is similar. □

Exercise (5.15) (Nine Lemma). — Consider this commutative diagram:

0  0  0
↓  ↓  ↓
0 ⟶ L′ ⟶ L ⟶ L′′ ⟶ 0
↓  ↓  ↓
0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0
↓  ↓  ↓
0 ⟶ N′ ⟶ N ⟶ N′′ ⟶ 0
↓  ↓  ↓
0  0  0

Assume all the columns are exact and the middle row is exact. Prove that the first row is exact if and only if the third is.

Solution: The first row is exact if the third is owing to the Snake Lemma (5.12) applied to the bottom two rows. The converse is proved similarly. □

Exercise (5.16). — Consider this commutative diagram with exact rows:

M′ ⟶^{β} M ⟶^{γ} M′′

α′↓  ^{α}↓ _{α′′}↓

N′ ⟶^{β′} N ⟶^{γ′} N′′

Assume α′ and γ are surjective. Given n ∈ N and m′′ ∈ M′′ with α′′(m′′) = γ′(n), show that there is m ∈ M such that α(m) = n and γ(m) = m′′.

Solution: Since γ is surjective, there is m_{1} ∈ M with γ(m_{1}) = m′′. Then γ′(n−α(m_{1})) = 0 as α′′(m′′) = γ′(n) and as the right-hand square is commutative. So by exactness of the bottom row, there is n′ ∈ N′ with β′(n′) = n−α(m_{1}). Since α′ is surjective, there is m′ ∈ M′ with α′(m′) = n′. Set m := m_{1} + β(m′). Then γ(m) = m′′ as γβ = 0. Further, α(m) = α(m_{1})+β′(n′) = n as the left-hand square is commutative. Thus m works. □

Exercise (5.21). — Show that a free module R^{⊕Λ} is projective.

Solution: Given β: M → → N and α: R^{⊕Λ} → N, use the UMP of (4.10) to define γ: R^{⊕Λ} → M by sending the standard basis vector e_{λ} to any lift of α(e_{λ}), that is, any m_{λ} ∈ M with β(m_{λ}) = α(e_{λ}). (The Axiom of Choice permits a simultaneous choice of all m_{λ} if Λ is infinite.) Clearly α = βγ. Thus R^{⊕Λ} is projective. □

Exercise (5.24). — Let R be a ring, and 0 → L → R^{n} → M → 0 an exact sequence. Prove M is finitely presented if and only if L is finitely generated.

Solution: Assume M is finitely presented; say R^{l} → R^{m} → M → 0 is a finite presentation. Let L′ be the image of R^{l}. Then L′ ⊕ R^{n} ≃ L ⊕ R^{m} by Schanuel’s Lemma (5.23). Hence L is a quotient of R^{l} ⊕ R^{n}. Thus L is finitely generated.

Conversely, assume L is generated by ℓ elements. They yield a surjection R^{ℓ} → → L by (4.10)(1). It yields a sequence R^{ℓ} → R^{n} → M → 0. The latter is, plainly, exact. Thus M is finitely presented. □

Exercise (5.25). — Let R be a ring, X_{1}, X_{2}, … infinitely many variables. Set P := R[X_{1}, X_{2}, …] and M := P/⟨X_{1}, X_{2}, … ⟩. Is M finitely presented? Explain.

Solution: No, otherwise by (5.24), the ideal ⟨X_{1}, X_{2}, … ⟩ would be generated by some f_{1}, …, f_{n} ∈ P, so also by X_{1}, …, X_{m} for some m, but plainly it isn’t. □

Exercise (5.27). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence with M finitely generated and N finitely presented. Prove L is finitely generated.

Solution: Let R be the ground ring. Say M is generated by m elements. They yield a surjection μ: R^{m} → → M by (4.10)(1). As in (5.26), μ induces the following commutative diagram, with λ surjective:

0 ⟶ K ⟶ R^{m} ⟶ N ⟶ 0

λ↓ ^{μ}↓ ^{1_{N}} ↓

0 ⟶ L ⟶^{α} M ⟶^{β} N ⟶ 0

By (5.24), K is finitely generated. Thus L is too, as λ is surjective. □

6. Direct Limits

Exercise (6.3). — (1) Show that the condition (6.2)(1) is equivalent to the commutativity of the corresponding diagram:

Hom_{𝒞} (B, C) ⟶ Hom_{𝒞′} (F (B), F (C))
↓  ↓
Hom_{𝒞} (A, C) ⟶ Hom_{𝒞′} (F (A), F (C))

(2) Given γ: C → D, show (6.2)(1) yields the commutativity of this diagram:

Hom_{𝒞} (B, C) ⟶ Hom_{𝒞′} (F (B), F (C))
↓  ↓
Hom_{𝒞} (A, D) ⟶ Hom_{𝒞′} (F (A), F (D))

Solution: The left-hand vertical map is given by composition with α, and the right-hand vertical map is given by composition with F (α). So the composition of the top map and the right-hand map sends β to F (β)F (α), whereas the composition of the left-hand map with the bottom map sends β to F (βα). These two images are always equal if and only if the diagram commutes. Thus (1) holds if and only if the diagram commutes.

As to (2), the argument is similar. □

Exercise (6.5). — Let 𝒞 and 𝒞′ be categories, F: 𝒞 → 𝒞′ and F′: 𝒞′ → 𝒞 an adjoint pair. Let φ_{A,A′}: Hom_{𝒞′} (F A, A′) ⟶^{∼} Hom_{𝒞} (A, F′A′) denote the natural bijection, and set η_{A} := φ_{A,F A}(1_{F A}). Do the following:

(1) Prove η_{A} is natural in A; that is, given g: A → B, the induced square

A ⟶^{η_{A}} F′F A

g↓  ↓^{F′F g}

B ⟶^{η_{B}} F′F B

is commutative. We call the natural transformation A ↦ η_{A} the unit of (F, F′). (2) Given f′: F A → A′, prove φ_{A,A′}(f′) = F′f′ ∘ η_{A}.

(3) Prove the natural map η_{A}: A → F′F A is universal from A to F′; that is, given f: A → F′A′, there is a unique map f′: F A → A′ with F′f′ ∘ η_{A} = f.

(4) Conversely, instead of assuming (F, F′) is an adjoint pair, assume given a natural transformation η: 1_{𝒞} → F′F satisfying (1) and (3). Prove the equation in (2) defines a natural bijection making (F, F′) an adjoint pair, whose unit is η.

(5) Identify the units in the two examples in (6.4): the “free module” functor and the “polynomial ring” functor.

(Dually, we can define a counit ε: F F′ → 1_{𝒞′} , and prove similar statements.)

Solution: For (1), form this canonical diagram, with horizontal induced maps:

Hom_{𝒞′} (F A, F A) ⟶^{(F g)_{∗}} Hom_{𝒞′} (F A, F B) ⟵^{(F g)^{∗}} Hom_{𝒞′} (F B, F B)

φ_{A, F A}↓  ^{φ_{A, F B}} ↓  ^{φ_{B, F B}} ↓

Hom_{𝒞} (A, F′F A) ⟶^{(F′F g)_{∗}} Hom_{𝒞} (A, F′F B) ⟵^{g^{∗}} Hom_{𝒞} (B, F′F B)

It commutes since φ is natural. Follow 1_{F A} out of the upper left corner to find F′F g ∘ η_{A} = φ_{A, F B}(g) in Hom_{𝒞} (A, F′F B). Follow 1_{F B} out of the upper right corner to find φ_{A, F B}(g) = η_{B} ∘ g in Hom_{𝒞} (A, F′F B). Thus (F′F g) ∘ η_{A} = η_{B} ∘ g.

For (2), form this canonical commutative diagram:

Hom_{𝒞′} (F A, F A) ⟶^{f′_{∗}} Hom_{𝒞′} (F A, A′)

φ_{A, F A}↓  ^{φ_{A,A′}} ↓

Hom_{𝒞} (A, F′F A) ⟶^{(F′f′)_{∗}} Hom_{𝒞} (A, F′A′)

Follow 1_{F A} out of the upper left-hand corner to find φ_{A,A′}(f′) = F′f′ ∘ η_{A}.

For (3), given an f′, note that (2) yields φ_{A,A′}(f′) = f; whence, f′ = φ_{A,A′}^{−1}(f). Thus f′ is unique. Further, an f′ exists: just set f′ := φ_{A,A′}^{−1}(f).

For (4), set ψ_{A,A′}(f′) := F′f′ ∘ η_{A}. As η_{A} is universal, given f: A → F′A′, there is a unique f′: F A → A′ with F′f′ ∘ η_{A} = f. Thus ψ_{A,A′} is a bijection:

ψ_{A,A′}: Hom_{𝒞′} (F A, A′) ⟶^{∼} Hom_{𝒞} (A, F′A′).

Also, ψ_{A,A′} is natural in A, as η_{A} is natural in A and F′ is a functor. And, ψ_{A,A′} is natural in A′, as F′ is a functor. Clearly, ψ_{A, F A}(1_{F A}) = η_{A}. Thus (4) holds.

For (5), use the notation of (6.4). Clearly, if F is the “free module” functor, then η_{Λ} : Λ → R^{⊕Λ} carries an element of Λ to the corresponding standard basis vector. Further, if F is the “polynomial ring” functor and if A is the set of variables X_{1}, …, X_{n}, then η_{A}(X_{i}) is just X_{i} viewed in R[X_{1}, …, X_{n}]. □

Exercise (6.9). — Let α: L → M and β: L → N be two maps. Their pushout is defined as the universal example of an object P equipped with a pair of maps γ: M → P and δ: N → P such that γα = δβ. In terms of the definitions, express the pushout as a direct limit. Show directly that, in ((Sets)), the pushout is the disjoint union M ⊔ N modulo the smallest equivalence relation ∼ with m ∼ n if there is ℓ ∈ L with α(ℓ) = m and β(ℓ) = n. Show directly that, in ((R-mod)), the pushout is the direct sum M ⊕ N modulo the image of L under the map (α, −β).

Solution: Let Λ be the category with three objects λ, μ, and ν and two nonidentity maps λ → μ and λ → ν. Define a functor λ ↦ M_{λ} by M_{λ} := L, M_{μ} := M, M_{ν} := N, α_{μ}^{λ} := α, and α_{ν}^{λ} := β. Set Q := \underrightarrow{lim} M_{λ}. Then writing

\begin{matrix}N ⟵^{β} L ⟶^{α} M \\ ^{η_{ν}} ↓ & ^{η_{λ}}↓ & ^{η_{μ}}↓ \\ Q ⟵^{1_{R}} Q ⟶^{1_{R}} Q \end{matrix}  as \begin{matrix}L ⟶^{α} M \\ ^{β}↓ & ^{η_{μ}}↓ \\ N ⟶^{η_{ν}} Q \end{matrix}

we see that Q is equal to the pushout of α and β; here γ = η_{μ} and δ = η_{ν}.

In ((Sets)), take γ and δ to be the inclusions followed by the quotient map. Clearly γα = δβ. Further, given P and maps γ′: M → P and δ′: N → P, they define a unique map M ⊔ N → P, and it factors through the quotient if and only if γ′α = δ′β. Thus (M ⊔ N)/ ∼ is the pushout.

In ((R-mod)), take γ and δ to be the inclusions followed by the quotient map. Then for all ℓ ∈ L, clearly ι_{M}α(ℓ) − ι_{N}β(ℓ) = (α(ℓ), −β(ℓ)). So ι_{M}α(ℓ) − ι_{N}β(ℓ) is in Im(L); hence, ι_{M}α(ℓ) and ι_{N}β(ℓ) have the same image in the quotient. Thus γα = δβ. Given γ′: M → P and δ′: N → P , let φ: M ⊕ N → P be the induced map. Clearly φ factors through the quotient if and only if with γ′α = δ′β. Thus (M ⊕ N)/ Im(L) is the pushout. □

Exercise (6.16). — Let 𝒞 be a category, Σ and Λ small categories.

(1) Prove 𝒞^{Σ×Λ} = (𝒞^{Λ})^{Σ} with (σ, λ) ↦ M_{σ,λ} corresponding to σ ↦ (λ ↦ M_{σλ}).

(2) Assume 𝒞 has direct limits indexed by Σ and by Λ. Prove that 𝒞 has direct limits indexed by Σ × Λ and that \underrightarrow{lim}_{λ∈Λ} \underrightarrow{lim}_{σ∈Σ} = \underrightarrow{lim}_{(σ,λ)∈Σ×Λ}.

Solution: In Σ × Λ, a map (σ, λ) → (τ, μ) factors in two ways:

(σ, λ) → (τ, λ) → (τ, μ) and (σ, λ) → (σ, μ) → (τ, μ).

So, given a functor (σ, λ) ↦ M_{σ,λ}, there is a commutative diagram like (6.13.1). It shows that the map σ → τ in Σ induces a natural transformation from λ ↦ M_{σ,λ} to λ ↦ M_{τ,λ}. Thus the rule σ ↦ (λ ↦ M_{σλ}) is a functor from Σ to 𝒞^{Λ}.

A map from (σ, λ) ↦ M_{σ,λ} to a second functor (σ, λ) ↦ N_{σ,λ} is a collection of maps θ_{σ,λ}: M_{σ,λ} → N_{σ,λ} such that, for every map (σ, λ) → (τ, μ), the square

M_{σλ} ⟶ M_{τμ}
θ_{σ,λ}↓  ↓^{θ_{τ,μ}}
N_{σλ} ⟶ N_{τμ}

is commutative. Factoring (σ, λ) → (τ, μ) in two ways as above, we get a commutative cube. It shows that the θ_{σ,λ} define a map in (𝒞^{Λ})^{Σ}.

This passage from 𝒞^{Σ×Λ} to (𝒞^{Λ})^{Σ} is reversible. Thus (1) holds.

Assume 𝒞 has direct limits indexed by Σ and Λ. Then 𝒞^{Λ} has direct limits indexed by Σ by (6.13). So the functors \underrightarrow{lim}_{λ∈Λ} : 𝒞^{Λ} → 𝒞 and \underrightarrow{lim}_{σ∈Σ} : (𝒞^{Λ})^{Σ} → 𝒞^{Λ} exist, and they are the left adjoints of the diagonal functors 𝒞 → 𝒞^{Λ} and 𝒞^{Λ} → (𝒞^{Λ})^{Σ} by (6.6). Hence the composition \underrightarrow{lim}_{λ∈Λ} \underrightarrow{lim}_{σ∈Σ} is the left adjoint of the composition of the two diagonal functors. But the latter is just the diagonal 𝒞 → 𝒞^{Σ×Λ} owing to (1). So this diagonal has a left adjoint, which is necessarily \underrightarrow{lim}_{(σ,λ)∈Σ×Λ} owing to the uniqueness of adjoints. Thus (2) holds. □

Exercise (6.17). — Let λ ↦ M_{λ} and λ ↦ N_{λ} be two functors from a small category Λ to ((R-mod)), and {θ_{λ}: M_{λ} → N_{λ}} a natural transformation. Show

dlim Coker(θ_{λ}) = Coker(\underrightarrow{lim} M_{λ} → \underrightarrow{lim} N_{λ}).

Show that the analogous statement for kernels can be false by constructing a counterexample using the following commutative diagram with exact rows:

ℤ ⟶^{μ_{2}} ℤ ⟶ ℤ/⟨2⟩ ⟶ 0

↓^{μ_{2}}  ↓^{μ_{2}}  ↓^{μ_{2}}

ℤ ⟶^{μ_{2}} ℤ ⟶ ℤ/⟨2⟩ ⟶ 0

Solution: By (6.8), the cokernel is a direct limit, and by (6.14), direct limits commute; thus, the asserted equation holds.

To construct the desired counterexample using the given diagram, view its rows as expressing the cokernel ℤ/⟨2⟩ as a direct limit over the category Λ of (6.8). View the left two columns as expressing a natural transformation {θ_{λ}}, and view the third column as expressing the induced map between the two limits. The latter map is 0, so its kernel is ℤ/⟨2⟩. However, Ker(θ_{λ}) = 0 for λ ∈ Λ; so \underrightarrow{lim} Ker(θ_{λ}) = 0. □

7. Filtered direct limits

Exercise (7.2). — Let R be a ring, M a module, Λ a set, M_{λ} a submodule for each λ ∈ Λ. Assume ⋃ M_{λ} = M. Assume, given λ, μ ∈ Λ, there is ν ∈ Λ such that M_{λ}, M_{μ} ⊂ M_{ν}. Order Λ by inclusion: λ ≤ μ if M_{λ} ⊂ M_{μ}. Prove that M = \underrightarrow{lim} M_{λ}.

Solution: Let us prove that M has the UMP characterizing \underrightarrow{lim} M_{λ}. Given homomorphisms β_{λ}: M_{λ} → P with β_{λ} = β_{ν}|M_{λ} when λ ≤ ν, define β: M → P by β(m) := β_{λ}(m) if m ∈ M_{λ}. Such a λ exists as ⋃ M_{λ} = M. If also m ∈ M_{μ} and M_{λ}, M_{μ} ⊂ M_{ν}, then β_{λ}(m) = β_{ν}(m) = β_{μ}(m); so β is well defined. Clearly, β: M → P is the unique set map such that β|M_{λ} = β_{λ}. Further, given m, n ∈ M and x ∈ R, there is ν such that m, n ∈ M_{ν}. So β(m+n) = β_{ν}(m+n) = β(m)+β(n) and β(xm) = β_{ν}(xm) = xβ(m). Thus β is R-linear. Thus M = \underrightarrow{lim} M_{λ}. □

Exercise (7.3). — Show that every module M is the filtered direct limit of its finitely generated submodules.

Solution: Every element m ∈ M belongs to the submodule generated by m; hence, M is the union of all its finitely generated submodules. Any two finitely generated submodules are contained in a third, for example, their sum. So the assertion results from (7.2) with Λ the set of all finite subsets of M. □

Exercise (7.4). — Show that every direct sum of modules is the filtered direct limit of its finite direct subsums.

Solution: Consider an element of the direct sum. It has only finitely many nonzero components. So it lies in the corresponding finite direct subsum. Thus the union of the subsums is the whole direct sum. Now, given any two finite direct subsums, their sum is a third. Thus the finite subsets of indices form a directed partially ordered set Λ. So the assertion results from (7.2). □

Exercise (7.6). — Keep the setup of (7.5). For each n ∈ Λ, set N_{n} := ℤ/⟨n⟩; if n = ms, define α_{n}^{m}: N_{m} → N_{n} by α_{n}^{m}(x) := xs (mod n). Show \underrightarrow{lim} N_{n} = ℚ/ℤ.

Solution: For each n ∈ Λ, set Q_{n} := M_{n}/ℤ ⊂ ℚ/Z. If n = ms, then clearly Diagram (7.5.1) induces this one:

N_{m} ⟶^{α_{n}^{m}} N_{n}

γ_{m}↓^{≃}  ^{γ_{n}}↓^{≃}

Q_{m} ⟶^{η_{n}^{m}} Q_{n}

where η_{n}^{m} is the inclusion. Now, ⋃ Q_{n} = ℚ/ℤ and Q_{n}, Q_{n′} ⊂ Q_{nn′}. So (7.2) yields ℚ/ℤ = \underrightarrow{lim} M_{n}. Thus \underrightarrow{lim} N_{n} = ℚ/ℤ. □

Exercise (7.9). — Let R be a filtered direct limit of rings R_{λ}. Show R = 0 if and only if R_{λ} = 0 for some λ. Show R is a domain if R_{λ} is a domain for every λ.

Solution: If R_{λ} = 0, then 1 = 0 in R_{λ}; so 1 = 0 in R as α_{λ}: R_{λ} → R carries 1 to 1 and 0 to 0; hence, R = 0 by (1.1). Conversely, assume R = 0. Then 1 = 0 in R. So α_{λ}1 = 0 for any given λ. Hence, by (7.8)(3) with ℤ for R, there is α_{μ}^{λ} such that α_{μ}^{λ}1 = 0. But α_{μ}^{λ}1 = 1. Thus 1 = 0 in R_{μ}, and so R_{μ} = 0 by (1.1).

Suppose every R_{λ} is a domain. Given x, y ∈ R with xy = 0, we can lift x, y back to x_{λ}, y_{λ} ∈ R_{λ} for some λ by (7.8)(1) and (7.1)(1). Then x_{λ}y_{λ} maps to 0 ∈ R. Hence, by (7.8)(3), there is a transition map α_{μ}^{λ} with α_{μ}^{λ}(x_{λ}y_{λ}) = 0 in R_{μ}. However, α_{μ}^{λ}(x_{λ}y_{λ}) = α_{μ}^{λ}(x_{λ})α_{μ}^{λ}(y_{λ}), and R_{μ} is a domain. Hence either α_{μ}^{λ}(x_{λ}) = 0 or α_{μ}^{λ}(y_{λ}) = 0. Therefore, either x = 0 or y = 0. Thus R is a domain. □

Exercise (7.11). — Let M := \underrightarrow{lim} M_{λ} be a filtered direct limit of modules, and N ⊂ M a submodule. For each λ, let α_{λ}: M_{λ} → M be the insertion, and set N_{λ} := α_{λ}^{−1}N ⊂ M_{λ}. Prove that N = \underrightarrow{lim} N_{λ}.

Solution: The given functor λ ↦ M_{λ} induces a functor λ ↦ N_{λ}, and the insertions α_{λ}: M_{λ} → M induce maps β_{λ}: N_{λ} → N. So there is β: \underrightarrow{lim} N_{λ} → N with βα_{λ} = β_{λ}. By (7.10), \underrightarrow{lim} N_{λ} → M is injective; so β is too. Further, for any m ∈ M, there is an m_{λ} ∈ M_{λ} such that m = α_{λ}m_{λ}, and if m ∈ N, then m_{λ} ∈ N_{λ} since N_{λ} := α_{λ}^{−1}N. Thus β is surjective, so an isomorphism. □

Exercise (7.13). — Let Λ and Λ′ be small categories, C: Λ′ → Λ a functor. Assume Λ′ is filtered. Assume C is cofinal; that is,

(1) given λ ∈ Λ, there is a map λ → Cλ′ for some λ′ ∈ Λ′, and

(2) given ψ, φ: λ ⇉ Cλ′, there is χ: λ′ → λ′_{1} with (Cχ)ψ = (Cχ)φ. Let λ ↦ M_{λ} be a functor from Λ to 𝒞 whose direct limit exists. Show that

\underrightarrow{lim}_{λ′∈Λ′} M_{Cλ′} = \underrightarrow{lim}_{λ∈Λ} M_{λ};

more precisely, show that the right side has the UMP characterizing the left.

Solution: Let P be an object of 𝒞. For λ′ ∈ Λ′, take maps γ_{λ′}: M_{Cλ′} → P compatible with the transition maps M_{Cλ′} → M_{Cμ′}. Given λ ∈ Λ, choose a map λ → Cλ′, and define β_{λ}: M_{λ} → P to be the composition

β_{λ}: M_{λ} ⟶ M_{Cλ′} ⟶^{γ_{λ′}} P.

Let’s check that β_{λ} is independent of the choice of λ → Cλ′.

Given a second choice λ → Cλ′′, there are maps λ′′ → μ′ and λ′ → μ′ for some μ′ ∈ Λ′ since Λ′ is filtered. So there is a map μ′ → μ′_{1} such that the compositions λ → Cλ′ → Cμ′ → Cμ′_{1} and λ → Cλ′′ → Cμ′ → Cμ′_{1} are equal since C is cofinal. Therefore, λ → Cλ′′ gives rise to the same β_{λ}, as desired.

Clearly, the β_{λ} are compatible with the transition maps M_{κ} → M_{λ}. So the β_{λ} induce a map β: \underrightarrow{lim} M_{λ} → P with βα_{λ} = β_{λ} for every insertion α_{λ}: M_{λ} → \underrightarrow{lim} M_{λ}. In particular, this equation holds when λ = Cλ′ for any λ′ ∈ Λ′, as required. □

Exercise (7.14). — Show that every R-module M is the filtered direct limit over a directed set of finitely presented modules.

Solution: By (5.19), there is a presentation R^{⊕Φ_{1}} ⟶^{α} R^{⊕Φ_{2}} ⟶ M → 0. For i = 1, 2, let Λ_{i} be the set of finite subsets Ψ_{i} of Φ_{i}, and order Λ_{i} by inclusion. Clearly, an inclusion Ψ_{i} ↪ Φ_{i} yields an injection R^{⊕Ψ_{i}} ↪ R^{⊕Φ_{i}}, which is given by extending vectors by 0. Hence (7.2) yields \underrightarrow{lim} R^{⊕Ψ_{i}} = R^{⊕Φ_{i}}.

Let Λ ⊂ Λ_{1} × Λ_{2} be the set of pairs λ := (Ψ_{1}, Ψ_{2}) such that α induces a map α_{λ}: R^{⊕Ψ_{1}} → R^{⊕Ψ_{2}}. Order Λ by componentwise inclusion. Clearly, Λ is directed. For λ ∈ Λ, set M_{λ} := Coker(α_{λ}). Then M_{λ} is finitely presented.

For i = 1, 2, the projection C_{i}: Λ → Λ_{i} is surjective, so cofinal. Hence, (7.13) yields \underrightarrow{lim}_{λ∈Λ} R^{⊕C_{i}λ} = \underrightarrow{lim}_{Ψ_{i}∈Λ_{i}} R^{⊕Ψ_{i}}. Thus (6.17) yields \underrightarrow{lim} M_{λ} = M. □

8. Tensor Products

Exercise (8.6). — Let R be a domain. Set K := Frac(R). Given a nonzero submodule M ⊂ K, show that M ⊗_{R} K = K.

Solution: Define a map β: 𝔞×K → K by β(x, y) := xy. It is clearly R-bilinear. Given any R-bilinear map α: 𝔞×K → P, fix a nonzero z ∈ 𝔞, and define an R-linear map γ: K → P by γ(y) := α(z, y/z). Then α = γβ as

α(x, y) = α(xz, y/z) = α(z, xy/z) = γ(xy) = γβ(x, y).

Clearly, β is surjective. So γ is unique with this property. Thus the UMP implies that K = 𝔞 ⊗_{R} K. (Also, as γ is unique, γ is independent of the choice of z.)

Alternatively, form the linear map φ: 𝔞⊗K → K induced by the bilinear map β. Since β is surjective, so is φ. Now, given any w ∈ 𝔞 ⊗ K, say w = ∑ a_{i} ⊗ x_{i}/x with all x_{i} and x in R. Set a := ∑ a_{i}x_{i} ∈ 𝔞. Then w = a ⊗ (1/x). Hence, if φ(w) = 0, then a/x = 0; so a = 0 and so w = 0. Thus φ is injective, so bijective. □

Exercise (8.8). — Let R be a ring, R′ an R-algebra, M, N two R′-modules. Show there is a canonical R-linear map τ: M ⊗_{R} N → M ⊗_{R′} N.

Let K ⊂ M ⊗_{R} N denote the R-submodule generated by all the differences (x′m) ⊗ n − m ⊗ (x′n) for x′ ∈ R′ and m ∈ M and n ∈ N. Show K = Ker(τ). Show τ is surjective, and is an isomorphism if R′ is a quotient of R.

Solution: The canonical map β′: M × N → M ⊗_{R′} N is R′-bilinear, so Rbilinear. Hence, by (8.3), it factors: β′ = τβ where β: M × N → M ⊗_{R} N is the canonical map and τ is the desired map.

Set Q := (M ⊗_{R} N)/K. Then τ factors through a map τ′: Q → M ⊗_{R′} N since each generator (x′m) ⊗ n − m ⊗ (x′n) of K maps to 0 in M ⊗_{R′} N.

By (8.7), there is an R′-structure on M ⊗_{R} N with y′(m ⊗ n) = m ⊗ (y′n), and so by (8.5)(1), another one with y′(m ⊗ n) = (y′m) ⊗ n. Clearly, K is a submodule for each structure, so Q is too. But on Q the two structures coincide. Further, the canonical map M × N → Q is R′-bilinear. Hence the latter factors through M ⊗_{R′} N, furnishing an inverse to τ′. So τ′: Q ⟶^{∼} M ⊗_{R′} N. Hence Ker(τ) is equal to K, and τ is surjective.

Finally, suppose R′ is a quotient of R. Then every x′ ∈ R′ is the residue of some x ∈ R. So each (x′m) ⊗ n − m ⊗ (x′n) is equal to 0 in M ⊗_{R} N as x′m = xm and x′n = xn. Hence Ker(τ) vanishes. Thus τ is an isomorphism. □

Exercise (8.13). — Let R be a ring, 𝔞 and 𝔟 ideals, and M a module. (1) Use (8.11) to show that (R/𝔞) ⊗ M = M/𝔞M.

(2) Use (1) to show that (R/𝔞) ⊗ (R/𝔟) = R/(𝔞 + 𝔟).

Solution: To prove (1), view R/𝔞 as the cokernel of the inclusion 𝔞 → R. Then (8.11) implies that (R/𝔞)⊗M is the cokernel of 𝔞⊗M → R⊗M. Now, R⊗M = M and x ⊗ m = xm by (8.5)(2). Correspondingly, 𝔞 ⊗ M → M has 𝔞M as image. The assertion follows. (Caution: 𝔞 ⊗ M → M needn’t be injective; if it’s not, then 𝔞 ⊗ M ≠ 𝔞M. For example, take R := ℤ, take 𝔞 := ⟨2⟩, and take M := ℤ/⟨2⟩; then 𝔞 ⊗ M → M is just multiplication by 2 on ℤ/⟨2⟩, and so 𝔞M = 0.)

