BOUNDS on the NUMBER of GENERATORS of a MODULE By

BOUNDS ON THE NUMBER OF GENERATORS OF A MODULE

by PEYTON MORRIS MARTIN EVANS, COMMITTEE CHAIR MARTYN DIXON JON CORSON RICHARD BORIE

A THESIS

Submitted in partial fulﬁllment of the requirements for the degree of Master of Arts in the Department of Mathematics in the Graduate School of The University of Alabama

TUSCALOOSA, ALABAMA

The aim of this work is to present the Forster-Swan bound on the number of generators of a module. It is not our intention to present a novel finding or research discovery. Instead, we will develop the commutative algebra necessary to understand its proof, as well as the significance of the result. As a consequence, we will not present merely a series of prerequisites to the proof of the Forster-Swan bound, but rather everything which is strictly necessary to obtain a genuine un- derstanding of the nature of the theorem. With this goal in mind, we will develop not only many of the fundamental results of commutative algebra, but in addition present results which are in a more profound way related to or even go beyond the Forster-Swan theorem itself. In section one, we describe the basic properties of commutative rings and modules over them. In particular, we will give the standard results on prime and maximal ideals, finitely generated modules, exact sequences, tensor products and flatness. In section two we define the notion of the spectrum of a ring, show that this is a geometric object associated with the ring, and present a few examples of spectra of rings. In section three, the theory of localization, the driving force be- hind the Forster-Swan theorem, is developed and several local-global principles of commutative algebra are demonstrated. In section four, the theory of Noetherian rings and modules, the dimension of rings, the concept of an algebraic variety, and Hilbert’s basis and zero-locus theorems are presented. Section five contains the full proof of the Forster-Swan theorem. We conclude in section 6 with a discussion of topics surrounding Serre’s problem on projective modules and the Eisenbud-Evans conjectures which give an improvement on the Forster-Swan bound.

ii DEDICATION

To Diane, Edward, and Bill. When we meet again, I really hope this isn’t all I have to tell you about.

iii LIST OF ABBREVIATIONS AND SYMBOLS

N The natural numbers. Z The integers. Q The rational numbers. R The real numbers. C The complex numbers.

In The n × n identity matrix.

Mn×n(R) The ring of m × n matrices over a ring R.

iv ACKNOWLEDGMENTS

If one can ﬁnd anything of value in this thesis, this is due to a few truly exceptional people it has been my good fortune to have on my side. All that belongs to me are the countless mistakes I made along the way. I don’t know how I have the merit to enjoy such tremendous support. I could try to list every member of my family who helped me get here, but that would be gratu- itous. There isn’t a single one, out of so many, to whom I am not eternally grateful. It would only be fair to thank all of them here. Martin Evans, my advisor, helped to substantially improve the clarity of both content and form of exposition. Throughout the majority of my time as an undergraduate, I have had the pleasure of learning mathematics from Dr. Evans. When I leave here, this is without a doubt what I will miss the most. To the other members of my committee, I must also express my gratitude for your time and valuable input on my work. I would also like to thank the entire faculty of the department of mathematics for the time and en- ergy they have invested in me, and for providing such an enjoyable and stimulating environment for doing mathematics. I would particularly like to thank Bulent Tosun, Kyungyong Lee, David Cruz-Uribe, and Tim Ferguson. Lastly, I would like to thank my fellow students Tucker Ervin, Jeremy Cummings, Katelyn Isbell, and Connor Malin who, whether they were aware of it or not, helped me on my way.

v CONTENTS

ABSTRACT ...... ii

DEDICATION ...... iii

LIST OF ABBREVIATIONS AND SYMBOLS ...... iv

ACKNOWLEDGMENTS ...... v

LIST OF FIGURES ...... viii

CHAPTER 1 Preliminaries ...... 1

1.1 Commutative Rings ...... 1

1.2 Ring Homomorphisms, Ideals ...... 2

1.3 Prime and Maximal Ideals, the Nilradical and Jacobson Radical ...... 5

1.4 Modules ...... 8

1.5 Nakayama’s Lemma ...... 10

1.6 Exact Sequences, Tensor Products, and Flatness ...... 13

CHAPTER 2 The Spectrum of a Ring ...... 23

2.1 The Zariski Topology ...... 23

2.2 Examples of Spectra of Rings ...... 28

CHAPTER 3 Localization; Passage from Local to Global ...... 31

3.1 Rings and Modules of Fractions ...... 33

3.2 The Passage from Local to Global ...... 42

CHAPTER 4 Noetherian Rings, Modules, and Spaces; Aﬃne Varieties; Dimension Theory 47

vi 4.1 Noetherian/Artinian Rings and Modules ...... 49

4.2 Algebraic Varieties; The Nullstellensatz ...... 56

4.3 Dimension of a ring ...... 62

CHAPTER 5 The Forster-Swan Theorem ...... 65

CHAPTER 6 Projective Modules, Quillen’s Local-Global Principle, The Eisenbud- Evans Conjecture, and Serre’s Problem ...... 73

6.1 Finitely Presentable Modules ...... 77

6.2 Quillen’s Local-Global Principle ...... 86

6.3 Projective Modules and Serre’s Problem ...... 91

6.4 The Eisenbud-Evans Conjectures ...... 100

REFERENCES ...... 104

vii LIST OF FIGURES

2.1 Mumford’s famous depiction of Spec(Z[x]), from “The Red Book.” ...... 29

4.1 Some pictures of aﬃne varieties, illustrating the concept of dimension...... 47

viii CHAPTER 1

PRELIMINARIES

We assume the reader is familiar with the rudiments of group theory. For a set S , an object (ideal,module,etc.) generated S is frequently denoted by (S ). When the presence of additional parentheses renders this notation ambiguous, we instead use hS i to denote an object generated by S .

1.1 Commutative Rings

Deﬁnition 1.1. A ring R is a set with two binary operations + and ×, called addition and multiplication, such that

1) R is an abelian group with respect to addition. We denote the additive identity of R by 0.

2) Multiplication is associative: for all a, b, c ∈ R

a × (b × c) = (a × b) × c.

3) The distributive laws hold: for all a, b, c ∈ R

(a + b) × c = (a × c) + (b × c) and a × (b + c) = (a × b) + (a × c).

We will only be concerned with rings satisfying two additional axioms

1 4) There is a multiplicative identity in R (denoted by 1): there is a 1 ∈ R such that for all a ∈ R

a × 1 = 1 × a = a.

5) Multiplication is commutative: for all a, b ∈ R

a × b = b × a.

From here on, unless explicit mention is made to the contrary, every ring is assumed to satisfy (4) and (5) above, and we will adopt the shorthand ab for a × b. A subset S of a ring R is said to be a subring of R if 1 ∈ S and S is closed under addition and multiplication. In a ring R, x ∈ R is a unit if and only if there is a y ∈ R such that xy = 1. Moreover, this y is uniquely determined by x, so we write y = x−1, and for all a, b ∈ R we have that ab is a unit if and only if a and b are units. Thus the units of a ring form a multiplicative abelian group, denoted R×.

1.2 Ring Homomorphisms, Ideals

Deﬁnition 1.2. A ring homomorphism is a map f : R → S from a ring R into a ring S such that

1) For all a, b ∈ R we have f (a + b) = f (a) + f (b).

2) For all a, b ∈ R we have f (ab) = f (a) f (b).

3) The map takes the multiplicative identity in R to the multiplicative identity in S : f (1) = 1.

Given two ring homomorphisms, their composition is easily seen to be a ring homomorphism. A homomorphism f : R → S is an isomorphism if there is a homomorphism g : S → R such

2 that for all x ∈ R and y ∈ S we have (g ◦ f )(x) = x and ( f ◦ g)(y) = y. If f is an isomorphism

−1 then clearly f is a bijection. If f is a bijection let g = f , then f (g(1S )) = 1S = f (1R) and since f is injective, g(1) = 1. Moreover, f (g(xy)) = xy = f (g(x)g(y)) which gives g(xy) = g(x)g(y) and similarly g(x + y) = g(x) + g(y), so that g = f −1 is a ring homomorphism which makes f an isomorphism. Thus, a ring homomorphism is an isomorphism if and only if it is a bijection.

Given any two rings R and S with exactly one element, i.e. where 1R = 0R and 1S = 0S , there is a

unique ring isomorphism given by 0R 7→ 0S . A ring R is the zero ring, i.e. R = {0}, if and only if 1 = 0. An ideal I of a ring R is a subset of R which is both an additive subgroup and is closed under multiplication by elements of R, i.e. for all x ∈ R and y ∈ I we have xy ∈ I. Whenever it is convenient, we will use I C R to indicate that I is an ideal of R. Since I is closed under the multiplication in R, there is an induced multiplication in the quotient group R/I, and equipped with this multiplication we call R/I a quotient ring. Its elements are the cosets of I in R, and the canonical map φ : R → R/I taking each x ∈ R to the coset x + I ∈ R/I is a surjective ring homomorphism.

Proposition 1.1. There is a one-to-one inclusion-preserving correspondence between the ideals J of R with I ⊂ J, and the ideals J of R/I given by J = φ−1(J¯).

Proof. Let I be an ideal of R. We already have the correspondence for subgroups of R. Let S =

{J ⊂ R : J C R and J ⊃ I} and T = {J¯ ⊂ R/I : J¯ C R/I}. For all J¯ ∈ T , we must have {0} = φ(I) ⊂ J¯ so that I ⊂ φ−1(J¯). Let x + I ∈ J¯. For all r + I ∈ R/I, there are r0, x0 ∈ R such that φ(r0) = r + I and φ(x0) = x + I, since φ is surjective. We have (r + I)(x + I) = φ(r)φ(x) ∈ J¯ so Rφ−1(J¯) ⊂ φ−1(J¯), and thus φ−1(J¯) ∈ S. Let J ∈ S, let y ∈ φ(J), and let x ∈ J be such that y = φ(x) = x + I. For all r + I ∈ R/I, since J is an ideal in R we have (r + I)y = (r + I)(x + I) = rx + I = φ(rx) ∈ φ(J) so that φ(J) is an ideal, and thus φ(J) ∈ T . Let F : S → T and G : T → S be such that F(J) = φ(J) and G(J¯) = φ−1(J¯). We have

3 (F ◦ G)(J¯) = φ(φ−1(J¯)) = J¯ and (G ◦ F)(J) = φ−1(φ(J)) = J. Moreover, if J¯ ⊂ T¯ and if x ∈ R is such that φ(x) ∈ J¯, then φ(x) ∈ T¯ so that G(J¯) = φ−1(J¯) ⊂ G(T¯) = φ−1(T¯). Thus, G gives the desired bijection.

For any ring homomorphism f : R → S , the kernel of f is the set ker f = {x ∈ R : f (x) = 0}, and it’s clear that ker f is an ideal of R. A ring homomorphism is injective if and only if ker f = {0}.

Moreover, f (R) is a subring of S and R/ ker f f (R). A zero divisor in a ring R is an element x ∈ R such that there is a non-zero y ∈ R with xy = 0. A non-zero ring with no zero-divisors is called an integral domain. For a ring R and for all x ∈ R the set (x) = {rx : r ∈ R} is an ideal, and such ideals are distinguished by referring to them as principal ideals. We have (x) = R if and only if x is a unit. An element x ∈ R is nilpotent if there is a positive integer n such that xn = 0. A non-zero ring is a ﬁeld if and only if every nonzero element is a unit.

Proposition 1.2. Let R , {0} be a ring. The following are equivalent

1) R is a ﬁeld

2) the only ideals in R are (0) and (1)

3) every homomorphism of F into a non-zero ring S is injective

Proof.

1) =⇒ 2) Let I be a non-zero ideal of R. There is an non-zero x ∈ I, but x is a unit so that (x) = (1) = R ⊂ I. Thus R = (1) = I.

2) =⇒ 3) If f : R → S is a homomorphism, then f (1) , 0 so that ker f , R is a proper ideal, which implies that ker f = (0) and so f is injective.

3) =⇒ 1) Suppose x ∈ R is not a unit. Since x is not a unit, R , (x) and thus R/(x) , 0. Now the canonical map φ : R → R/(x) is injective, so ker φ = (x) = (0), whence x = 0.

4 If I and J are ideals in R, we can deﬁne their sum, product, and the radical r(I) of an ideal

I + J = {a + b : a ∈ I, b ∈ J}

X IJ = aibi : ai ∈ I, bi ∈ J, almost all ai, bi = 0 i r(I) = {a ∈ R : an ∈ I for some positive integer n}

all of which are easily seen to be ideals. Moreover, I ∩ J is also an ideal. We have IJ ⊂ I ∩ J ⊂ T I, J ⊂ I + J. For any family of ideals {Iα}α∈A we have α∈A Iα an ideal, and whenever an ideal J T is such that J ⊂ Iα for all α ∈ A we have J ⊂ α∈A Iα. For any subset S ⊂ R if {Iα}α∈A is the set T of all ideals with S ⊂ Iα, then α∈A Iα = (S ) is the smallest ideal containing S , which we call the ideal generated by S . Now, IJ ⊂ I ∩ J suggests approaching (S ) from below by the product of all P principal ideals (x) where x ∈ S , and in fact (S ) = { i ri xi : ri ∈ R, xi ∈ S }. An ideal is said to be ﬁnitely generated if it possesses a ﬁnite generating set. If I + J = (1) we have IJ = I ∩ J, and in this case the ideals I and J are said to be coprime. The ideals I and J are coprime if and only if there are x ∈ I and y ∈ J such that x + y = 1.