To prove (2), apply (1) with M := R/𝔟. Note 𝔞(R/𝔟) = (𝔞 + 𝔟)/𝔟. Hence

R/𝔞 ⊗ R/𝔟 = (R/𝔟)/((𝔞 + 𝔟)/𝔟).

The latter is equal to R/(𝔞 + 𝔟) by (4.8). □

Exercise (8.14). — Let k be a field, M and N nonzero vector spaces. Prove that M ⊗ N ≠ 0.

Solution: Since k is a field, M and N are free; say M = k^{⊕Φ} and N = k^{⊕Ψ}. Then (8.11) yields M ⊗ N = k^{⊕(Φ×Ψ)} as k ⊗ k = k by (8.5)(2). Thus M ⊗ N ≠ 0. □

Exercise (8.16). — Let F: ((R-mod)) → ((R-mod)) be a linear functor. Show that F always preserves finite direct sums. Show that θ(M): M ⊗ F (R) → F (M) is surjective if F preserves surjections and M is finitely generated, and that θ(M) is an isomorphism if F preserves cokernels and M is finitely presented.

Solution: The first assertion follows immediately from the characterization of finite direct sum in terms of maps (4.15), since F preserves the stated relations.

The second assertion follows from the first via the second part of the proof of Watt’s Theorem (8.15), but with Σ and Λ finite. □

Exercise (8.21). — Let X be a variable, ω a complex cubic root of 1, and \sqrt[3]{2} the real cube root of 2. Set k := ℚ(ω) and K := k[\sqrt[3]{2}]. Show K = k[X]/⟨X^{3} − 2⟩ and then K ⊗_{k} K = K × K × K.

Solution: Note ω is a root of X^{2} + X + 1, which is irreducible over ℚ; hence, [k : ℚ] = 2. But the three roots of X^{3} − 2 are \sqrt[3]{2} and ω\sqrt[3]{2} and ω^{2} \sqrt[3]{2}. Therefore, X^{3} −2 has no root in k. So X^{3} −2 is irreducible over k. Thus k[X]/⟨X^{3} −2⟩ ⟶^{∼} K.

Note K[X] = K ⊗_{k} k[X] as k-algebras by (8.20). So (8.5)(2) and (8.10) and (8.13)(1) yield

k[X]/⟨X^{3} − 2⟩ ⊗_{k} K = k[X]/⟨X^{3} − 2⟩ ⊗_{k[X]} (k[X] ⊗_{k} K)
= k[X]/⟨X^{3} − 2⟩ ⊗_{k[X]} K[X] = K[X]/⟨X^{3} − 2⟩.

However, X^{3} − 2 factors in K as follows:

X^{3} − 2 = (X − \sqrt[3]{2})(X − ω\sqrt[3]{2})(X − ω^{2} \sqrt[3]{2}).

So the Chinese Remainder Theorem, (1.13), yields

K[X]/⟨X^{3} − 2⟩ = K × K × K,

because K[X]/⟨X − ω^{i} \sqrt[3]{2}⟩ ⟶^{∼} K for any i by (1.7). □

9. Flatness

Exercise (9.7). — Let R be a ring, R′ a flat algebra, and P a flat R′-module. Show that P is a flat R-module.

Solution: Cancellation (8.10) yields • ⊗_{R} P = (• ⊗_{R} R′) ⊗_{R′} P. But • ⊗_{R} R′ and • ⊗_{R′} P are exact. Hence, •_{R}P is too. Thus P is R-flat. □

Exercise (9.8). — Let R be a ring, M a flat module, and R′ an algebra. Show that M ⊗_{R} R′ is a flat R′-module.

Solution: Cancellation (8.10) yields (M ⊗_{R} R′) ⊗_{R′} • = M ⊗_{R} •. Therefore, (M ⊗_{R} R′) ⊗_{R′} • is exact. Thus M ⊗_{R} R′ is R′-flat. □

Exercise (9.9). — Let R be a ring, 𝔞 an ideal. Assume that R/𝔞 is R-flat. Show that 𝔞 = 𝔞^{2}.

Solution: Since R/𝔞 is flat, tensoring it with the inclusion 𝔞 ↪ R yields an injection 𝔞 ⊗_{R} (R/𝔞) ↪ R ⊗_{R} (R/𝔞). But the image vanishes: a ⊗ r = 1 ⊗ ar = 0. Further, 𝔞 ⊗_{R} (R/𝔞) = 𝔞/𝔞^{2} by (8.13). Hence 𝔞/𝔞^{2} = 0. Thus 𝔞 = 𝔞^{2}. □

Exercise (9.10). — Let R be a ring, R′ a flat algebra with structure map φ. Then R′ is said to be faithfully flat if for every R-module M, the map M → M ⊗ R′ given by x ↦ x ⊗ 1 is injective. Show that the following conditions are equivalent:

(1) R′ is faithfully flat.

(2) Every ideal 𝔞 of R is the contraction of its extension, or 𝔞 = φ^{−1}(𝔞R′).

(3) Every prime 𝔭 of R is the contraction of some prime 𝔮 of R′, or 𝔭 = φ^{−1}𝔮 .

(4) Every maximal ideal 𝔪 of R extends to a proper ideal, or 𝔪R′ ≠ R′.

(5) Every nonzero R-module M remains nonzero when tensored with R′, or M ⊗_{R} R′ ≠ 0.

Solution: Assume (1). Then R/𝔞 → (R/𝔞) ⊗ R′ is injective. Hence (8.13)(1) implies R/𝔞 → R′/𝔞R′ is injective. Thus (2) holds.

Assume (2). Let 𝔭 be a prime ideal of R. Then (2) yields 𝔭 = φ^{−1}(𝔭R′). Thus (3.10) yields (3).

Assume (3). Let 𝔪 be a maximal ideal of R. By (3), there is a prime ideal 𝔫 of R′ with φ^{−1}(𝔫) = 𝔪. So 𝔪R′ ⊂ 𝔫. Thus (4) holds.

Assume (4). Take a nonzero m ∈ M; set M′ := Rm. As R′ is flat, the inclusion M′ ↪ M yields an injection M′⊗R′ ↪ M ⊗R′. So it suffices to show M′⊗_{R}R′ ≠ 0.

Note M′ = R/𝔞 for some 𝔞 by (4.7). So M′ ⊗_{R} R′ = R′/𝔞R′ by (8.13)(1). Take a maximal ideal 𝔪 ⊃ 𝔞. Then 𝔞R′ ⊂ 𝔪R′. But 𝔪R′ ⫋ R′ by (4). Hence M′ ⊗_{R} R′ ≠ 0. Thus (5) holds.

Assume (5). Set K := Ker(M → M ⊗_{R}R′). Since R′ is flat, the induced sequence

0 → K ⊗_{R} R′ → M ⊗_{R} R′ ⟶^{α} M ⊗_{R} R′ ⊗_{R} R′

is exact. But α has a retraction, namely, m ⊗ x ⊗ y ↦ m ⊗ xy; hence, α is injective. Thus K ⊗_{R} R′ = 0. Hence (5) yields K = 0. Thus (1) holds. □

Exercise (9.11). — Let A and B be local rings, 𝔪 and 𝔫 their maximal ideals. Let φ: A → B be a local homomorphism; that is, φ(𝔪) ⊂ 𝔫. Assume φ is flat. Show that φ is faithfully flat.

Solution: The assertion results from (9.10), as (4) holds since φ(𝔪) ⊂ 𝔫. □

Exercise (9.15). — Let R be a ring, R′ an algebra, M and N modules. Show that there is a canonical map

σ: Hom_{R}(M, N) ⊗_{R} R′ → Hom_{R′}(M ⊗_{R} R′, N ⊗_{R} R′).

Assume R′ is flat over R. Show that if M is finitely generated, then σ is injective, and that if M is finitely presented, then σ is an isomorphism.

Solution: Simply put R′ := R and P := R′ in (9.14), put P := N ⊗_{R} R′ in the second equation in (8.10), and combine the two results. □

Exercise (9.19) (Equational Criterion for Flatness). — Prove that Condition (9.18)(4) can be reformulated as follows: For every relation ∑_{i} x_{i}y_{i} = 0 with x_{i} ∈ R and y_{i} ∈ M, there are x_{ij} ∈ R and y′_{j} ∈ M such that

∑_{j} x_{ij}y′_{j} = y_{i} for all i and ∑_{i} x_{ij}x_{i} = 0 for all j.  (9.19.1)

Solution: Assume (9.18)(4) holds. Let e_{1}, …, e_{m} be the standard basis of R^{m}. Given a relation ∑_{1}^{m} x_{i}y_{i} = 0, define α: R^{m} → M by α(e_{i}) := y_{i} for each i. Set k := ∑ x_{i}e_{i}. Then α(k) = 0. So (9.18)(4) yields a factorization α: R^{m}⟶^{φ} R^{n} ⟶^{β}M with φ(k) = 0. Let e′_{1}, …, e′_{n} be the standard basis of R^{n}, and set y′_{j} := β(e′_{j}) for each j. Let (x_{ij}) be the n × m matrix of φ; that is, φ(e_{i}) = ∑ x_{ji}e′_{j}. Then y_{i} = ∑ x_{ji}y′_{j}. Now, φ(k) = 0; hence, ∑_{i,j} x_{ji}x_{i}e′_{j} = 0. Thus (9.19.1) holds.

Conversely, given α: R^{m} → M and k ∈ Ker(α), write k = ∑ x_{i}e_{i}. Assume (9.19.1). Let φ: R^{m} → R^{n} be the map with matrix (x_{ij}); that is, φ(e_{i}) = ∑ x_{ji}e′_{j}. Then φ(k) = ∑ x_{i}x_{ji}e′_{j} = 0. Define β: R^{n} → M by β(e′_{j}) := y′_{j}. Then βφ(e_{i}) = y_{i}; hence, βφ = α. Thus (9.18)(4) holds. □

Exercise (9.22). — Let R be a domain, M a module. Prove that, if M is flat, then M is torsion free; that is, μ_{x}: M → M is injective for all nonzero x ∈ R. Prove that, conversely, if R is a PID and M is torsion free, then M is flat.

Solution: Since R is a domain, μ_{x}: R → R is injective. So if M is flat, then μ_{x} ⊗ M: R ⊗ M → R ⊗ M is injective too. But R ⊗ M = M by (8.5).

Conversely, assume R is a PID and M is torsion free. Let 𝔞 be a nonzero ideal, say 𝔞 = ⟨x⟩. Define α: R → 𝔞 by α(y) := xy. Then α is injective as R is a domain and x ≠ 0. Further, α is surjective as 𝔞 = ⟨x⟩. So α is bijective.

Consider the composition

β: M = R ⊗ M ⟶^{α⊗M} 𝔞 ⊗ M → M.

Clearly, β = μ_{x}. So β is injective since M is torsion free. Hence 𝔞 ⊗ M → M is injective too. So M is flat by the Ideal Criterion (9.20). □

10. Cayley–Hamilton Theorem

Exercise (10.6). — Let R be a ring, 𝔞 an ideal. Assume 𝔞 is finitely generated and idempotent (or 𝔞 = 𝔞^{2}). Prove there is a unique idempotent e with ⟨e⟩ = 𝔞.

Solution: By (10.3) with 𝔞 for M, there is e ∈ 𝔞 such that (1 − e)𝔞 = 0. So for all x ∈ 𝔞, we have (1 − e)x = 0, or x = ex. Thus 𝔞 = ⟨e⟩ and e = e^{2}.

Finally, e is unique by (1.16)(2). □

Exercise (10.8). — Prove the following conditions on a ring R are equivalent:

(1) R is absolutely flat; that is, every module is flat.

(2) Every finitely generated ideal is a direct summand of R.

(3) Every finitely generated ideal is idempotent.

(4) Every principal ideal is idempotent.

Solution: Assume (1). Let 𝔞 be a finitely generated ideal. Then R/𝔞 is flat by hypotheses. So 𝔞 is a direct summand of R by (10.7). Thus (2) holds.

Conditions (2) and (3) are equivalent by (10.7).

Trivially, if (3) holds, then (4) does. Conversely, assume (4). Given a finitely generated ideal 𝔞, say 𝔞 = ⟨x_{1}, …, x_{n}⟩. Then each ⟨x_{i}⟩ is idempotent by hypothesis. So ⟨x_{i}⟩ = ⟨f_{i}⟩ for some idempotent f_{i} by (1.16)(2). Then 𝔞 = ⟨f_{1}, …, f_{n}⟩. Hence 𝔞 is idepotent by (1.16)(5), (1). Thus (3) holds.

Assume (2). Let M be a module, and 𝔞 a finitely generated ideal. Then 𝔞 is a direct summand of R by hypothesis. So R/𝔞 is flat by (9.5). Hence 𝔞 ⊗ M ⟶^{∼} 𝔞M by (9.12)(1). So M is flat by (9.20). Thus (1) holds. □

Exercise (10.9). — Let R be a ring.

(1) Assume R is Boolean. Prove R is absolutely flat.

(2) Assume R is absolutely flat. Prove any quotient ring R′ is absolutely flat.

(3) Assume R is absolutely flat. Prove every nonunit x is a zerodivisor.

(4) Assume R is absolutely flat and local. Prove R is a field.

Solution: In (1), as R is Boolean, every element is idempotent. Hence every principal ideal is idempotent by (1.15)(1). Thus (10.8) yields (1).

For (2), let 𝔟 ⊂ R′ be principal, say 𝔟 = ⟨\overline{x}⟩. Let x ∈ R lift \overline{x}. Then ⟨x⟩ is idempotent by (10.8). Hence 𝔟 is also idempotent. Thus (10.8) yields (2).

For (3) and (4), take a nonunit x. Then ⟨x⟩ is idempotent by (10.8). So x = ax^{2} for some a. Then x(ax−1) = 0. But x is a nonunit. So ax−1 ≠ 0. Thus (3) holds.

Suppose R is local, say with maximal ideal 𝔪. Since x is a nonunit, x ∈ 𝔪. So ax ∈ 𝔪. So ax − 1 ∉ 𝔪. So ax − 1 is a unit. But x(ax − 1) = 0. So x = 0. Thus 0 is the only nonunit. Thus (4) holds. □

Exercise (10.12). — Let R be a ring, 𝔪 ⊂ rad(R) an ideal. Let α, β: M → N be two maps of finitely generated modules. Assume α is surjective and β(M) ⊂ 𝔪N. Set γ := α + β. Show that γ is an isomorphism.

Solution: As α is surjective, given n ∈ N, there is m ∈ M with α(m) = n. So

n = α(m) + β(m) − β(m) ∈ γ(M) + 𝔪N.

Hence γ(M) = N by (10.11). So γ is an isomorphism by (10.4). □

Exercise (10.13). — Let A be a local ring, 𝔪 the maximal ideal, M a finitely generated A-module, and m_{1}, …, m_{n} ∈ M. Set k := A/𝔪 and M′ := M/𝔪M, and write m′_{i} for the image of m_{i} in M′. Prove that m′_{1}, …, m′_{n} ∈ M′ form a basis of the k-vector space M′ if and only if m_{1}, …, m_{n} form a minimal generating set of M (that is, no proper subset generates M), and prove that every minimal generating set of M has the same number of elements.

Solution: By (10.11), reduction mod 𝔪 gives a bijective correspondence between generating sets of M as an A-module, and generating sets of M′ as an A-module, or equivalently by (4.5), as an k-vector space. This correspondence preserves inclusion. Hence, a minimal generating set of M corresponds to a minimal generating set of M′, that is, to a basis. But any two bases have the same number of elements. □

Exercise (10.14). — Let A be a local ring, k its residue field, M and N finitely generated modules. (1) Show that M = 0 if and only if M ⊗_{A} k = 0. (2) Show that M ⊗_{A} N ≠ 0 if M ≠ 0 and N ≠ 0.

Solution: Let 𝔪 be the maximal ideal. Then M ⊗ k = M/𝔪M by (8.13)(1). So (1) is nothing but a form of Nakayama’s lemma (10.10).

In (2), M ⊗ k ≠ 0 and N ⊗ k ≠ 0 by (1). So (M ⊗ k) ⊗ (N ⊗ k) ≠ 0 by (8.14) and (8.8). But (M ⊗ k) ⊗ (N ⊗ k) = (M ⊗ N) ⊗ (k ⊗ k) by the associative and commutative laws. Finally, k ⊗ k = k by (8.13)(1). □

Exercise (10.17). — Let G be a finite group acting on a domain R, and R′ the ring of invariants. Show every x ∈ R is integral over R′, in fact, over the subring R′′ generated by the elementary symmetric functions in the conjugates gx for g ∈ G.

Solution: Given an x ∈ R, form F (X) := ∏_{g∈G}(X −gx). Then the coefficients of F (X) are the elementary symmetric functions in the conjugates gx for g ∈ G; hence, they are invariant under the action of G. So F (x) = 0 is a relation of integral dependence for x over R′, in fact, over its subring R′′. □

Exercise (10.19). — Let k be a field, P := k[X] the polynomial ring in one variable, f ∈ P. Set R := k[X^{2}] ⊂ P. Using the free basis 1, X of P over R, find an explicit equation of integral dependence of degree 2 on R for f.

Solution: Write f = f_{e} + f_{o}, where f_{e} and f_{o} are the polynomials formed by the terms of f of even and odd degrees. Say f_{o} = gX. Then the matrix of μ_{f} is ( \begin{matrix}f_{e} & gX^{2} \\ g & f_{e} \end{matrix} ). Its characteristic polynomial is T^{2} − 2f_{e}T + f_{e}^{2} − f_{o}^{2}. So the Cayley– Hamilton Theorem (10.1) yields f^{2} − 2f_{e}f + f_{e}^{2} − f_{o}^{2} = 0. □

Exercise (10.24). — Let R_{1}, …, R_{n} be R-algebras that are integral over R. Show that their product ∏ R_{i} is a integral over R.

Solution: Let y = (y_{1}, …, y_{n}) ∈ ∏_{i=1}^{n} R_{i}. Since R_{i}/R is integral, R[y_{i}] is a module-finite R-subalgebra of R_{i}. Hence ∏_{i=1}^{n} R[y_{i}] is a module-finite R-subalgebra of ∏_{i=1}^{n} R_{i} by (4.14) and induction on n. Now, y ∈ ∏_{i=1}^{n} R[y_{i}]. Therefore, y is integral over R. Thus ∏_{i=1}^{n} R_{i} is integral over R. □

Exercise (10.26). — For 1 ≤ i ≤ r, let R_{i} be a ring, R′_{i} an extension of R_{i}, and x_{i} ∈ R′_{i}. Set R := ∏ R_{i}, set R′ := ∏ R′_{i}, and set x := (x_{1}, …, x_{r}). Prove

(1) x is integral over R if and only if x_{i} is integral over R_{i} for each i;

(2) R is integrally closed in R′ if and only if each R_{i} is integrally closed in R′_{i}.

Solution: Assume x is integral over R. Say x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0 with a_{j} ∈ R. Say a_{j} =: (a_{1j}, …, a_{rj}). Fix i. Then x_{i}^{n} + a_{i1}x^{n−1} + ⋯ + a_{in} = 0. So x_{i} is integral over R_{i}.

Conversely, assume each x_{i} is integral over R_{i}. Say x_{i}^{n_{i}} +a_{i1}x_{i}^{n_{i}−1}+⋯+a_{in_{i}} = 0. Set n := max n_{i}, set a_{ij} := 0 for j > n_{i}, and set a_{j} := (a_{1j}, …, a_{rj}) ∈ R for each j. Then x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0. Thus x is integral over R. Thus (1) holds.

Assertion (2) is an immediate consequence of (1). □

Exercise (10.30). — Let k be a field, X and Y variables. Set

R := k[X, Y]/⟨Y^{2} − X^{2} − X^{3}⟩,

and let x, y ∈ R be the residues of X, Y. Prove that R is a domain, but not a field. Set t := y/x ∈ Frac(R). Prove that k[t] is the integral closure of R in Frac(R).

Solution: As k[X, Y] is a UFD and Y^{2}−X^{2}−X^{3} is irreducible, ⟨Y^{2}−X^{2}−X^{3}⟩ is prime by (2.6); however, it is not maximal by (2.27). Hence R is a domain by (2.9), but not a field by (2.17).

Note y^{2} − x^{2} − x^{3} = 0. Hence x = t^{2} − 1 and y = t^{3} − t. So k[t] ⊃ k[x, y] = R. Further, t is integral over R; so k[t] is integral over R by (2)⇒(1) of (10.23).

Finally, k[t] has Frac(R) as fraction field. Further, Frac(R) ≠ R, so x and y cannot be algebraic over k; hence, t must be transcendental. So k[t] is normal by (10.29)(1). Thus k[t] is the integral closure of R in Frac(R). □

11. Localization of Rings

Exercise (11.2). — Let R be a ring, S a multiplicative subset. Prove S^{−1}R = 0 if and only if S contains a nilpotent element.

Solution: By (1.1), S^{−1}R = 0 if and only if 1/1 = 0/1. But by construction, 1/1 = 0/1 if and only if 0 ∈ S. Finally, since S is multiplicative, 0 ∈ S if and only if S contains a nilpotent element. □

Exercise (11.3). — Let R be a ring, S a multiplicative subset, \overline{S} its saturation. Set T := (S^{−1}R)^{×}. Show T = \{ x/s | x ∈ \overline{S} and s ∈ S \}. Show φ_{S}^{−1}T = \overline{S}.

Solution: First, given x ∈ \overline{S} and s ∈ S, take y ∈ R such that xy ∈ S. Then x/s · sy/xy = 1 in S^{−1}R. Thus x/s ∈ T. Conversely, say x/s · y/t = 1 in S^{−1}R with x, y ∈ R and s, t ∈ S. Then there’s u ∈ S with xyu = stu in R. But stu ∈ S. Thus x ∈ \overline{S}. Thus the first assertion holds.

Set U := φ_{S}^{−1}T. Then U is saturated multiplicative by (3.12). Further, U ⊃ S by (11.1). Thus (1)(c) of (3.14) yields U ⊃ \overline{S}. Conversely, take x ∈ U. Then x/1 ∈ T. So the first assertion yields x/1 = y/s with y ∈ \overline{S} and s ∈ S. So there’s t ∈ S with xst = yt in R. But \overline{S} ⊃ S by (1)(a) of (3.14), and \overline{S} is multiplicative by (1)(b); so yt ∈ \overline{S}. But \overline{S} is saturated by (1)(b). Thus x ∈ \overline{S}. Thus U = \overline{S}. □

Exercise (11.5). — Find all intermediate rings ℤ ⊂ R ⊂ ℚ, and describe each R as a localization of ℤ. As a starter, prove ℤ[2/3] = S^{−1}ℤ where S = {3^{i} | i ≥ 0}.

Solution: Clearly ℤ[2/3] ⊂ ℤ[1/3] as 2/3 = 2·(1/3). But the opposite inclusion holds as 1/3 = 1 − (2/3). Obviously, S^{−1}ℤ = ℤ[1/3].

Let P ⊂ ℤ be the set of all prime numbers that appear as factors of the denominators of elements of R in lowest terms; recall that x = r/s ∈ ℚ is in lowest terms if r and s have no common prime divisor. Let S be the multiplicative subset generated by P, that is, the smallest multiplicative subset containing P. Clearly, S is equal to the set of all products of elements of P.

First note that, if p ∈ P, then 1/p ∈ R. Indeed, take an element x = r/ps ∈ R in lowest terms. Then sx = r/p ∈ R. Also the Euclidean algorithm yields m, n ∈ ℤ such that mp + nr = 1. Then 1/p = m + nsx ∈ R, as desired. Hence S^{−1}ℤ ⊂ R. But the opposite inclusion holds because, by the very definition of S, every element of R is of the form r/s for some s ∈ S. Thus S^{−1}ℤ = R. □

Exercise (11.8). — Let R′ and R′′ be rings. Consider R := R′ × R′′ and set S := { (1, 1), (1, 0) }. Prove R′ = S^{−1}R.

Solution: Let’s show that the projection map π: R′ × R′′ → R′ has the UMP of (11.6). First, note that πS = {1} ⊂ R^{′×}. Let ψ: R′ × R′′ → B be a ring map such that ψ(1, 0) ∈ B^{×}. Then in B,

ψ(1, 0) · ψ(0, x) = ψ((1, 0) · (0, x)) = ψ(0, 0) = 0 in B.

Hence ψ(0, x) = 0 for all x ∈ R′′. So ψ factors uniquely through π by (1.5). □

Exercise (11.9). — Take R and S as in (11.8). On R × S, impose this relation:

(x, s) ∼ (y, t) if xt = ys.

Prove that it is not an equivalence relation.

Solution: Observe that, for any z ∈ R′′, we have

((1, z), (1, 1)) ∼ ((1, 0), (1, 0)).

However, if z ≠ 0, then

((1, z), (1, 1)) ≁ ((1, 0), (1, 1)).

Thus although ∼ is reflexive and symmetric, it is not transitive if R′′ ≠ 0. □

Exercise (11.15). — Let R be a ring, S a multiplicative subset, 𝔞 and 𝔟 ideals. Show (1) if 𝔞 ⊂ 𝔟, then 𝔞^{S} ⊂ 𝔟^{S}; (2) (𝔞^{S})^{S} = 𝔞^{S}; and (3) (𝔞^{S}𝔟^{S})^{S} = (𝔞𝔟)^{S}.

Solution: For (1), take x ∈ 𝔞^{S}. Then there is s ∈ S with sx ∈ 𝔞. If 𝔞 ⊂ 𝔟, then sx ∈ 𝔟, and so x ∈ 𝔟^{S}. Thus (1) holds.

To show (2), proceed by double inclusion. First, note 𝔞^{S} ⊃ 𝔞 by (11.14)(2). So (𝔞^{S})^{S} ⊃ 𝔞^{S} again by (11.14)(2). Conversely, given x ∈ (𝔞^{S})^{S}, there is s ∈ S with sx ∈ 𝔞^{S}. So there is t ∈ S with tsx ∈ a. But ts ∈ S. So x ∈ 𝔞^{S}. Thus (2) holds.

To show (3), proceed by double inclusion. First, note 𝔞 ⊂ 𝔞^{S} and 𝔟 ⊂ 𝔟^{S} by (11.14)(2). So 𝔞𝔟 ⊂ 𝔞^{S}𝔟^{S}. Thus (11.14)(2) yields (𝔞𝔟)^{S} ⊂ (𝔞^{S}𝔟^{S})^{S}.

Conversely, given x ∈ 𝔞^{S}𝔟^{S}, say x := ∑ y_{i}z_{i} with y_{i} ∈ 𝔞^{S} and z_{i} ∈ 𝔟^{S}. Then there are s_{i}, t_{i} ∈ S such that s_{i}y_{i} ∈ 𝔞 and t_{i}z_{i} ∈ 𝔟. Set u := ∏ s_{i}t_{i}. Then u ∈ S and ux ∈ 𝔞𝔟. So x ∈ (𝔞𝔟)^{S}. Thus 𝔞^{S}𝔟^{S} ⊂ (𝔞𝔟)^{S}. So (𝔞^{S}𝔟^{S})^{S} ⊂ ((𝔞𝔟)^{S})^{S} by (1). But ((𝔞𝔟)^{S})^{S} = (𝔞𝔟)^{S} by (2). Thus (3) holds. □

Exercise (11.16). — Let R be a ring, S a multiplicative subset. Prove that

nil(R)(S^{−1}R) = nil(S^{−1}R).

Solution: Proceed by double inclusion. Given an element of nil(R)(S^{−1}R), put it in the form x/s with x ∈ nil(R) and s ∈ S using (11.12)(1). Then x^{n} = 0 for some n ≥ 1. So (x/s)^{n} = 0. So x/s ∈ nil(S^{−1}R). Thus nil(R)(S^{−1}R) ⊂ nil(S^{−1}R).

Conversely, take x/s ∈ nil(S^{−1}R). Then (x/s)^{m} = 0 with m ≥ 1. So there’s t ∈ S with tx^{m} = 0 by (11.14)(1). Then (tx)^{m} = 0. So tx ∈ nil(R). But tx/ts = x/s. So x/s ∈ nil(R)(S^{−1}R) by (11.12)(1). Thus nil(R)(S^{−1}R) ⊃ nil(S^{−1}R). □

Exercise (11.23). — Let R′/R be an integral extension of rings, and S a multiplicative subset of R. Show that S^{−1}R′ is integral over S^{−1}R.

Solution: Given x/s ∈ S^{−1}R′, let x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{0} = 0 be an equation of integral dependence of x on R. Then

(x/s)^{n} + (a_{n−1}/1)(1/s)(x/s)^{n−1} + ⋯ + a_{0}(1/s)^{n} = 0

is an equation of integral dependence of x/s on S^{−1}R, as required. □

Exercise (11.24). — Let R be a domain, K its fraction field, L a finite extension field, and \overline{R} the integral closure of R in L. Show that L is the fraction field of \overline{R}. Show that, in fact, every element of L can be expressed as a fraction b/a where b is in \overline{R} and a is in R.

Solution: Let x ∈ L. Then x is algebraic (integral) over K, say

x^{n} + y_{1}x^{n−1} + ⋯ + y_{n} = 0

with y_{i} ∈ K. Write y_{i} = a_{i}/a with a_{1}, …, a_{n}, a ∈ R. Then

(ax)^{n} + (aa_{1})(ax)^{n−1} + ⋯ + a^{n}a_{0} = 0.