1.3 Prime and Maximal Ideals, the Nilradical and Jacobson Radical

An ideal p in R is prime if 1 < p and if xy ∈ p implies that x ∈ p or y ∈ p. An ideal m is maximal if 1 < m and there is no ideal I such that m ( I ( R.

Proposition 1.3. Let R be a ring and let p, m be ideals in R. We have that p prime if and only if R/p is an integral domain and m maximal if and only if R/m is a ﬁeld.

Proof. Suppose p is prime, and suppose to get a contradiction x + p , 0 is a zero divisor, then there is a y + p , 0 such that xy + p = 0, i.e. xy ∈ p, which is impossible since x, y < p. Now suppose R/p is an integral domain. Let x, y ∈ R \ p, and suppose that xy ∈ p. Since x + p , 0 and

5 y + p , 0, but xy + p = 0 this contradicts our hypothesis that R/p has no zero-divisors. Now m is a maximal ideal if and only if there are no proper ideals I such that m ( I, which is true

if and only if the only ideals in R/m are (0) and (1), which is true if and only if R/m is a ﬁeld.

As an immediate corollary, every maximal ideal is prime, since every ﬁeld is an integral domain.

Proposition 1.4. Every non-zero ring R has at least one maximal ideal.

Proof. Let S = {I C R : I , R} and order S by inclusion. Let {Iα}α∈A be a chain in S , so that for S all α, β ∈ A either Iα ⊂ Iβ or Iβ ⊂ Iα. Let I = α∈A Iα. Let x, y ∈ I then x, y ∈ Iα for some α ∈ A, so that x − y ∈ Iα and thus x − y ∈ I. Similarly, let x ∈ I, then x ∈ Iα for some α ∈ A so for all T r ∈ R we have rx ∈ Iα and thus rx ∈ I for all r ∈ R. Since 1 ∈ α∈A R \ Iα we have 1 < I and thus

I , R. It follows that I is an ideal in S , and Iα ⊂ I for all α ∈ A, so that I is an upper bound for the chain {Iα}α∈A. Thus, by Zorn’s lemma, there is a maximal element m in S , and m satisﬁes the axioms for a maximal ideal.

Given a proper ideal I of R, by applying the previous result to the quotient ring R/I, we must have that every ideal I , R is contained in a maximal ideal. A fortiori, every non-unit is contained in a maximal ideal. If a ring R has exactly one maximal ideal, we call R a local ring, and we let k denote its residue ﬁeld R/m. Such rings do in fact exist, and are of extreme importance in what follows. We deﬁne the nilradical N of R to be the set of all nilpotent elements of R. Now, if x is nilpotent then clearly for all r ∈ R we have rx nilpotent. If x and y are nilpotent, there are positive integers n and m such that xn = 0 = ym. By the binomial theorem, (x + y)n+m−1 is a sum of products of the form xrys with r, s ≥ 0 and r + s = m + n − 1, so that either r ≥ m or s ≥ n. Thus x + y is nilpotent. This gives the following proposition.

Proposition 1.5. Let R be a ring. Its nilradical is an ideal.

Suppose that a < N, and let S = {I C R : an < I for all n > 0}. Since a is not nilpotent, S (0) ∈ S . Order S by inclusion. Let {Iα}α∈A be a chain in S , and let I = α∈A Iα. I is an ideal. Since

6 n T a ∈ α∈A R\Iα for all n > 0 we have I ∈ S and Iα ⊂ I for all α ∈ A. By Zorn’s lemma, S contains a maximal element.

Proposition 1.6. The nilradical of a ring R is the intersection of all the prime ideals of R.

Proof. Let N denote the intersection of all the prime ideals of R. Let a ∈ R and let p be a prime ideal, then if a ∈ N we must have an n > 0 such that an = 0 ∈ p, so that a ∈ p by the primality of p. Hence N ⊂ N.

Suppose a ∈ R is not nilpotent. Let p be a maximal element in S = {I C R : an < I for all n > 0}. Suppose x, y < p, then (x, p), (y, p) ) p so (x, p), (y, p) < S . Hence, there is an m > 0 and an n > 0 such that am ∈ (x, p) and an ∈ (y, p). We then have am+n ∈ (x, p) · (y, p) = (xy, p) so that (xy, p) < S and so xy < p. Thus, p is a prime ideal with a < p, so a < N.

Given an ideal I of R, let φ : R → R/I denote the canonical map. Clearly, we must have r(I) =

−1 φ (NR/I) and so proposition 1.6 applied to the ring R/I yields the following result.

Corollary 1.1. The radical of an ideal I is the intersection of the prime ideals which contain I.

The proof of the previous proposition highlights an interesting principle in commutative algebra, namely that ideals which are maximal with respect to some condition have a tendency to be prime. The Jacobson radical R of a ring R is deﬁned to be the intersection of all the maximal ideals of R.

Proposition 1.7. For a ring R, its Jacobson radical is completely characterized by the property that x ∈ R if and only if 1 − xy is a unit in R for all y ∈ R.

Proof. Suppose x ∈ R. If 1 − xy is a non-unit for some y ∈ R, then 1 − xy ∈ m for some maximal ideal m. But x ∈ m and so xy ∈ m and thus 1 ∈ m, which is impossible. Let x ∈ R be ﬁxed. Now suppose that 1 − xy is a unit for all y ∈ R. If there is a maximal ideal m of R such that x < m, then (m, x) = (1) so that there is a u ∈ m and a y ∈ R with u + xy = 1, i.e. u = 1 − xy ∈ m so that m contains a unit, which is impossible.

7 Let f : R → S be a ring homomorphism. For an ideal I in S , f −1(I) is always an ideal. For an ideal J in R, f (J) need not be an ideal. However, the set f (J) still generates an ideal, and we will denote this ideal by Ie. On the other hand, it is also convienent to write Ic for the ideal f −1(I). For a prime ideal of S , pc is always prime, but for a maximal ideal m of S we need not have mc a maximal ideal.

1.4 Modules

The following generalization of the concept of a vector space is of fundamental importance.

Deﬁnition 1.3. For a ring R, an R-module is an abelian group M for which there exists a mapping • : A × M → M such that for all a, b ∈ R and x, y ∈ M

1) •(a, x + y) = •(a, x) + •(a, y)

2) •(a + b, x) = •(a, x) + •(b, x)

3) •(ab, x) = •(a, bx)

4) •(1, x) = x.

We adpot the shorthand ax for •(a, x). Let End(M) denote the set of all group homomorphisms f : M → M. For f, g ∈ End(M) and m ∈ M define the “pointwise addition” of f and g by ( f + g)(m) = f (m) + g(m), then the End(M) is a ring with respect to the operations of pointwise addition and composition. Moreover, φ : R → End(M) is a ring homomorphism if and only if defining rx = φ(r)(x) for all r ∈ R and x ∈ M gives M the structure of an R-module, i.e. the map • : A × M → M with •(r, x) = φ(r)(x) satisfies the axioms above.

Deﬁnition 1.4. Given R-modules M, N a mapping f : M → N is an R-module homomorphism (or R-linear) if for all a ∈ R and x, y ∈ M

1) f (x + y) = f (x) + f (y)

8 2) f (ax) = a f (x)

Let HomR(M, N) denote the set of all R-module homomorphisms. Deﬁning addition in HomR(M, N)

pointwise and scalar multiplication by (a f )(x) = a( f (x)) for all a ∈ R and f ∈ HomR(M, N) gives

HomR(M, N) the structure of an R-module. We also have a free object.

Deﬁnition 1.5. For an R-module F,we say that a subset B ⊂ F is a basis for F if for all nonzero

x ∈ F there are unique, nonzero r1,..., rn ∈ R and unique, nonzero a1,..., an ∈ B such that

x = r1a1 + r2a2 + ··· + rnan.

If F has a basis it is called a free R-module. For R-modules M and N we can form their direct sum M ⊕ N = {(x, y): x ∈ M, y ∈ N} which is an R-module when we define addition and scalar multiplication componentwise. For a family of R-modules {Mi}i∈I their direct sum is the collection of families {xi}i∈I such that for xi ∈ Mi for each i ∈ I and almost all of the xi are zero. Hence, an R-module F is free if and only if F ⊕i∈IR for some index set I. Given an R-module M, we define a submodule of M as a subgroup M0 of M which is closed under multiplication by elements of R. The scalar multiplication on M induces an R-module structure on the abelian group M/M0. We call M/M0 the quotient (module) of M by M0. It’s not hard to see that the canonical map M → M/M0 is R-linear. Moreover, there is an inclusion preserving bijective correspondence between submodules of M containing M0 and the submodules of M/M0. Given an R-module homomorphism f : M → N the kernel and image of f are submodules of M and N, respectively. As usual, we have an isomorphism M/ ker f im f . We define the cokernel of f to be Coker( f ) = N/im f . Let M be an R-module. If x ∈ M then the set (x) = {rx : r ∈ R, x ∈ M} is a submodule of

M. A subset S = {xi}i∈I ⊂ M such that M = (xi)i∈I is called a set of generators or a generat-

9 ing system for M. If there is a finite generating system for M we say that M is finitely generated. P P Let {Mi}i∈I be a family of submodules of M. We define their sum to be i∈I Mi = { i xi : xi ∈

Mi, almost all xi = 0}, which is a submodule. Their intersection is also a submodule. Given an ideal I of R we can define the product IM to be the submodule generated by all of the products ax P (a ∈ I, x ∈ M), i.e. the set of all finite sums IM = { i ai xi : ai ∈ I, xi ∈ Mi}. Moreover, M/IM is an R/I-module in the obvious way. We can also define the annihilator of M, Ann(M) = {r ∈ R : for all m ∈ M, rm = 0}, or equivalently as the kernel of the ring homomorphism R → End(M). Thus, Ann(M) is an ideal of R. If F is a free R-module, then Ann(F) = {0}. Clearly, if K is a field, then every K-module is a vector space. Since every vector space has a basis, every K-module is free. Suppose that R is not a field, then there exists a non-zero, proper ideal I and R/I , 0. Since Ann(R/I) = I, the R-module R/I is not free. Hence, R is a field if and only if every R-module is free. It would be beneficial to point out a key difference between the theory of modules and vector spaces. In a (finite dimensional) vector space, one can always express its elements, the “vectors”, uniquely in terms of coordinates with respect to a basis. However, generating sets for (finitely generated) modules need not be linearly independent. To be more precise, even if two generators are linearly dependent it may not be possible to express one in terms of the other. As anyone who remembers the proof from linear algebra could tell you, this is due to the (potential) lack of multiplicative inverses. Hence, the expression of an element of a module in terms of coordinates with respect to a generating system may no longer be unique, and the notion of dimension does not carry over in a well-defined manner. This is the fundamental issue one has to contend with to extend results of linear algebra to the category of R-modules.

1.5 Nakayama’s Lemma

We could also express the fact that we no longer have a unique system of coordinates as follows.

10 Proposition 1.8. M is a ﬁnitely generated R-module if and only if M is isomorphic (as an R- module) to a quotient of Rn for some n > 0.

n Proof. If M is ﬁnitely generated, let {x1,..., xn} be a generating system. Then φ : R → M taking

n (a1,..., an) to a1 x1 + ··· + an xn is clearly R-linear and surjective. Thus M R / ker φ. Now if M is isomorphic a quotient of Rn, by N say, there is an R-linear homomorphism φ : An →

n M with kernel N, and so if {x1,..., xn} is a basis for R then the φ(x1), . . . , φ(xn) generate M.

Let f : R → S be a ring homomorphism. If a ∈ R and b ∈ S , deﬁne a product ab = f (a)b. This makes S into an R-module, and the ring S equipped with this R-module structure is called

P i an R-algebra. The prototypical example of an R-algebra is the polynomial ring R[x] = { i ai x : ai ∈ R, almost all ai = 0}, with the ring homomorphism f : R → R[x] given by f (r) = xr.

Clearly, with this R-module structure, R[x1,..., xn] is the free module of rank n. We can extend the Cayley-Hamilton theorem.

Proposition 1.9. Let M be a ﬁnitely generated R-module, I an ideal of R, and φ : M → M an R-linear map with im φ ⊂ IM. Then φ satisﬁes an equation of the form

n n−1 φ + a1φ + ··· + an = 0

i where a1,..., an ∈ I and φ is the i-fold composition of φ with itself.

Proof. Let {x1,..., xn} be a generating system for M. Now for each i we have φ(xi) ∈ IM, so

Pn there are ai j ∈ I such that φ(xi) = j=1 ai j x j. We can view M as an R[t]-module by deﬁning a multiplication ti x = φi(x), and so Xn (tδi j − ai j)x j = 0 j=1 where δi j is the Kronecker delta. Let A be the n × n matrix whose entries are the ai j, In the n × n

11 identity matrix, and v the column vector whose entries are the x j, so that

(tIn − A)v = 0

and multiplying on the left by the adjoint of tIn − A we have

det(tIn − A)Inv = 0

so that det(tIn − A)x j = 0 for each j. Let p(t) = det(tIn − A). Then, p(φ) = det(φIn − A) is the zero endomorphism of M, and by expanding the determinant we have an equation of the desired

form.

The celebrated Nakayama lemma is an easy consequence of this result. There are actually several related results which are occasionally referred to as Nakayama’s lemma.

Corollary 1.2. (Nakayama’s lemma) Let M be a ﬁnitely generated R-module

1) If I is an ideal of R such that IM = M, then there is an x ∈ R such that x − 1 ∈ I and xM = 0.

2) If I is an ideal of R contained in the Jacobson radical R and IM = M, then M = 0.

3) If N is a submodule of M and I is an ideal of R contained in the Jacobson radical R with IM + N = M, then M = N.

Proof.

1) Take φ = 1 in the Cayley-Hamilton theorem, and let x = 1 + a1 + ··· + an. Then for all m ∈ M, x · φ(m) = xm = 0.

2) By (1), there is an x such that x − 1 ∈ R and xM = 0. Thus, x is a unit, and so M = x−1 xM = 0.

12 3) By (2), since I(M/N) = (IM + N)/N = M/N we must have M/N = 0, i.e. M = N.

Now let R be a local ring with maximal ideal m. If M is a ﬁnitely generated R-module, then M/mM

is annihilated by m, so it is naturally a R/m-module, and thus a vector space. Suppose x1,..., xn

are elements of M whose images form a basis for this vector space. Let N = (x1,..., xn), then the canonical map N → M/mM is surjective, so that (N + mM)/mM = M/mM, which implies

that N + mM = M. By Nakayama’s lemma, N = M, i.e. the x1,..., xn generate M. We have just shown the following.

Corollary 1.3. Let R be a local ring with maximal ideal m and residue ﬁeld k = R/m. If M is a

ﬁnitely generated R-module and x1,..., xn ∈ M are such that their images in M/mM generate the

k-vector space M/mM, we must have that M = (x1,..., xn), i.e. the xi generate M.

Clearly, if M = (x1,..., xn) we must have the images of the xi generators for M/mM. We then have that the number of generators in a minimal generating system for M is given by the dimen-

sion of the k-vector space M/mM, and {x1,..., xn} is a minimal generating system for M if and

only if the set of images {x¯1,..., x¯n} forms a basis for M/mM.

1.6 Exact Sequences, Tensor Products, and Flatness

Deﬁnition 1.6. A sequence of R-modules and R-linear homomorphisms

fi fi+1 · · · → Mi−1 −→ Mi −−→ Mi+1 → ...

is said to be exact at Mi if im fi = ker fi+1. The sequence is exact if it is exact at each Mi. Given R-modules M, M0, and M00 and R-linear f, g, we have: f 0 → M0 −→ M exact if and only if f is injective, g M −→ M00 → 0 exact if and only if g is surjective, f g 0 → M0 −→ M −→ M00 → 0 exact if and only if f is injective, g is surjective, and g induces an

13 isomorphism Coker( f ) M00. Moreover, for any R-linear homomorphism f : M → N

f 0 → ker f → M −→ im f → 0

is exact, and for any submodule N ≤ M

0 → N → M → M/N → 0

is exact. As a special case of the latter

0 → im f → N → Coker( f ) → 0

is exact.

Proposition 1.10. The sequence of R-modules and R-module homomorphisms

u v M0 −→ M −→ M00 → 0

is exact if and only if for all R-modules N

v¯ u¯ 0 → Hom(M00, N) −→ Hom(M, N) −→ Hom(M0, N)

is exact, whereu ¯( f ) = f ◦ u andv ¯( f ) = f ◦ v. In other words, the functor HomR(, N) is left exact.

Proof. Suppose u v M0 −→ M −→ M00 → 0

14 is exact. Let N be any R-module. Consider

v¯ u¯ 0 → Hom(M00, N) −→ Hom(M, N) −→ Hom(M0, N) and let f ∈ Hom(M00, N), then if f ∈ kerv ¯ we must have f ◦ v = 0. Since v is surjective, f maps every element of M00 to 0, i.e. f = 0. Hence,v ¯ is injective. Now since im u = ker v and u¯ ◦ v¯( f ) = f ◦ v ◦ u we must have imv ¯ ⊂ keru ¯. Let f ∈ Hom(M, N), then if f ∈ keru ¯, we have f ◦ u = 0 so ker v = im u ⊂ ker f . Since v is surjective, M00 = M/ ker v = M/im u. For x ∈ M letx ˜ = x + im u be its image in M/im u. We can

00 00 associate eachx ˜ to a yx ∈ M and deﬁne an R-linear map φ : M → N by φ(yx) = f (x). This is

well-deﬁned, for if y1 = y2 then x1 − x2 ∈ im u ⊂ ker f so f (x1) = f (x2) and thus φ(y1) = φ(y2). Moreover, φ ◦ v = v¯(φ) = f , so that f ∈ imv ¯. Hence imv ¯ = keru ¯, and the sequence is exact. Suppose that the sequence

v¯ u¯ 0 → Hom(M00, N) −→ Hom(M, N) −→ Hom(M0, N) is exact for every R-module N. Sincev ¯ is injective for all N, if we take N = M there is an f ∈ Hom(M00, M) such that f ◦ v = 1, so that v is surjective. Now let N = M00. Sinceu ¯ ◦ v¯ = 0 we have v ◦ u ◦ f = 0 for all f ∈ Hom(M00, M00), so that in particular, for f = 1 we have v ◦ u = 0. Thus im u ⊂ ker v. Now let N = Coker(u) = M/im u and let φ : M → Coker(u) take x to its image in the cokernel. Since φ ∈ keru ¯ = imv ¯ there is a ψ : M00 → Coker(u) with φ = ψ ◦ v. Then u v ker v ⊂ ker φ = im u. Thus v is surjective and ker v = im u, so that M0 −→ M −→ M00 → 0 is exact.

15 Recall that for vector spaces V, W, and X a map f : V × W → X is bilinear if it is linear in both arguments, i.e. for each v ∈ V the map w 7→ f (v, w) is linear and for each w ∈ W the map v 7→ f (v, w) is linear. For a ring R, the notion of an R-bilinear map between R-modules carries over mutatis mutandis. Now having already built up a few methods to work with modules and their linear maps, it would be prudent to explain how to work with bilinear or even multilinear maps. Fortunately, we can construct a module, called the tensor product, which brings us back to the setting of linear maps. To be more speciﬁc, for a ring R and R-modules M, N, and P, there is a universal property of the tensor product M ⊗ N wherein bilinear maps from M × N to P are in a

natural one-to-one correspondence with the R-linear maps from M ⊗R N to P.

Proposition 1.11. Let M and N be R-modules. Then there exists a pair (T, g) consisting of an R-module T and an R-bilinear mapping g : M × N → T with the following property: Given any R-module P and any R-bilinear mapping f : M × N → P there exists a unique R-linear mapping f 0 : T → P such that f = f 0 ◦ g. Moreover, if (T, g) and (T 0, g0) are two pairs with this property, then there exists a unique isomorphism j : T → T 0 such that j◦g = g0. This is expressed by the commutativity of the following diagrams.

g M × N T f f 0 P

g M × N T g0 j ∼ T 0

Proof.

Uniqueness: Since T 0 is an R-module and g0 is R-bilinear, by the universal property for (T, g), there is a unique j : T → T 0 such that g0 = j ◦ g. On the other hand, T is also an R-module and g is R-bilinear, so by the universal property for (T 0, g0) there is a unique j0 : T 0 → T such that

16 g = j0 ◦ g0. Since j ◦ j0 ◦ g0 = j ◦ g = g0 and j0 ◦ j ◦ g = j0 ◦ g0 = g, we must have j ◦ j0 = 1 = j0 ◦ j.

Existence: Let C denote the free R-module with basis M × N. Let D be the submodule of C generated by all the elements (which one should think of as “relations”) of the following form:

(x + x0, y) − (x, y) − (x0, y)

(x, y + y0) − (x, y) − (x, y0)

(ax, y) − a · (x, y)

(x, ay) − a · (x, y) and let T = C/D. For each (x, y) ∈ M × N let x ⊗ y denote its image in T. Then the quotient map g : M × N → T is such that g(x, y) = x ⊗ y. Since M × N is a basis for C, T is generated by the x ⊗ y. We have defined D in such a way that g must be an R-bilinear map. Let an R-module P and map f : M × N → P be given, then since f is defined on the basis for C we can extend f linearly to an R-module homomorphism f¯ : C → P. If f is R-bilinear, then f¯ = 0 on D. Since f¯ vanishes on D, if we define f 0 : T → P by f 0(a + D) = f¯(a), for a ∈ C, we get a well-defined R-linear homomorphism. Moreover, f 0(x ⊗ y) = f¯((x, y)) = f (x, y), and f 0 is uniquely defined by this condition. Thus, for any R-module P and any R-bilinear f : M × N → P

we must have f = f 0 ◦ g with f 0 unique.

The R-module T is the tensor product, and it is denoted by M ⊗R N. By construction, if M and N

are ﬁnitely generated R-modules, so too is M⊗R N. A general element of M⊗R N is not of the form x ⊗ y, but rather a ﬁnite sum whose terms are of the form x ⊗ y. Moreover, such a representation need not be unique. Clearly, the map sending elements (x, y) of M ×N to their tensor product x⊗y is R-bilinear. The universal property of tensor products gives us a few very helpful isomorphisms.

17 Proposition 1.12. Let M,N, and P be R-modules. There are unique isomorphisms

1) M ⊗ N → N ⊗ M

2)( M ⊗ N) ⊗ P → M ⊗ (N ⊗ P) → M ⊗ N ⊗ P

3)( M ⊕ N) ⊗ P → (M ⊗ P) ⊕ (N ⊗ P)

4) M ⊗ R → M.

Proof.

1) Deﬁne f : M × N → N ⊗ M by f (x, y) = y⊗ x. Then f is R-bilinear and we get a unique R-linear map φ : M ⊗ N → N ⊗ M such that φ(x ⊗ y) = y ⊗ x. Similarly, by deﬁning g : N × M → M ⊗ N by g(y, x) = x ⊗ y we can get a unique R-linear map ψ : N ⊗ M → M ⊗ N such that ψ(y ⊗ x) = x ⊗ y. Clearly, φ ◦ ψ = 1 and ψ ◦ φ = 1. Thus φ is an isomorphism.

2) Fix z ∈ P. From the R-bilinear map (x, y) 7→ x ⊗ (y ⊗ z), with x ∈ M and y ∈ N, we get a unique

R-linear map fz : M ⊗ N → M ⊗ (N ⊗ P) with fz(x ⊗ y) = x ⊗ (y ⊗ z). For t ∈ M ⊗ N, z ∈ P, the map (t, z) 7→ fz(t) is R-bilinear and induces a homomorphism φ :(M ⊗ N) ⊗ P → M ⊗ (N ⊗ P) such that φ((x ⊗ y) ⊗ z) = x ⊗ (y ⊗ z). Fix x ∈ M. From the R-bilinear map (y, z) 7→ (x ⊗ y) ⊗ z, with y ∈ N and z ∈ P, we get a unique

R-linear map gx : N ⊗ P → (M ⊗ N) ⊗ P with gx(y ⊗ z) = (x ⊗ y) ⊗ z. For s ∈ N ⊗ P, x ∈ M, the map (x, t) 7→ gx(t) is R-bilinear and induces a homomorphism ψ : M ⊗ (N ⊗ P) → (M ⊗ N) ⊗ P) such that ψ(x(⊗y ⊗ z)) = (x ⊗ y) ⊗ z. Clearly, φ ◦ ψ = 1 and ψ ◦ φ = 1. Thus φ is an isomorphism. By a similar argument, we can deﬁne an isomorphism π :(M ⊗ N) ⊗ P → M ⊗ N ⊗ P by π((x ⊗ y) ⊗ z) = x ⊗ y ⊗ z.

3) The bilinear map f :(M ⊕ N) × P → (M ⊗ P) ⊕ (N ⊗ P) with f ((x, y), z) = (x ⊗ y, y ⊗ z) induces a unique R-linear map φ :(M ⊕ N) ⊗ P → (M ⊗ P) ⊕ (N ⊗ P) with φ((x, y) ⊗ z) = (x ⊗ z, y ⊗ z).