Set b := ax. Then b ∈ \overline{R} and x = b/a. □

Exercise (11.25). — Let R ⊂ R′ be domains, K and L their fraction fields. Assume that R′ is a finitely generated R-algebra, and that L is a finite dimensional K-vector space. Find an f ∈ R such that R′_{f} is module finite over R_{f}.

Solution: Let x_{1}, …, x_{n} generate R′ over R. Using (11.24), write x_{i} = b_{i}/a_{i} with b_{i} integral over R and a_{i} in R. Set f := ∏ a_{i}. The x_{i} generate R′_{f} as an R_{f}-algebra; so the b_{i} do too. Thus R′_{f} is s module finite over R_{f} by (10.23). □

Exercise (11.28). — Let R be a ring, S and T multiplicative subsets. (1) Set T′ := φ_{S}(T) and assume S ⊂ T. Prove

T^{−1}R = T^{′−1}(S^{−1}R) = T^{−1}(S^{−1}R).

(2) Set U := {st ∈ R | s ∈ S and t ∈ T }. Prove

T^{−1}(S^{−1}R) = S^{−1}(T^{−1}R) = U^{−1}R.

(3) Let S′ := {t′ ∈ R | t′t ∈ S for some t ∈ R}. Prove S^{′−1}R = S^{−1}R.

Solution: A proof similar to that of (11.26) shows T^{−1}R = T^{′−1}(S^{−1}R). By (11.22), T^{′−1}(S^{−1}R) = T^{−1}(S^{−1}R). Thus (1) holds.

As 1 ∈ T, obviously S ⊂ U. So (1) yields U^{−1}R = U^{−1}(S^{−1}R). Now, clearly U^{−1}(S^{−1}R) = T^{−1}(S^{−1}R). Similarly, U^{−1}R = S^{−1}(T^{−1}R). Thus (2) holds.

Finally, in any ring, a product is a unit if and only if each factor is a unit. So a homomorphism φ: R → R′ carries S′ into R′^{×} if and only if φ carries S into R′^{×}. Thus S^{′−1}R and S^{−1}R are universal examples of R-algebras that satisfy equivalent conditions. Thus (3) holds. □

Exercise (11.31) (Localization and normalization commute). — Given a domain R and a multiplicative subset S with 0 ∉ S. Show that the localization of the normalization S^{−1}\overline{R} is equal to the normalization of the localization \overline{S^{−1}R}.

Solution: Since 0 ∉ S, clearly Frac(S^{−1}R) = Frac(R) owing to (11.4). Now, S^{−1}\overline{R} is integral over S^{−1}R by (11.23). Thus S^{−1}\overline{R} ⊂ \overline{S^{−1}R}.

Conversely, given x ∈ \overline{S^{−1}R}, consider an equation of integral dependence:

x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0.

Say a_{i} = b_{i}/s_{i} with b_{i} ∈ R and s_{i} ∈ S; set s := ∏ s_{i}. Multiplying by s^{n} yields

(sx)^{n} + sa_{1}(sx)^{n−1} + ⋯ + s^{n}a_{n} = 0.

Hence sx ∈ \overline{R}. So x ∈ S^{−1}\overline{R}. Thus S^{−1}\overline{R} ⊃ \overline{S^{−1}R}, as desired. □

12. Localization of Modules

Exercise (12.4). — Let R be a ring, S a multiplicative subset, and M a module. Show that M = S^{−1}M if and only if M is an S^{−1}R-module.

Solution: If M = S^{−1}M, then obviously M is an S^{−1}R-module. Conversely, if M is an S^{−1}R-module, then M equipped with the identity map has the UMP that characterizes S^{−1}M; whence, M = S^{−1}M. □

Exercise (12.5). — Let R be a ring, S ⊂ T multiplicative subsets, M a module. Set T_{1} := φ_{S}(T) ⊂ S^{−1}R. Show T^{−1}M = T^{−1}(S^{−1}M) = T_{1}^{−1}(S^{−1}M).

Solution: Let’s check that both T^{−1}(S^{−1}M) and T_{1}^{−1}(S^{−1}M) have the UMP characterizing T^{−1}M. Let ψ: M → N be an R-linear map into an T^{−1}R-module. Then the multiplication map μ_{s}: N → N is bijective for all s ∈ T by (12.1), so for all s ∈ S since S ⊂ T. Hence ψ factors via a unique S^{−1}R-linear map ρ: S^{−1}M → N by (12.3) and by (12.1) again.

Similarly, ρ factors through a unique T^{−1}R-linear map ρ′: T^{−1}(S^{−1}M) → N. Hence ψ = ρ′φ_{T} φ_{S}, and ρ′ is clearly unique, as required. Also, ρ factors through a unique T_{1}^{−1}(S^{−1}R)-linear map ρ′_{1} : T_{1}^{−1}(S^{−1}M) → N. Hence ψ = ρ′_{1}φ_{T_{1}}φ_{S}, and ρ′_{1} is clearly unique, as required. □

Exercise (12.6). — Let R be a ring, S a multiplicative subset. Show that S becomes a filtered category when equipped as follows: given s, t ∈ S, set

Hom(s, t) := {x ∈ R | xs = t}.

Given a module M, define a functor S → ((R-mod)) as follows: for s ∈ S, set M_{s} := M; to each x ∈ Hom(s, t), associate μ_{x}: M_{s} → M_{t}. Define β_{s}: M_{s} → S^{−1}M by β_{s}(m) := m/s. Show the β_{s} induce an isomorphism \underrightarrow{lim} M_{s} ⟶^{∼} S^{−1}M.

Solution: Clearly, S is a category. Now, given s, t ∈ S, set u := st. Then u ∈ S; also t ∈ Hom(s, u) and s ∈ Hom(t, u). Given x, y ∈ Hom(s, t), we have xs = t and ys = t. So s ∈ Hom(t, u) and xs = ys in Hom(s, u). Thus S is filtered.

Further, given x ∈ Hom(s, t), we have β_{t}μ_{x} = β_{s} since m/s = xm/t as xs = t. So the β_{s} induce a homomorphism β: \underrightarrow{lim} M_{s} → S^{−1}M. Now, every element of S^{−1}M is of the form m/s, and m/s =: β_{s}(m); hence, β is surjective.

Each m ∈ \underrightarrow{lim} M_{s} lifts to an m′ ∈ M_{s} for some s ∈ S by (7.8)(1). Assume βm = 0. Then β_{s}m′ = 0 as the β_{s} induce β. But β_{s}m′ = m′/s. So there is t ∈ S with tm′ = 0. So μ_{t}m′ = 0 in M_{st}, and μ_{t}m′ ↦ m. So m = 0. Thus β is injective, so an isomorphism. □

Exercise (12.7). — Let R be a ring, S a multiplicative subset, M a module. Prove S^{−1}M = 0 if Ann(M) ∩ S ≠ ∅. Prove the converse if M is finitely generated.

Solution: Say f ∈ Ann(M)∩S. Let m/t ∈ S^{−1}M. Then f/1·m/t = fm/t = 0. Hence m/t = 0. Thus S^{−1}M = 0.

Conversely, assume S^{−1}M = 0, and say m_{1}, … m_{n} generate M. Then for each i, there is f_{i} ∈ S with f_{i}m_{i} = 0. Then ∏ f_{i} ∈ Ann(M) ∩ S, as desired. □

Exercise (12.11). — Let R be a ring, S a multiplicative subset, P a projective module. Then S^{−1}P is a projective S^{−1}R-module.

Solution: By (5.22), there is a module K such that F := K ⊕ P is free. So (12.9) yields that S^{−1}F = S^{−1}P ⊕ S^{−1}L and that S^{−1}F is free over S^{−1}R. Hence S^{−1}P is a projective S^{−1}R-module again by (5.22). □

Exercise (12.13). — Let R be a ring, S a multiplicative subset, M and N modules. Show S^{−1}(M ⊗_{R}N) = S^{−1}M ⊗_{R}N = S^{−1}M ⊗_{S^{−1}R}S^{−1}N = S^{−1}M ⊗_{R}S^{−1}N.

Solution: By (12.12), S^{−1}(M ⊗_{R} N) = S^{−1}R ⊗_{R} (M ⊗_{R} N). The latter is equal to (S^{−1}R ⊗_{R} M) ⊗_{R} N by associativity (8.9). Again by (12.12), the latter is equal to S^{−1}M ⊗_{R} N. Thus the first equality holds.

By cancellation (8.10), S^{−1}M ⊗_{R} N = S^{−1}M ⊗_{S^{−1}R} (S^{−1}R ⊗_{R} N), and the latter is equal to S^{−1}M ⊗_{S^{−1}R} S^{−1}N by (12.12). Thus the second equality holds.

Finally by (8.8), the kernel of the map S^{−1}M ⊗_{R} S^{−1}N → S^{−1}M ⊗_{S^{−1}R} S^{−1}N is generated by elements (xm/s) ⊗ (n/1) − (m/1) ⊗ (xn/s) with m ∈ M, n ∈ N, x ∈ R, and s ∈ S. Those elements are zero because μ_{s} is an isomorphism on the S^{−1}R-module S^{−1}M ⊗_{R} S^{−1}N. Thus the third equality holds. □

Exercise (12.24). — Set R := ℤ and S = ℤ − ⟨0⟩. Set M := ⨁_{n≥2} ℤ/⟨n⟩ and N := M. Show that the map σ of (12.21) is not injective.

Solution: Given m > 0, let e_{n} be the nth standard basis element for some n > m. Then m · e_{n} ≠ 0. Hence μ_{R}: R → Hom_{R}(M, M) is injective. But S^{−1}M = 0, as any x ∈ M has only finitely many nonzero components; so kx = 0 for some nonzero integer k. So Hom(S^{−1}M, S^{−1}M) = 0. Thus σ is not injective. □

13. Support

Exercise (13.2). — Let R be a ring, 𝔭 ∈ Spec(R). Show that 𝔭 is a closed point — that is, {𝔭} is a closed set — if and only if 𝔭 is a maximal ideal.

Solution: If 𝔭 is maximal, then V(𝔭) = {𝔭}; so 𝔭 is closed.

Conversely, suppose 𝔭 is not maximal. Then 𝔭 ⫋ 𝔪 for some maximal ideal 𝔪. If 𝔭 ∈ V(𝔞), then 𝔪 ∈ V(𝔞) too. So {𝔭} ≠ V(𝔞). Thus {𝔭} is not closed. □

Exercise (13.3). — Let R be a ring, R′ a flat algebra with structure map φ. Show that R′ is faithfully flat if and only if Spec(φ) is surjective.

Solution: Owing to the definition of Spec(φ) in (13.1), the assertion amounts to the equivalence of (1) and (3) of (9.10). □

Exercise (13.5). — Let R be a ring, X := Spec(R), and U an open subset. Show U is quasi-compact if and only if X − U = V (𝔞) where 𝔞 is finitely generated.

Solution: Assume U is quasi-compact. By (13.1), U = ⋃_{λ} D(f_{λ}) for some f_{λ}. Hence U = ⋃_{1}^{n} D(f_{i}) for some f_{i}. Thus X − U = ⋂ V(f_{i}) = V(⟨f_{1}, …, f_{n}⟩).

Conversely, assume X −U = V(⟨f_{1}, …, f_{n}⟩). Then U = ⋃_{i=1}^{n} D(f_{i}). By (13.4), each D(f_{i}) is quasi-compact. Thus U is quasi-compact. □

Exercise (13.6). — Let B be a Boolean ring, and set X := Spec(B). Show X is a compact Hausdorff space. (Following Bourbaki, “quasi-compact” is shortened to “compact” when the space is Hausdorff.) Further, show a subset U ⊂ X is both open and closed if and only if U = D(f) for some f ∈ B.

Solution: Let f ∈ B. Then D(f) ∪ D(1 − f) = X whether B is Boolean or not; indeed, if 𝔭 ∈ X − D(f), then f ∈ 𝔭, so 1 − f ∉ 𝔭, so 𝔭 ∈ D(1 − f). Further, D(f) ∩ D(1 − f) = ∅; indeed, if 𝔭 ∈ D(f), then f ∉ 𝔭, but f(1 − f) = 0 as B is Boolean, so 1 − f ∈ 𝔭, so 𝔭 ∉ D(1 − f). Thus X − D(f) = D(1 − f). Thus D(f) is closed as well as open.

Let 𝔭, 𝔮 be prime ideals with 𝔭 ≠ 𝔮. Then there is f ∈ 𝔭 − 𝔮. So 𝔭 ∉ D(f), but 𝔮 ∈ D(f). By the above, D(f) is both open and closed. Thus X is Hausdorff. By (13.4), X is quasi-compact, so compact as it is Hausdorff.

Finally, let U ⊂ X be open and closed. Then U is quasi-compact, as U is closed and X is quasi-compact. So X − U = V (𝔞) where 𝔞 is finitely generated by (13.5). Since B is Boolean, 𝔞 = ⟨f⟩ for some f ∈ B by (1.16) (5). Thus U = D(f). □

Exercise (13.7) (Stone’s Theorem). — Show every Boolean ring B is isomorphic to the ring of continuous functions from a compact Hausdorff space X to 𝔽_{2} with the discrete topology. Equivalently, show B is isomorphic to the ring R of open and closed subsets of X; in fact, X := Spec(B), and B ⟶^{∼} R is given by f ↦ D(f).

Solution: The two statements are equivalent by (1.2). Further, X := Spec(B) is compact Hausdorff, and its open and closed subsets are precisely the D(f) by (13.6). Thus f ↦ D(f) is a well defined function, and is surjective.

This function preserves multiplication owing to (13.1.1). To show it preserves addition, we must show that, for any f, g ∈ B,

D(f + g) = (D(f) − D(g)) ∪ (D(g) − D(f)).  (13.7.1)

Fix a prime 𝔭. There are four cases. First, if f ∉ 𝔭 and g ∈ 𝔭, then f + g ∉ 𝔭. Second, if g ∉ 𝔭 but f ∈ 𝔭, then again f + g ∉ 𝔭. In both cases, 𝔭 lies in the open sets on both sides of (13.7.1).

Third, if f ∈ 𝔭 and g ∈ 𝔭, then f + g ∈ 𝔭. The first three cases do not use the hypothesis that B is Boolean. The fourth does. Suppose f ∉ 𝔭 and g ∉ 𝔭. Now, B/𝔭 = 𝔽_{2} by (2.18). So the residues of f and g are both equal to 1. But 1 + 1 = 0 ∈ 𝔽_{2}. So again f + g ∈ 𝔭. Thus in both the third and fourth cases, 𝔭 lies in neither side of (13.7.1). Thus (13.7.1) holds.

Finally, to show that f ↦ D(f) is injective, suppose that D(f) is empty. Then f ∈ nil(B). But nil(B) = ⟨0⟩ by (3.23). Thus f = 0. □

Exercise (13.14). — Let R be a ring, M a module, 𝔭 ∈ Supp(M). Prove

V(𝔭) ⊂ Supp(M).

Solution: Let 𝔮 ∈ V(𝔭). Then 𝔮 ⊃ 𝔭. So M_{𝔭} = (M_{𝔮})_{𝔭} by (11.28)(1). Now, 𝔭 ∈ Supp(M). So M_{𝔭} ≠ 0. Hence M_{𝔮} ≠ 0. Thus 𝔮 ∈ Supp(M). □

Exercise (13.15). — Let ℤ be the integers, ℚ the rational numbers, and set M := ℚ/ℤ. Find Supp(M), and show that it is not Zariski closed.

Solution: Let 𝔭 ∈ Spec(R). Then M_{𝔭} = ℚ_{𝔭}/ℤ_{𝔭} since localization is exact by (12.16). Now, ℚ_{𝔭} = ℚ by (12.4) and (12.1) since ℚ is a field. If 𝔭 ≠ ⟨0⟩, then ℤ_{𝔭} ≠ ℚ_{𝔭} since 𝔭ℤ_{𝔭} ∩ ℤ = 𝔭 by (11.17). If 𝔭 = ⟨0⟩, then ℤ_{𝔭} = ℚ_{𝔭}. Thus Supp(M) consists of all the nonzero primes of ℤ.

Finally, suppose Supp(M) = V(𝔞). Then 𝔞 lies in every nonzero prime; so 𝔞 = ⟨0⟩. But ⟨0⟩ is prime. Hence ⟨0⟩ ∈ V(𝔞) = Supp(M), contradicting the above. Thus Supp(M) is not closed. □

Exercise (13.17). — Let R be a ring, P a module, and M, N submodules. Show M = N if M_{𝔪} = N_{𝔪} for every maximal ideal 𝔪. First assume M ⊂ N.

Solution: If M ⊂ N, then (12.16) yields (N/M)_{𝔪} = N_{𝔪}/M_{𝔪} = 0 for each 𝔪; so N/M = 0 by (13.16). The general case follows by replacing N by M + N owing to (12.15)(4), (5). □

Exercise (13.18). — Prove these three conditions on a ring R are equivalent:

(1) R is reduced.

(2) S^{−1}R is reduced for all multiplicatively closed sets S.

(3) R_{𝔪} is reduced for all maximal ideals 𝔪.

Solution: Assume (1) holds. Then nil(R) = 0. But nil(R)(S^{−1}R) = nil(S^{−1}R) by (11.16). Thus (2) holds. Trivially (2) implies (3).

Assume (3) holds. Then nil(R_{𝔪}) = 0. Hence nil(R)_{𝔪} = 0 by (11.16) and (12.2). So nil(R) = 0 by (13.16). Thus (1) holds. □

Exercise (13.19). — Let R be a ring, Σ the set of minimal primes. Prove this:

(1) If R_{𝔭} is a domain for any prime 𝔭, then the 𝔭 ∈ Σ are pairwise comaximal.

(2) R_{𝔭} is a domain for any prime 𝔭 and Σ is finite if and only if R = ∏_{i=1}^{n} R_{i} where R_{i} is a domain. If so, then R_{i} = R/𝔭_{i} with {𝔭_{1}, …, 𝔭_{n}} = Σ.

Solution: Consider (1). Suppose 𝔭, 𝔮 ∈ Σ are not comaximal. Then 𝔭 + 𝔮 lies in some maximal ideal 𝔪. Hence R_{𝔪} contains two minimal primes, 𝔭R_{𝔪} and 𝔮R_{𝔪}, by (11.18). However, R_{𝔪} is a domain by hypothesis, and so ⟨0⟩ is its only minimal prime. Hence 𝔭R_{𝔪} = 𝔮R_{𝔪}. So 𝔭 = 𝔮. Thus (1) holds.

Consider (2). Assume R_{𝔭} is a domain for any 𝔭. Then R is reduced by (13.18). Assume, also, Σ is finite. Form the canonical map φ: R → ∏_{𝔭∈Σ} R/𝔭; it is injective by (3.28), and surjective by (1) and the Chinese Remainder Theorem (1.13). Thus R is a finite product of domains.

Conversely, assume R = ∏_{i=1}^{n} R_{i} where R_{i} is a domain. Let 𝔭 be a prime of R. Then R_{𝔭} = ∏(R_{i})_{𝔭} by (12.10). Each (R_{i})_{𝔭} is a domain by (11.4). But R_{𝔭} is local. So R_{𝔭} = (R_{i})_{𝔭} for some i by (2.5). Thus R_{𝔭} is a domain. Further, owing to (2.12), each 𝔭_{i} ∈ Σ has the form 𝔭_{i} = ∏ 𝔞_{j} where, after renumbering, 𝔞_{i} = ⟨0⟩ and 𝔞_{j} = R_{j} for j ≠ i. Thus the ith projection gives R/𝔭_{i} ⟶^{∼} R_{i}. Thus (2) holds. □

Exercise (13.21). — Let R be a ring, M a module. Prove elements m_{λ} ∈ M generate M if and only if, at every maximal ideal 𝔪, their images m_{λ} generate M_{𝔪}.

Solution: The m_{λ} define a map α: R^{⊕{λ}} → M. By (13.20), it is surjective if and only if α_{𝔪}: (R^{⊕{λ}})_{𝔪} → M_{𝔪} is surjective for all 𝔪. But (R^{⊕{λ}})_{𝔪} = R_{𝔪}^{⊕{λ}} by (12.10). Hence (4.10)(1) yields the assertion. □

Exercise (13.24). — Let R be a ring, R′ a flat algebra, 𝔭′ a prine in R′, and 𝔭 its contraction in R. Prove that R′_{𝔭′} is a faithfully flat R_{𝔭}-algebra.

Solution: First, R′_{𝔭} is flat over R_{𝔭} by (13.23). Next, R′_{𝔭′} is flat over R′_{𝔭} by (12.17) and (11.28) as R − 𝔭 ⊂ R′ − 𝔭′. Hence R′_{𝔭′} is flat over R_{𝔭} by (9.7). But a flat local homomorphism is faithfully flat by (9.10)(4). □

Exercise (13.28). — Given n, prove an R-module P is locally free of rank n if and only if P is finitely generated and P_{𝔪} ≃ R_{𝔪}^{n} holds at each maximal ideal 𝔪.

Solution: If P is locally free of rank n, then P is finitely generated by (13.27). Also, for any 𝔭 ∈ Spec(R), there’s f ∈ R − 𝔭 with P_{f} ≃ R_{f}^{n}; so P_{𝔭} ≃ R_{𝔭}^{n} by (12.5).

As to the converse, given any prime 𝔭, take a maximal ideal 𝔪 containing it. Assume P_{𝔪} ≃ R_{𝔪}^{n}. Take a free basis p_{1}/f_{1}^{k_{1}}, …, p_{n}/f_{n}^{k_{n}} of P_{𝔪} over R_{𝔪}. The p_{i} define a map α: R^{n} → P, and α_{𝔪}: R_{𝔪}^{n} → P_{𝔪} is bijective, so surjective.

Assume P is finitely generated. Then (12.20)(1) provides f ∈ R − 𝔪 such that α_{f}: R_{f}^{n} → P_{f} is surjective. Hence α_{𝔮}: R_{𝔮}^{n} → P_{𝔮} is surjective for every 𝔮 ∈ D(f) by (12.5) and (12.16). Assume P_{𝔮} ≃ R_{𝔮}^{n} if also 𝔮 is maximal. So α_{𝔮} is bijective by (10.4). Clearly, α_{𝔮} = (α_{f})_{(𝔮R_{f})}. Hence α_{f}: R_{f}^{n} → P_{f} is bijective owing to (13.20) with R_{f} for R, as desired. □

Exercise (13.29). — Let A be a semilocal ring, P a locally free module of rank n. Show that P is free of rank n.

Solution: As P is locally free, P is finitely presented by (13.27), and P_{𝔪} ≃ A_{𝔪}^{n} at the maximal 𝔪 by (13.28). But A is semilocal. So P ≅ A^{n} by (13.22). □

14. Krull–Cohen–Seidenberg Theory

Exercise (14.4). — Let R ⊂ R′ be an integral extension of rings, and 𝔭 a prime of R. Suppose R′ has just one prime 𝔭′ over 𝔭. Show (a) that 𝔭′R′_{𝔭} is the only maximal ideal of R′_{𝔭}, (b) that R′_{𝔭′} = R′_{𝔭}, and (c) that R′_{𝔭′} is integral over R_{𝔭}.

Solution: Since R′ is integral over R, the localization R′_{𝔭} is integral over R_{𝔭} by (11.23). Moreover, R_{𝔭} is a local ring with unique maximal ideal 𝔭R_{𝔭} by (11.20). Hence, every maximal ideal of R′_{𝔭} lies over 𝔭R_{𝔭} by (14.3)(1). But every maximal ideal of R′_{𝔭} is the extension of some prime 𝔮′ ⊂ R′ by (11.18)(2), and therefore 𝔮′ lies over 𝔭 in R. So, by hypothesis, 𝔮′ = 𝔭′. Thus 𝔭′R′_{𝔭} is the only maximal ideal of R′_{𝔭}; that is, (a) holds. So R′_{𝔭} − 𝔭′R′_{𝔭} consists of units. Hence (11.28) and (11.7) yield (b). But R′_{𝔭} is integral over R_{𝔭}; so (c) holds too. □

Exercise (14.5). — Let R ⊂ R′ be an integral extension of domains, and 𝔭 a prime of R. Suppose R′ has at least two distinct primes 𝔭′ and 𝔮′ lying over 𝔭. Show that R′_{𝔭′} is not integral over R_{𝔭}. Show that, in fact, if y lies in 𝔮′, but not in 𝔭′, then 1/y ∈ R′_{𝔭′} is not integral over R_{𝔭}.

Solution: Suppose 1/y is integral over R_{𝔭}. Say

(1/y)^{n} + a_{1}(1/y)^{n−1} + ⋯ + a_{n} = 0

with n ≥ 1 and a_{i} ∈ R_{𝔭}. Multiplying by y^{n−1}, we obtain

1/y = −(a_{1} + ⋯ + a_{n}y^{n−1}) ∈ R′_{𝔭}.

However, y ∈ 𝔮′, so y ∈ 𝔮′R′_{𝔭}. Hence 1 ∈ 𝔮′R′_{𝔭}. So 𝔮′ ∩ (R − 𝔭) ≠ ∅ by (11.17)(3). But 𝔮′ ∩ R = 𝔭, a contradiction. So 1/y is not integral over R_{𝔭}. □

Exercise (14.6). — Let k be a field, and X an indeterminate. Set R′ := k[X], and Y := X^{2}, and R := k[Y]. Set 𝔭 := (Y − 1)R and 𝔭′ := (X − 1)R′. Is R′_{𝔭′} integral over R_{𝔭}? Explain.

Solution: Note that R′ is a domain, and that the extension R ⊂ R′ is integral as R′ is generated by 1 and X as an R-module.

Suppose the characteristic is not 2. Set 𝔮′ := (X + 1)R′. Then both 𝔭′ and 𝔮′ contain Y − 1, so lie over the maximal ideal 𝔭 of R. Further X + 1 lies in 𝔮′, but not in 𝔭′. Hence R′_{𝔭′} is not integral over R_{𝔭} by (14.5).

Suppose the characteristic is 2. Then (X − 1)^{2} = Y − 1. Let 𝔮′ ⊂ R′ be a prime over 𝔭. Then (X − 1)^{2} ∈ 𝔮′. So 𝔭′ ⊂ 𝔮′. But 𝔭′ is maximal. So 𝔮′ = 𝔭′. Thus R′ has just one prime 𝔭′ over 𝔭. Hence R′_{𝔭′} is integral over R_{𝔭} by (14.4). □

Exercise (14.12). — Let R be a reduced ring, Σ the set of minimal primes. Prove that z.div(R) = ⋃_{𝔭∈Σ} 𝔭 and that R_{𝔭} = Frac(R/𝔭) for any 𝔭 ∈ Σ.

Solution: If 𝔭 ∈ Σ, then 𝔭 ⊂ z.div(R) by (14.10). Thus z.div(R) ⊃ ⋃_{𝔭∈Σ} 𝔭. Conversely, say xy = 0. If x ∉ 𝔭 for some 𝔭 ∈ Σ, then y ∈ 𝔭. So if x ∉ ⋃_{𝔭∈Σ} 𝔭,

then y ∈ ⋂_{𝔭∈Σ} 𝔭. But ⋂_{𝔭∈Σ} 𝔭 = ⟨0⟩ by the Scheinnullstellensatz (3.22) and (3.11). So y = 0. Hence if x ∉ ⋃_{𝔭∈Σ} 𝔭, then x ∉ z.div(R). Thus z.div(R) ⊂ ⋃_{𝔭∈Σ} 𝔭. Thus z.div(R) = ⋃_{𝔭∈Σ} 𝔭.

Fix 𝔭 ∈ Σ. Then R_{𝔭} is reduced by (13.18). Further, R_{𝔭} has only one prime, namely 𝔭R_{𝔭}, by (11.18)(2). Hence R_{𝔭} is a field, and 𝔭R_{𝔭} = ⟨0⟩. But by (12.19), R_{𝔭}/𝔭R_{𝔭} = Frac(R/𝔭). Thus R_{𝔭} = Frac(R/𝔭). □

Exercise (14.13). — Let R be a ring, Σ the set of minimal primes, and K the total quotient ring. Assume Σ is finite. Prove these three conditions are equivalent:

(1) R is reduced.

(2) z.div(R) = ⋃_{𝔭∈Σ} 𝔭, and R_{𝔭} = Frac(R/𝔭) for each 𝔭 ∈ Σ.

(3) K/𝔭K = Frac(R/𝔭) for each 𝔭 ∈ Σ, and K = ∏_{𝔭∈Σ} K/𝔭K.

Solution: Assume (1) holds. Then (14.12) yields (2).

Assume (2) holds. Set S := R − z.div(R). Let 𝔮 be a prime of R with 𝔮 ∩ S = ∅. Then 𝔮 ⊂ ⋃_{𝔭∈Σ} 𝔭. But Σ is finite. So 𝔮 ⊂ 𝔭 for some 𝔭 ∈ Σ by Prime Avoidance (3.15). Hence 𝔮 = 𝔭 since 𝔭 is minimal. But K = S^{−1}R. Therefore, by (11.18)(2), for 𝔭 ∈ Σ, the extensions 𝔭K are the only primes of K, and they all are both maximal and minimal.

Fix 𝔭 ∈ Σ. Then K/𝔭K = S^{−1}(R/𝔭) by (12.18). So S^{−1}(R/𝔭) is a field. But clearly S^{−1}(R/𝔭) ⊂ Frac(R/𝔭). Therefore, K/𝔭K = Frac(R/𝔭) by (2.3). Further, S ⊂ R−𝔭. Hence (11.18)(2) yields 𝔭 = φ_{S}^{−1}(𝔭K). Therefore, φ_{S}^{−1}(K−𝔭K) = R−𝔭. So K_{𝔭K} = R_{𝔭} by (11.26). But R_{𝔭} = Frac(R/𝔭) by hypothesis. Thus K has only finitely many primes, the 𝔭K; each 𝔭K is minimal, and each K_{𝔭K} is a domain.

Therefore, (13.19)(2) yields K = ∏_{𝔭∈Σ} K/𝔭K. Thus (3) holds.

Assume (3) holds. Then K is a finite product of fields, and fields are reduced. But clearly, a product of reduced ring is reduced. Further, R ⊂ K, and trivially, a subring of a reduced ring is reduced. Thus (1) holds. □

Exercise (14.14). — Let A be a reduced local ring with residue field k and a finite set Σ of minimal primes. For each 𝔭 ∈ Σ, set K(𝔭) := Frac(A/𝔭). Let P be a finitely generated module. Show that P is free of rank r if and only if dim_{k}(P ⊗_{A} k) = r and dim_{K(𝔭)}(P ⊗_{A} K(𝔭)) = r for each 𝔭 ∈ Σ.

Solution: If P is free of rank r, then dim(P ⊗ k) = r and dim(P ⊗ K(𝔭)) = r owing to (8.11).

Conversely, suppose dim(P ⊗ k) = r. As P is finitely generated, (10.13) implies P is generated by r elements. So (5.19) yields an exact sequence

0 → M ⟶^{α} A^{r} → P → 0.

Momentarily, fix a 𝔭 ∈ Σ. Since A is reduced, K(𝔭) = R_{𝔭} by (14.12). So K(𝔭) is flat by (12.17). So the induced sequence is exact:

0 → M ⊗ K(𝔭) → K(𝔭)^{r} → P ⊗ K(𝔭) → 0.

Suppose dim(P ⊗ K(𝔭)) = r too. It then follows that M ⊗_{A} K(𝔭) = 0. Let K be the total quotient ring of A, and form this commutative square:

M ⟶^{α} A^{r}

↓^{φ_{M}}  ↓^{φ_{A^{r}}}

M ⊗ K ⟶ K^{r}

Here α is injective. And φ_{A^{r}} is injective as φ_{A}: A → K is. Hence, φ_{M} is injective.

By hypothesis, A is reduced and Σ is finite; so K = ∏_{𝔭∈Σ} K(𝔭) by (14.13). So M ⊗ K = ∏(M ⊗ K(𝔭)). But M ⊗_{A} K(𝔭) = 0 for each 𝔭 ∈ Σ. So M ⊗ K = 0. But φ_{M}: M → M ⊗ K is injective. So M = 0. Thus A^{r} ⟶^{∼} P, as desired. □

Exercise (14.15). — Let A be a reduced local ring with residue field k and a finite set of minimal primes. Let P be a finitely generated module, B an A-algebra with Spec(B) → Spec(A) surjective. Show that P is a free A-module of rank r if and only if P ⊗ B is a free B-module of rank r.

Solution: If P is a free A-module of rank r, then P ⊗ B is a free B-module of rank r owing to (8.11).

Conversely, let 𝔭 ⊂ A be a prime. Since Spec(B) → Spec(A) is surjective, there is a prime 𝔮 ⊂ B whose trace is 𝔭. Set K := Frac(A/𝔭) and L := Frac(B/𝔮). Then the structure map A → B induces a map K → L. Moreover,

(P ⊗_{A} B) ⊗_{B} L = (P ⊗_{A} K) ⊗_{K} L.  (14.15.1)

Suppose P ⊗ B is a free B-module of rank r. Then dim_{L}((P ⊗_{A} B) ⊗_{B} L) = r owing to (8.11). Hence (14.15.1) implies dim_{K}(P ⊗_{A} K) = r. But 𝔭 is arbitrary. Thus P is a free A-module of rank r by (14.14), as desired. □

Exercise (14.17). — Let R be a ring, 𝔭_{1} …, 𝔭_{r} all its minimal primes, and K the total quotient ring. Prove that these three conditions are equivalent:

(1) R is normal.

(2) R is reduced and integrally closed in K.

(3) R is a finite product of normal domains R_{i}. If so, then the R_{i} are equal to the R/𝔭_{j} up to order.

Solution: Assume (1). Then R is reduced by (13.18). Let x ∈ K be integral over R, and 𝔪 any maximal ideal. Then x/1 is integral over R_{𝔪}. So x/1 ∈ R_{𝔪} by hypothesis. Hence (R[x]/R)_{𝔪} = 0. Therefore, R[x]/R = 0 by (13.16). So x ∈ R Thus (2) holds.

Assume (2). Set R_{i} := R/𝔭_{i} and K_{i} := Frac(R_{i}). Then K = ∏ K_{i} by (14.13). Let R′_{i} be the normalization of R_{i}. Then R ⊂ ∏ R_{i} ⊂ ∏ R′_{i}. Further, the first extension is integral by (10.24), and the second, by (10.26); whence, R ⊂ ∏ R′_{i} is integral by the tower property (10.22). However, R is integrally closed in K by hypothesis. Hence R = ∏ R_{i} = ∏ R′_{i}. Thus (3) and the last assertion hold.

Assume (3). Let 𝔭 be any prime of R. Then R_{𝔭} = ∏(R_{i})_{𝔭} by (12.10), and each (R_{i})_{𝔭} is normal by (11.31). But R_{𝔭} is local. So R_{𝔭} = (R_{i})_{𝔭} for some i by (3.5). Hence R_{𝔭} is a normal domain. Thus (1) holds. □

15. Noether Normalization

Exercise (15.2). — Let k := 𝔽_{q} be the finite field with q elements, and k[X, Y] the polynomial ring. Set f := X^{q}Y − XY^{q} and R := k[X, Y]/⟨f⟩. Let x, y ∈ R be the residues of X, Y. For every a ∈ k, show that R is not module finite over P := k[y−ax]. (Thus, in (15.1), no k-linear combination works.) First, take a = 0.

Solution: Take a = 0. Then P = k[y]. Any algebraic relation over P satisfied by x is given by a polynomial in k[X, Y], which is a multiple of f. However, no multiple of f is monic in X. So x is not integral over P. By (10.18), R is not module finite over P.

Consider an arbitrary a. Since a^{q} = a, after the change of variable Y′ := Y −aX, our f still has the same form. Thus, we have reduced to the previous case. □

Exercise (15.3). — Let k be a field, and X, Y, Z variables. Set

R := k[X, Y, Z]/⟨X^{2} − Y^{3} − 1, XZ − 1⟩,

and let x, y, z ∈ R be the residues of X, Y, Z. Fix a, b ∈ k, and set t := x + ay + bz and P := k[t]. Show that x and y are integral over P for any a, b and that z is integral over P if and only if b ≠ 0.

Solution: To see x is integral, notice xz = 1, so x^{2}−tx+b = −axy. Raising both sides of the latter equation to the third power, and using the equation y^{3} = x^{2} − 1, we obtain an equation of integral dependence of degree 6 for x over P. Now, y^{3} − x^{2} − 1 = 0, so y is integral over P [x]. Hence, the Tower Property, (10.22), implies that y too is integral over P.

If b ≠ 0, then z = b^{−1}(t − x − ay) ∈ P [x, y], and so z is integral over P by (10.23).

Assume b = 0 and z is integral over P. Now, P ⊂ k[x, y]. So z is integral over k[x, y] as well. But y^{3} − x^{2} + 1 = 0. So y is integral over k[x]. Hence z is too. However, k[x] is a polynomial ring, so integrally closed in its fraction field k(x) by (10.29)(1). Moreover, z = 1/x ∈ k(x). Hence, 1/x ∈ k[x], which is absurd. Thus z is not integral over P if b = 0. □

Exercise (15.7). — Let k be a field, K an algebraically closed extension field. (So K contains a copy of every finite extension field.) Let P := k[X_{1}, …, X_{n}] be the polynomial ring, and f, f_{1}, …, f_{r} ∈ P. Assume f vanishes at every zero in K^{n} of f_{1}, …, f_{r}; in other words, if (a) := (a_{1}, …, a_{n}) ∈ K^{n} and f_{1}(a) = 0, …, f_{r}(a) = 0, then f(a) = 0 too. Prove that there are polynomials g_{1}, …, g_{r} ∈ P and an integer N such that f^{N} = g_{1}f_{1} + ⋯ + g_{r}f_{r}.

Solution: Set 𝔞 := ⟨f_{1}, …, f_{r}⟩. We have to show f ∈ \sqrt{𝔞}. But, by the Hilbert Nullstellensatz, \sqrt{𝔞} is equal to the intersection of all the maximal ideals 𝔪 containing 𝔞. So given an 𝔪, we have to show that f ∈ 𝔪.

Set L := P/𝔪. By the weak Nullstellensatz, L is a finite extension field of k. So we may embed L/k as a subextension of K/k. Let a_{i} ∈ K be the image of the variable X_{i} ∈ P, and set (a) := (a_{1}, …, a_{n}) ∈ K^{n}. Then f_{1}(a) = 0, …, f_{r}(a) = 0. Hence f(a) = 0 by hypothesis. Therefore, f ∈ 𝔪, as desired. □

Exercise (15.10). — Let R be a domain of (finite) dimension r, and 𝔭 a nonzero prime. Prove that dim(R/𝔭) < r.

Solution: Every chain of primes of R/𝔭 is of the form 𝔭_{0}/𝔭 ⫋ ⋯ ⫋ 𝔭_{s}/𝔭 where 0 ⫋ 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{s} is a chain of primes of R. So s < r. Thus dim(R/𝔭) < r. □

Exercise (15.11). — Let R′/R be an integral extension of rings. Prove that dim(R) = dim(R′).

Solution: Let 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} be a chain of primes of R. Set 𝔭′_{−1} := 0. Given 𝔭′_{i−1} for 0 ≤ i ≤ r, Going up, (14.3)(4), yields a prime 𝔭′_{i} of R′ with 𝔭′_{i−1} ⊂ 𝔭′_{i} and 𝔭′_{i} ∩ R = 𝔭_{i}. Then 𝔭′_{0} ⫋ ⋯ ⫋ 𝔭′_{r} as 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r}. Thus dim(R) ≤ dim(R′).

Conversely, let 𝔭′_{0} ⫋ ⋯ ⫋ 𝔭′_{r} be a chain of primes of R′. Set 𝔭_{i} := 𝔭′_{i} ∩ R. Then 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} by Incomparability, (14.3)(2). Thus dim(R) ≥ dim(R′). □

Exercise (15.16). — Let k be a field, R a finitely generated k-algebra, f ∈ R nonzero. Assume R is a domain. Prove that dim(R) = dim(R_{f}).

Solution: Note that R_{f} is a finitely generated R-algebra by (11.11), as R_{f} is, by (11.11), obtained by adjoining 1/f. So since R is a finitely generated k-algebra, R_{f} is one too. Moreover, R and R_{f} have the same fraction field K. Hence both dim(R) and dim(R_{f}) are equal to tr. deg_{k}(K) by (15.12). □

Exercise (15.17). — Let k be a field, P := k[f] the polynomial ring in one variable f. Set 𝔭 := ⟨f⟩ and R := P_{𝔭}. Find dim(R) and dim(R_{f}).

Solution: In P, the chain of primes 0 ⊂ 𝔭 is of maximal length by (2.6) and (2.23) or (15.12). So ⟨0⟩ and 𝔭R are the only primes in R by (11.18). Thus dim(R) = 1.

Set K := Frac(P). Then R_{f} = K since, if a/(bf^{n}) ∈ K with a, b ∈ P and f ∤ b, then a/b ∈ R and so (a/b)/f^{n} ∈ R_{f}. Thus dim(R_{f}) = 0. □

Exercise (15.18). — Let R be a ring, R[X] the polynomial ring. Prove

1 + dim(R) ≤ dim(R[X]) ≤ 1 + 2 dim(R).

Solution: Let 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{n} be a chain of primes in R. Then

𝔭_{0}R[X] ⫋ ⋯ ⫋ 𝔭_{n}R[X] ⫋ 𝔭_{n}R[X] + ⟨X⟩

is a chain of primes in R[X] by (2.10). Thus 1 + dim(R) ≤ dim(R[X]).

Let 𝔭 be a prime of R, and 𝔮_{0} ⫋ ⋯ ⫋ 𝔮_{r} be a chain of primes of R[X] with 𝔮_{i}∩R = 𝔭 for each i. Then (1.8) yields a chain of primes of length r in R[X]/𝔭R[X]. Further, as 𝔮_{i} ∩ R = 𝔭 for each i, the latter chain gives rise to a chain of primes of length r in k(𝔭)[X] where k(𝔭) = (R/𝔭)_{𝔭} by (11.29) and (11.18). But k(𝔭)[X] is a PID. Hence r ≤ 1.

Take any chain 𝔓_{0} ⫋ ⋯ ⫋ 𝔓_{m} of primes in R[X]. Then it contracts to a chain 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{n} in R. But at most two 𝔓_{j} can contract to a given 𝔭_{i} by the above discussion. Thus m ≤ 2n + 1. □

Exercise (15.22). — Let X be a topological space. We say a subset Y is locally closed if Y is the intersection of an open set and a closed set; equivalently, Y is open in its closure \overline{Y}; equivalently, Y is closed in an open set containing it.

We say a subset X_{0} of X is very dense if X_{0} meets every nonempty locally closed subset Y. We say X is Jacobson if its set of closed points is very dense.

Show that the following conditions on a subset X_{0} of X are equivalent:

(1) X_{0} is very dense.

(2) Every closed set F of X satisfies \overline{F ∩ X_{0}} = F.

(3) The map U ↦ U ∩ X_{0} from the open sets of X to those of X_{0} is bijective.

Solution: Assume (1). Given a closed set F, take any x ∈ F, and let U be an open neighborhood of x in X. Then F ∩ U is locally closed, so meets X_{0}. Hence x ∈ \overline{F ∩ X_{0}}. Thus F ⊂ \overline{F ∩ X_{0}}. The opposite inclusion is trivial. Thus (2) holds.

Assume (2). In (3), the map is trivially surjective. To check it’s injective, suppose U ∩X_{0} = V ∩X_{0}. Then (X −U)∩X_{0} = (X −V)∩X_{0}. So (2) yields X −U = X −V. So U = V. Thus (3) holds.

Assume (3). Then the map F ↦ F ∩ X_{0} of closed sets is bijective too; whence, so is the map Y ↦ Y ∩ X_{0} of locally closed sets. In particular, if a locally closed set Y is nonempty, then so is Y ∩ X_{0}. Thus (1) holds. □

Exercise (15.23). — Let R be a ring, X := Spec(R), and X_{0} the set of closed points of X. Show that the following conditions are equivalent:

(1) R is a Jacobson ring.

(2) X is a Jacobson space.

(3) If y ∈ X is a point such that {y} is locally closed, then y ∈ X_{0}.

Solution: Assume (1). Let F ⊂ X be closed. Trivially, F ⊃ \overline{F ∩ X_{0}}. To prove F ⊂ \overline{F ∩ X_{0}}, say F = V(𝔞) and \overline{F ∩ X_{0}} = V(𝔟). Then F ∩ X_{0} is the set of maximal ideals 𝔪 containing 𝔞 by (13.2), and every such 𝔪 contains 𝔟. So (1) implies 𝔟 ⊂ \sqrt{𝔞}. But V(\sqrt{𝔞}) = F. Thus F ⊂ \overline{F ∩ X_{0}}. Thus (15.22) yields (2).

Assume (2). Let y ∈ X be a point such that {y} is locally closed. Then {y} ⋂ X_{0} is nonempty by (2). So ({y} ⋂ X_{0}) ∋ y. Thus (3) holds.

Assume (3). Let 𝔭 be a prime ideal of R such that 𝔭R_{f} is maximal for some f ∉ 𝔭. Then {𝔭} is closed in D(f) by (13.1). So {𝔭} is locally closed in X. Hence {𝔭} is closed in X by (3). Thus 𝔭 is maximal. Thus (15.21) yields (1). □

Exercise (15.25). — Let P := ℤ[X_{1}, …, X_{n}] be the polynomial ring. Assume f ∈ P vanishes at every zero in K^{n} of f_{1}, …, f_{r} ∈ P for every finite field K; that is, if (a) := (a_{1}, …, a_{n}) ∈ K^{n} and f_{1}(a) = 0, …, f_{r}(a) = 0 in K, then f(a) = 0 too. Prove there are g_{1}, …, g_{r} ∈ P and N ≥ 1 such that f^{N} = g_{1}f_{1} + ⋯ + g_{r}f_{r}.

Solution: Set 𝔞 := ⟨f_{1}, …, f_{r}⟩. Suppose f ∉ \sqrt{𝔞}. Then f lies outside some maximal ideal 𝔪 containing 𝔞 by (15.24)(2) and (15.19). Set K := P/𝔪. Then K is a finite extension of 𝔽_{p} for some prime p by (15.24)(1). So K is finite. Let a_{i} be the residue of X_{i}, set (a) := (a_{1}, …, a_{n}) ∈ K^{n}. Then f_{1}(a) = 0, …, f_{r}(a) = 0. So f(a) = 0 by hypothesis. Thus f ∈ 𝔪, a contradiction. Thus f ∈ \sqrt{𝔞}. □

Exercise (15.26). — Let R be a ring, R′ an algebra. Prove that if R′ is integral over R and R is Jacobson, then R′ is Jacobson.

Solution: Given an ideal 𝔞′ ⊂ R′ and an f outside \sqrt{𝔞}, set R′′ := R[f]. Then R′′ is Jacobson by (15.24). So R′′ has a maximal ideal 𝔪′′ that avoids f and contains 𝔞′ ∩ R′′. But R′ is integral over R′′. So R′ contains a prime 𝔪′ that contains 𝔞′ and that contracts to 𝔪′′ by Going Up (14.3)(4). Then 𝔪′ avoids f as 𝔪′′ does, and 𝔪′ is maximal by Maximality, (14.3)(1). Thus R′ is Jacobson. □

Exercise (15.27). — Let R be a Jacobson ring, S a multiplicative subset, f ∈ R. True or false: prove or give a counterexample to each of the following statements:

(1) The localized ring R_{f} is Jacobson.

(2) The localized ring S^{−1}R is Jacobson.

(3) The filtered direct limit \underrightarrow{lim} R_{λ} of Jacobson rings R_{λ} is Jacobson.

Solution: (1) True: R_{f} = R[1/f]; so R_{f} is Jacobson by (15.24).

(2) False: by (15.20), ℤ is Jacobson, but ℤ_{⟨p⟩} isn’t for any prime number p.

(3) False: Z_{⟨p⟩} isn’t Jacobson by (2), but Z_{⟨p⟩} = \underrightarrow{lim} ℤ by (12.6). □

Exercise (15.28). — Let R be a reduced Jacobson ring with a finite set Σ of minimal primes, and P a finitely generated module. Show that P is locally free of rank r if and only if dim_{R/𝔪}(P/𝔪P) = r for any maximal ideal 𝔪.

Solution: Suppose P is locally free of rank r. Then given any maximal ideal 𝔪, there is an f ∈ R − 𝔪 such that P_{f} is a free R_{f}-module of rank r by (13.25). But P_{𝔪} is a localization of P_{f} by (12.5). So P_{𝔪} is a free R_{𝔪}-module of rank r by (12.10). But P_{𝔪}/𝔪P_{𝔪} = (P/𝔪P)_{𝔪} by (12.18). Also R_{𝔪}/𝔪R_{𝔪} = R/𝔪 by (12.19). Thus dim_{R/𝔪}(P/𝔪P) = r.

Consider the converse. Given a 𝔭 ∈ Σ, set K := Frac(R/𝔭). Then P ⊗_{R} K is a K-vector space, say of dimension n. Since R is reduced, K = R_{𝔭} by (14.12). So by (12.20), there is an h ∈ R −𝔭 with P_{h} free of rank n. As R is Jacobson, there is a maximal ideal 𝔪 avoiding h, by (15.19). Hence, as above, dim_{R/𝔪}(P/𝔪P) = n. But, by hypothesis, dim_{R/𝔪}(P/𝔪P) = r. Thus n = r.

Given a maximal ideal 𝔪, set A := R_{𝔪}. Then A is reduced by (13.18). Each minimal prime of A is of the form 𝔭A where 𝔭 ∈ Σ by (11.18)(2). Further, it’s not hard to see, essentially as above, that P_{𝔪} ⊗ Frac(A/𝔭A) = P ⊗ Frac(R/𝔭). Hence (14.14) implies P_{𝔪} is a free A-module of rank r. Finally, (13.28) implies P is locally free of rank r. □

16. Chain Conditions Exercise (16.2). — Let 𝔞 be a finitely generated ideal in an arbitrary ring. Show every set that generates 𝔞 contains a finite subset that generates 𝔞.

Solution: Say 𝔞 is generated by x_{1}, …, x_{r} and also by the y_{λ} for λ ∈ Λ. Write x_{i} = ∑_{j} z_{j}y_{λ_{ij}}. Then the y_{λ_{ij}} generate 𝔞. □

Exercise (16.8). — Let R be a ring, X a variable, R[X] the polynomial ring. Prove this statement or find a counterexample: if R[X] is Noetherian, then so is R.

Solution: It’s true. Since R[X] is Noetherian, so is R[X]/⟨X⟩ by (16.7). But the latter ring is isomorphic to R by (1.7); so R is Noetherian. □

Exercise (16.14). — Let 0 → L ⟶^{α} M ⟶^{β} N → 0 be a short exact sequence of R-modules, and M_{1}, M_{2} two submodules of M. Prove or give a counterexample to this statement: if β(M_{1}) = β(M_{2}) and α^{−1}(M_{1}) = α^{−1}(M_{2}), then M_{1} = M_{2}.

Solution: The statement is false: form the exact sequence

0 → ℝ ⟶^{α} ℝ ⊕ ℝ ⟶^{β} ℝ → 0

with α(r) := (r, 0) and β(r, s) := s, and take

M_{1} := {(t, 2t) | t ∈ ℝ} and M_{2} := {(2t, t) | t ∈ ℝ}.

(Geometrically, we can view M_{1} as the line determined by the origin and the point (1, 2), and M_{2} as the line determined by the origin and the point (2, 1). Then β(M_{1}) = β(M_{2}) = ℝ, and α^{−1}(M_{1}) = α^{−1}(M_{2}) = 0, but M_{1} ≠ M_{2} in ℝ ⊕ ℝ.) □

Exercise (16.17). — Let R be a ring, 𝔞_{1}, …, 𝔞_{r} ideals such that each R/𝔞_{i} is a Noetherian ring. Prove (1) that ⨁ R/𝔞_{i} is a Noetherian R-module, and (2) that, if ⋂ 𝔞_{i} = 0, then R too is a Noetherian ring.

Solution: Any R-submodule of R/𝔞_{i} is an ideal of R/𝔞_{i}. Since R/𝔞_{i} is a Noetherian ring, such an ideal is finitely generated as an (R/𝔞_{i})-module, so as an Rmodule as well. Thus R/𝔞_{i} is a Noetherian R-module. So ⨁ R/𝔞_{i} is a Noetherian R-module by (16.16). Thus (1) holds.

To prove (2), note that the kernel of the natural map R → ⨁ R/𝔞_{i} is ⋂ 𝔞_{i}, which is 0 by hypothesis. So R can be identified with a submodule of the Noetherian R-module ⨁ R/𝔞_{i}. Hence R itself is a Noetherian R-module by (16.15)(2). So R is a Noetherian ring by (16.12). □

Exercise (16.20). — Let G be a finite group acting on a domain R, and R′ the subring of invariants. Let k ⊂ R′ be a field. Using (10.17), prove this celebrated theorem of E. Noether (1926): if R is algebra finite over k, then so is R′.

Solution: By (10.17), R is integral over R′′. But it’s algebra finite. So it’s module finite by (10.23). Hence (16.19) yields the assertion. □

Exercise (16.24). — Let k be a field, R an algebra. Assume that R is finite dimensional as a k-vector space. Prove that R is Noetherian and Artinian.

Solution: View R as a vector space, and ideals as subspaces. Now, by a simple dimension argument, any ascending or descending chain of subspaces of R stabilizes. Thus R is Noetherian by (16.5) and is Artinian by definition. □

Exercise (16.25). — Let p be a prime number, and set M := ℤ[1/p]/ℤ. Prove that any ℤ-submodule N ⊂ M is either finite or all of M. Deduce that M is an Artinian ℤ-module, and that it is not Noetherian.

Solution: Given q ∈ N, write q = n/p^{e} where n is relatively prime to p. Then there is an m ∈ ℤ with nm ≡ 1 (mod p^{e}). Hence N ∋ m(n/p^{e}) = 1/p^{e}, and so 1/p^{r} = p^{e−r}(1/p^{e}) ∈ N for any 0 ≤ r ≤ e. Therefore, either N = M, or there is a largest integer e ≥ 0 with 1/p^{e} ∈ N. In the second case, N is finite.

Let M ⊋ N_{1} ⊃ N_{2} ⊃ ⋯ be a descending chain. By what we just proved, each N_{i} is finite, say with n_{i} elements. Then the sequence n_{1} ≥ n_{2} ≥ ⋯ stabilizes; say n_{i} = n_{i+1} = · · ·. But N_{i} ⊃ N_{i+1} ⊃ ⋯, so N_{i} = N_{i+1} = · · ·. Thus M is Artinian.

Finally, suppose m_{1}, …, m_{r} generate M, say m_{i} = n_{i}/p^{e_{i}}. Set e := max e_{i}. Then 1/p^{e} generates M, a contradiction since 1/p^{e+1} ∈ M. Thus M is not finitely generated, and so not Noetherian. □

Exercise (16.26). — Let R be an Artinian ring. Prove that R is a field if it is a domain. Deduce that in general every prime ideal 𝔭 of R is maximal.

Solution: Take any nonzero element x ∈ R, and consider the chain of ideals ⟨x⟩ ⊃ ⟨x^{2}⟩ ⊃ · · ·. Since R is Artinian, the chain stabilizes; so ⟨x^{e}⟩ = ⟨x^{e+1}⟩ for some e. Hence x^{e} = ax^{e+1} for some a ∈ R. If R is a domain, then we can cancel to get 1 = ax; thus R is then a field.

In general, R/𝔭 is Artinian by (16.23)(2). Now, R/𝔭 is also a domain by (2.9). Hence, by what we just proved, R/𝔭 is a field. Thus 𝔭 is maximal by (2.17). □

17. Associated Primes

Exercise (17.6). — Given modules M_{1}, …, M_{r}, set M := M_{1} ⊕ ⋯ ⊕ M_{r}. Prove

Ass(M) = Ass(M_{1}) ∪ ⋯ ∪ Ass(M_{r}).

Solution: Set N := M_{2} ⊕ ⋯ ⊕ M_{r}. Then N, M_{1} ⊂ M. Also, M/N = M_{1}. So (17.5) yields

Ass(N), Ass(M_{1}) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(M_{1}).

So Ass(M) = Ass(N) ∪ Ass(M_{1}). The assertion follows by induction on r. □

Exercise (17.7). — Take R := ℤ and M := ℤ/⟨2⟩ ⊕ ℤ. Find Ass(M) and find two submodules L, N ⊂ M with L + N = M but Ass(L) ∪ Ass(N) ⫋ Ass(M).

Solution: First, we have Ass(M) = {⟨0⟩, ⟨2⟩} by (17.6) and (17.4)(2). Next, take L := R·(1, 1) and N := R·(0, 1). Then the canonical maps ℤ → L and ℤ → N are isomorphisms. Hence both Ass(L) and Ass(N) are {⟨0⟩} by (17.4)(2). Finally, L + N = M because (a, b) = a · (1, 1) + (b − a) · (0, 1). □

Exercise (17.10). — Let R be a ring, and suppose R_{𝔭} is a domain for every prime 𝔭. Prove every associated prime of R is minimal.

Solution: Let 𝔭 ∈ Ass(R). Then 𝔭R_{𝔭} ∈ Ass(R_{𝔭}) by (17.9). By hypothesis, R_{𝔭} is a domain. So 𝔭R_{𝔭} = ⟨0⟩ by (17.4). Hence 𝔭 is a minimal prime of R by (11.18)(2).

Alternatively, say 𝔭 = Ann(x) with x ∈ R. Then x/1 ≠ 0 in R_{𝔭}; otherwise, there would be some s ∈ R − 𝔭 such that sx = 0, contradicting 𝔭 = Ann(x). However, for any y ∈ 𝔭, we have xy/1 = 0 in R_{𝔭}. Since R_{𝔭} is a domain and since x/1 ≠ 0, we must have y/1 = 0 in R_{𝔭}. So there exists some t ∈ R − 𝔭 such that ty = 0. Now, 𝔭 ⊃ 𝔮 for some minimal prime 𝔮 by (3.11). Suppose 𝔭 ≠ 𝔮. Then there is some y ∈ 𝔭 − 𝔮. So there exists some t ∈ R − 𝔭 such that ty = 0 ∈ 𝔮, contradicting the primeness of 𝔮. Thus 𝔭 = 𝔮; that is, 𝔭 is minimal. □

Exercise (17.15). — Let R be a Noetherian ring, M a module, N a submodule, x ∈ R. Show that, if x ∉ 𝔭 for any 𝔭 ∈ Ass(M/N), then xM ∩ N = xN.