18 By a similar argument we can obtain unique R-linear maps ψ0 : M ⊗ P → (M ⊕ N) ⊗ P and ψ00 : N ⊗ P → (M ⊕ N) ⊗ P with ψ0(x ⊗ z) = (x, 0) ⊗ z and ψ00(y ⊗ z) = (0, y) ⊗ z. Deﬁne ψ :(M ⊗ P) ⊕ (N ⊗ P) → (M ⊕ N) ⊗ P by ψ((x ⊗ z, y ⊗ z) = ψ0(x ⊗ z) + ψ00(y ⊗ z), then ψ is unique and R-linear because ψ0 and ψ00 are. Moreover, ψ ◦ φ = 1 and φ ◦ ψ = 1,so φ is an isomorphism.

4) We have a bilinear map f : M × R → M given by f (x, r) = rx. Hence, there is a unique R-linear map φ : M ⊗ R → M given by φ(x ⊗ r) = rx. The map ψ : M → M ⊗ R with ψ(x) = x ⊗ 1 is R-linear, and (ψ ◦ φ)(x ⊗ r) = ψ(rx) = rx ⊗ 1 = x ⊗ r and (φ ◦ ψ)(m) = φ(m ⊗ 1) = m. Thus φ is an

isomorphism.

We can also consider the tensor product of R-module homomorphisms. For if f : M → N and f 0 : M0 → N0 are R-linear then the map M × M0 → N ⊗ N0 which sends (x, x0) to f (x) ⊗ f 0(x0) is R-bilinear, so that there is a unique R-linear homomorphism f ⊗ f 0: M ⊗ M0 → N ⊗ N0 with ( f ⊗ f 0)(x ⊗ x0) = f (x) ⊗ f 0(x0). We also have the following exactness property of the tensor product

Proposition 1.13. For all R-modules M, N, and P

Hom(M ⊗ N, P) Hom(M, Hom(N, P)).

Moreover, for every exact sequence of R-modules and R-module homomorphisms and any R- module N u v M0 −→ M −→ M00 → 0

the sequence u⊗1 v⊗1 M0 ⊗ N −−→ M ⊗ N −−→ M00 ⊗ N → 0

is exact. In other words, tensor products are right exact.

19 Proof. First for R-modules M, N, and P, let f : M × N → P be R-bilinear. Then for each x

the maps fx given by y 7→ f (x, y) are R-linear, i.e. fx ∈ Hom(N, P) for each x ∈ M. Thus each

f : M × N → P gives rise to a φ ∈ Hom(M, Hom(N, P)) given by φ(x) = fx. On the other hand, if φ ∈ Hom(M, Hom(N, P)) the map (x, y) 7→ φ(x)(y) is bilinear. It follows that the set of R-bilinear maps is in bijective correspondence with Hom(M, Hom(N, P)). But the set of R-bilinear maps is in bijective correspondence with Hom(M ⊗ N, P) by the universal property of tensor products. In other words, we can associate each f ∈ Hom(M ⊗ N, P) with a F ∈ Hom(M, Hom(N, P)) where f (x ⊗ y) = F(x)(y). The maps sending f to F and F to f are clearly inverses and R-linear. Hence, Hom(M ⊗ N, P) Hom(M, Hom(N, P)). Fix an R-module N. Now if u v M0 −→ M −→ M00 → 0 is exact, for all R-modules P, by proposition 1.10 the sequence

0 → Hom(M0, Hom(N, P)) → Hom(M, Hom(N, P)) → Hom(M00, Hom(N, P)) is exact. Thus

0 → Hom(M0 ⊗ N, P) → Hom(M ⊗ N, P) → Hom(M00 ⊗ N, P) is exact. But again by proposition 1.10, this means that

M0 ⊗ N → M ⊗ N → M00 ⊗ N → 0 is exact. The result follows.

Proposition 1.14. For a ring R, let I be an ideal and let M be an R-module. Then R/I ⊗R M M/IM.

20 Proof. We have an exact sequence 0 → I → R → R/I → 0 and so I⊗M → R⊗M → R/I⊗M → 0

is exact. Thus R/I ⊗ M (R ⊗ M)/(im i ⊗ R) (R ⊗ M)/(I ⊗ R) where i : I → R is the inclusion map. Since the unique isomorphism R ⊗ M → M sending r ⊗ x to rx maps I ⊗ R to IM, we have

(R ⊗ M)/(I ⊗ R) M/IM. Hence R/I ⊗ M M/IM.

Corollary 1.4. For a local ring R, let M and N be ﬁnitely generated R-modules. If M ⊗ N = 0 then M = 0 or N = 0.

Proof. Let m be the maximal ideal of R and let k = R/m denote its residue ﬁeld. Now by proposition 1.14 k ⊗R M is isomorphic to the k-vector space M/mM, and similarly for k ⊗R N. Moreover, since M ⊗ N = 0 = k ⊗R (M ⊗ N) and

k ⊗R (M ⊗R N) (k ⊗k k) ⊗ (M ⊗R N) (M ⊗R k) ⊗k (N ⊗R k) M/mM ⊗k N/mN = 0

Since M/mM, N/mN are k-vector spaces, the tensor product of their basis elements are basis elements of the tensor product, and so M/mM ⊗k N/mN = 0 implies that we must have one of them zero. Without loss we may assume M/mM = 0, then by Nakayama’s lemma M = 0.

If the sequence M0 → M → M00 is exact, we need not have M0 ⊗ N → M ⊗ N → M00 ⊗ N f exact for any R-module N. For example, given a non-zero integer n, the sequence 0 → Z −→ Z, with f (x) = nx for all x ∈ Z, is exact. Let N = Z/nZ, then for all x ⊗ y ∈ Z ⊗ N we have f ⊗1 ( f ⊗ 1)(x ⊗ y) = f (x) ⊗ y = nx ⊗ y = x ⊗ ny = x ⊗ 0 = 0. Hence 0 → Z ⊗ N −−−→ Z ⊗ N is not exact, since f is not injective. To say that we are interested in R-modules where this is true would be a massive understatement.

Deﬁnition 1.7. We say that an R-module N is ﬂat if for all exact sequences of R-modules

M0 → M → M00

21 the sequence M0 ⊗ N → M ⊗ N → M00 ⊗ N is exact.

Proposition 1.15. The following are equivalent, for an R-module N:

1) N is ﬂat.

2) If 0 → M0 → M → M00 → 0 is any exact sequence of R-modules, then 0 → M0 ⊗ N → M ⊗ N → M00 ⊗ N → 0 is exact.

3) If f : M0 → M is injective, then f ⊗ 1 : M0 ⊗ N → M ⊗ N is injective.

fi fi+1 Proof. Any sequence · · · → Mi−1 −→ Mi −−→ Mi+1 → ... can be split up into

· · · → Ni → 0 → Ni → Mi → Ni+1 → 0 → Ni+1 → ...

where Ni = im fi = ker fi+1 for each i. Thus (1) ⇐⇒ (2) is clear. Now proposition 1.13 gives (2) =⇒ (3), while (3) =⇒ (2) is clear.

Mumford, in the red book of varieties and schemes, writes that “the concept of ﬂatness is a riddle that comes out of algebra, but which technically is the answer to many prayers.” Up to here, this is probably the best available explanation for why ﬂat modules are important. Flatness will be lurking in the background of a lot of what follows.

22 CHAPTER 2

THE SPECTRUM OF A RING

It is often useful to associate a ring with a certain geometric object. Perhaps surprising at first, this object is the set of all prime ideals of the ring. Under suitable conditions, the prime ideals can be shown to correspond to points, curves, surfaces, and so on. However, to even begin to jus- tify the claim that the prime ideals of a ring contain information of any geometric significance, we must first show that the collection of all prime ideals is endowed with some type of structure, namely a topology.

2.1 The Zariski Topology

The set of all prime ideals of a ring R is called the spectrum of R, and is denoted by Spec(R). For any subset S of R we set V(S ) = {p ∈ Spec(R): S ⊂ p}.

Let I be the ideal of R generated by S , then it’s easy to see that V(S ) = V(I) = V(r(I)). For any pair of ideals I and J, we have p ∈ V(I) ∪ V(J) if and only if p ⊃ I or p ⊃ J if and only if p ⊃ I ∩ J if and only if p ∈ V(I ∩ J), so that V(I) ∪ V(J) = V(I ∩ J). For any family of ideals T S {Ik}k∈I we have p ∈ k∈I V(Ik) if and only if Ik ⊂ p for all k ∈ I if and only if k∈I Ik ⊂ p if S T S and only if p ∈ V( k∈I Ik), so that k∈I V(Ik) = V( k∈I Ik). We also clearly have V(R) = 0 and V(0) = Spec(R). Thus the sets of the form V(I) with I an ideal of R satisfy the axioms for closed sets in a topo-

23 logical space. It follows that there is a unique topology on Spec(R) where the closed sets are the V(I).

Deﬁnition 2.1. Let R be a ring. We deﬁne the Zariski topology on X = Spec(R) to be the topology whose closed sets are of the form V(S ) for some S ⊂ R.

For f ∈ R, we adopt the shorthand V( f ) for V(( f )). Let X = Spec(R). For each f ∈ R let X f =

X \ V( f ). The X f are called the basic open sets of Spec(R).

Proposition 2.1. For a ring R, with f, g ∈ R, we have the following.

(1) The basic open sets form a basis for the Zariski topology on Spec(R).

(2) X f ∩ Xg = X f g

(3) X f = ∅ if and only if f is nilpotent

(4) X f = X if and only if f is a unit

(5) X f = Xg if and only if r(( f )) = r((g))

(6) X is compact

Proof.

(1) Assume for the moment items (2) - (4). Let B = {X f } f ∈R. Now R contains 1 and 0, X0 = ∅ ∈

B and X1 = X ∈ B. Suppose that p ∈ X f ∩ Xg, then p ∈ X f g ∈ B. Thus B forms a basis.

(2) We have V( f ) ∪ V(g) = V( f g) since f g ∈ p if and only if f ∈ p or g ∈ p. Thus X f g =

X \ (V( f ) ∪ V(g)) = X \ V( f ) ∩ X \ V(g) = X f ∩ Xg.

(3) Suppose X f = ∅, then V( f ) = X so that p ⊂ ( f ) for all primes p. Hence, by proposition 1.6,

T nil nil ( f ) ⊂ p∈X p = R so f ∈ R .

nil If f ∈ R then r( f ) = r(0) so that V( f ) = X and thus X f = ∅.

24 (4) If X f = X then V( f ) = ∅. Suppose f is a non-unit, then ( f ) ⊂ m for some maximal ideal m. But then m ∈ V( f ), a contradiction. Hence f is a unit.

If f is a unit, it’s clear that since ( f ) = (1) = R we have V( f ) = ∅, so X f = X. T T (5) Suppose X f = Xg, then V( f ) = V(g) so that p⊃( f ) p = p⊃(g) p which, by corollary 1.1, implies that r( f ) = r(g).

If r( f ) = r(g) then V(r( f )) = V( f ) = V(r(g)) = V(g) so that X f = Xg. S (6) Let {Uk}k∈I be a family of open sets such that X = k∈I Uk. Now since the basic open sets form S a basis for the topology on X, for each k ∈ I the open set Uk = f ∈S X f for some subset S ⊂ R. S Hence X = k∈I X fk and its enough to ﬁnd a sub-cover for the {X fk }. Now since

[ [ X = X fk = X \ V( fk) = X \ (∩k∈IV( fK)) = X \ (V(∪k∈I( fk))) k∈I k∈I

we have V(∪k∈I( fk)) = ∅. Let E = ∪k∈I( fk) and let I = (E) be the ideal generated by E. Then I P contains a unit, as V(I) = ∅, and thus I = (1). Hence, 1 ∈ I = k∈I( fk) which implies that there is

a ﬁnite set J ⊂ I and gk ∈ R such that X gk fk = 1 k∈J P P Hence k∈J( fk) = (1) and so if we let L = h∪k∈J( fk)i then L = k∈J( f j) = (1), so that

\ V(L) = V(∪k∈J( f j)) = V( fk) = ∅ k∈J

and thus [ [ X \ V(L) = X \ (∩k∈JV( fk)) = X \ V( fk) = X fk = X k∈J k∈J

Now, as one example, the spectrum of Z consists of the zero ideal and the principal ideals generated by prime numbers. Notice that (0) ∈ Spec(Z) is a point whose closure V(0) = Spec(Z) is the

25 whole space. Such a point is called a generic point. In fact, for an integral domain the zero ideal will always be a generic point. What’s more, for a general commutative ring R, since every prime ideal must contain all of the nilpotent elements of the ring, i.e. each q ∈ Spec(R) contains Rnil, the closure of any generic point p must equal the closure of the nilradical Rnil. However, the nilradical is the intersection of all the prime ideals of R, and since p ⊂ q for every q ∈ Spec(R), we must

T nil have p = q∈Spec(R) q = R , so the nilradical is a prime ideal. On the other hand, if the nilradical is prime, it’s closure must be all of Spec(R). It follows that Spec(R) has a generic point if and only if the nilradical is prime, and moreover the generic point, if it exists, is uniquely determined by the nilradical. There is a special geometric condition which is connected to the existence of a generic point in a general topological space. Namely if X has no generic point there must exist closed, proper subsets X1 and X2 such that X = X1 ∪ X2, in which case we call X reducible. Now if X = Spec(R) is such that there are no proper, closed subsets X1, X2 with X = X1 ∪ X2, we claim that X has a generic point, i.e. the nilradical is prime. Suppose to the contrary, then there are points x, y ∈ R such that xy ∈ Rnil but x < Rnil and y < Rnil. Thus, since V( f ) ∪ V(g) = V( f g) = X but V( f ), V(g) , X by part (3) of proposition 2.1, we obtain a contradiction. Thus, there is a generic point in Spec(R) if and only if Spec(R) is irreducible. The geometric intuition here is that while one can obtain the entire x or y axis from the lines x = 0 and y = 0 respectively, the set deﬁned by xy = 0 is obtained from the union of the two lines x = 0 and y = 0. Thus, subsets analogous to lines have generic points, while those built up of subsets analogous to lines need not. Let φ : A → B be a ring homomorphism. Let X = Spec(A) and let Y = Spec(B), then for all p ∈ Y we have φ−1(p) a prime ideal of A, so that p 7→ φ−1(p) deﬁnes a mapping φ∗ : Y → X.

Proposition 2.2. Let φ : A → B be a ring homomorphism. Let X = Spec(A) and let Y = Spec(B).

∗ −1 ∗ (1) If f ∈ A then (φ ) (X f ) = Yφ( f ) and hence φ is continuous.

(2) If φ is surjective, then φ∗ is a homeomorphism of Y onto the closed subset V(ker φ) of X.

26 Proof.

∗ −1 (1) We have (φ ) (X f ) = {y ∈ Y : y 2 φ( f )} = Y \ V(φ( f )) = Yφ( f ) and since the basic open

sets X f form a basis for the topology on X, any open subset U of X is such that U = ∪ f ∈I X f

∗ −1 S for some I ⊂ A, and thus (φ ) (U) = f ∈I Yφ( f ) is an open set in Y.

(2) Suppose φ is surjective, then for p ∈ X, φ(p) ∈ Y. Moreover, deﬁne φ∗−1(p) = φ(p) and notice that φ∗−1(φ∗(p)) = φ(φ−1(p)) = p. Hence, φ∗ has a left inverse and is therefore injective. If we corestrict to φ∗ : Y → Im φ∗ we get a bijection. Yet

Im φ∗ = {p ∈ X : p = qc for some q ∈ Y} = {p ∈ X : p = pec}.

Suppose p ∈ X satisﬁes p = φ−1(φ(p)) = pec, so that p ∈ Im φ∗. Since φ(p) is an ideal (0) ⊂ φ(p), whence p = pec ⊃ φ−1(0) = ker φ, i.e. p ∈ V(ker φ).

Suppose p ∈ V(ker φ), then p corresponds to an ideal p of A/ ker φ such that p = φ−1(p) and so φ(p) = p is prime, and p = pc with p ∈ Y so that p ∈ Im φ∗.

We conclude that Im φ∗ = V(ker φ). Thus φ∗ : Y → V(ker φ) is a bijection. Now let φ∗−1 : V(ker φ) → Y be given by φ∗−1(p) = φ(p). Let U be an open subset of Y, so that S U = f ∈I Y f for some I ⊂ B.Then

∗−1 −1 [ −1 [ (φ ) Y f = {p ∈ V(ker φ): p 2 φ ( f ) for some f ∈ I} = V(ker φ) ∩ Xφ−1( f ) f ∈I f ∈I

is open. Hence φ∗ is a homeomorphism onto V(ker φ).

Note that a singleton set {x} ⊂ Spec(R) is closed if and only if x ∈ Spec(R) is a maximal ideal. For {x} is closed if and only if {x} = V(I) for some ideal of R if and only if x is the only prime ideal containing I if and only if x is maximal.

27 2.2 Examples of Spectra of Rings

(1) Let k be a ﬁeld. Let X = Spec(k[x]). The prime ideal (0) is the generic point of X. Since k[x] is a PID, every prime ideal is maximal, and thus the non-generic points of

X are all closed. Suppose V(I) is a proper closed subset of X, then I , 0. Suppose q ∈

k[x] generates I, and let q1,..., qt be the irreducible factors of q. Then the only prime

ideals containing I are those generated by divisors of q, so that V(I) = {(q1),..., (qt)}. Thus the proper closed subsets of X are all ﬁnite. This spectrum is often referred to as the aﬃne line over k.

(2) Since Z is a PID we may again view Spec(Z) as a line. At every prime number there is a closed point, and (0) is the generic point. The proper closed subsets are again ﬁ- nite sets.

(3) How about X = Spec(Z[x])? To tackle this example, we’ll need to assume a result about localization, proved later. Let φ : Z → Z[x] be the inclusion map, which induces φ∗ : X → Spec(Z). We have

[ X = (φ∗)−1(0) ∪ (φ∗)−1(pZ) . p prime

Let p ∈ X. If p ∈ (φ∗)−1(pZ) for some p prime then we must have p ∩ Z = pZ, so that pZ[x] ⊂ p. However, we have a bijective correspondence between ideals containing pZ[x] and ideals of Z[x]/pZ[x] = Z/pZ[x]. But then by part (2) of the proposition we have (φ∗)−1(pZ) homeomorphic to Spec(Z/pZ[x]). Now if the image of p in Z/pZ[x] is zero we must have p = (p). Otherwise, there is a monic, irreducible polynomial q over Z/pZ[x] such that p = (q(x)) + (p) = (q(x), p).

If p ∈ (φ∗)−1(0) we have p ∩ Z = (0). Let S = Z \{0}, then we have p ∩ S = ∅.

Now, in section 3 we establish that for any ring R we can form a ring S −1R called the

28 localization at S , which introduces fractions whose denominator belongs to S . For example, here S −1Z = Q, and this is just the same as the formation of the ﬁeld of fractions. Localization has a homomorphism R → S −1R which gives a homeomorphism from Spec(S −1R) onto {p ∈ Spec(R): S ∩ p = ∅}, by part (2) of proposition 2.2. The prime ideals of S −1R are in one-to-one correspondence with the prime ideals of R satisfying p ∩ S = ∅.

Back to the issue at hand, we have a bijective correspondence between the set of primes p ∈ (φ∗)−1(0) and the prime ideals of S −1(Z[x]) = Q[x]. We thus have(φ∗)−1(0) homeomorphic to Spec(Q[x]). Thus if p ∈ (φ∗)−1(0) we must either have p = (0) or p = (q) where q is a monic irreducible polynomial over Q[x].

We can then summarize the above as saying that for X we have a generic point at (0) and prime ideals of the form:

i)( p) where p is a prime number

ii)( q(x)) where q is a monic irreducible polynomial of any degree over Q[x] iii)( f (x)) + (p) where p is prime and f is a monic irreducible integral polynomial modulo p.

Figure 2.1: Mumford’s famous depiction of Spec(Z[x]), from “The Red Book.”

Thus X can be visualized as a family of lines parametrized by points of Spec(Z) over ﬁelds of various characteristics.

29 (4) We can carry out a similar process for Z[i] using the inclusion Z → Z[i]. We have (φ∗)−1(0) = (0) and for a prime p,(φ∗)−1(p) consists of principal prime ideals of Z[i] generated by elements of the form a + bi which divide p.

In particular, if p ≡ 1 mod 4 then p can be written as a sum of squares p = a2 + b2 = (a + bi)(a − bi), so that in this case there are two such prime ideals in (φ∗)−1(p).

If p = 2 or p ≡ 3 mod 4 then a + bi divides p only when |a| = p and b = 0 or |b| = p and a = 0, so in these cases there is only one such prime ideal. Thus, Spec(Z[i]) can be viewed as a sort of 2 stranded braid with knots corresponding to the primes p = 2

or p ≡ 3 mod 4. Since Z[i] Z[x]/(x2 + 1) the picture of Spec(Z[x]) given in Figure 1 contains an image of Spec(Z[x]).

It is occasionally useful to concentrate our attention on certain subspaces of Spec(R).

Deﬁnition 2.2. For a ring R we deﬁne

a) J(R) to be the set of all prime ideals p ∈ Spec(R) such that p can be written as the intersection of maximal ideals

b) Max(R) to be the set of all maximal ideals of R.

These two subsets of Spec(R) inherit the subspace topology from Spec(R). Another useful piece of notation is the “vanishing ideal” of a subset A of Spec(R)

\ I(A) = p p∈A and we claim that V(I(A)) = A, the closure of A in Spec(R). Now if p ∈ A we must have T p ⊃ p∈A p = I(A), so that p ∈ V(I(A)). If V(I) ⊃ A for some ideal I, then p ⊃ I for all p ∈ A, T so I ⊂ p∈A p = I(A) and thus V(I(A)) ⊂ V(I). It then follows that V(I(A)) is the smallest closed subset of Spec(R) containing A, i.e. V(I(A)) = A.

30 CHAPTER 3

LOCALIZATION; PASSAGE FROM LOCAL TO GLOBAL

Consider the equation V : X2 + Y2 − 3Z2 = 0.

Suppose one wants to know if there exist non-trivial integer solutions to V. If by passing to Q we obtain a non-trivial solution, then by canceling common denominators and dividing by the greatest common divisors of the numerators we obtain a non-trivial integer solution (x, y, z) with gcd(x, y, z) = 1. However, since x2 + y2 = 3z2 we must have x2 + y2 ≡ 0 mod 3, but squares of integers are congruent to either 0 or 1 modulo 3 so that 3 divides x and y. Since 32 divides x2 + y2 = 3z2, it follows immediately that 3 divides z, which contradicts the fact that gcd(x, y, z) = 1. Hence, V admits no non-trivial rational solutions.

Rather than ﬁrst looking for solutions over Q = {a/s : a, s ∈ Z, s < (0)}, we could have started by looking for solutions in Z(3) = {a/s : a, s ∈ Z, s < (3)}. Now, R = Z(3) is a local ring with maximal ideal m = (3), and by passing to the vector space R/mR we have transformed the equation V into a congruence modulo 3. In this case, doing so was enough to show that there were no non-trivial solutions. The equation V and the prime 3 are particularly special here. For some other equation, this process may not be suﬃcient. However, for an arbitrary prime p, a natural extension of this method is to consider not just congruences modulo p, but congruences modulo pr for any r > 0. This will give increasingly more precise information, where letting r get arbitrarily large allows one to get closer and closer to the original equation. This corresponds to passing to the p-adic completion of Z(p), i.e. the ring of p-adic integers.

31 Now Z(p) is known as the localization of Z at the prime ideal (p). This process of passing to a p- adic completion can be done for any Diophantine problem, and so it is suﬃcient to show that just one prime p and r > 0 are such that no non-trivial solutions modulo pr exist in order to conclude that there are no non-trivial rational solutions. The converse need not hold in general, if for all prime ideals (p) there is a non-trivial solution upon reduction modulo p, it may not be possible to piece these solutions together to obtain a rational one. This is a prototypical example of the problem of passing from the local to the global, which consists of identifying appropriate local properties and the conditions under which this local information constrains the properties of the global structure under consideration.

The juxtaposition of Q with Z(3) may seem contrived in the case of the ring Z, but consider instead an arbitrary commutative ring R. If R is no longer an integral domain, or equivalently if (0) is no longer prime, then we can no longer form the field of fractions as we did for Z. So in this situation, introducing the “ring of fractions” Rp = {a/s : a, s ∈ R, s < p} allows us to very naturally apply a similar method to problems on R-modules as we did for the Diophantine problem. Multiplying in a naive fashion, the set of fractions of the form a/s with a ∈ p should form an ideal in Rp, while every fraction of the form b/t with b < p should have a multiplicative inverse t/b. With these notions properly defined, we will see that this suffices to show that Rp is a local ring. Although the formation of fractions and the process of localization are perhaps the single most important tools of commutative algebra, their introduction into the subject seems to have oc- curred rather late. A ring of fractions corresponding to an integral domain was only first system- atically defined in 1926 by Grell , whereas its extension to Noetherian rings possibly containing zero divisors, no doubt motivated by the work of Krull, was made 18 years later by Chevalley. Fi- nally, in 1948 Uzkov showed that one could free himself of the Noetherian restriction and work over any commutative ring. It is in Krull that we first find examples of the local-global method of proving a property of an integral domain by exhibiting it for the localizations Rp at each prime

32 ideal p. The extension of such methods to arbitrary commutative rings, their modules, and even the quite frequent ability to weaken to localizations at each maximal ideal is seemingly due to Serre.

3.1 Rings and Modules of Fractions

Let R be a ring. A multiplicatively closed subset of R is a subset S of R which contains 1 and is closed under multiplication. If we define a relation ≡ on R × S by (a, s) ≡ (b, t) if and only if there is a u ∈ S such that u(at − bs) = 0 in R, we obtain an equivalence relation. That ≡ is symmetric and reflexive is immediately clear. If (a, s) ≡ (b, t) and (b, t) ≡ (c, u) then we have v, w ∈ S such that v(at − bs) = 0 = w(bu − ct) so that vat = vbs and tvw(au − cs) = vs(wbu − wct) = vs(w(bu − ct)) = 0. The fraction a/s is then used to denote the equivalence class of (a, s) under ≡, and the set S −1R is defined as the set of all equivalence classes under ≡ in R × S . Define addition and multiplication on S −1R by a/t + b/t = (at + bs)/st

(a/s)(b/t) = ab/st.