Solution: Trivially, xN ⊂ xM ⋂ N. Conversely, take m ∈ M with xm ∈ N. Let m′ be the residue of m in M/N. Then xm′ = 0. By (17.14), x ∉ z.div(M/N). So m′ = 0. So m ∈ N. So xm ∈ xN. Thus xM ⋂ N ⊂ xN, as desired. □

Exercise (17.21). — Let R be a Noetherian ring, 𝔞 an ideal. Prove the primes minimal containing 𝔞 are associated to 𝔞. Prove such primes are finite in number.

Solution: Since 𝔞 = Ann(R/𝔞), the primes in question are the primes minimal in Supp(R/𝔞) by (13.9)(3). So they are associated to 𝔞 by (17.17), and they are finite in number by (17.20). □

Exercise (17.22). — Take R := ℤ and M := Z in (17.19). Determine when a chain 0 ⊂ M_{1} ⫋ M is acceptable, and show that then 𝔭_{2} ∉ Ass(M).

Solution: If the chain is acceptable, then M_{1} ≠ 0 as M_{1}/0 ≃ R/𝔭_{1}, and M_{1} is a prime ideal as M_{1} = Ann(M/M_{1}) = 𝔭_{2}. Conversely, the chain is acceptable if M_{1} is a nonzero prime ideal 𝔭, as then M_{1}/0 ≃ R/0 and M/M_{1} ≃ R/𝔭.

Finally, Ass(M) = 0 by (17.4). Further, as just observed, given any acceptable chain, 𝔭_{2} = M_{1} ≠ 0. So 𝔭_{2} ∉ Ass(M). □

Exercise (17.23). — Take R := ℤ and M := Z/⟨12⟩ in (17.19). Find all three acceptable chains, and show that, in each case, {𝔭_{i}} = Ass(M).

Solution: An acceptable chain in M corresponds to chain

⟨12⟩ ⊂ ⟨a_{1}⟩ ⊂ ⟨a_{2}⟩ ⊂ ⋯ ⊂ ⟨a_{n}⟩ = ℤ.

Here ⟨a_{1}⟩/⟨12⟩ ≃ ℤ/⟨p_{1}⟩ with p_{1} prime. So a_{1}p_{1} = 12. Hence the possibilities are p_{1} = 2, a_{1} = 6 and p_{1} = 3, a_{1} = 4. Further, ⟨a_{2}⟩/⟨a_{1}⟩ ≃ ℤ/⟨p_{2}⟩ with p_{2} prime. So a_{2}p_{2} = a_{1}. Hence, if a_{1} = 6, then the possibilities are p_{2} = 2, a_{2} = 3 and p_{2} = 3, a_{2} = 2; if a_{1} = 4, then the only possibility is p_{2} = 2 and a_{2} = 2. In each case, a_{2} is prime; hence, n = 3, and these three chains are the only possibilities. Conversely, each of these three possibilities, clearly, does arise.

In each case, {𝔭_{i}} = {⟨2⟩, ⟨3⟩}. Hence (17.19.1) yields Ass(M) ⊂ {⟨2⟩, ⟨3⟩}. For any M, if 0 ⊂ M_{1} ⊂ ⋯ ⊂ M is an acceptable chain, then (17.5) and (17.4)(2) yield Ass(M) ⊃ Ass(M_{1}) = {𝔭_{1}}. Here, there’s one chain with 𝔭_{1} = ⟨2⟩ and another with 𝔭_{1} = ⟨3⟩; hence, Ass(M) ⊃ {⟨2⟩, ⟨3⟩}. Thus Ass(M) = {⟨2⟩, ⟨3⟩}. □

18. Primary Decomposition Exercise (18.6). — Let R be a ring, and 𝔭 = ⟨p⟩ a principal prime generated by a nonzerodivisor p. Show every positive power 𝔭^{n} is 𝔭-primary, and conversely, if R is Noetherian, then every 𝔭-primary ideal 𝔮 is equal to some power 𝔭^{n}.

Solution: Let’s proceed by induction. Form the exact sequence

0 → 𝔭^{n}/𝔭^{n+1} → R/𝔭^{n+1} → R/𝔭^{n} → 0.

Consider the map R → 𝔭^{n}/𝔭^{n+1} given by x ↦ xp^{n}. It is surjective, and its kernel is 𝔭 as p is a nonzerodivisor. Hence R/𝔭 ⟶^{∼} 𝔭^{n}/𝔭^{n+1}. But Ass(R/𝔭) = {𝔭} by (17.4)(2). Hence (17.5) yields Ass(R/𝔭^{n}) = {𝔭} for every n ≥ 1, as desired.

Conversely, 𝔭 = \sqrt{𝔮} by (18.5). So p^{n} ∈ 𝔮 for some n; take n minimal. Then 𝔭^{n} ⊂ 𝔮. Suppose there is an x ∈ 𝔮 − 𝔭^{n}. Say x = yp^{m} for some y and m ≥ 0. Then m < n as x ∉ 𝔭^{n}. Take m maximal. Now, p^{m} ∉ 𝔮 as n is minimal. So (18.5) yields y ∈ 𝔮 ⊂ 𝔭. Hence y = zp for some z. Then x = zp^{m+1}, contradicting the maximality of m. Thus 𝔮 = 𝔭^{n}. □

Exercise (18.7). — Let k be a field, and k[X, Y] the polynomial ring. Let 𝔞 be the ideal ⟨X^{2}, XY ⟩. Show 𝔞 is not primary, but \sqrt{𝔞} is prime. Show 𝔞 satisfies this condition: ab ∈ 𝔞 implies a^{2} ∈ 𝔞 or b^{2} ∈ 𝔞.

Solution: First, ⟨X⟩ is prime by (2.11). But ⟨X^{2}⟩ ⊂ 𝔞 ⊂ ⟨X⟩. So \sqrt{𝔞} = ⟨X⟩ by (3.26). On the other hand, XY ∈ 𝔞, but X ∉ 𝔞 and Y ∉ \sqrt{𝔞}; thus 𝔞 is not primary by (18.5). If ab ∈ 𝔞, then X | a or X | b, so a^{2} ∈ 𝔞 or b^{2} ∈ 𝔞. □

Exercise (18.8). — Let φ: R → R′ be a homomorphism of Noetherian rings, and 𝔮 ⊂ R′ a 𝔭-primary ideal. Show that φ^{−1}𝔮 ⊂ R is φ^{−1}𝔭-primary. Show that the converse holds if φ is surjective.

Solution: Let xy ∈ φ^{−1}𝔮, but x ∉ φ^{−1}𝔮. Then φ(x)φ(y) ∈ 𝔮, but φ(x) ∉ 𝔮. So φ(y)^{n} ∈ 𝔮 for some n ≥ 1 by (18.5). Hence, y^{n} ∈ φ^{−1}𝔮. So φ^{−1}𝔮 is primary by (18.5). Its radical is φ^{−1}𝔭 as 𝔭 = \sqrt{𝔮}, and taking the radical commutes with taking the inverse image by (3.27). The converse can be proved similarly. □

Exercise (18.16). — Let k be a field, R := k[X, Y, Z] be the polynomial ring. Set 𝔞 := ⟨XY, X − Y Z⟩, set 𝔮_{1} := ⟨X, Z⟩ and set 𝔮_{2} := ⟨Y^{2}, X − Y Z⟩. Show that 𝔞 = 𝔮_{1} ∩𝔮_{2} holds and that this expression is an irredundant primary decomposition.

Solution: First, XY = Y (X −Y Z)+Y^{2}Z ∈ 𝔮_{2}. Hence 𝔞 ⊂ 𝔮_{1} ∩𝔮_{2}. Conversely, take F ∈ 𝔮_{1} ∩ 𝔮_{2}. Then F ∈ 𝔮_{2}, so F = GY^{2} + H(X − Y Z) with G, H ∈ R. But F ∈ 𝔮_{1}, so G ∈ 𝔮_{1}; say G = AX + BZ with A, B ∈ R. Then

F = (AY + B)XY + (H − BY)(X − ZY) ∈ 𝔞.

Thus 𝔞 ⊃ 𝔮_{1} ∩ 𝔮_{2}. Thus 𝔞 = 𝔮_{1} ∩ 𝔮_{2} holds.

Finally, 𝔮_{1} is prime by (2.11). Now, using (18.8), let’s show 𝔮_{2} is ⟨X, Y ⟩- primary. Form φ: k[X, Y, Z] → k[Y, Z] with φ(X) := Y Z. Clearly, 𝔮_{2} = φ^{−1}⟨Y^{2}⟩ and ⟨X, Y ⟩ = φ^{−1}⟨Y ⟩; also, ⟨Y^{2}⟩ is ⟨Y ⟩-primary by (18.2). Thus 𝔞 = 𝔮_{1} ∩ 𝔮_{2} is a primary decomposition. It is irredundant as 𝔮_{1} and ⟨X, Y ⟩ are distinct. □

Exercise (18.17). — Let R := R′ × R′′ be a product of two domains. Find an irredundant primary decomposition of ⟨0⟩.

Solution: Set 𝔭′ := ⟨0⟩ × R′′ and 𝔭′′ := R′′ × ⟨0⟩. Then 𝔭′ and 𝔭′′ are prime by (2.12), so primary by (17.4)(2). Clearly ⟨0⟩ = 𝔭′ ∩ 𝔭′′. Thus this representation is a primary decomposition; it is irredundant as both 𝔭′ and 𝔭′′ are needed. □

Exercise (18.21). — Let R be a Noetherian ring, 𝔞 an ideal, and M a finitely generated module. Consider the following submodule of M:

Γ_{𝔞}(M) := ⋃_{n≥1}{m ∈ M | 𝔞^{n}m = 0 for some n ≥ 1}.

(1) For any decomposition 0 = ⋂ Q_{i} with Q_{i} 𝔭_{i}-primary, show Γ_{𝔞}(M) = ⋂_{𝔞⊄𝔭_{i}} Q_{i}.

(2) Show Γ_{𝔞}(M) is the set of all m ∈ M such that m/1 ∈ M_{𝔭} vanishes for every prime 𝔭 with 𝔞 ⊄ 𝔭. (Thus Γ_{𝔞}(M) is the set of all m whose support lies in V(𝔞).)

Solution: For (1), given m ∈ Γ_{𝔞}(M), say 𝔞^{n}m = 0. Given i with 𝔞 ⊄ 𝔭_{i}, take a ∈ 𝔞 − 𝔭_{i}. Then a^{n}m = 0 ∈ Q_{i}. Hence m ∈ Q_{i} by (18.4). Thus m ∈ ⋂_{𝔞⊄𝔭_{i}} Q_{i}.

Conversely, given m ∈ ⋂_{𝔞⊄𝔭_{i}} Q_{i}, take any j with 𝔞 ⊂ 𝔭_{j}. Now, 𝔭_{j} = nil(M/Q_{j}) by (18.3). So there is n_{j} with 𝔞^{n_{j}}m ∈ Q_{j}. Set n := max{n_{j}}. Then a^{n}m ∈ Q_{i} for all i, if 𝔞 ⊂ 𝔭_{i} or not. Hence a^{n}m ∈ ⋂ Q_{i} = 0. Thus m ∈ Γ_{𝔞}(M).

For (2), given m ∈ Γ_{𝔞}(M), say 𝔞^{n}m = 0. Given a prime 𝔭 with 𝔞 ⊄ 𝔭, take a ∈ 𝔞 − 𝔭. Then a^{n}m = 0 and a^{n} ∉ 𝔭. So m/1 ∈ M_{𝔭} vanishes.

Conversely, given an m ∈ M such that m/1 ∈ M_{𝔭} vanishes for every prime 𝔭 with 𝔞 ⊄ 𝔭, consider a decomposition 0 = ⋂ Q_{i} with Q_{i} 𝔭_{i}-primary; one exists by (18.20). By (1), it suffices to show m ∈ Q_{i} if 𝔞 ⊄ 𝔭_{i}. But m/1 ∈ M_{𝔭_{i}} vanishes. So there’s an a ∈ R − 𝔭_{i} with am = 0 ∈ Q_{i}. So (18.4) yields m ∈ Q_{i}, as desired. □

Exercise (18.25). — Let R be a Noetherian ring, M a finitely generated module, N a submodule. Prove N = ⋂_{𝔭∈Ass(M/N)} φ_{𝔭}^{−1}(N_{𝔭}).

Solution: (18.20) yields an irredundant primary decomposition N = ⋂_{1}^{r} Q_{i}. Say Q_{i} is 𝔭_{i}-primary. Then {𝔭_{i}}_{1}^{r} = Ass(M/N) by (18.19). Also, (18.23) yields φ_{𝔭_{i}}^{−1}(N_{𝔭_{i}}) = ⋂_{𝔭_{j}⊂𝔭_{i}} Q_{j}. Thus ⋂_{1}^{r} φ_{𝔭_{i}}^{−1}(N_{𝔭_{i}}) = ⋂_{1}^{r}(⋂_{𝔭_{j}⊂𝔭_{i}} Q_{j}) = ⋂_{1}^{r} Q_{i} = N. □

Exercise (18.26). — Let R be a Noetherian ring, 𝔭 a prime. Its nth symbolic power 𝔭^{(n)} is defined as the saturation (𝔭^{n})^{S} where S := R − 𝔭.

(1) Show 𝔭^{(n)} is the 𝔭-primary component of 𝔭^{n}.

(2) Show 𝔭^{(m+n)} is the 𝔭-primary component of 𝔭^{(n)}𝔭^{(m)}.

(3) Show 𝔭^{(n)} = 𝔭^{n} if and only if 𝔭^{n} is 𝔭-primary.

(4) Given a 𝔭-primary ideal 𝔮, show 𝔮 ⊃ 𝔭^{(n)} for all large n.

Solution: Clearly, 𝔭 is minimal in V(𝔭^{n}). But V(𝔭^{n}) = Supp(R/𝔭^{n}) by (13.9). Hence 𝔭 is minimal in Ass(R/𝔭^{n}) by (17.17) and (17.3). Thus (18.24) yields (1).

Clearly, (11.15)(3) yields (𝔭^{(m)}𝔭^{(n)})^{S} = 𝔭^{(m+n)}. Thus (18.24) yields (2).

If 𝔭^{(n)} = 𝔭^{n}, then 𝔭^{n} is 𝔭-primary by (1). Conversely, if 𝔭^{n} is 𝔭-primary, then 𝔭^{n} = 𝔭^{(n)} because primary ideals are saturated by (18.22). Thus (3) holds.

For (4), recall 𝔭 = \sqrt{𝔮} by (18.5). So 𝔮 ⊃ 𝔭^{n} for all large n by (3.26). Hence 𝔮^{S} ⊃ 𝔭^{(n)}. But 𝔮^{S} = 𝔮 by (18.22) since 𝔭 ∩ (R − 𝔭) = ∅. Thus (4) holds. □

Exercise (18.27). — Let R be a Noetherian ring, ⟨0⟩ = 𝔮_{1}∩⋯∩𝔮_{n} an irredundant primary decomposition. Set 𝔭_{i} := \sqrt{𝔮_{i}} for i = 1, …, n.

(1) Suppose 𝔭_{i} is minimal for some i. Show 𝔮_{i} = 𝔭_{i}^{(r)} for all large r.

(2) Suppose 𝔭_{i} is not minimal for some i. Show that replacing 𝔮_{i} by 𝔭_{i}^{(r)} for large r gives infinitely many distinct irredundant primary decompositions of ⟨0⟩.

Solution: Set A := R_{𝔭_{i}} and 𝔪 := 𝔭_{i}A. Then A is Noetherian by (16.7). Suppose 𝔭_{i} is minimal. Then 𝔪 is the only prime in A. So 𝔪 = \sqrt{⟨0⟩} by the Scheinnullstellensatz (3.22). So 𝔪^{r} = 0 for all large r by (3.25). So p^{(r)} = 𝔮_{i} by the Second Uniqueness Theorem (18.24) and Lemma (18.22). Thus (1) holds.

Suppose 𝔭_{i} is not minimal. Then the 𝔪^{r} are distinct. Otherwise, 𝔪^{r} = 𝔪^{r+1} for some r. So 𝔪^{r} = 0 by Nakayama’s Lemma (10.10). But then 𝔪 is minimal; so 𝔭_{i} is too, contrary to hypothesis. So (11.17)(1) implies the 𝔭_{i}^{(r)} are distinct.

However, 𝔮_{i} ⊃ 𝔭_{i}^{(r)} for all large r by (18.26)(4). Hence ⟨0⟩ = 𝔭_{i}^{(r)} ∩ ⋂_{j≠i} 𝔮_{j}.

But 𝔭_{i}^{(r)} is 𝔭_{i}-primary by (18.26)(1). Thus replacing 𝔮_{i} by 𝔭_{i}^{(r)} for large r gives infinitely many distinct primary decompositions of ⟨0⟩.

These decompositions are irredundant owing to two applications of (18.18). A first yields {𝔭_{i}} = Ass(R) as ⟨0⟩ = 𝔮_{1} ∩ ⋯ ∩ 𝔮_{n} is irredundant. So a second yields the desired irredundancy. □

Exercise (18.29). — Let R be a Noetherian ring, 𝔪 ⊂ rad(R) an ideal, M a finitely generated module, and M′ a submodule. Considering M/N, show that

M′ = ⋂_{n≥0}(𝔪^{n}M + M′).

Solution: Set N := ⋂_{n≥0} 𝔪^{n}(M/M′). Then by (18.28), there is x ∈ 𝔪 such that (1 + x)N = 0. By (3.2), 1 + x is a unit since 𝔪 ⊂ rad(R). Therefore, N = (1 + x^{−1})(1 + x)N = ⟨0⟩. However, 𝔪^{n}(M/M′) = (𝔪^{n}M + M′)/M′. Thus ⋂(𝔪^{n}M + M′)/M′ = 0, as desired. □

19. Length

Exercise (19.2). — Let R be a ring, M a module. Prove these statements: (1) If M is simple, then any nonzero element m ∈ M generates M.

(2) M is simple if and only if M ≃ R/𝔪 for some maximal ideal 𝔪, and if so, then 𝔪 = Ann(M).

(3) If M has finite length, then M is finitely generated.

Solution: Obviously, Rm is a nonzero submodule. So it is equal to M, because M is simple. Thus (1) holds.

Assume M is simple. Then M is cyclic by (1). So M ≃ R/𝔪 for 𝔪 := Ann(M) by (4.7). Since M is simple, 𝔪 is maximal owing to the bijective correspondence of (1.8). By the same token, if, conversely, M ≃ R/𝔪 with 𝔪 maximal, then M is simple. Thus (2) holds.

Assume ℓ(M) < ∞. Let M = M_{0} ⊃ M_{1} ⊃ ⋯ ⊃ M_{m} = 0 be a composition series. If m = 0, then M = 0. Assume m ≥ 1. Then M_{1} has a composition series of length m − 1. So, by induction on m, we may assume M_{1} is finitely generated. Further, M/M_{1} is simple, so finitely generated by (1). Hence M is finitely generated by (16.15)(1). Thus (3) holds. □

Exercise (19.4). — Let R be a Noetherian ring, M a finitely generated module. Prove the equivalence of the following three conditions:

(1) that M has finite length;

(2) that Supp(M) consists entirely of maximal ideals;

(3) that Ass(M) consists entirely of maximal ideals.

Prove that, if the conditions hold, then Ass(M) and Supp(M) are equal and finite.

Solution: If (1) holds, then (2) holds owing to (19.3). If (2) holds, then (1) holds owing to (17.19) and (19.2)(2). Finally, (17.16) and (17.20) imply that (2) and (3) are equivalent and that the last assertion holds. □

Exercise (19.5). — Let R be a Noetherian ring, 𝔮 a 𝔭-primary ideal. Consider chains of primary ideals from 𝔮 to 𝔭. Show (1) all such chains have length at most ℓ(A) where A := (R/𝔮)_{𝔭} and (2) all maximal chains have length exactly ℓ(A).

Solution: There is a natural bijective correspondence between the 𝔭-primary ideals containing 𝔮 and the (𝔭/𝔮)-primary ideals of R/𝔮, owing to (18.8). In turn, there is one between the latter ideals and the ideals of A primary for its maximal ideal 𝔪, owing to (18.8) again and also to (18.22) with M := A.

However, 𝔭 = \sqrt{𝔮} by (18.5). So 𝔪 = \sqrt{⟨0⟩}. Hence every ideal of A is 𝔪-primary by (18.10). Further, 𝔪 is the only prime of A; so ℓ(A) is finite by (19.4) with M := A. Hence (19.3) with M := A yields (1) and (2). □

Exercise (19.8). — Let k be a field, and R a finitely generated k-algebra. Prove that R is Artinian if and only if R is a finite-dimensional k-vector space.

Solution: Since k is Noetherian by (16.1) and since R is a finitely generated k-algebra, R is Noetherian by (16.11). Assume R is Artinian. Then ℓ(R) < ∞ by (19.6). So R has a composition series. The successive quotients are isomorphic to residue class fields by (19.2)(2). These fields are finitely generated k-algebras, since R is so. Hence these fields are finite extension fields of k by the Weak Nullstellensatz. Thus R is a finite-dimensional k-vector space. The converse holds by (16.24). □

Exercise (19.10). — Let k be a field, A a local k-algebra. Assume the map from k to the residue field is bijective. Given an A-module M, prove ℓ(M) = dim_{k}(M).

Solution: If M = 0, then ℓ(M) = 0 and dim_{k}(M) = 0. If M ≅ k, then ℓ(M) = 1 and dim_{k}(M) = 1. Assume 1 ≤ ℓ(M) < ∞. Then M has a submodule M′ with M/M′ ≅ k. So Additivity of Length, (19.9), yields ℓ(M′) = ℓ(M) − 1 and dim_{k}(M′) = dim_{k}(M) − 1. Hence ℓ(M′) = dim_{k}(M′) by induction on ℓ(M). Thus ℓ(M) = dim_{k}(M).

If ℓ(M) = ∞, then for every m ≥ 1, there exists a chain of submodules,

M = M_{0} ⫌ M_{1} ⫌ ⋯ ⫌ M_{m} = 0.

Hence dim_{k}(M) = ∞. □

Exercise (19.12). — Prove these conditions on a Noetherian ring R equivalent: (1) that R is Artinian;

(2) that Spec(R) is discrete and finite;

(3) that Spec(R) is discrete.

Solution: Condition (1) holds, by (19.11), if and only if Spec(R) consists of finitely points and each is a maximal ideal. But a prime 𝔭 is a maximal ideal if and only if {𝔭} is closed in Spec(R) by (13.2). It follows that (1) and (2) are equivalent.

Trivially, (2) implies (3). Conversely, (3) implies (2), since Spec(R) is quasicompact by (13.4). Thus all three conditions are equivalent. □

Exercise (19.13). — Let R be an Artinian ring. Show that rad(R) is nilpotent.

Solution: Set 𝔪 := rad(R). Then 𝔪 ⊃ 𝔪^{2} ⊃ ⋯ is a descending chain. So 𝔪^{r} = 𝔪^{r+1} for some r. But R is Noetherian by Akizuki’s Theorem (19.11). So 𝔪 is finitely generated. Thus Nakayama’s Lemma (10.10) yields 𝔪^{r} = 0. □

Exercise (19.16). — Let R be a ring, 𝔭 a prime ideal, and R′ a module-finite R-algebra. Show that R′ has only finitely many primes 𝔭′ over 𝔭, as follows: reduce to the case that R is a field by localizing at 𝔭 and passing to the residue rings.

Solution: First note that, if 𝔭′ ⊂ R′ is a prime lying over 𝔭, then 𝔭′R′_{𝔭} ⊂ R′_{𝔭} is a prime lying over the maximal ideal 𝔭R_{𝔭}. Hence, by (11.18)(2), it suffices to show that R′_{𝔭} has only finitely many such primes. Note also that R′_{𝔭} is module-finite over R_{𝔭}. Hence we may replace R and R′ by R_{𝔭} and R′_{𝔭} , and thus assume that 𝔭 is the unique maximal ideal of R. Similarly, we may replace R and R′ by R/𝔭 and R′/𝔭R′, and thus assume that R is a field.

There are a couple of ways to finish. First, R′ is now Artinian by (19.15) or by (16.24); hence, R′ has only finitely many primes by (19.11). Alternatively, every prime is now minimal by incomparability (14.3)(2). Further, R′ is Noetherian by (16.11); hence, R′ has only finitely many minimal primes by (17.21). □

Exercise (19.18). — Let R be a Noetherian ring, and M a finitely generated module. Prove the following four conditions are equivalent:

(1) that M has finite length;

(2) that M is annihilated by some finite product of maximal ideals ∏ 𝔪_{i};

(3) that every prime 𝔭 containing Ann(M) is maximal;

(4) that R/Ann(M) is Artinian.

Solution: Assume (1) holds. Let M = M_{0} ⊃ ⋯ ⊃ M_{m} = 0 be a composition series, and set 𝔪_{i} := Ann(M_{i−1}/M_{i}). Then 𝔪_{i} is maximal by (19.2)(2). Further, 𝔪_{i}M_{i−1} ⊂ M_{i}. Hence 𝔪_{i} ⋯ 𝔪_{1}M_{0} ⊂ M_{i}. Thus (2) holds.

If (2) holds, then (3) does too. Indeed, if 𝔭 ⊃ Ann(M) ⊃ ∏ 𝔪_{i}, then 𝔭 ⊃ 𝔪_{i} for some i by (2.2) as 𝔭 is prime, and so 𝔭 = 𝔪_{i} as 𝔪_{i} is maximal.

Assume (3) holds. Then dim(R/Ann(M)) = 0. But, by (16.7), any quotient of R is Noetherian. Hence Akizuki’s Theorem (19.11) yields (4).

If (4) holds, then (19.14) yields (1), because M is a finitely generated module over R/ Ann(M) owing to (4.5). □

20. Hilbert Functions

Exercise (20.5). — Let k be a field, k[X, Y] the polynomial ring. Show ⟨X, Y^{2}⟩ and ⟨X^{2}, Y^{2}⟩ have different Hilbert Series, but the same Hilbert Polynomial.

Solution: Set 𝔪 := ⟨X, Y ⟩ and 𝔞 := ⟨X, Y^{2}⟩ and 𝔟 := ⟨X^{2}, Y^{2}⟩. They are graded by degree. So ℓ(𝔞_{1}) = 1, and ℓ(𝔞_{n}) = ℓ(𝔪_{n}) for all n ≥ 2. Further, ℓ(𝔟_{1}) = 0, ℓ(𝔟_{2}) = 2, and ℓ(𝔟_{n}) = ℓ(𝔪_{n}) for n ≥ 3. Thus the three ideals have the same Hilbert Polynomial, namely h(n) = n + 1, but different Hilbert Series. □

Exercise (20.6). — Let R = ⨁ R_{n} be a graded ring, M = ⨁ M_{n} a graded Rmodule. Let N = ⨁ N_{n} be a homogeneous submodule; that is, N_{n} = N ∩ M_{n}. Assume R_{0} is Artinian, R is a finitely generated R_{0}-algebra, and M is a finitely generated R-module. Set

N′ := { m ∈ M | there is k_{0} such that R_{k}m ∈ N for all k ≥ k_{0} }.

(1) Prove that N′ is a homogeneous submodule of M with the same Hilbert Polynomial as N, and that N′ is the largest such submodule.

(2) Let N = ⋂ Q_{i} be a decomposition with Q_{i} 𝔭_{i}-primary. Set R_{+} := ⨁_{n>0} R_{n}. Prove that N′ = ⋂_{𝔭_{i}⊅R_{+}} Q_{i}.

Solution: Given m = ∑ m_{i} ∈ N′, say R_{k}m ⊂ N. Then R_{k}m_{i} ⊂ N since N is homogeneous. Hence m_{i} ∈ N′. Thus N′ is homogeneous.

By (19.11) and (16.11), R is Noetherian. So N′ is finitely generated by (16.18). Let n_{1}, …, n_{r} be homogeneous generators of N′ with n_{i} ∈ N_{k_{i}}; set k′ := max{k_{i}}. There is k such that R_{k}n_{i} ∈ N for all i. Given ℓ ≥ k + k′, take n ∈ N′_{ℓ}, and write n = ∑ y_{i}n_{i} with y_{i} ∈ R_{ℓ−k_{i}}. Then y_{i}n_{i} ∈ N_{ℓ} for all i. So n ∈ N_{ℓ}. Thus N′_{ℓ} = N_{ℓ} for all ℓ ≥ k + k′. Thus N and N′ have the same Hilbert polynomial.

Say N′′ ⊃ N, and both have the same Hilbert Polynomial. Then there is k_{0} with ℓ(N′′_{k}) = ℓ(N_{k}) for all k ≥ k_{0}. So N′′_{k} = N_{k} for all k ≥ k_{0}. So, if n ∈ N′′, then R_{k}n ∈ N for all k ≥ k_{0}. Thus N′′ ⊂ N′. Thus (1) holds.

To prove (2), note 0 = ⋂(Q_{i}/N) in M/N. By (18.21),

Γ_{R_{+}}(M/N) = ⋂_{𝔭_{i}⊅R_{+}}(Q_{i}/N).

But clearly Γ_{R_{+}}(M/N) = N′/N. Thus N′ = ⋂_{𝔭_{i}⊅R_{+}} Q_{i}. □

Exercise (20.9). — Let k be a field, P := k[X, Y, Z] the polynomial ring in three variables, f ∈ P a homogeneous polynomial of degree d ≥ 1. Set R := P/⟨f⟩. Find the coefficients of the Hilbert Polynomial h(R, n) explicitly in terms of d.

Solution: Clearly, the following sequence is exact:

0 → P (−d) ⟶^{μ_{f}} P → R → 0.