The following proposition is elementary, but familiarity with the manipulation of these fractions will aﬀord the reader the conﬁdence to tacitly apply the rules later on when it becomes more convenient.

Proposition 3.1. Both addition and multiplication are well-deﬁned and S −1R is a ring.

Proof. Suppose (a, s) ≡ (a0, s0) and (b, t) ≡ (b0, t0). There are u, v ∈ S such that u(as0 − a0 s) = 0 =

33 v(bt0 − b0t). Thus

uv((at + bs)s0t0 − (a0t0 + b0 s0)st) = vt0t(uas0 − ua0 s) + us0 s(vbt0 − vb0t) = 0

so that a/s + b/t = (at + bs)/st = (a0t0 + b0 s0)/s0t0 = a0/s0 + b0/t0. Similarly,

uv((ab)s0t0 −(a0b0)st) = uv(abs0t0 −a0bst0 +a0bst0 −a0b0 st) = vbt0(uas0 −ua0 s)+ua0 s(vbt0 −vb0t) = 0

so that (a/s)(b/t) = ab/st = a0b0/s0t0 = (a0/s0)(b0/t0). Thus the two operations are in fact well- deﬁned. Now since 1 ∈ S and R is associative we have, for any t ∈ S , at/st = a/s. The commutativity of addition and multiplication is directly inherited from the commutativity of R. We have, by associativity and distributivity in R,

(a/s+b/t)+c/u = (at+bs)/st+c/u = (a(tu)+(bu+ct)s)/stu = a/s+(bu+ct)/tu = a/s+(b/t+c/u) so that addition is associative. Now 0/v = 0v/1v = 0/1 for all v ∈ S . Since 0/1 + a/s = a/s we have an additive identity. Since a/s + (−a)/s = 0/s2 = 0/1, the set S −1R forms an abelian group. The associativity of multiplication is directly inherited from associativity in R, and 1/1 is easily seen to be an identity for multiplication. The distributive law holds

a/s(b/t + c/u) = a/s(bu + ct/tu) = (abu + act)/stu = (absu + acst)/s2tu

= absu/s2tu + acst/s2tu = ab/st + ac/su = (a/s)(b/t) + (a/s)(c/u) so that S −1R indeed forms a ring.

Let S = R \ p, then 1 ∈ S and for x, y < p we have xy < p, so S is multiplicatively closed. Thus

34 −1 S R = Rp is indeed a ring, and our previous remarks are enough to show that it must be a local

ring. For if I is an ideal such that pRp ( I, then I contains a unit b/t with b ∈ S . So m = pRp is maximal, and every element of R \ m is a unit.

One may reasonably ask if it’s legitimate to regard pRp as an ideal of Rp. However, there is a canonical ring homomorphism f : R → S −1R given by f (x) = x/1, which is injective if and only if S does not contain zero divisors since ker( f ) = {a ∈ R : there is an s ∈ S such that as = 0}. Now R\p may very well contain zero divisors, but if f (x) = f (y) then there is a u ∈ R\p such that

ux = uy ∈ p, so that y ∈ p. So we can identify pRp with f (p)Rp and not worry about potentially multiplying by elements of R \ p in the numerator.

In fact, by the same token R is embedded in Rp, and then we see that pRp ∩ R = p. Deﬁne φ :

R/p → Rp/pRp by φ(a + p) = a + pRp := a¯. If a < p then 1/a < pRp so 1/a is a unit in the ﬁeld

Rp/pRp. Thusa ¯(1/a) = φ(a + p)(1/a) = 1,¯ so that ker φ = 0. Hence, we may view Rp/pRp as the ﬁeld of fractions of R/p. Using the canonical homomorphism f : R → S −1R, it can be shown that the ring of fractions respects the following universal property:

Theorem 3.1. Universal property of localization

Given a ring homomorphism g : R → T with g(s) a unit for all s ∈ S , there exists a unique ring homomorphism h : S −1R → T such that g = h ◦ f .

Proof. Deﬁne h(a/s) = g(a)g(s)−1. We must check that h is well-deﬁned. If a/s = a0/s0 then there is a u ∈ S such that u(as0 − a0 s) = 0, so that g(u(as0 − a0 s)) = g(u)(g(a)g(s0) − g(a0)g(s)) = 0. But since u ∈ S we have g(u) a unit in S by hypothesis, so that h(a/s) = g(a)g(s)−1 = g(a0)g(s0)−1 = h(a0/s0). If h0 is also such that g = h0 ◦ f , then h0(a/1) = g(a) for all a ∈ R

and h0(1/s) = h0((s/1)−1) = g(s)−1 so that h0(a/s) = h0(a/1)h0(1/s) = g(a)g(s)−1 = h(a/s).

Examples:

35 1) We have already mentioned the local ring Rp obtained from the multiplicative set S = R \ p. For our purposes, it is the single most important example. This example, and the name localization, originated in algebraic geometry. Given a variety V defined over a field k and a point p ∈ V, one may wish to study the variety locally around this point. If we let R = k[V], the coordinate ring of V, then the ideal I(a) = { f ∈ k[V]: f (a) = 0} is maximal in k[V] since it is the kernel of the map k[V] → k which sends f to f (a). Then by localizating at I(a) we obtain a local ring of rational functions defined at a. Such functions f /g will

also be deﬁned on small open neighborhoods of a in the Zariski topology on which g , 0. In this sense, the localization at I(a) is representative of the germ of functions near a. (See section 4.)

2) The ring S −1R = 0 if and only if 0 ∈ S .

n −1 3) If f ∈ R and S = { f }n≥0, then S is multiplicatively closed and we write R f for S R. We

have R f = 0 if and only if f is nilpotent.

4) Suppose R is a local ring with maximal ideal m. Then S = R \ m consists entirely of units,

so that Rm = R.

P∞ n 5) Consider the formal power series ring R[[x]] = { f = n=0 an x : an ∈ R}. Now x ∈ R[[x]],

n P∞ n and if S = {x }n≥0, we have R[[x]]x = { n=−m an x : an ∈ R} = R((x)), the ring of formal Laurent series.

6) (How to use the universal property) Let Zp denote the ring of p-adic integers, where p is

P∞ n prime and Zp = { n=0 an p : an ∈ {0, 1,..., p − 1}}. The homomorphism g : Z → Zp

P∞ n sending m to n=0 mp is injective. If p does not divide m, it is an elementary fact of number theory that m is in the group of units of Z/piZ for all i > 0. Hence if p does not di-

vide m, then g(m) is a unit in Zp. By the universal property of localization, there is a unique

−1 ring homomorphism h : Z(p) → Zp given by h(a/s) = g(a)g(s) . Now Zp is uncount-

36 able, whereas Z(p) ⊂ Q is countable, so that h cannot be surjective. Since g is injective and h(a/s) = g(a)g(s)−1 = 0 only if g(a) = 0, we must have h injective.

Implicit in our proof of the universal property of localization is that f sends elements of S to units in S −1R, ker( f ) = {a ∈ R : there is an s ∈ S such that as = 0}, and every element of S −1R is of the form f (a) f (s)−1.

Corollary 3.1. Let g : R → T be a homomorphism with g(s) a unit for all s ∈ S . If

(1) g(a) = 0 implies that as = 0 for some s ∈ S

(2) every element of T is of the form g(a)g(s)−1 then the homomorphism h : S −1R → T is an isomorphism.

Proof. Since h(a/s) = g(a)g(s)−1, h is obviously surjective. If h(a/s) = g(a)g(s)−1 = 0 then

g(a) = 0 so a/s = 0/1.

Remark: In example 6 above, g : Z → Zp satisfies (1) , but not (2). In general, if g satisfies (1) then h is injective, and if g satisfies (2) then h is surjective. We can also carry out the formation of fractions for an R-module M. Given a multiplicatively closed subset S of R, define an equivalence relation on M × S by (m, s) ≡ (m0, s0) if and only if there is a t ∈ S such that t(sm0 − s0m) = 0. Then S −1 M is the set of all equivalence classes m/s, and forms an S −1R-module. It is also sometimes convenient to view S −1 M as an R-module, with multiplication a · m/s := f (a) · m/s. Note that S −1 M = 0 if and only if for each m ∈ M there is an s ∈ S such that sm = 0. Throughout the remainder of this section S will denote a multiplicatively closed subset of a ring R. For an R-module homomorphism u : M → N, there is a canonical S −1R-homomorphism

−1 −1 uS : S M → S N given by uS (m/s) = u(m)/s.

Proposition 3.2. S −1 is an exact functor of modules.

37 u v u v Proof. Let M0 −→ M −→ M00 be exact. Consider the sequence S −1 M0 −→S S −1 M −→S S −1 M00. Since

v◦u = 0 and (v◦u)S = vS ◦us, we have im uS ⊂ ker(vS ). If m/s ∈ ker(vS ), then there is a t ∈ S such that tv(m) = 0. Thus v(tm) = 0 so tm ∈ ker(v) = im u, so there is a n ∈ M0 such that sm = u(n). u v −1 0 S −1 S −1 00 Thus tm/st = u(n)/st ∈ im uS . Hence ker(vS ) = im vS , and so S M −→ S M −→ S M is exact.

Corollary 3.2. Let M be an R-module and N, P ≤ M. Let I, J be ideals of R. We have

(1) S −1(N + P) = S −1N + S −1P

(2) S −1(N ∩ P) = S −1N ∩ S −1P

(3) S −1(M/N) S −1 M/S −1N

(4) S −1r(I) = r(S −1I)

(5) S −1(IJ) = S −1IS −1 J

Proof. (1) Since (n + p)/s = n/s + p/s and n/s + p/t = (nt + ps)/st for all n ∈ N, p ∈ P, and s, t ∈ S .

(2) For n ∈ N, s ∈ S , if n/s = p/t for some p ∈ P and t ∈ S , then there is a u ∈ S such that utn = usp ∈ N ∩ P, so n/s = utn/uts ∈ S −1(N ∩ P). If n ∈ N ∩ P and s ∈ S then n/s ∈ S −1N ∩ S −1P

ι π (3) Since N ≤ M we have 0 → N −→ M −→ M/N → 0, with ι inclusion and π projection, exact. ι π −1 S −1 S −1 −1 −1 Thus 0 → S N −→ S M −−→ S (M/N) → 0 is exact. Now coker(ιS ) = S M/S N but

−1 we have coker(ιS ) im πS = S (M/N) by exactness.

(4) Let a/s ∈ S −1r(I), then there is an n > 0 such that an ∈ I, so (a/s)n = an/sn ∈ S −1I. If a/s ∈ r(S −1(I)), there is an n > 0 such that an/sn = b/t for some b ∈ I, t ∈ S . Hence there is a u ∈ S such that uant = busn ∈ I so (uat)n = (bus)n ∈ I, uat ∈ r(I) and thus a/s = uat/sut ∈ S −1r(I).

38 (5) For ab/s ∈ S −1(IJ), since bs ∈ J we have ab/s = abs/s2 = (a/s)(bs/s) ∈ S −1IS −1 J. For (a/s)(b/t) ∈ S −1IS −1 J, clearly (a/s)(b/t) = ab/st ∈ S −1(IJ).

The extension to ﬁnite sums, intersections, and products is immediate.

−1 −1 −1 Theorem 3.2. For an R-module M, we have S M M ⊗R S R as S R-modules.

−1 −1 Proof. Deﬁne φ : M ⊗R S R → S M by taking φ(m ⊗ a/s) = am/s. Since the map (m, a/s) 7→ am/s is bilinear, the universal property of the tensor product guarantees that φ is well-deﬁned.

−1 −1 Deﬁne ψ : S M → M ⊗R S R by ψ(m/s) = m ⊗ 1/s. To show this is well-deﬁned, suppose m/s = m0/s0, then there is a u ∈ S such that ums0 = um0 s and so

m ⊗ 1/s = m ⊗ u/su = m ⊗ u · (1/su) = mu ⊗ 1/su = ums0 ⊗ s0 · (1/su)

= um0 s ⊗ s0/su = um0 s ⊗ 1/ss0u = m0 ⊗ (us) · (1/ss0u) = m0 ⊗ 1/s0.

Now ψ(φ(m⊗a/s)) = ψ(am/s) = am⊗1/s = m⊗a·(1/s) = m⊗a/s and φ(ψ(am/s)) = φ(m⊗a/s) = am/s. Thus, φ is an isomorphism.

As a direct consequence of theorem 3.2 and the fact that S −1 is an exact functor, we conclude that S −1R is ﬂat as an R-module. Hence, if f : M0 → M is injective, then f ⊗ 1 : S −1 M0 → S −1 M is injective.

Lemma 3.1. Let R, S be rings, let M be an R-module, P a S -module, and let N be both simulta- neously an R-module and a S -module such that a(xb) = (ax)b for all a ∈ R, x ∈ N, b ∈ S . In this

case, N is also called an (R, S )-bimodule. Then M ⊗R N is a S -module and N ⊗S P is an R-module, with

(M ⊗R N) ⊗S P M ⊗R (N ⊗S P).

39 Proof. Let z ∈ P be ﬁxed. The map fz : M×N → M⊗R (N ⊗S P) given by fz(x, y) = x⊗(y⊗z) is R- bilinear. Then by the universal property of tensor products, there is an induced R-homomorphism

0 0 fz : M⊗R N → M⊗R (N ⊗S P) such that fz (x⊗y) = x⊗(y⊗z). Let g :(M⊗R N)×P → M⊗R (N ⊗S P)

0 be such that g(x ⊗ y, z) = fz (x ⊗ y). By deﬁnition g is an R-homomorphism, and its easy to see that

0 g is S -bilinear. Thus g induces a bimodule homomorphism g :(M ⊗R N) ⊗S P → M ⊗R (N ⊗S P) given by g0((x ⊗ y) ⊗ z) = x ⊗ (y ⊗ z). This same argument carries over mutatis mutandi to show

0 the existence of a bimodule homomorphism h : M ⊗R (N ⊗S P) → (M ⊗R N) ⊗S P such that h0(x ⊗ (y ⊗ z)) = (x ⊗ y) ⊗ z, and thus g0 is an isomorphism.

−1 −1 −1 Proposition 3.3. For all R-modules M, N we have S M ⊗S −1R S N S (M ⊗R N).

Proof. By theorem 3.2 and lemma 3.1, we have

−1 −1 −1 −1 −1 −1 −1 S M ⊗S −1R S N (M ⊗R S R) ⊗S −1R S N M ⊗R (S R ⊗S −1R S N) M ⊗R S N

and yet

−1 −1 −1 −1 M ⊗R S N M ⊗R (N ⊗R S R)) (M ⊗R N) ⊗R S R S (M ⊗R N).

In particular, we have Mp ⊗Rp Np (M ⊗R N)p.

Deﬁnition 3.1. Let R be an integral domain. For an R-module M, the torsion submodule T(M) is

the set of all m ∈ M such that Ann(m) , 0. If T(M) = 0 then M is called torsion-free. M is called a torsion module if T(M) = M. Now if f : M → S −1 M is the canonical homomorphism, we have ker f = T(M). Hence, T(M) is in fact a submodule. Moreover, M is torsion-free if and only if f is injective.

Proposition 3.4.

40 Let R be an integral domain.

(1) For an R-module M, M/T(M) is torsion-free.

(2) For an R-module homomorphism f : M → N, f (T(M)) ⊂ T(N).

(3) If 0 → M0 → M → M00 is exact, then 0 → T(M0) → T(M) → T(M00) is exact.

(4) T(M) is the kernel of the map x 7→ 1 ⊗ x of M into K ⊗R M where K is the ﬁeld of fractions of R.

Proof.

(1) Lets ¯ ∈ M/T(M). Supposes ¯ is torsion, then there is an a , 0 such that as¯ = as + T(M) = 0 so that as ∈ T(M) and therefore s ∈ T(M), a contradiction.

(2) For s ∈ T(M), we have an a , 0 such that as = 0, so that f (as) = a f (s) = 0, and thus f (s) ∈ T(N).

(3) By exactness, f : M0 → M is injective, and we have by (2) that T(M0) → T(M) is the restriction to T(M0). Similarly, the composition T(M0) → T(M) → T(M00) is zero since it is the restriction of M0 → M → M00. If x is in the kernel of T(M) → T(M00), then x is in T(M) and the kernel of g : M → M00 and vice versa. By exactness, x ∈ im f , so there is a y ∈ M0 such that

x = f (y). Since x ∈ T(M) there is an a , 0 such that ax = a f (y) = f (ay) = 0, so ay = 0 and y ∈ T(M0). It follows that 0 → T(M0) → T(M) → T(M00) is exact.

− −1 (4) Let S = R \ 0, then S M S R ⊗R M = K ⊗R M. Moreover, the isomorphism is given by a/s ⊗ m 7→ am/s so that 1 ⊗ m maps to zero only if m/1 = 0/1, i.e. if there is an a , 0 such that am = 0. Thus by taking the composition we see that the kernel of the map m 7→ 1 ⊗ m is precisely

the set of m ∈ M such that am = 0 for some a , 0, i.e. T(M).

41 3.2 The Passage from Local to Global

A local-global statement concerning a ring R or an R-module M is any statement of the form R

(respectively M) has property P if and only if Rp (respectively Mp) has property P for all prime ideals p.

Deﬁnition 3.2. Let R be any ring. For an R-module M, the support of M is deﬁned to be the set

Supp(M) of prime ideals p such that Mp , 0.

Proposition 3.5. If M is ﬁnitely generated, then Supp(M) = V(Ann(M)) = {p ∈ Spec(R): p ⊃Ann(M)}

Proof. Let m1,..., mn generate M. If p < Supp(M) then Mp = (m1,..., mn)p = 0. Hence there

are s1,..., sn ∈ R \ p such that simi = 0 for each i. Now s1 s2 ... sn annihilates M but s1 s2 ... sn is not in p. Hence p 2 Ann(M). If Mp , 0, suppose there is an s ∈ Ann(M) such that s < p. Then

sm = 0 for all m ∈ M, so Mp = 0, a contradiction.

Proposition 3.6.

For an R-module M, the following are equivalent

(1) M = 0

(2) Mp = 0 for all primes p

(3) Mm = 0 for every maximal ideal m

Proof. That (1) implies (2) and (2) implies (3) is obvious. Suppose Mm = 0 for every maximal

ideal m. If M , 0 there is an x , 0 ∈ M. Since Ann(x) is a proper ideal, it is contained in a maximal ideal m. Yet f (x) = 0/1, so there is an s ∈ R \ m such that sx = 0, but s < Ann(x).

As an immediate corollary, M , 0 if and only if Supp(M) , ∅.

42 Proposition 3.7. For an R-module homomorphism φ : M → N, the following are equivalent

(1) φ is injective (respectively, surjective)

(2) φp is injective (respectively, surjective) for all primes p

(3) φm is injective (respectively, surjective) for all maximal ideals m

Proof.

(1) =⇒ (2) If φ is injective, 0 → M → N is exact, so 0 → Mp → Np is exact for each prime p. If

φ is surjective then Mp → Np → 0 is exact for each prime p. Thus, (1) implies (2). It’s clear that (2) implies (3).

(3) =⇒ (1) We have 0 → ker φ → M → N exact and M → N → N/im φ → 0 exact. Hence

0 → (ker φ)m → Mm → Nm and Mm → Nm → Nm/(im φ)m → 0 are exact.

Suppose φm is injective for every maximal ideal m. We have ker φm = (ker φ)m and since ker φm =

0 for every maximal ideal m, we have ker(φ) = 0. Similarly im φm = (im φ)m shows surjectivity.

Proposition 3.8. For an R-module M, the following are equivalent

(1) M is ﬂat

(2) Mp is ﬂat for all primes p

(3) Mm is ﬂat for all maximal ideals m

Proof.

(1) =⇒ (2) Suppose M is ﬂat, and let 0 → N0 → N → N00 → 0 be exact. Then 0 → N0 ⊗ M →

00 N ⊗ M → N ⊗ M → 0 is exact. Recall that by proposition 3.3 Mp ⊗ Np (M ⊗ N)p. Let p be prime, then 0 → (N0 ⊗ M)p → (N ⊗ M)p → (N00 ⊗ M)p → 0 is exact, so that by proposition 1.15

Mp is ﬂat. It’s clear that (2) implies (3).

43 0 (3) =⇒ (1) Suppose Mm is ﬂat for all maximal ideals m. Suppose N → N is injective. Then

0 Nm → Nm is injective for all maximal ideals m by proposition 3.7. Since Mm is ﬂat for all maxi-

0 mal ideals m, by proposition 1.15 we have Nm ⊗ Mm → Nm ⊗ Mm injective for all maximal ideals

0 0 m. Thus (N ⊗ M)m → (N ⊗ M)m is injective for all maximal ideals m, so that N ⊗ M → N ⊗ M

is injective and M is ﬂat.

Proposition 3.9.

Let R be an integral domain. For an R-module M, the following are equivalent

(1) M is torsion-free

(2) Mp is torsion-free for all primes p

(3) Mm is torsion-free for all maximal ideals m

Proof.

(1) =⇒ (2) For a , 0, deﬁne the R-module homomorphism µa : M → M by µ(m) = am. Suppose

M is torsion-free. Then µa is injective, so that by proposition 3.7 (µa)p is injective for each prime

p. Since this holds for each a , 0, Mp is torsion-free for each prime p. It’s clear that (2) implies (3).

(3) =⇒ (1) If Mm is torsion-free for each maximal ideal m, then for each a , 0 we have (µa)m

injective for each maximal ideal m, so that for each a , 0 by proposition 3.7 µa is injective, and M is torsion-free.

Not every property can pass from the local to the global. Counterexample

If Rp is an integral domain for all primes p in R, we need not in general have R an integral domain.

44 Let R = Z/2Z × Z/2Z. We have (0, 1) and (1, 0) zero-divisors in R. The only prime ideals are

p1 = Z/2Z × {0} and p2 = {0} × Z/2Z. We let 0 = (0, 0), a = (0, 1), b = (1, 0), and 1 = (1, 1).Note

2 2 that a = a and b = b. We have R \ p1 = {1, a} and R \ p2 = {1, b}. Localizing at p1, since a(a − 1) = a2 − a = a − a = 0 we have

a/1 = 1/a = a/a = 1/1 so that b/1 = (b/1)(1/a) = b/a = (b/1)(a/1) = ba/1 = 0/1.

Since we obtain a symmetric result by localizing at p2, we have

Rp1 = {0/1, 1/1} and

Rp2 = {0/1, 1/1} both of which are isomorphic to Z/2Z, a ﬁeld. We end with a characterization of the prime ideals in S −1R. Recall that for a ring homomorphism f : R → S and ideals I of R and J of S , we denote the contraction f −1(J) of J by Jc and the extension f (I) of I by Ie. Recall that the contraction of an ideal is always an ideal, and moreover the contraction of a prime ideal is always prime. For ideals I, J of a ring R let (I : J) = {x ∈ R : xJ ⊂ I}.

Proposition 3.10. The prime ideals of S −1R are in one-to-one correspondence with the prime ideals p of R with p ∩ S = ∅.

Proof. Suppose J is an ideal in S −1R, and let x/s ∈ J. Let f : R → S −1R be the canonical map. Since x/1 = f (x) ∈ J we have x ∈ Jc, and thus x/s ∈ Jce. Since J ⊃ Jce = f ( f −1(I)) we have

45 J = Jce. Thus every ideal of S −1R is extended. Now let I be an ideal of R. Its extension is S −1I. We have Iec = ( f (I))c = (S −1I)c. Moreover

(S −1I)c = f −1(S −1I) = {x ∈ R : x/1 ∈ S −1I} = {x ∈ R : x/1 = a/s, for some a ∈ I, s ∈ S } so that x ∈ Iec = (S −1I)c if and only if x/1 = a/s for some a ∈ I and s ∈ S , which is true if and only if there is a t ∈ S such that t(xs − a) = 0. Yet, this holds if and only if txs = ta ∈ I, and since ts is always in the ideal (s), this is true if and only if x(s) ⊂ I. We conclude x ∈ Iec if and only if

ec S e x ∈ (I :(s)) for some s ∈ S . Hence, I = s∈S (I :(s)). In particular, I = (1) if and only if S s∈S (I :(s)) = (1), which is true if and only if I ∩ S , ∅. Let q be a prime ideal of S −1R, then qc is a prime ideal in R, say qc = p, and q = qce , (1). Thus, by the above, p ∩ S = qc ∩ S = ∅. Suppose p is a prime ideal in R with p ∩ S = ∅. Then S ⊂ R \ p. Let T = R \ p, then we can view S as a multiplicatively closed subset of T. Moreover, S −1T = {t/s : t ∈ T, s ∈ S } is a multiplicatively closed subset of S −1R, and since S −1R \ S −1T = S −1p we claim S −1p is

prime. Suppose x < S −1p and y < S −1p, then x, y ∈ S −1T. Since S −1T is multiplicatively closed, xy ∈ S −1T, so that xy < S −1p. Thus S −1p = pe is prime. The bijection is then given by the map p 7→ pe from the set {p ∈ Spec(R): p ∩ S = ∅} to Spec(S −1R).

46 CHAPTER 4

NOETHERIAN RINGS, MODULES, AND SPACES; AFFINE VARIETIES; DIMENSION THEORY

2 Given a family of polynomials { fk}k∈I in R[x, y], let Z( fk)k∈I denote the set of all points (x, y) ∈ R

2 3 such that fk(x, y) = 0 for all k ∈ I. One has the descending chain R ) Z(xy) ) Z(y) ) Z(y, x − x) ) Z(y, x2 − x) ) Z(y, x) ) ∅ which of course terminates.