Hence, Additivity of Length, (19.9), yields h(R, n) = h(P, n) − h(P (−d), n). But P (−d)_{n} = P (n − d), so h(P (−d), n) = h(P, n − d). Therefore, (20.4) yields

h(R, n) = \binom{2+n}{2} − \binom{2−d+n}{2} = dn − (d − 3)d/2.  □

Exercise (20.10). — Under the conditions of (20.8), assume there is a homogeneous nonzerodivisor f ∈ R with M_{f} = 0. Prove deg(h(R, n)) > deg(h(M, n)); start with the case M := R/⟨f^{k}⟩.

Solution: Suppose M := R/⟨f^{k}⟩. Set c := k deg(f). Form the exact sequence 0 → R(−c) ⟶^{μ} R → M → 0 where μ is multiplication by f^{k}. Then Additivity of Length (19.9) yields h(M, n) = h(R, n) − h(R, n − c). But

h(R, n) = \frac{e(1)}{(d−1)!}n^{d−1} + ⋯ and h(R, n − c) = \frac{e(1)}{(d−1)!}(n − c)^{d−1} + · · · .

by (20.8). Thus deg(h(R, n)) > deg(h(M, n)).

In the general case, there is k with f^{k}M = 0 by (12.7). Set M′ := R/⟨f^{k}⟩. Then generators m_{i} ∈ M_{c_{i}} for 1 ≤ i ≤ r yield a surjection ⨁_{i} M′(−c_{i}) → → M. Hence ∑_{i} ℓ(M′_{n−c_{i}}) ≥ ℓ(M_{n}) for all n. But deg(h(M′(−c_{i}), n)) = deg(h(M′, n)). Hence deg(h(M′, n)) ≥ deg(h(M, n)). But deg(h(R, n)) > deg(h(M′, n)) by the first case. Thus deg(h(R, n)) > deg(h(M, n)). □

Exercise (20.15). — Let R be a Noetherian ring, 𝔮 an ideal, and M a finitely generated module. Assume ℓ(M/𝔮M) < ∞. Set 𝔪 := \sqrt{𝔮}. Show

deg p_{𝔪}(M, n) = deg p_{𝔮}(M, n).

Solution: There is an m such that 𝔪 ⊃ 𝔮 ⊃ 𝔪^{m} by (3.25). Hence

𝔪^{n}M ⊃ 𝔮^{n}M ⊃ 𝔪^{mn}M

for all n ≥ 0. Dividing into M and extracting lengths yields

ℓ(M/𝔪^{n}M) ≤ ℓ(M/𝔮^{n}M) ≤ ℓ(M/𝔪^{mn}M).

Therefore, for large n, we get

p_{𝔪}(M, n) ≤ p_{𝔮}(M, n) ≤ p_{𝔪}(M, nm).

The two extremes are polynomials in n with the same degree, say d, (but not the same leading coefficient). Dividing by n^{d} and letting n → ∞, we conclude that the polynomial p_{𝔮}(M, n) also has degree d. □

Exercise (20.19). — Derive the Krull Intersection Theorem, (18.28), from the Artin–Rees Lemma, (20.18).

Solution: In the notation of (18.28), we must prove that N = 𝔞N. So apply the Artin–Rees Lemma to N and the 𝔞-adic filtration of M; we get an m such that 𝔞(N ∩ 𝔞^{m}M) = N ∩ 𝔞^{m+1}M. But N ∩ 𝔞^{n}M = N for all n ≥ 0. Thus N = 𝔞N. □

20. Appendix: Homogeneity

Exercise (20.22). — Let R = ⨁ R_{n} be a graded ring, M = ⨁_{n≥n_{0}} M_{n} a graded module, 𝔞 ⊂ ⨁_{n>0} R_{n} a homogeneous ideal. Assume M = 𝔞M. Show M = 0.

Solution: Suppose M ≠ 0; say M_{n_{0}} ≠ 0. Note M = 𝔞M ⊂ ⨁_{n>n_{0}} M_{n}; hence M_{n_{0}} = 0, a contradiction. Thus M = 0. □

Exercise (20.23). — Let R = ⨁ R_{n} be a Noetherian graded ring, M = ⨁ M_{n} a finitely generated graded R-module, N = ⨁ N_{n} a homogeneous submodule. Set

N′ := { m ∈ M | R_{n}m ∈ N for all n ≫ 0 }.

Show that N′ is the largest homogeneous submodule of M containing N and having, for all n ≫ 0, its degree-n homogeneous component N′_{n} equal to N_{n}.

Solution: Given m, m′ ∈ N′, say R_{n}m, R_{n}m′ ∈ N for n ≫ 0. Let x ∈ R. Then R_{n}(m + m′), R_{n}xm ∈ N for n ≫ 0. So N′ ⊂ M is a submodule. Trivially N ⊂ N′. Let m_{i} be a homogeneous component of m. Then R_{n}m_{i} ∈ N for n ≫ 0 as N is homogeneous. Thus N′ ⊂ M is a homogeneous submodule containing N.

Since R is Noetherian and M is finitely generated, N′ is finitely generated, say by g, g′, …, g^{(r)}. Then there is n_{0} with R_{n}g, R_{n}g′, …, R_{n}g^{(r)} ∈ N for n ≥ n_{0}. Replace g, g′, …, g^{(r)} by their homogeneous components. Say g, g′, …, g^{(r)} are now of degrees d, d′, …, d^{(r)} with d ≥ d′ ≥ ⋯ ≥ d^{(r)}. Set n_{1} := d + n_{0}.

Given m ∈ N′_{n} with n ≥ n_{1}, say m = xg + x′g′ + ⋯ with x ∈ R_{n−d} and x′ ∈ R_{n−d′} and so on. Then n_{0} ≤ n − d ≤ n − d′ ≤ · · ·. Hence m ∈ N_{n}. Thus N′_{n} ⊂ N_{n}. But N′ ⊃ N. Thus N′_{n} = N_{n} for n ≥ n_{1}, as desired.

Let N′′ = ⨁ N′′_{n} ⊂ M be homogeneous with N′′_{n} = N_{n} for n ≥ n_{2}. Let m ∈ N′′ and p ≥ n_{2}. Then R_{p}m ∈ ⨁_{n≥n_{2}} N′′_{n} ⊂ N. So m ∈ N′. Thus N′′ ⊂ N′. □

Exercise (20.25). — Let R be a graded ring, 𝔞 a homogeneous ideal, and M a graded module. Prove that \sqrt{𝔞} and Ann(M) and nil(M) are homogeneous.

Solution: Take x = ∑_{i≥r}^{r+n} x_{i} ∈ R with the x_{i} the homogeneous components. First, suppose x ∈ \sqrt{𝔞}. Say x^{k} ∈ 𝔞. Either x_{r}^{k} vanishes or it is the initial

component of x^{k}. But 𝔞 is homogeneous. So x_{r}^{k} ∈ 𝔞. So x_{r} ∈ \sqrt{𝔞}. So x − x_{r} ∈ \sqrt{𝔞} by (3.24). So all the x_{i} are in \sqrt{𝔞} by induction on n. Thus \sqrt{𝔞} is homogeneous.

Second, suppose x ∈ Ann(M). Let m ∈ M. Then 0 = xm = ∑ x_{i}m. If m is homogeneous, then x_{i}m = 0 for all i, since M is graded. But M has a set of homogeneous generators. Thus x_{i} ∈ Ann(M) for all i, as desired.

Finally, nil(M) is homogeneous, as nil(M) = \sqrt{Ann(M)} by (13.10). □

Exercise (20.26). — Let R be a Noetherian graded ring, M a finitely generated graded module, Q a submodule. Let Q^{∗} ⊂ Q be the submodule generated by the homogeneous elements of Q. Assume Q is primary. Then Q^{∗} is primary too.

Solution: Let x ∈ R and m ∈ M be homogeneous with xm ∈ Q^{∗}. Assume x ∉ nil(M/Q^{∗}). Then, given ℓ ≥ 1, there is m′ ∈ M with x^{ℓ}m′ ∉ Q^{∗}. So m′ has a homogeneous component m′′ with x^{ℓ}m′′ ∉ Q^{∗}. Then x^{ℓ}m′′ ∉ Q by definition of Q^{∗}. Thus x ∉ nil(M/Q). Since Q is primary, m ∈ Q by (18.4). Since m is homogeneous, m ∈ Q^{∗}. Thus Q^{∗} is primary by (20.24). □

Exercise (20.30). — Under the conditions of (20.8), assume that R is a domain and that its integral closure \overline{R} in Frac(R) is a finitely generated R-module.

(1) Prove that there is a homogeneous f ∈ R with R_{f} = \overline{R}_{f}.

(2) Prove that the Hilbert Polynomials of R and \overline{R} have the same degree and same leading coefficient.

Solution: Let x_{1}, …, x_{r} be homogeneous generators of \overline{R} as an R-module. Write x_{i} = a_{i}/b_{i} with a_{i}, b_{i} ∈ R homogeneous. Set f := ∏ b_{i}. Then fx_{i} ∈ R for each i. So \overline{R}_{f} = R_{f}. Thus (1) holds.

Consider the short exact sequence 0 → R → \overline{R} → \overline{R}/R → 0. Then (\overline{R}/R)_{f} = 0 by (12.16). So deg(h(\overline{R}/R, n)) < deg(h(\overline{R}, n)) by (20.10) and (1). But

h(\overline{R}, n) = h(R, n) + h(\overline{R}/R, n)

by (19.9) and (20.8). Thus (2) holds. □

21. Dimension

Exercise (21.8). — Let R be a Noetherian ring, and 𝔭 be a prime minimal containing x_{1}, …, x_{r}. Given r′ with 1 ≤ r′ ≤ r, set R′ := R/⟨x_{1}, …, x_{r′}⟩ and 𝔭′ := 𝔭/⟨x_{1}, …, x_{r′}⟩. Assume ht(𝔭) = r. Prove ht(𝔭′) = r − r′.

Solution: Let x′_{i} ∈ R′ be the residue of x_{i}. Then 𝔭′ is minimal containing x′_{r′+1}, … x′_{r} by (1.8) and (2.7). So ht(𝔭′) ≤ r − r′ by (21.7).

On the other hand, R′_{𝔭′} = R′_{𝔭} by (11.22), and R′_{𝔭} = R_{𝔭}/⟨x_{1}/1, …, x_{r′}/1⟩ by (12.18) Hence dim(R′_{𝔭′}) ≥ dim(R_{𝔭}) − r′ by repeated application of (21.5). So ht(𝔭′) ≥ r − r′ by (21.6.1), as required. □

Exercise (21.10). — Let R be a Noetherian ring, 𝔭 a prime ideal with ht(𝔭) ≥ 2. Prove 𝔭 is the union of infinitely many distinct prime ideals 𝔮 with ht(𝔮) = 1.

Solution: Dividing R by a minimal prime ideal contained in 𝔭, we may, plainly, assume R is a domain. Given a nonzero x ∈ 𝔭, let 𝔮_{x} ⊂ 𝔭 be a prime ideal minimal containing x. Then ht(𝔮_{x}) = 1 by the Krull Principal Theorem (21.9). Plainly ⋃ 𝔮_{x} = 𝔭. Finally, if there were only finitely many distinct 𝔮_{x}, then by Prime Avoidance (3.15), one 𝔮_{x} would be equal to 𝔭, a contradiction. □

Exercise (21.11). — Let R be a Noetherian ring with only finitely many prime ideals. Show dim(R) ≤ 1.

Solution: By (21.10), there’s no prime 𝔭 with ht(𝔭) ≥ 2. So dim(R) ≤ 1. □

Exercise (21.12). — Let R be a domain. Prove that, if R is a UFD, then every height-1 prime is principal, and that the converse holds if R is Noetherian.

Solution: Let 𝔭 be a height-1 prime. Then there’s a nonzero x ∈ 𝔭. Factor x. One prime factor p must lie in 𝔭 as 𝔭 is prime. Clearly, ⟨p⟩ is a prime ideal as p is a prime element. But ⟨p⟩ ⊂ 𝔭 and ht(p) = 1. Thus, ⟨p⟩ = 𝔭.

Conversely, assume every height-1 prime is principal and assume R is Noetherian. To prove R is a UFD, it suffices to prove every irreducible element p is prime (see [1, Ch. 11, Sec. 2, pp. 392–396]). Let 𝔭 be a prime minimal containing p. By Krull’s Principal Ideal Theorem, ht(𝔭) = 1. So 𝔭 = ⟨x⟩ for some x. Then x is prime by (2.6). And p = xy for some y as p ∈ 𝔭 . But p is irreducible. So y is a unit. Thus p is prime, as desired. □

Exercise (21.13). — (1) Let A be a Noetherian local ring with a principal prime 𝔭 of height at least 1. Prove that A is a domain.

(2) Let k be a field, P := k[[X]] the formal power series ring in one variable. Set R := P × P. Prove that P is Noetherian and semilocal, and that P contains a principal prime 𝔭 of height 1, but that P is not a domain.

Solution: To prove (1), say 𝔭 = ⟨x⟩, and let 𝔮 ⊂ 𝔭 be a minimal prime. Take y ∈ 𝔮. Then y = ax for some a. But x ∉ 𝔮 since ht 𝔭 ≥ 1. Hence a ∈ 𝔮. Thus 𝔮 = 𝔮x. But x lies in the maximal ideal of the local ring A, and 𝔮 is finitely generated since A is Noetherian. Hence Nakayama’s Lemma (10.10) yields 𝔮 = ⟨0⟩. Thus ⟨0⟩ is prime, and so A is a domain.

Alternatively, as a ∈ 𝔮, also a = a_{1}x with a_{1} ∈ 𝔮. Repeating yields an ascending chain of ideals ⟨a⟩ ⊂ ⟨a_{1}⟩ ⊂ ⟨a_{2}⟩ ⊂ · · ·. It must stabilize as A is Noetherian: there’s a k such that a_{k} ∈ ⟨a_{k−1}⟩. Then a_{k} = ba_{k−1} = ba_{k}x for some b. So a_{k}(1 − bx) = 0. But 1 − bx is a unit by (3.4) as A is local. So a_{k} = 0. Hence y = 0 and so 𝔮 = ⟨0⟩. Thus A is a domain.

As to (2), every nonzero ideal of P is of the form ⟨X^{n}⟩ by (3.8). Hence P is Noetherian. Thus R is Noetherian by (16.16).

The primes of R are of the form 𝔮×P or P ×𝔮 where 𝔮 is a prime of P by (2.11). Further, 𝔪 := ⟨X⟩ is the unique maximal ideal by (3.7). Hence R has just two maximal ideals 𝔪 × P and P × 𝔪. Thus R is semilocal.

Set 𝔭 := ⟨(X, 1)⟩. Then 𝔭 = 𝔪 × P. So 𝔭 is a principal prime. Further, 𝔭 contains just one other prime 0 × P. Thus ht(𝔭) = 1.

Finally, R is not a domain as (1, 0) · (0, 1) = 0. □

Exercise (21.14). — Let R be a finitely generated algebra over a field. Assume R is a domain of dimension r. Let x ∈ R be neither 0 nor a unit. Set R′ := R/⟨x⟩. Prove that r − 1 is the length of any chain of primes in R′ of maximal length.

Solution: A chain of primes in R′ of maximal length lifts to a chain of primes 𝔭_{i} in R of maximal length with ⟨x⟩ ⊆ 𝔭_{1} ⫋ ⋯ ⫋ 𝔭_{d}. As x is not a unit, d ≥ 1. As x ≠ 0, also 𝔭_{1} ≠ 0. But R is a domain. So Krull’s Principal Ideal Theorem, (21.8), yields ht 𝔭_{1} = 1. So 0 ⫋ 𝔭_{1} ⫋ ⋯ ⫋ 𝔭_{r} is of maximal length in R. But R is a finitely generated algebra over a field. Hence d = dim R by (15.8). □

Exercise (21.16). — Let R be a Noetherian ring. Show that

dim(R[X]) = dim(R) + 1.

Solution: Let 𝔓 be a prime ideal of R[X], and 𝔭 its contraction in R. Then R_{𝔭} → R[X]_{𝔓} is a flat local homomorphism by (13.24). Hence (21.15) yields

dim(R[X]_{𝔓}) = dim(R_{𝔭}) + dim(R[X]_{𝔓}/𝔭R[X]_{𝔓}).  (21.16.1)

But R[X]_{𝔓}/𝔭R[X]_{𝔓} = k(𝔭)[X]_{𝔓}, owing to (1.6) and (11.29). So its dimension is 1, as k(𝔭)[X] is a PID. Hence if 𝔓 has dim(R[X]_{𝔓}) = dim(R[X]), then (21.16.1) yields dim(R[X]) ≤ dim(R)+1. Thus the desired equality follows from (15.18). □

Exercise (21.17). — Let A be a Noetherian local ring of dimension r. Let 𝔪 be the maximal ideal, and k := A/𝔪 the residue class field. Prove that

r ≤ dim_{k}(𝔪/𝔪^{2}),

with equality if and only if 𝔪 is generated by r elements.

Solution: By (21.4), dim(A) is the smallest number of elements that generate a parameter ideal. But 𝔪 is a parameter ideal, and the smallest number of generators of 𝔪 is dim_{k}(𝔪/𝔪^{2}) by (10.11)(2). The assertion follows. □

Exercise (21.21). — Let A be a Noetherian local ring of dimension r, and x_{1}, …, x_{s} ∈ A with s ≤ r. Set 𝔞 := ⟨x_{1}, …, x_{s}⟩ and B := A/𝔞. Prove these two conditions are equivalent:

(1) A is regular, and there are x_{s+1}, …, x_{r} ∈ A with x_{1}, …, x_{r} a regular sop.

(2) B is regular of dimension r − s.

Solution: Assume (1). Then x_{1}, …, x_{r} generate the maximal ideal 𝔪 of A. So the residues of x_{s+1}, …, x_{r} generate that 𝔫 of B. Hence dim(B) ≥ r − s by (21.4). But dim(B) ≥ r − s by (21.5). So dim(B) = r − s. Thus (2) holds.

Assume (2). Then 𝔫 is generated by r − s elements, say by the residues of x_{s+1}, …, x_{r} ∈ A. Hence 𝔪 is generated by x_{1}, …, x_{r}. Thus (1) holds. □

22. Completion

Exercise (22.3). — In the 2-adic integers, evaluate the sum 1 + 2 + 4 + 8 + · · ·.

Solution: In the 2-adic integers, 1 + 2 + 4 + 8 + ⋯ = 1/(1 − 2) = −1. □

Exercise (22.4). — Let R be a ring, 𝔞 an ideal, and M a module. Prove the following three conditions are equivalent:

(1) κ: M → M̂ is injective;  (2) ⋂ 𝔞^{n}M = ⟨0⟩;  (3) M is separated.

Solution: Clearly, Ker(κ) = ⋂ 𝔞^{n}M; so (1) and (2) are equivalent. Moreover,

(2) and (3) were proved equivalent in (22.1). □

Exercise (22.8). — Let A be a Noetherian semilocal ring, and 𝔪_{1}, …, 𝔪_{m} all its maximal ideals. Prove that Â = ∏ Â_{𝔪_{i}}.

Solution: Set 𝔪 := rad(R). Fix n ≥ 0. Then A/𝔪^{n} is Noetherian of dimension 0; so it’s Artinian by (19.18). Hence (19.17) yields

A/𝔪^{n} = ∏_{i}(A/𝔪^{n})_{(𝔪_{i}/𝔪^{n})}.

However, (A/𝔪^{n})_{(𝔪_{i}/𝔪^{n})} is equal to (A/𝔪^{n})_{𝔪_{i}} by (11.22), so to A_{m_{i}}/𝔪^{n}A_{m_{i}} by Exactness of Localization (12.16). Furthermore, 𝔪^{n} = (∏ 𝔪_{i})^{n} = ⋂ 𝔪_{i}^{n} by (1.13). Now, 𝔪_{i}^{n} is 𝔪_{i}-primary by (18.10). Hence 𝔪^{n}A_{m_{i}} = 𝔪_{i}^{n}A_{m_{i}} by (18.23). Therefore, A/𝔪^{n} = ∏_{i}(A_{𝔪_{i}}/𝔪_{i}^{n}A_{𝔪_{i}}). Taking inverse limits, we obtain the assertion, because inverse limit commutes with finite product by the construction of the limit. □

Exercise (22.9). — Let R be a ring, M a module, M = M_{0} ⊃ M_{1} ⊃ ⋯ a filtration, and N ⊂ M a submodule. Filter N by N_{n} := N ∩ M_{n}. Assume N ⊃ M_{n} for n ≥ n_{0} for some n_{0}. Prove N̂ ⊂ M̂ and M̂/N̂ = M/N and G(M̂) = G(M).

Solution: For each n ≥ n_{0}, form this commutative diagram with exact rows:

0 ⟶ N/M_{n+1} ⟶ M/M_{n+1} ⟶ M/N ⟶ 0
↓  ↓  ↓
0 ⟶ N/M_{n} ⟶ M/M_{n} ⟶ M/N ⟶ 0

The left vertical map is surjective; the right is the identity. So the induced sequence

0 → N̂ → M̂ → M/N → 0

is exact by (22.6) and (22.7). Thus N̂ ⊂ M̂ and M̂/N̂ = M/N.

In particular, M̂/M̂_{n} = M/M_{n} for each n. Therefore, M̂_{n}/M̂_{n+1} = M_{n}/M_{n+1}. Thus G(M̂) = G(M). □

Exercise (22.10). — (1) Let R be a ring, 𝔞 an ideal. If G_{𝔞}(R) is a domain, show R̂ is an domain. If also ⋂_{n≥0} 𝔞^{n} = 0, show R is a domain.

(2) Use (1) to give an alternative proof that a regular local ring A is a domain.

Solution: Consider (1). Let x, y ∈ R̂ be nonzero. Since R̂ is separated there are positive integers r and s with x ∈ 𝔞̂^{r} −𝔞̂^{r+1} and y ∈ 𝔞̂^{s} −𝔞̂^{s+1}. Let x′ ∈ Gr ; 𝔞̂(R̂) and y′ ∈ Gs ; 𝔞̂(R̂) denote the images of x and y. Then x′ ≠ 0 and y′ ≠ 0. Now, G𝔞̂(R̂) = G_{𝔞}(R) by (22.9). Assume G_{𝔞}(R) is a domain. Then x′y′ ≠ 0. Hence x′y′ ∈ Gr+s ; 𝔞̂ is the image of xy ∈ 𝔞̂^{r+s}. Hence xy ≠ 0. Thus R̂ is a domain.

If ⋂_{n≥0} 𝔞^{n} = 0, then R ⊂ R̂ by (22.4); so R is a domain if R̂ is. Thus (1) holds. As to (2), denote the maximal ideal of A by 𝔪. Then ⋂_{n≥0} 𝔪^{n} = ⟨0⟩ by the

Krull Intersection Theorem (18.28), and G_{𝔪}(A) is a polynomial ring by (21.20), so a domain. Hence A is a domain, by (1). Thus (2) holds. □

Exercise (22.12). — Let A be a semilocal ring, 𝔪_{1}, …, 𝔪_{m} all its maximal ideals, and set 𝔪 := rad(A). Prove that Â is a semilocal ring, that 𝔪̂_{1}, …, 𝔪̂_{m} are all its maximal ideals, and that 𝔪̂ = rad(Â).

Solution: First, (22.9) yields Â/𝔪̂ = A/𝔪 and Â/𝔪̂_{i} = A/𝔪_{i}. So 𝔪̂_{i} is maximal. By hypothesis, 𝔪 = ⋂ 𝔪_{i}; so A/𝔪 ⊂ ∏(A/𝔪_{i}). Hence Â/𝔪̂ ⊂ ∏(Â/𝔪̂_{i}); so 𝔪̂ = ⋂ 𝔪̂_{i}. So 𝔪̂ ⊃ rad(Â). But 𝔪̂ ⊂ rad(Â) by (22.2). Thus 𝔪̂ = rad(Â).

Finally, let 𝔪′ be any maximal ideal of Â. Then 𝔪′ ⊃ rad(Â) = ⋂ 𝔪̂_{i}. Hence 𝔪′ ⊃ 𝔪̂_{i} for some i by (2.2). But 𝔪̂_{i} is maximal. So 𝔪′ = 𝔪̂_{i}. Thus 𝔪̂_{1}, …, 𝔪̂_{m} are all the maximal ideals of Â, and so Â is semilocal. □

Exercise (22.15). — Let A be a Noetherian ring, x ∈ A, and x̂ ∈ Â its image. Prove x̂ is a nonzerodivisor if x is. Prove the converse holds if A is semilocal.

Solution: Assume x is a nonzerodivisor. Then the multiplication map μ_{x} is injective on A. So by Exactness of Completion, the induced map μ̂_{x} is injective on Â. But μ̂_{x} = μx̂. Thus x̂ is a nonzerodivisor.

Conversely, assume x̂ is a nonzerodivisor and A is semilocal. Then μ̂_{x} is injective on Â. So its restriction is injective on the image of the canonical map A → Â. But this map is injective, as the completion is taken with respect to the Jacobson radical; further, μ̂_{x} induces μ_{x}. Thus x is a nonzerodivisor. □

Exercise (22.16). — Let p ∈ ℤ be prime. For n > 0, define a ℤ-linear map

α_{n}: ℤ/⟨p⟩ → ℤ/⟨p^{n}⟩ by α_{n}(1) = p^{n−1}.

Set A := ⨁_{n≥1} ℤ/⟨p⟩ and B := ⨁_{n≥1} ℤ/⟨p^{n}⟩. Set α := ⨁ α_{n}; so α: A → B. (1) Show that the p-adic completion Â is just A.

(2) Show that, in the topology on A induced by the p-adic topology on B, the completion \overline{A} is equal to ∏_{n=1}^{∞} ℤ/⟨p⟩.

(3) Show that the natural sequence of p-adic completions

0 → Â \underrightarrow{α̂} B̂ \underrightarrow{κ̂} (B/A)

is not exact at B̂. (Thus p-adic completion is not right-exact on ((ℤ-mod)).)

Solution: For (1), note pA = 0. So every Cauchy sequence is constant. Hence Â = A. Thus (1) holds.

For (2), set A_{k} := α^{−1}(p^{k}B). These A_{k} are the fundamental open neighborhoods of 0 in the topology induced from the p-adic topology of B. So

A_{k} = α^{−1}(0 ⊕ ⋯ ⊕ 0 ⊕ ⨁_{n>k}⟨p^{k}⟩/⟨p^{n}⟩) = (0 ⊕ ⋯ ⊕ 0 ⊕ ⨁_{n>k} ℤ/⟨p⟩).

Hence A/A_{k} = ⨁_{i=1}^{k} ℤ/⟨p⟩ = ∏_{n=1}^{k} ℤ/⟨p⟩. But by (22.7), in the induced topology, the completion \overline{A} is equal to \underleftarrow{lim}_{k≥1} A/A_{k}. Thus

\overline{A} = \underleftarrow{lim}_{k≥1} ∏_{n=1}^{k} ℤ/⟨p⟩.

In general, let M_{1}, M_{2}, … be a sequence of modules, π_{k}^{k+1} : ∏_{n=1}^{k+1} M_{n} → ∏_{n=1}^{k} M_{n} the projections. Then (22.6) yields \underleftarrow{lim}_{k≥1} ∏_{n=1}^{k} M_{n} = ∏_{n=1}^{∞} M_{n}. Thus (2) holds.

For (3), note that, by (2) and (22.6.2), the following sequence is exact:

0 → \overline{A} → B̂ \underrightarrow{κ̂} (B/A) .

But Â = A by (1), and A ≠ \overline{A} as A is countable yet \overline{A} is not. Thus Im(α̂) ≠ Ker(κ̂); that is, (3) holds. □

Exercise (22.18). — Let R be a ring, 𝔞 an ideal. Show that M ↦ M̂ preserves surjections, and that R̂ ⊗ M → M̂ is surjective if M is finitely generated.

Solution: The first part of the proof of (22.14) shows that M ↦ M̂ preserves surjections. So (8.16) yields the desired surjectivity. □

Exercise (22.21). — Let R be a Noetherian ring, and 𝔞 and 𝔟 ideals. Assume 𝔞 ⊂ rad(R), and use the 𝔞-adic toplogy. Prove 𝔟 is principal if 𝔟R̂ is.

Solution: Since R is Noetherian, 𝔟 is finitely generated. But 𝔞 ⊂ rad(R). Hence, 𝔟 is principal if 𝔟/𝔞𝔟 is cyclic by (10.11)(2). But 𝔟/𝔞𝔟 = 𝔟̂/(𝔞𝔟) by (22.9), and 𝔟̂ = 𝔟R̂ by (22.19)(2). Hence, if 𝔟R̂ is principal, then 𝔟/𝔞𝔟 is cyclic, as desired. □

Exercise (22.24) (Nakayama’s Lemma for a complete ring). — Let R be a ring, 𝔞 an ideal, and M a module. Assume R is complete, and M separated. Show m_{1}, …, m_{n} ∈ M generate if their images in M/𝔞M generate.

Solution: Note that the images of m_{1}, …, m_{n} in G(M) generate over G(R). Therefore, m_{1}, …, m_{n} ∈ M generate over R by the proof of (22.23).

Alternatively, M is finitely generated over R and complete by the statement of (22.23). Since M is also separated, M = M̂. Hence M is also an R̂-module. Since R is complete, κ_{R}: R → R̂ is surjective. Now, 𝔞 is closed by (22.1); so 𝔞 is complete; whence, κ_{𝔞}: 𝔞 → 𝔞̂ is surjective too. Hence 𝔞M = 𝔞̂M. Thus M/𝔞M = M/𝔞̂M. So the m_{i} generate M/𝔞̂M. But 𝔞̂ ⊂ rad(R̂) by (22.2). So by Nakayama’s Lemma (10.11)(2), the m_{i} generate M over R̂, so also over R as κ_{R} is surjective. □

Exercise (22.28). — Let A be a Noetherian local ring, 𝔪 the maximal ideal. Prove (1) that Â is a Noetherian local ring with 𝔪̂ as maximal ideal, (2) that dim(A) = dim(Â), and (3) that A is regular if and only if Â is regular.

Solution: First, Â is Noetherian by (22.26), and local with 𝔪̂ as maximal ideal by (22.8); thus (1) holds.

Second, A/𝔪^{n} = Â/𝔪̂^{n} by (22.9). So d(A) = d(Â) by (20.13). Thus (2) holds by (21.4).

Third, 𝔪/𝔪^{2} = 𝔪̂/𝔪̂^{2} by (22.9). So 𝔪 and 𝔪̂ have the same number of generators by (10.13). Thus (3) holds. □

23. Discrete Valuation Rings

Exercise (23.6). — Let R be a ring, M a module, and x, y ∈ R.

(1) Prove that, if x, y form an M-sequence, then, given any m, n ∈ M such that xm = yn, there exists p ∈ M such that m = yp and n = xp.

(2) Prove the converse of (1) if R is local, and x, y lie in its maximal ideal 𝔪, and M is Noetherian.

Solution: Consider (1). Let n_{1} be the residue of n in M_{1} := M/xM. Then yn_{1} = 0, but y ∉ z.div(M_{1}). Hence n_{1} = 0. So there exists p ∈ M such that n = xp. So x(m − yp) = 0. But x ∉ z.div(M). Thus m = yp.

Consider (2). Given m ∈ M such that xm = 0, take n := 0. Then xm = yn; so there exists p ∈ M such that m = yp and n = xp. Repeat with p in place of m, obtaining p_{1} ∈ M such that p = yp_{1} and 0 = xp_{1}. Induction yields p_{i} ∈ M for i ≥ 2 such that p_{i−1} = yp_{i} and 0 = xp_{i}.

Then Rp_{1} ⊂ Rp_{2} ⊂ ⋯ is an ascending chain. It stabilizes as M is Noetherian. Say Rp_{n} = Rp_{n+1}. So p_{n+1} = zp_{n} for some z ∈ R. Then p_{n} = yp_{n+1} = yzp_{n}. So (1 − yz)p_{n} = 0. But y ∈ 𝔪. So 1 − yz is a unit. Hence p_{n} = 0. But m = y^{n+1}p_{n}. Thus m = 0. Thus x ∉ z.div(M).

Given n_{1} ∈ M_{1} := M/xM such that yn_{1} = 0, take n ∈ M with n_{1} as residue. Then yn = xm for some m ∈ M. So there exists p ∈ M such that m = yp and n = xp. Thus n_{1} = 0. Thus y ∉ z.div(M_{1}). Thus x, y form an M-sequence. □

Exercise (23.7). — Let R be a local ring, 𝔪 its maximal ideal, M a Noetherian module, x_{1}, …, x_{n} ∈ 𝔪, and σ a permutation of 1, …, n. Assume x_{1}, …, x_{n} form an M-sequence, and prove x_{σ1}, …, x_{σn} do too; first, say σ transposes i and i + 1.

Solution: Say σ transposes i and i + 1. Set M_{j} := M/⟨x_{1}, …, x_{j}⟩. Then x_{i}, x_{i+1} form an M_{i−1}-sequence; so x_{i+1}, x_{i} do too owing to (23.6). So

x_{1}, …, x_{i−1}, x_{i+1}, x_{i}

form an M-sequence. But M/⟨x_{1}, …, x_{i−1}, x_{i+1}, x_{i}⟩ = M_{i+1}. Hence x_{σ1}, …, x_{σn} form an M-sequence. In general, σ is a composition of transpositions of successive integers; hence, the general assertion follows. □

Exercise (23.8). — Prove that a Noetherian local ring A of dimension r ≥ 1 is regular if and only if its maximal ideal 𝔪 is generated by an A-sequence.

Solution: Assume A is regular. Given a regular sop x_{1}, …, x_{r}, let’s show it’s an A-sequence. Set A_{1} := A/⟨x_{1}⟩. Then A_{1} is regular of dimension r − 1 by (21.21). So x_{1} ≠ 0. But A is a domain by (21.22). So x_{1} ∉ z.div(A). Further, if r ≥ 2, then the residues of x_{2}, …, x_{r} form a regular sop of A_{1}; so we may assume they form an A_{1}-sequence by induction on r. Thus x_{1}, …, x_{r} is an A-sequence.

Conversely, if 𝔪 is generated by an A-sequence x_{1}, …, x_{n}, then n ≤ depth(A) ≤ r by (23.4) and (23.5)(3), and n ≥ r by (21.17); thus n = r, and A is regular. □

Exercise (23.10). — Let A be a DVR with fraction field K, and f ∈ A a nonzero nonunit. Prove A is a maximal proper subring of K. Prove dim(A) ≠ dim(A_{f}).

Solution: Let R be a ring, A ⫋ R ⊂ K. Then there’s an x ∈ R − A. Say x = ut^{n} where u ∈ A^{×} and t is a uniformizing parameter. Then n < 0. Set y := u^{−1}t^{−n−1}. Then y ∈ A. So t^{−1} = xy ∈ R. Hence wt^{m} ∈ R for any w ∈ A^{×} and m ∈ ℤ. Thus R = K, as desired.

Since f is a nonzero nonunit, A ⫋ A_{f} ⊂ K. Hence A_{f} = K by the above. So dim(A_{f}) = 0. But dim(A) = 1 by (23.9). □

Exercise (23.11). — Let k be a field, P := k[X, Y] the polynomial ring in two variables, f ∈ P an irreducible polynomial. Say f = ℓ(X, Y) + g(X, Y) with ℓ(X, Y) = aX + bY for a, b ∈ k and with g ∈ ⟨X, Y ⟩^{2}. Set R := P/⟨f⟩ and 𝔭 := ⟨X, Y ⟩/⟨f⟩. Prove that R_{𝔭} is a DVR if and only if ℓ ≠ 0. (Thus R_{𝔭} is a DVR if and only if the plane curve C : f = 0 ⊂ k^{2} is nonsingular at (0, 0).)

Solution: Set A := R_{𝔭} and 𝔪 := 𝔭A. Then (12.18) and (12.4) yield

A/𝔪 = (R/𝔭)_{𝔭} = k and 𝔪/𝔪^{2} = 𝔭/𝔭^{2}.

First, assume ℓ ≠ 0. Now, the k-vector space 𝔪/𝔪^{2} is generated by the images x and y of X and Y in A. Clearly, the image of f is 0 in 𝔪/𝔪^{2}. Also, g ∈ (X, Y)^{2}; so its image in 𝔪/𝔪^{2} is also 0. Hence, the image of ℓ is 0 in 𝔪/𝔪^{2}; that is, x and y are linearly dependent. Now, f cannot generate ⟨X, Y ⟩, so 𝔪 ≠ 0; hence, 𝔪/𝔪^{2} ≠ 0 by Nakayama’s Lemma, (10.10). Therefore, 𝔪/𝔪^{2} is 1-dimensional over k; hence, 𝔪 is principal by (10.11)(2). Now, since f is irreducible, A is a domain. Hence, A is a DVR by (23.9).

Conversely, assume ℓ = 0. Then f = g ∈ (X, Y)^{2}. So

𝔪/𝔪^{2} = 𝔭/𝔭^{2} = ⟨X, Y ⟩/⟨X, Y ⟩^{2}.

Hence, 𝔪/𝔪^{2} is 2-dimensional. Therefore, A is not a DVR by (23.10). □

Exercise (23.12). — Let k be a field, A a ring intermediate between the polynomial ring and the formal power series ring in one variable: k[X] ⊂ A ⊂ k[[X]]. Suppose that A is local with maximal ideal ⟨X⟩. Prove that A is a DVR. (Such local rings arise as rings of power series with curious convergence conditions.)

Solution: Let’s show that the ideal 𝔞 := ⋂_{n≥0}⟨X^{n}⟩ of A is zero. Clearly, 𝔞 is a subset of the corresponding ideal ⋂_{n≥0}⟨X^{n}⟩ of k[[X]], and the latter ideal is clearly zero. Hence (23.3) implies A is a DVR. □

Exercise (23.13). — Let L/K be an algebraic extension of fields, X_{1}, …, X_{n} variables, P and Q the polynomial rings over K and L in X_{1}, …, X_{n}.

(1) Let 𝔮 be a prime of Q, and 𝔭 its contraction in P. Prove ht(𝔭) = ht(𝔮).

(2) Let f, g ∈ P be two polynomials with no common prime factor in P. Prove that f and g have no common prime factor q ∈ Q.

Solution: Since L/K is algebraic, Q/P is integral. Furthermore, P is normal, and Q is a domain. Hence we may apply the Going Down Theorem (14.9). So given any chain of primes 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} = 𝔭, we can proceed by descending induction on i for 0 ≤ i ≤ r, and thus construct a chain of primes 𝔮_{0} ⫋ ⋯ ⫋ 𝔮_{r} = 𝔮 with 𝔮_{i} ∩ P = 𝔭_{i}. Thus ht 𝔭 ≤ ht 𝔮. Conversely, any chain of primes 𝔮_{0} ⫋ ⋯ ⫋ 𝔮_{r} = 𝔮 contracts to a chain of primes 𝔭_{0} ⊂ ⋯ ⊂ 𝔭_{r} = 𝔭, and 𝔭_{i} ≠ 𝔭_{i+1} by Incomparability, (14.3); whence, ht 𝔭 ≥ ht 𝔮. Hence ht 𝔭 = ht 𝔮. Thus (1) holds.

Alternatively, by (15.13), ht(𝔭) + dim(P/𝔭) = n and ht(𝔮) + dim(Q/𝔮) = n as both P and Q are polynomial rings in n variables over a field. However, by (15.12), dim P/𝔭 = tr. deg_{K} Frac(P/𝔭) and dim Q/𝔮 = tr. deg_{L} Frac(Q/𝔮), and these two transcendence degrees are equal as Q/P is an integral extension. Thus again, (1) holds.

Suppose f and g have a common prime factor q ∈ Q, and set 𝔮 := Qq. Then the maximal ideal 𝔮Q_{𝔮} of Q_{𝔮} is principal and nonzero. Hence Q_{𝔮} is a DVR by (23.9). Thus ht(𝔮) = 1. Set 𝔭 := 𝔮 ∩ P. Then 𝔭 contains f; whence, 𝔭 contains some prime factor p of f. Then 𝔭 ⊇ P p, and P p is a nonzero prime. Hence 𝔭 = P p since ht 𝔭 = 1 by (1). However, 𝔭 contains g too. Therefore, p | g, contrary to the hypothesis. Thus (2) holds. (Caution: if f := X_{1} and g := X_{2}, then f and g have no common factor, yet there are no φ and ψ such that φf + ψg = 1.) □

Exercise (23.15). — Let R be a Noetherian ring. Show that R is reduced if and only if (R_{0}) and (S_{1}) hold.

Solution: Assume (R_{0}) and (S_{1}) hold. Consider an irredundant primary decomposition ⟨0⟩ = ⋂ 𝔮_{i}. Set 𝔭_{i} := \sqrt{𝔮_{i}}. Then 𝔭_{i} is minimal by (S_{1}), and 𝔭_{i} = 𝔮_{i} by (R_{0}) and (18.22). So ⟨0⟩ = ⋂ 𝔭_{i} = \sqrt{⟨0⟩}. Thus R is reduced.

Conversely, assume R is reduced. Then R_{𝔭} is reduced for any prime 𝔭 by (13.18). So if 𝔭 is minimal, then R_{𝔭} is a field. Thus (R_{0}) holds. But ⟨0⟩ = ⋂_{𝔭 minimal} 𝔭. So 𝔭 is minimal whenever 𝔭 ∈ Ass(R) by (18.19). Thus R satisfies (S_{1}). □

Exercise (23.20). — Prove that a Noetherian domain R is normal if and only if, given any prime 𝔭 associated to a principal ideal, 𝔭R_{𝔭} is principal.

Solution: Assume R normal. Say 𝔭 ∈ Ass(R/⟨x⟩). Then 𝔭R_{𝔭} ∈ Ass(R_{𝔭}/⟨x/1⟩) by (17.9). So depth(R_{𝔭}) = 1. But R_{𝔭} is normal by (11.31). Hence 𝔭R_{𝔭} is principal by (23.9).

Conversely, assume that, given any prime 𝔭 associated to a principal ideal, 𝔭R_{𝔭} is principal. Given any prime 𝔭 of height 1, take a nonzero x ∈ 𝔭. Then 𝔭 is minimal containing ⟨x⟩. So 𝔭 ∈ Ass(R/⟨x⟩) by (17.17). So, by hypothesis, 𝔭R_{𝔭} is principal. So R_{𝔭} is a DVR by (23.9). Thus R satisfies (R_{1}).

Given any prime 𝔭 with depth(R_{𝔭}) = 1, say 𝔭R_{𝔭} ∈ Ass(R_{𝔭}/⟨x/s⟩) with x ≠ 0. Then ⟨x/s⟩ = ⟨x/1⟩ ⊂ R_{𝔭}. So 𝔭 ∈ Ass(R/⟨x⟩) by (17.9). So, by hypothesis, 𝔭R_{𝔭} is principal. So dim(R_{𝔭}) = 1 by (23.9). Thus R also satisfies (S_{2}). So R is normal by Serre’s Criterion, (23.18). □

Exercise (23.21). — Let R be a Noetherian ring, K its total quotient ring, Set

Φ := { 𝔭 prime | ht(𝔭) = 1 } and Σ := { 𝔭 prime | depth(R_{𝔭}) = 1 }.

Assuming (S_{1}) holds in R, prove Φ ⊂ Σ, and prove Φ = Σ if and only if (S_{2}) holds. Further, without assuming (S_{1}) holds, prove this canonical sequence is exact:

R → K → ∏_{𝔭∈Σ} K_{𝔭}/R_{𝔭}.  (23.21.1)

Solution: Assume (S_{1}) holds. Then, given 𝔭 ∈ Φ, there exists a nonzerodivisor x ∈ 𝔭. Clearly, 𝔭 is minimal containing ⟨x⟩. So 𝔭 ∈ Ass(R/⟨x⟩) by (17.17). Hence depth(R_{𝔭}) = 1 by (23.5)(2). Thus Φ ⊂ Σ.

However, as (S_{1}) holds, (S_{2}) holds if and only if Φ ⊃ Σ. Thus Φ = Σ if and only if R satisfies (S_{2}).

Further, without assuming (S_{1}), consider (23.21.1). Trivially, the composition is zero. Conversely, take an x ∈ K that vanishes in ∏_{𝔭∈Σ} K_{𝔭}/R_{𝔭}. Say x = a/b with a, b ∈ R and b a nonzerodivisor. Then a/1 ∈ bR_{𝔭} for all 𝔭 ∈ Σ. But b/1 ∈ R_{𝔭} is, clearly, a nonzerodivisor for any prime 𝔭. Hence, if 𝔭 ∈ Ass(R_{𝔭}/bR_{𝔭}), then 𝔭 ∈ Σ by (23.5)(2). Therefore, a ∈ bR by (18.25). Thus x ∈ R. Thus (23.21.1) is exact. □

Exercise (23.22). — Let R be a Noetherian ring, and K its total quotient ring. Set Φ := { 𝔭 prime | ht(𝔭) = 1 }. Prove these three conditions are equivalent:

(1) R is normal.

(2) (R_{1}) and (S_{2}) hold.

(3) (R_{1}) and (S_{1}) hold, and R → K → ∏_{𝔭∈Φ} K_{𝔭}/R_{𝔭} is exact.

Solution: Assume (1). Then R is reduced by (14.17). So (23.15) yields (R_{0}) and (S_{1}). But R_{𝔭} is normal for any prime 𝔭 by (14.16). Thus (2) holds by (23.9).

Assume (2). Then (R_{1}) and (S_{1}) hold trivially. Thus (23.21) yields (3).

Assume (3). Let x ∈ K be integral over R. Then x/1 ∈ K is integral over R_{𝔭} for any prime 𝔭. Now, R_{𝔭} is a DVR for all 𝔭 of height 1 as R satisfies (R_{1}). Hence, x/1 ∈ R_{𝔭} for all 𝔭 ∈ Φ. So x ∈ R by the exactness of the sequence in (3). But R is reduced by (23.15). Thus (14.17) yields (1). □

24. Dedekind Domains

Exercise (24.5). — Let R be a domain, S a multiplicative subset.

(1) Assume dim(R) = 1. Prove dim(S^{−1}R) = 1 if and only if there is a nonzero prime 𝔭 with 𝔭 ∩ S = ∅.

(2) Assume dim(R) ≥ 1. Prove dim(R) = 1 if and only if dim(R_{𝔭}) = 1 for every nonzero prime 𝔭.

Solution: Consider (1). Suppose dim(S^{−1}R) = 1. Then there’s a chain of primes 0 ⫋ 𝔭′ ⊂ S^{−1}R. Set 𝔭 := 𝔭′ ∩ R. Then 𝔭 is as desired by (11.18)(2).

Conversely, suppose there’s a nonzero 𝔭 with 𝔭 ∩ S = ∅. Then 0 ⫋ 𝔭S^{−1}R is a chain of primes by (11.18)(2); so dim(S^{−1}R) ≥ 1. Now, given a chain of primes 0 = 𝔭′_{0} ⫋ ⋯ ⫋ 𝔭′_{r} ⊂ S^{−1}R, set 𝔭_{i} := 𝔭′_{i} ∩ R. Then 0 = 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} ⊂ R is a chain of primes by (11.18)(2). So r ≤ 1 as dim(R) = 1. Thus dim(S^{−1}R) = 1.

Consider (2). If dim(R) = 1, then (1) yields dim(R_{𝔭}) = 1 for every nonzero 𝔭. Conversely, let 0 = 𝔭_{0} ⫋ ⋯ ⫋ 𝔭_{r} ⊂ R be a chain of primes. Set 𝔭′_{i} := 𝔭_{i}R_{𝔭_{i}}.

Then 0 = 𝔭′_{0} ⫋ ⋯ ⫋ 𝔭′_{r} is a chain of primes by (11.18)(2). So if dim(R_{𝔭_{i}}) = 1, then r ≤ 1. Thus, if dim(R_{𝔭}) = 1 for every nonzero 𝔭, then dim(R) ≤ 1, as desired. □

Exercise (24.6). — Let R be a Dedekind domain, S a multiplicative subset. Prove S^{−1}R is a Dedekind domain if and only if there’s a nonzero prime 𝔭 with 𝔭 ∩ S = ∅.

Solution: Suppose there’s a prime nonzero 𝔭 with 𝔭 ∩ S = ∅. Then 0 ∉ S. So S^{−1}R is a domain by (11.4). And S^{−1}R is normal by (11.31). Further, S^{−1}R is Noetherian by (16.7). Also, dim(S^{−1}R) = 1 by (24.5)(1). Thus S^{−1}R is Dedekind.

The converse results directly from (24.5)(1). □

Exercise (24.8). — Let R be a Dedekind domain, and 𝔞, 𝔟, 𝔠 ideals. By first reducing to the case that R is local, prove that

𝔞 ∩ (𝔟 + 𝔠) = (𝔞 ∩ 𝔟) + (𝔞 ∩ 𝔠),
𝔞 + (𝔟 ∩ 𝔠) = (𝔞 + 𝔟) ∩ (𝔞 + 𝔠).

Solution: By (13.17), it suffices to establish the two equations after localizing at each maximal ideal 𝔭. But localization commutes with sum and intersection by (12.15)(4), (5). So the localized equations look like the original ones, but with 𝔞, 𝔟, 𝔠 replaced by 𝔞_{𝔭}, 𝔟_{𝔭}, 𝔠_{𝔭}. Thus we may replace R by R_{𝔭}, and so assume R is a DVR.

Referring to (23.1), take a uniformizing parameter t, and say 𝔞 = ⟨t^{i}⟩ and 𝔟 = ⟨t^{j}⟩ and 𝔠 = ⟨t^{k}⟩. Then the two equations in questions are equivalent to these two:

max{i, min{j, k}} = min{max{i, j}, max{i, k}},
min{i, max{j, k}} = max{min{i, j}, min{i, k}}.

However, these two equations are easy to check for any integers i, j, k. □

Exercise (24.12). — Prove that a semilocal Dedekind domain A is a PID. Begin by proving that each maximal ideal is principal.

Solution: Let 𝔭_{1}, …, 𝔭_{r} be the maximal ideals of A. Let’s prove they are principal, starting with 𝔭_{1}. By Nakayama’s lemma (10.10), 𝔭_{1}A_{𝔭_{1}} ≠ 𝔭_{1}^{2}A_{𝔭_{1}}; so 𝔭_{1} ≠ 𝔭_{1}^{2}. Take y ∈ 𝔭_{1}−𝔭_{1}^{2}. The ideals 𝔭_{1}^{2}, 𝔭_{2}, …, 𝔭_{r} are pairwise comaximal because no two of them lie in the same maximal ideal. Hence, by the Chinese Remainder Theorem, (1.13), there is an x ∈ A with x ≡ y mod 𝔭_{1}^{2} and x ≡ 1 mod 𝔭_{i} for i ≥ 2.

The Main Theorem of Classical Ideal Theory, (24.10), yields ⟨x⟩ = 𝔭_{1}^{n_{1}}𝔭_{2}^{n_{2}} ⋯ 𝔭_{r}^{n_{r}} with n_{i} ≥ 0. But x ∉ 𝔭_{i} for i ≥ 2; so n_{i} = 0 for i ≥ 2. Further, x ∈ 𝔭_{1} − 𝔭_{1}^{2}; so n_{1} = 1. Thus 𝔭_{1} = ⟨x⟩. Similarly, all the other 𝔭_{i} are principal.

Finally, let 𝔞 be any nonzero ideal. Then the Main Theorem, (24.10), yields 𝔞 = ∏ 𝔭_{i}^{m_{i}} for some m_{i}. Say 𝔭_{i} = ⟨x_{i}⟩. Then 𝔞 = ∏ x_{i}^{m_{i}}, as desired. □

Exercise (24.13). — Let R be a Dedekind domain, 𝔞 and 𝔟 two nonzero ideals. Prove (1) every ideal in R/𝔞 is principal, and (2) 𝔟 is generated by two elements.

Solution: To prove (1), let 𝔭_{1}, …, 𝔭_{r} be the associated primes of 𝔞, and set S := ⋂_{i}(R − 𝔭_{i}). Then S is multiplicative. Set R′ := S^{−1}R. Then R′ is Dedekind by (24.6). Let’s prove R′ is semilocal.

Let 𝔮 be a maximal ideal of R′, and set 𝔭 := 𝔮 ∩ R. Then 𝔮 = 𝔭R′ by (11.18). So 𝔭 is nonzero, whence maximal since R has dimension 1. Suppose 𝔭 is distinct from all the 𝔭_{i}. Then 𝔭 and the 𝔭_{i} are pairwise comaximal. So, by the Chinese Remainder Theorem, (1.13), there is a u ∈ R that is congruent to 0 modulo 𝔭 and to 1 modulo each 𝔭_{i}. Hence, u ∈ 𝔭 ∩ S, but 𝔮 = 𝔭R′, a contradiction. Thus 𝔭_{1}R′, …, 𝔭_{r}R′ are all the maximal ideals of R′.

So R′ is a PID by (24.12); so every ideal in R′/𝔞R′ is principal. But by (12.18), R′/𝔞R′ = S^{−1}(R/𝔞). Finally, S^{−1}(R/𝔞) = R/𝔞 by (11.7) because every u ∈ S maps to a unit in R/𝔞 since the image lies in no maximal ideal of R/𝔞. Thus (1) holds.

Alternatively, we can prove (1) without using (24.12), as follows. The Main Theorem of Classical Ideal Theory, (24.10), yields 𝔞 = 𝔭_{1}^{n_{1}} ⋯ 𝔭_{k}^{n_{k}} for distinct maximal ideals 𝔭_{i}. The 𝔭_{i}^{n_{i}} are pairwise comaximal. So, by the Chinese Remainder Theorem, (1.13), there’s a canonical isomorphism:

R/𝔞 ⟶^{∼} R/𝔭_{1}^{n_{1}} × ⋯ × R/𝔭_{k}^{n_{k}}.

Next, let’s prove each R/𝔭_{i}^{n_{i}} is a Principal Ideal Ring (PIR); that is, every ideal is principal. Set S := R − 𝔭_{i}. Then S^{−1}(R/𝔭_{i}^{n_{i}}) = R_{𝔭_{i}}/𝔭_{i}^{n_{i}}R_{𝔭_{i}}, and the latter ring is a PIR because R_{𝔭_{i}} is a DVR. However, R/𝔭_{i}^{n_{i}} = S^{−1}(R/𝔭_{i}^{n_{i}}) by (11.7), because every u ∈ S maps to a unit in R/𝔭_{i}^{n_{i}} since 𝔭/𝔭_{i}^{n_{i}} is the only prime in R/𝔭_{i}^{n_{i}}.

Finally, given finitely many PIRs R_{1}, …, R_{k}, we must prove their product is a PIR. Consider an ideal 𝔟 ⊂ R_{1} × ⋯ × R_{k}. Then 𝔟 = 𝔟_{1} × ⋯ × 𝔟_{k} where 𝔟_{i} ⊂ R_{i} is an ideal by (1.15). Say 𝔟_{i} = ⟨a_{i}⟩. Then 𝔟 = ⟨(a_{1}, …, a_{k})⟩. Thus again, (1) holds.

Consider (2). Let x ∈ 𝔟 be nonzero. By (1), there is a y ∈ 𝔟 whose residue generates 𝔟/⟨x⟩. Then 𝔟 = ⟨x, y⟩. □

25. Fractional Ideals

Exercise (25.2). — Let R be a domain, M and N nonzero fractional ideals. Prove that M is principal if and only if there exists some isomorphism M ≃ R. Construct the following canonical surjection and canonical isomorphism:

π: M ⊗ N → → MN and φ: (M : N) ⟶^{∼} Hom(N, M).

Solution: If M ≃ R, let x correspond to 1; then M = Rx. Conversely, assume M = Rx. Then x ≠ 0 as M ≠ 0. Form the map R → M with a ↦ ax. It’s surjective as M = Rx. It’s injective as x ≠ 0 and M ⊂ Frac(R).

Form the canonical M × N → MN with (x, y) ↦ xy. It’s bilinear. So it induces a map π: M ⊗ N → MN, and clearly π is surjective.

Define φ as follows: given z ∈ (M : N), define φ(z): N → M by φ(z)(y) := yz. Clearly, φ is R-linear. Say y ≠ 0. Then yz = 0 implies z = 0; thus, φ is injective.

Finally, given θ: N → M, fix a nonzero n ∈ N, and set z := θ(n)/n. Given y ∈ N, say y = a/b and n = c/d with a, b, c, d ∈ R. Then bcy = adn. So bcθ(y) = adθ(n). Hence θ(y) = yz. Thus, φ is surjective, as desired. □

Exercise (25.6). — Let R be a domain, M and N fractional ideals. Prove that the map π: M ⊗ N → MN is an isomorphism if M is locally principal.

Solution: By (13.20), it suffices to prove that, for each maximal ideal 𝔪, the localization π_{𝔪}: (M ⊗ N)_{𝔪} → (MN)_{𝔪} is bijective. But (M ⊗ N)_{𝔪} = M_{𝔪} ⊗ N_{𝔪} by (12.13), and (MN)_{𝔪} = M_{𝔪}N_{𝔪} by (25.4). By hypothesis, M_{𝔪} = R_{𝔪}x for some x. Clearly R_{𝔪}x ≃ R_{𝔪}. And R_{𝔪} ⊗ N_{𝔪} = N_{𝔪} by (8.5)(2). Thus π_{𝔪} ≃ 1_{N_{𝔪}}. □

Exercise (25.11). — Let R be a UFD. Show that a fractional ideal M is invertible if and only if M is principal and nonzero.

Solution: By (25.7), a nonzero principal ideal is always invertible.

Conversely, assume M is invertible. Then trivially M ≠ 0. Say 1 = ∑ m_{i}n_{i} with m_{i} ∈ M and n_{i} ∈ M^{−1}. Fix a nonzero m ∈ M.

Then m = ∑ m_{i}n_{i}m. But n_{i}m ∈ R as m ∈ M and n_{i} ∈ M^{−1}. Set

d := gcd{n_{i}m} ∈ R and x := ∑(n_{i}m/d)m_{i} ∈ M.

Then m = dx.

Given m′ ∈ M, write m′/m = a/b where a, b ∈ R are relatively prime. Then

d′ := gcd{n_{i}m′} = gcd{n_{i}ma/b} = a gcd{n_{i}m}/b = ad/b.

So m′ = (a/b)m = (ad/b)x = d′x. But d′ ∈ R. Thus M = Rx. □

Exercise (25.14). — Show that a ring is a PID if and only if it’s a Dedekind domain and a UFD.

Solution: A PID is Dedekind by (24.2), and is a UFD by (2.23).

Conversely, let R be a Dedekind UFD. Then every nonzero fractional ideal is invertible by (25.3) and (25.13), so is principal by (25.11). Thus R is a PID.

Alternatively and more directly, every nonzero prime is of height 1 as dim R = 1, so is principle by (21.12). But, by (24.10), every nonzero ideal is a product of nonzero prime ideals. Thus again, R is a PID. □

Exercise (25.16). — Let R be an ring, M an invertible module. Prove that M is finitely generated, and that, if R is local, then M is free of rank 1.

Solution: Say α: M ⊗ N ⟶^{∼} R and 1 = α(∑ m_{i} ⊗ n_{i}) with m_{i} ∈ M and n_{i} ∈ N. Given m ∈ M, set a_{i} := α(m ⊗ n_{i}). Form this composition:

β: M = M ⊗ R ⟶^{∼} M ⊗ M ⊗ N = M ⊗ N ⊗ M ⟶^{∼} R ⊗ M = M.

Then β(m) = ∑ a_{i}m_{i}. But β is an isomorphism. Thus the m_{i} generate M.

Suppose R is local. Then R − R^{×} is an ideal. So u := α(m_{i} ⊗ n_{i}) ∈ R^{×} for some i. Set m := u^{−1}m_{i} and n := n_{i}. Then α(m ⊗ n) = 1. Define ν: M → R by ν(m′) := α(m′ ⊗ n). Then ν(m) = 1; so ν is surjective. Define μ: R → M by μ(x) := xm. Then μν(m′) = ν(m′)m = β(m′), or μν = β. But β is an isomorphism. So ν is injective. Thus ν is an isomorphism, as desired. □

Exercise (25.17). — Show these conditions on an R-module M are equivalent:

(1) M is invertible.

(2) M is finitely generated, and M_{𝔪} ≃ R_{𝔪} at each maximal ideal 𝔪.

(3) M is locally free of rank 1.

Assuming the conditions, show M is finitely presented and M ⊗ Hom(M, R) = R.

Solution: Assume (1). Then M is finitely generated by (25.16). Further, say M ⊗ N ≃ R. Let 𝔪 be a maximal ideal. Then M_{𝔪} ⊗ N_{𝔪} ≃ R_{𝔪}. Hence M_{𝔪} ≃ R_{𝔪} again by (25.16). Thus (2) holds.

Conditions (2) and (3) are equivalent by (13.28).

Assume (3). Then (2) holds; so M_{𝔪} ≃ R_{𝔪} at any maximal ideal 𝔪. Also, M is finitely presented by (13.27); so Hom_{R}(M, R)_{𝔪} = Hom_{R_{𝔪}}(M_{𝔪}, R_{𝔪}) by (12.21).

Consider the evaluation map

ev(M, R): M ⊗ Hom(M, R) → R defined by ev(M, R)(m, α) := α(m).

Clearly ev(M, R)_{𝔪} = ev(M_{𝔪}, R_{𝔪}). Clearly ev(R_{𝔪}, R_{𝔪}) is bijective. Hence ev(M, R) is bijective by (13.20). Thus the last assertions hold; in particular, (1) holds. □

26. Arbitrary Valuation Rings

Exercise (26.3). — Let V be a domain. Show that V is a valuation ring if and only if, given any two ideals 𝔞 and 𝔟, either 𝔞 lies in 𝔟 or 𝔟 lies in 𝔞.

Solution: First, suppose V is a valuation ring. Suppose also 𝔞 ⊄ 𝔟; say x ∈ 𝔞, but x ∉ 𝔟. Take y ∈ 𝔟. Then x/y ∉ V; else x = (x/y)y ∈ 𝔟. So y/x ∈ V. Hence y = (y/x)x ∈ 𝔞. Thus 𝔟 ⊂ 𝔞.

Conversely, let x, y ∈ V − {0}, and suppose x/y ∉ V. Then ⟨x⟩ ⊄ ⟨y⟩; else, x = wy with w ∈ V. Hence ⟨y⟩ ⊂ ⟨x⟩ by hypothesis. So y = zx for some z ∈ V; in other words, y/x ∈ V. Thus V is a valuation ring. □

Exercise (26.4). — Let V be a valuation ring, 𝔪 its maximal ideal, and 𝔭 ⊂ 𝔪 another prime ideal. Prove that V_{𝔭} is a valuation ring, that its maximal ideal 𝔭V_{𝔭} is equal to 𝔭, and that V/𝔭 is a valuation ring of the field V_{𝔭}/𝔭.

Solution: First, set K := Frac(V_{𝔭}). So K = Frac(V). Let x ∈ K − V_{𝔭}. Then 1/x ∈ V ⊂ V_{𝔭}. Thus V_{𝔭} is a valuation ring.

Second, let r/s ∈ 𝔭V_{𝔭} where r ∈ 𝔭 − {0} and s ∈ V − 𝔭. Then s/r ∉ V, else s = (s/r)r ∈ 𝔭. Hence r/s ∈ V. Now, (r/s)s = r ∈ 𝔭, but s ∉ 𝔭; since 𝔭 is prime, r/s ∈ 𝔭. Thus 𝔭V_{𝔭} = 𝔭.

Third, to prove V/𝔭 is a valuation ring of V_{𝔭}/𝔭V_{𝔭}, we need only show that, whenever x ∈ V_{𝔭}−V, then x^{−1} ∈ V. But, V is a valuation ring; hence, x^{−1} ∈ V. □

Exercise (26.5). — Prove that a valuation ring V is normal.

Solution: Set K := Frac(V), and let 𝔪 be the maximal ideal. Take x ∈ K integral over V, say x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} = 0 with a_{i} ∈ V. Then

1 + a_{1}x^{−1} + ⋯ + a_{n}x^{−n} = 0.  (26.5.1)

If x ∉ V, then x^{−1} ∈ 𝔪 by (26.2). So (26.5.1) yields 1 ∈ 𝔪, a contradiction. Hence x ∈ V. Thus V is normal. □

Exercise (26.10). — Let K be a field, 𝒮 the set of local subrings with fraction field K, ordered by domination. Show its maximal elements are the valuation rings.

Solution: Let V be maximal in 𝒮. By (26.9), V is dominated by a valuation ring V′ of K. By maximality, V = V′.

Conversely, let V be a valuation ring of K. Then V lies in 𝒮 by (26.2). Let V′ ∈ 𝒮 dominate V. Let 𝔪 and 𝔪′ be the maximal ideals of V and V′. Take any nonzero x ∈ V′. Then 1/x ∉ 𝔪′ as 1 ∉ 𝔪′; so also 1/x ∉ 𝔪. So x ∈ V by (26.2). Hence, V′ = V. Thus V is maximal in 𝒮. □

Exercise (26.15). — Let V be a valuation ring, such as a DVR, whose value group Γ is Archimedean; that is, given any nonzero α, β ∈ Γ, there’s n ∈ ℤ such that nα > β. Show that V is a maximal proper subring of its fraction field K.

Solution: Let R be a subring of K strictly containing V, and fix a ∈ R − V. Given b ∈ K, let α and β be the values of a and b. Then α < 0. So, as Γ is Archimedean, there’s n > 0 such that −nα > −β. Then v(b/a^{n}) > 0. So b/a^{n} ∈ V. So b = (b/a^{n})a^{n} ∈ R. Thus R = K. □

Exercise (26.16). — Let V be a valuation ring. Show that (1) every finitely generated ideal 𝔞 is principal, and

(2) V is Noetherian if and only if V is a DVR.

Solution: To prove (1), say 𝔞 = ⟨x_{1}, …, x_{n}⟩ with x_{i} ≠ 0 for all i. Let v be the valuation. Suppose v(x_{1}) ≤ v(x_{i}) for all i. Then x_{i}/x_{1} ∈ V for all i. So x_{i} ∈ ⟨x_{1}⟩. Hence 𝔞 = ⟨x_{1}⟩. Thus (1) holds.

To prove (2), first assume V is Noetherian. Then V is local by (26.2), and by (1) its maximal ideal 𝔪 is principal. Hence V is a DVR by (23.9). Conversely, assume V is a DVR. Then V is a PID by (23.1), so Noetherian. Thus (2) holds. □

References

[1] Artin, M., “Algebra,” Prentice-Hall, 1991.

[2] Atiyah, M., and Macdonald, I., “Introduction to Commutative Algebra,” Addison-Wesley, 1969.

[3] Eisenbud, D., “Commutative Algebra with a View Toward Algebraic Geomertry,” Springer-Verlag, 1999.

[4] Judson, T., “Abstract Algebra: theory and Applications,” Open source, Electronic Book,

[5] Lang, S., “Undergraduate Analysis,” Springer-Verlag, 1997.

[6] Lang, S., “Algebra” Graduate Texts in Mathematics 211, Springer-Verlag, 2002.

[7] Reid, M., “Undergraduate Commutative Algebra,” Cambridge University Press, 1995.

Index

[8] Stark H., “An Introduction to Number Theory,” MIT Press, 1978. algebra: (1.1), 1

algebra finite: (4.5), 16 algebra map: (1.1), 1 coproduct: (8.19), 42 faithfully flat: (9.4), 44 finitely generated: (4.5), 16 flat: (9.4), 44 group algebra: (26.13), 138 homomorphism: (1.1), 1 integral over a ring: (10.16), 52 localization: (11.22), 58 module finite: (10.16), 51 Rees Algebra: (20.16), 105 structure map: (1.1), 1 subalgebra: (4.5), 15

generated by: (4.5), 15 tensor product: (8.19), 41

canonical: (1.1), 1; (4.2), 14 category theory

coequalizer: (6.8), 30 colimit: (6.6), 29 coproduct: (6.7), 30 direct limit: (6.6), 29 dually: (5.2), 20 filtered direct limit: (7.1), 33 identity: (6.1), 26 inclusion: (6.7), 30 initial object: (6.7), 30 insertion: (6.6), 29 inverse: (6.1), 26 isomorphism: (6.1), 26 map: (6.1), 26 morphism: (6.1), 26 object: (6.1), 26 pushout: (6.9), 30 transition map: (6.6), 29

category: (6.1), 26 directed set: (7.1), 33 discrete: (6.7), 30 functor: (6.6), 29 has direct limits: (6.6), 29 product: (6.1), 26 small: (6.6), 29

diagram chase: (5.12), 21 commutative: (1.5), 3

element annihilator: (4.1), 14 Cauchy sequence: (22.1), 116 complementary idempotents: (1.10), 5 equation of integral dependence: (10.16), 52 free: (4.10), 17 generators: (1.4), 2; (4.10), 17 homogeneous: (20.1), 101 homogeneous component: (20.21), 107 homogeneous of degree n: (20.21), 107;

(20.28), 108 idempotent: (1.10), 5 initial component: (20.21), 107 integral over a ring: (10.16), 51 integrally dependent on a ring: (10.16), 51 irreducible: (2.6), 7 Kronecker delta function: (4.10), 18 limit: (22.1), 116 linear combination: (1.4), 2 linearly independent: (4.10), 17 lowest terms: (11.5), 164 multiplicative inverse: (1.1), 1 nilpotent: (3.18), 12; (13.10), 67 nonzerodivisor: (2.1), 6; (17.13), 88 orthogonal idempotents: (1.10), 5 p-adic integer: (22.1), 116 prime: (2.6), 7 reciprocal: (1.1), 1 relatively prime: (2.24), 8 residue of: (1.5), 3 restricted vectors: (4.10), 17; (4.13), 18 uniformizing parameter: (23.1), 122 unit: (1.1), 1 zerodivisor: (2.1), 6; (17.13), 88

field: (2.3), 6 discrete valuation: (23.1), 122 fraction field: (2.3), 6 rational functions: (2.3), 6 Trace Pairing: (24.15), 129 trace: (24.15), 129

functor: (6.2), 26 additive: (8.17), 41 adjoint: (6.4), 27

adjoint pair: (6.4), 27 counit: (6.5), 28 unit: (6.5), 28 universal: (6.5), 28

cofinal: (7.13), 35 constant: (6.6), 29 contravariant: (6.1), 27 covariant: (6.2), 26 diagonal: (6.6), 29 direct system: (6.6), 29 exact: (9.2), 43 forgetful: (6.2), 26 isomorphic: (6.2), 27 left adjoint: (6.4), 27 left exact: (9.2), 43 linear: (8.4), 38; (9.2), 43 natural bijection: (6.4), 27 natural transformation: (6.2), 27 right adjoint: (6.4), 27 right exact: (9.2), 43

ideal: (1.4), 2 associated prime: (17.1), 87 chain stabilizes: (16.3), 82 comaximal: (1.13), 5 contraction: (1.4), 3 extension: (1.4), 3 fractional: (25.1), 131

integral: (25.1), 131

invertible: (25.7), 132

locally principal: (25.5), 132 principal: (25.1), 131 product: (25.1), 131 quotient: (25.1), 131

generated: (1.4), 2 height: (21.6), 111 idempotent: (1.16), 5 intersection: (1.4), 3 length of chain: (15.9), 78 lie over: (14.2), 71 maximal: (2.13), 7 nested: (1.8), 4 nilradical: (3.18), 12 parameter: (21.2), 109 prime: (2.1), 6 principal: (1.4), 2 product: (1.4), 3 proper: (1.4), 3 radical: (3.18), 12 saturated: (11.13), 57 saturation: (11.13), 57 sum: (1.4), 3 symbolic power: (18.26), 95 variety: (13.1), 66

Lemma Artin–Rees: (20.18), 105 E. Artin: (24.14), 129 Equational Criterion for Flatness: (9.19),

47 Equational Criterion for Vanishing: (8.18),

41 Five: (5.14), 22 Ideal Criterion for Flatness: (9.20), 47;

(22.20), 120 Nakayama: (10.10), 50; (22.24), 120 Nine: (5.15), 22 Noether Normalization: (15.1), 75 Nonunit Criterion: (3.4), 10 Prime Avoidance: (3.15), 12; (21.4), 111;

(23.5), 123 Schanuel: (5.23), 24 Snake: (5.12), 21 Zorn’s: (2.28), 9; (3.9), 11; (16.9), 83;

(17.8), 88; (26.9), 137

map automorphism: (1.1), 1 bilinear: (8.1), 37 bimodule homomorphism: (8.7), 38 endomorphism: (1.1), 1; (4.4), 15 homogeneous: (20.21), 107 homomorphism: (1.1), 1; (4.2), 14 isomorphism: (1.1), 1; (4.2), 14 lift: (5.20), 23 Noether Isomorphisms: (4.8), 16 quotient map: (4.6), 16 retraction: (5.8), 21 section: (5.8), 21 trilinear: (8.9), 38

matrix of cofactors: (10.2), 49 module: (4.1), 14

𝔞-dic topology: (22.1), 115

ascending chain condition (acc): (16.12), 84 annihilator: (4.1), 14 Artinian: (16.22), 86 associated graded: (20.11), 103 associated prime: (17.1), 87 bimodule: (8.7), 38 bimodule homomorphism: (8.7), 38 chain stabilizes: (16.12), 84; (16.22), 86 characteristic polynomial: (10.1), 49 closed: (4.1), 14 Cohen–Macaulay: (23.4), 123 coimage: (4.9), 17 cokernel: (4.9), 17 complete: (22.1), 116 composition series: (19.1), 97 cyclic: (4.7), 16 depth: (23.4), 123 descending chain condition (dcc): (16.22),

86 dimension: (21.1), 109 direct product: (4.13), 18 direct sum: (4.10), 18; (4.13), 18 discrete: (22.1), 115 embedded prime: (17.1), 87 endomorphism: (4.4), 15 extension of scalars: (8.7), 38 faithful: (4.4), 15; (10.18), 52; (12.22), 65 filtration: (20.11), 103

Hilbert–Samuel Function: (20.11), 103 Hilbert–Samuel Polynomial: (20.11), 103 Hilbert–Samuel Series: (20.11), 103 𝔮-adic: (20.11), 103 𝔮-filtration: (20.11), 103 stable 𝔮-filtration: (20.11), 103 topology: (22.1), 115 finitely generated: (4.10), 17 finitely presented: (5.18), 23 flat: (9.4), 44 free: (4.10), 17 free basis: (4.10), 17 free of rank ℓ: (4.10), 17 generated: (4.10), 17 graded: (20.1), 101

homogeneous component: (20.1), 101 Hilbert Function: (20.3), 102 Hilbert Polynomial: (20.3), 102 Hilbert Series: (20.3), 102 shifting (20.1), 101

homogeneous component: (20.21), 107 homomorphism: (4.2), 14 image: (4.2), 14 inverse limit: (22.6), 116 invertible: (25.15), 133 isomorphism: (4.2), 14 kernel: (4.2), 14 length: (19.1), 97 localization: (12.2), 61 localization at f: (12.2), 61 localizaton at 𝔭: (12.2), 61 locally finitely generated: (13.25), 69 locally finitely presented: (13.25), 69 locally free: (13.25), 69 maximal condition (maxc): (16.12), 84 minimal condition (minc): (16.22), 86 minimal generating set: (10.13), 51 minimal prime: (17.1), 87 modulo: (4.6), 16 M-sequence: (23.4), 123 Noetherian: (16.12), 84 presentation: (5.18), 23 projective (5.20), 23 quotient: (4.6), 16 quotient map: (4.6), 16 R-linear map: (4.2), 14 radical: (21.2), 109 regular sequence: (23.4), 123 residue: (4.6), 16 restriction of scalars

left adjoint: (8.10), 39 restriction of scalars: (4.5), 15 restriction of scalars: (8.10), 39 saturated: (12.14), 63 saturation: (12.14), 63 scalar multiplication: (4.1), 14 semilocal: (21.2), 109 separated: (22.1), 115 separated completion: (22.1), 116 simple: (19.1), 97 standard basis: (4.10), 17 submodule: (4.1), 14

homogeneous: (20.6), 102 irredundant primary decomposition:

(18.13), 92 minimal primary decomposition: (18.13),

92 𝔭-primary: (18.1), 91 primary: (18.1), 91

primary decomposition: (18.13), 92 sum: (4.8), 17 support: (13.8), 67 system of parameters (sop): (21.2), 110 tensor product, see also torsion free: (9.22), 48

notation 𝔞 + 𝔟: (1.4), 3 M = N: (4.2), 14 R = R′: (1.1), 1 𝔞 ∩ 𝔟: (1.4), 3 𝔭^{(n)}: (18.26), 95 ∏ M_{λ}: (4.13), 18 R ≃ R′: (1.1), 1 𝔞𝔟: (1.4), 3 M ≃ N: (4.2), 14 ((R-alg)): (6.1), 26 ((R-mod)): (6.1), 26 ((Rings)): (6.1), 26; (13.1), 66 ((Sets)): (6.1), 26 ((Top spaces)): (13.1), 66 𝔞R′: (1.4), 3 α_{R} ⊗ α′: (8.4), 38 𝔞N: (4.1), 14 Ann(M): (4.1), 14 Ann(m): (4.1), 14 𝔞^{S}: (11.13), 57 Ass(M): (17.1), 87 Bil_{R}(M, M′; N): (8.1), 37 𝔟/𝔞: (1.8), 4 Coim(α): (4.9), 17 Coker(α): (4.9), 17 ℂ: (2.3), 6 ∐ M_{λ}: (6.7), 29 D(f): (13.1), 66 δ_{μλ}: (4.10), 18 depth(𝔞, M): (23.4), 123 depth(M): (23.4), 123 dim(M): (21.1), 109 dim(R): (15.9), 78 \underrightarrow{lim} M_{λ}: (6.6), 29 d(M): (21.2), 110 e_{μ}: (4.10), 18 End_{R}(M): (4.4), 15 ℱ(R) : (25.21), 135 𝔽_{2}: (1.1), 1 𝔽_{q}: (15.2), 76 Frac(R): (2.3), 6 Γ_{𝔞}(M): (18.21), 94 G(M): (20.11), 103 G_{𝔮}(M): (20.11), 103 G_{𝔮}(R): (20.11), 103 h(M, n): (20.3), 102 H(M, t): (20.3), 102 Hom(M, N): (4.2), 14 Im(α): (4.2), 14 \underleftarrow{lim} M_{λ}: (22.6), 116 ι_{κ}: (4.13), 18 Ker(α): (4.2), 14 ⟨a_{1}, …, a_{n}⟩: (1.4), 2 ℓ(M): (19.1), 97 S^{−1}R: (11.1), 55; (11.22), 58 L + M: (4.8), 17 M(m): (20.1), 101 (M : N) : (25.1), 131 M̂: (22.1), 116 M^{−1}: (25.8), 132 M_{f}: (12.2), 61 M_{𝔭}: (12.2), 61 M/N: (4.6), 16 MN : (25.1), 131 M ⊕ N: (4.13), 18 M ⊗_{R} N: (8.2), 37 m ⊗ n: (8.2), 37 μ_{R}: (4.4), 15 μ_{x}: (4.4), 15 nil(M): (13.10), 67 nil(R): (3.18), 12 1_{A}: (6.1), 26 1_{M}: (4.2), 15 p(M_{•}, n): (20.11), 103 P (M_{•}, t): (20.11), 103 𝒫(R) : (25.21), 135 (α_{κ}): (4.13), 18 (m_{λ}): (4.13), 18 (x_{λ}): (4.10), 17 φ_{𝔭}: (11.19), 58; (12.2), 61 φ_{f}: (11.10), 56; (12.2), 61 φ_{S}: (11.1), 55; (12.2), 61 π_{κ}: (4.13), 18 Pic(R) : (25.21), 134 p_{𝔮}(M, n): (20.11), 103 P_{𝔮}(M, t): (20.11), 103 ℚ: (2.3), 6 R/𝔞: (1.5), 3 R^{×}: (1.1), 1 R′ × R′′: (1.11), 5 R[[X_{1}, …, X_{n}]]: (3.7), 10 R[X_{1}, …, X_{n}]: (1.3), 2 rad(R): (3.1), 10 rad(M): (21.2), 109 ℝ: (2.3), 6 R_{f}: (11.10), 56 R_{𝔭}: (11.19), 58 R^{ℓ}: (4.10), 17 R^{⊕Λ}: (4.10), 17 ℛ(M_{•}): (20.16), 105 (R_{n}): (23.14), 125 R_{+}: (20.6), 102 ℛ(𝔮): (20.16), 105 R[x_{1}, …, x_{n}]: (4.5), 16 N^{S}: (12.14), 63 \overline{S}: (3.14), 12 s(M): (21.2), 110 S − T: (1.1), 2 (S_{n}): (23.14), 125 Spec(R): (13.1), 66 \sqrt{𝔞}: (3.18), 12 ⨁ M_{λ}: (4.13), 18 ∑ β_{κ}: (4.13), 19 ∑ 𝔞_{λ}: (1.4), 2 Supp(M): (13.8), 67 β: M → → N (5.20), 23 △: (1.1), 2 t: (24.15), 129 tr. deg: (15.8), 77 V(𝔞): (13.1), 66 v_{𝔭}: (24.10), 128 x/s: (11.1), 55; (11.22), 58 ℤ: (1.1), 1 z.div(M): (17.13), 88 z.div(R): (2.1), 6

ring: (1.1), 1 absolutely flat: (10.8), 50 algebra, see also Artinian: (16.22), 86; (16.26), 86; (19.8),

99; (19.11), 99; (19.13), 100 ascending chain condition (acc): (16.3), 82 associated graded: (20.11), 103 Boolean: (1.2), 2; (2.18), 8; (3.23), 13;

(13.6), 67 catenary: (15.14), 78 coefficient field: (22.30), 121 Dedekind domain: (24.1), 127 dimension: (15.9), 78 Discrete Valuation Ring (DVR): (23.1), 122 domain: (2.3), 6 dominates: (26.8), 137 extension: (14.1), 71 factor ring: (1.5), 3 field, see also formal power series ring: (3.7), 10 graded: (20.1), 101 homomorphism: (1.1), 1 Ideal Class Group: (25.21), 135 integral closure: (10.25), 53 integrally closed: (10.25), 53 Jacobson: (15.19), 80 Jacobson radical: (3.1), 10 kernel: (1.5), 3 Laurent series ring: (3.8), 11 local: (3.3), 10 local homomorphism: (9.11), 44 localization: (11.1), 55 localization at f: (11.10), 56 localizaton at 𝔭: (11.19), 58 map: (1.1), 1 maximal condition (maxc): (16.3), 82 modulo: (1.5), 3 Noetherian: (16.1), 82 nonzerodivisor: (2.1), 6 normal: (10.25), 53; (10.28), 53; (11.31), 60; (14.8), 72; (14.16), 74; (23.17), 125; (23.20), 126; (24.1), 127; (24.16), 129; (26.5), 136

normalization: (10.25), 53 p-adic integers: (22.1), 116; (22.6), 117 Picard Group: (25.21), 134 polynomial ring: (1.3), 2 Principal Ideal Domain (PID): (2.26), 8;

(2.23), 8; (3.8), 11; (9.22), 48; (16.1), 82; (23.1), 122; (24.2), 127; (24.12), 129; (25.14), 133

Principal Ideal Ring (PIR): (24.13), 198 product ring: (1.11), 5; (2.5), 6; (2.12), 7;

(3.5), 10; (4.14), 19; (10.24), 53; (14.17), 74; (18.17), 93; (19.17), 100

quotient map: (1.5), 3 quotient ring: (1.5), 3 radical: (3.1), 10 reduced: (3.18), 12 regular local: (21.18), 113; (21.21), 113;

(21.24), 114; (22.27), 121; (23.1), 122; (23.8), 124

regular system of parameters: (21.18), 113 residue ring: (1.5), 3 ring of fractions: (11.1), 55 semilocal: (3.3), 10 Serre’s Conditions: (23.14), 125 spectrum: (13.1), 66

compact: (13.6), 67 principal open set: (13.1), 66 quasi-compact: (13.4), 67 Zariski topology: (13.1), 66

subring: (1.1), 1 total quotient ring: (11.4), 55 Unique Factorization Domain (UFD): (2.6),

7; (2.23), 8; (10.28), 53; (16.1), 82; (23.1), 122; (25.11), 132; (25.14), 133; (25.21), 135

valuation: (26.1), 136

sequence Cauchy: (22.1), 116 exact: (5.1), 20 M-sequence: (23.4), 123 regular sequence: (23.4), 123 short exact: (5.3), 20 split exact: (5.8), 21

subset characteristic function: (1.2), 1 multiplicative: (2.1), 6

generated by: (11.5), 164 saturated: (3.12), 11 saturatation: (3.14), 12

symmetric difference: (1.2), 2 system of parameters (sop): (21.2), 110

regular (21.18), 113

tensor product: (8.2), 37 adjoint associativity: (8.9), 38 associative law: (8.9), 38 cancellation law: (8.10), 39 commutative law: (8.5), 38 unitary law: (8.5), 38

Theorem Additivity of Length: (19.9), 99 Akizuki: (19.11), 99 Artin’s Character: (24.14), 129 Characterization of DVRs: (23.9), 124 Cayley–Hamilton: (10.1), 49 Cohen: (16.9), 83 Cohen Structure: (22.30), 121 Determinant Trick: (10.2), 49 Dimension: (21.4), 110 Direct limits commute: (6.14), 32 Exactness of Localization: (12.16), 63 Exactness of Completion: (22.14), 119 Exactness of filtered direct limits: (7.10), 34 Finiteness of Integral Closure: (24.17), 129 First Uniqueness: (18.18), 94 Gauss: (10.28), 53 Generalized Hilbert Nullstellensatz:

(15.24), 81 Going down

for Flat Algebras: (14.11), 73 for integral extensions: (14.9), 72

Going up: (14.3), 71 Hilbert Basis: (16.11), 84 Hilbert Nullstellensatz: (15.6), 77 Hilbert–Serre: (20.7), 102 Incomparability: (14.3), 71 Jordan–Hölder: (19.3), 97 Krull Intersection: (18.28), 95; (20.19),

105 Krull Principal Ideal: (21.9), 112 Lasker–Noether: (18.20), 94 Lazard: (9.18), 46 Left Exactness of Hom: (5.17), 23 Lying over: (14.3), 71 Main of Classical Ideal Theory: (24.10),

128; (25.13), 133 Maximality: (14.3), 71 Noether: (24.20), 130 Scheinnullstellensatz: (3.22), 12 Second Uniqueness: (18.24), 95 Serre’s Criterion: (23.18), 125 Stone’s: (13.7), 67 Tower Law for Integrality: (10.22), 53 Watts: (8.15), 40 Weak Nullstellensatz: (15.4), 77

topological space closed point: (13.2), 66 Jacobson: (15.22), 80 locally closed subset: (15.22), 80 quasi-compact: (13.4), 67 very dense subset: (15.22), 80

topology 𝔞-adic: (22.1), 115 discrete: (1.2), 2 separated: (22.1), 115 Zariski: (13.1), 66

totally ordered group: (26.12), 137 Archimedean: (26.15), 139 value group: (26.12), 138

unitary: (6.1), 26 universal example: (1.3), 2 Universal Mapping Property (UMP)

coequalizer: (6.8), 30 cokernel: (4.9), 17 colimit: (6.6), 29 coproduct: (6.7), 29 direct limit: (6.6), 29 direct product: (4.13), 18 direct sum: (4.13), 18 Formal Power Series: (22.29), 121 fraction field: (2.3), 6 free module: (4.10), 18 inverse limit: (22.6), 116 localization: (11.6), 55; (12.3), 61 polynomial ring: (1.3), 2 pushout: (6.9), 30 residue module: (4.6), 16 residue ring: (1.5), 3 tensor product: (8.3), 37 universal example: (1.3), 2

valuation discrete: (26.12), 122 general: (26.12), 138 p-adic: (23.2), 123