Z(xy) Z(y) Z(y, x3 − x) Z(y, x2 − x)

Z(y, x)

Figure 4.1: Some pictures of aﬃne varieties, illustrating the concept of dimension.

∞ On the other hand, for every sequence of distinct real numbers {xn}n=1 we can form the polynomi- QN als pN(x) = n=1(x − xn) ∈ R[x, y] for all N ∈ N, so that if In = {k : 1 ≤ k ≤ n} we have an ascending chain ∅ ( Z(y, pk)k∈I1 ( Z(y, pk)k∈I2 ( Z(y, pk)k∈I3 ( ... which never terminates. It is rather straightforward to extend the above example of a descending chain to one of any ﬁnite length, but we will ultimately see that it is impossible to produce an inﬁnitely long descending

47 chain which never terminates. This expresses the fact that under a suitable, “Zariski-like” topol-

ogy, i.e. where the subsets of the form Z = Z( fk)k∈I are closed, affine 2-space over R is Noethe- rian (Definition 4.1). In fact, if we work over an algebraically closed field L, then by Hilbert’s Nullstellensatz (Theorem 4.2) the subsets of the form Z can be shown to correspond to certain ideals in what is called the coordinate ring (Definition 4.4) over Z, denoted by L[Z]. The above can then be used to show that R[Z] is Noetherian as a ring, but is not Artinian. Notice that in the chain R2 ) Z(xy) ) Z(y) ) Z(y, x3 − x) ) Z(y, x2 − x) ) Z(y, x) ) ∅ we pass from a plane to some finite collection of lines, and then from a line to some finite collection of points. Intuitively, it appears that in order to go further down the chain we must either lower the dimension by one or remove some component. If we restrict our attention to the non-empty irreducible subsets of a topological space, in the above example planes, lines, and distinct points, by going down such a chain we must reduce the dimension by one at each step. Then since the chain should terminate when the dimension reaches zero, it would seem reasonable to define the dimension of the space as one less than the maximal number of subsets appearing in such a chain. Now, the Nullstellensatz will allow us to pass back and forth from the nonempty irreducible

Z( fk)k∈I to prime ideals in the coordinate ring, at the cost of reversing inclusions (see the remarks following definition 4.4). In particular, singleton sets will correspond to maximal ideals in the coordinate ring. One can then use this to extend the definition of dimension to a ring R. It’s reasonable to ask whether the Noetherian and Artinian finiteness conditions might have some impact on the dimension of a ring. Suppose X is a space with a “Zariski-like” topology. When X is infinite we could just as easily form the polynomials pN as we did before, so any ring corresponding to X cannot be Artinian. When X consists of only finitely many distinct points every descending and ascending chain terminates, and so the corresponding ring ought to be Artinian. Since X consists of only finitely many points, the only prime ideals of R are maximal ideals, so that R is of dimension zero. Surprisingly perhaps, the Noetherian condition doesn’t bound the dimension of a ring at all, and thus it is quite rare for a Noetherian ring to be Artinian. This highlights the appar-

48 ent lack of symmetry between the Noetherian and Artinian conditions, which will become even more apparent when looking at modules.

4.1 Noetherian/Artinian Rings and Modules

We have the following result:

Proposition 4.1. Let R be a ring, and M and R-module, then the following are equivalent:

(1) Every submodule N of M is ﬁnitely generated

(2) Every ascending chain of submodules

M1 ⊂ M2 ⊂ · · · ⊂ Mn ⊂ ...

is such that there is a k with Mn = Mk for all n ≥ k.

(3) Every nonempty subset of the set of all submodules of M has a maximal element.

Proof.

(1) =⇒ (2) Suppose M1 ⊂ M2 ⊂ · · · ⊂ Mn ⊂ ... is an ascending chain of submodules of M. S∞ Let N = Mn, then N is a submodule of M and hence ﬁnitely generated. Let x1,..., xt gen- n=1 S∞ erate N. Since each xs ∈ Mn, there is an rs such that xs ∈ Mrs for each s = 1,..., t. Let n=1

k = max{r1,..., rt}, then for all n ∈ N we have Mn ⊂ N ⊂ Mk so that Mn = Mk for all n ≥ k.

(2) =⇒ (3) Suppose that a nonempty subset of the set of submodules of M, say Σ, is such that for

every submodule N ∈ Σ there is a submodule K ∈ Σ such that N ≤ K and N , K, i.e. Σ contains

no maximal element. Let N1 ∈ Σ be ﬁxed, then there is a submodule N2 ∈ Σ such that N1 N2.

Moreover, there is a N3 ∈ Σ such that N2 N3. By repeating this process, we contradict (2).

49 (3) =⇒ (1) Let N be a submodule of M. The set of ﬁnitely generated submodules of N has a

maximal element K. If K , N then there is an x ∈ N such that x < K. Let L be the submodule generated by x, then K ⊕ L properly contains K and is ﬁnitely generated, which contradicts the

maximality of K. Hence K = N, so that N is ﬁnitely generated.

Remark We have the following equivalent conditions

(20) Every descending chain of submodules

M1 ⊃ M2 ⊃ · · · ⊃ Mn ⊃ ...

is such that there is a k with Mn = Mk for all n ≥ k.

(30) Every nonempty subset of the set of all submodules of M has a minimal element.

where (20) → (30) follows by reversing the inclusions in the above proof, and to see that (30) →

0 (2 ) let M1 ⊃ M2 ⊃ · · · ⊃ Mn ⊃ ... be such a chain. The set {Mi : i ∈ N} contains a minimal

element Mk, so that Mn = Mk for all n ≥ k.

Definition 4.1. An R-module M satisfying any of conditions (1) − (3) of proposition 4.1 is called Noetherian. If M satisfies (20) or (30), it is called Artinian. A ring R is Noetherian (resp. Artinian) if R satisfies any of (1) − (3) (resp. (2) or (3)) as an R-module. Examples:

(1) The ring Z is a principal ideal domain, and so it is a Noetherian ring. However, for any nonzero integer n we have (n) ) (n2) ) (n3) ) ··· ) (nm) ) ... so that Z is not Artinian, even though every ideal of Z is ﬁnitely generated.

k (2) For a prime p ∈ Z, let Zp = {n/p : n ∈ Z, k ≥ 0} = M. We may regard M/Z as a Z-module, and 1 1 1 Z( ( + Z) ( ( + Z) ( ··· ( ( + Z) ( ... p p2 pk

50 a chain of submodules which never becomes stationary. Thus M/Z is not Noetherian. Sup- pose N is a proper submodule of M/Z,then there is an m ∈ Z and an n ≥ 1 such that p

m n does not divide m and pn + Z ∈ N. There are integers a and b such that ap + bm = 1, so m bm −apn 1 1 1 b( pn + Z) = pn + Z = pn + pn + Z = −a + pn + Z = pn + Z ∈ N. Then if k is the largest such m ∈ 1 ⊃ ⊃ · · · ⊃ ⊃ integer such that pn + Z N we have N = ( pk + Z). Thus if N1 N2 Nk ... is 1 a decreasing chain of submodules of M/Z we have N1 = ( ps + Z) for some s ≥ 0 and so we

must have Nk = Ns+1 for all k ≥ s + 1. Thus M/Z is Artinian, but not Noetherian. Moreover, M is not Noetherian or Artinian as a Z-module. M is not Artinian because it contains Z as a submodule, and it is not Noetherian as it is not ﬁnitely generated.

(3) Viewed as a Z-module, Z[x] is not Artinian because it contains Z as a submodule. It is also not Noetherian as a Z-module (later we will see that it is Noetherian as a ring) since the chain of submodules

(1) ( (1, t) ( (1, t, t2) ( ··· ( (1, t, t2,..., tn) ( ...

does not terminate.

Proposition 4.2.

(1) Every submodule N of a Noetherian R-module M is Noetherian

(2) If M is a Noetherian R-module and N ≤ M, then M/N is Noetherian.

(3) If M is an R-module, and N ≤ M is Noetherian with the property that M/N is also Noethe- rian, then M must be Noetherian.

Proof. (1) Since every submodule of N is a submodule of M, we have every submodule of N ﬁnitely generated and (1) follows.

51 (2) Let M1/N ⊂ M2/N ⊂ ... be an ascending chain of submodules of M/N. Since each Mk

corresponds to a submodule Mk of M, we have an ascending chain of submodules M1 ⊂

M2 ⊂ ... so that there is a k such that Mn = Mk for all n ≥ k. But since π : M → M/N is

surjective and Mn/N = π(Mn) for all n we must have Mn/N = Mk/N for all n ≥ k and (2) follows.

(3) Let K be a submodule of M. Since K ∩ N ≤ N it is ﬁnitely generated. Thus there are

x1,..., xn such that N ∩ K = (x1,..., xn). Since N/N ∩ K = N + K/N ≤ M/K there

are y1,..., ym such that N/N ∩ K = (y1,..., ym) so that K = (y1,..., ym, x1,..., xn). Thus M is Noetherian, since every submodule of M is ﬁnitely generated.

Corollary 4.1. Ever ﬁnitely generated free module over a Noetherian ring R is Noetherian

Proof. Let M = Rn. For n = 2 we have R and M/R R Noetherian. The result then follows by induction.

Corollary 4.2. Every ﬁnitely generated module M over a Noetherian ring R is Noetherian

Proof. Since every ﬁnitely generated module is a quotient of a free module, the result follows.

We claimed earlier that Z[x] is a Noetherian ring. We will now prove this and more.

Theorem 4.1. Hilbert’s Basis Theorem

For a Noetherian ring R, R[x] is a Noetherian ring.

Proof. Suppose to the contrary. Then there is an ideal I of R[x] which is not ﬁnitely generated.

Let f1 ∈ I be of least degree. Having chosen f1,..., fk, let fk+1 be of least degree in I \ ( f1,..., fk).

For each k ∈ N, let ak be the leading coeﬃcient and let nk be the degree of fk. Then by hypothesis nk ≤ nk+1 for each k and

(a1) ( (a1, a2) ( ...

52 is an ascending chain of ideals in R which does not terminate. For if (a1,..., ak) = (a1,..., ak+1)

Pk Pk nk+1−n j there are b1,..., bk ∈ R such that ak+1 = j=1 b ja j so that fk+1 − j=1 b j x f j ∈ I \ ( f1,..., fk) and is of smaller degree than fk+1.

Corollary 4.3. If R is Noetherian so is R[x1,..., xn]

Proof. If S = R[x1,..., xn−1] is Noetherian, then so is S [xn] = R[x1,..., xn]. The result follows by induction.

Corollary 4.4. For a Noetherian ring R, if S is a ﬁnitely generated R-algebra, then S is Noethe- rian.

Proof. We know that R[x1,..., xn] is Noetherian and by proposition 1.8 that S is an image of

R[x1,..., xn], so that there is an ideal I of R[x1,..., xn] such that S = R[x1,..., xn]/I, and thus S satisﬁes the ascending chain condition. Hence, S is Noetherian.

Suppose R[x] is Noetherian, then there is a surjective map φ : R[x] → R given by φ( f ) = f (0). Thus R = R[x]/ ker φ is Noetherian.

α β Proposition 4.3. Let 0 → M0 −→ M −→ M00 → 0 be an exact sequence of A-modules. Then M is Noetherian if and only if M0 and M00 are Noetherian.

Proof. Suppose M is Noetherian. Then M0 im α ≤ M, and so by proposition 4.2 we can conclude that M0 is Noetherian.

Since M/im α M00, we can conclude that M00 is Noetherian. Suppose M00 and M0 are Noetherian. Then M/im α and im α ≤ M are Noetherian. Hence M is

Noetherian.

n Corollary 4.5. If Mi is a Noetherian R-module for all i = 1,..., n, then ⊕i=1 Mi is Noetherian.

n−1 Proof. Since 0 → M2 → M1 ⊕ M2 → M1 → 0 is exact, M1 ⊕ M2 is Noetherian. If ⊕1 Mi is n n−1 n Noetherian,then since 0 → Mn → ⊕1 Mi → ⊕1 Mi → 0 is exact, ⊕1 Mi is Noetherian. The result follows by induction.

53 Lemma 4.1. Let {Mi}i∈I be a family of R-modules and let M = ⊕i∈I. Then M is ﬂat if and only if each Mi is ﬂat.

Proof. By proposition 1.12, for R-modules M, M0, M00 we have (M0 ⊕ M00) ⊗ M = (M0 ⊗ M) ⊕ (M00 ⊗ M). Suppose 0 → N0 → N → N00 → 0 is an exact sequence of R-modules.

0 0 We have N ⊗ M = ⊕i∈I(N ⊗ Mi). Moreover

0 00 0 → ⊕i∈I(N ⊗ Mi) → ⊕i∈I(N ⊗ Mi) → ⊕i∈I(N ⊗ Mi) → 0

is exact if and only if

0 00 0 → N ⊗ Mi → N ⊗ Mi → N ⊗ Mi → 0

is exact for all i ∈ I. Thus M is exact if and only if Mi is exact for all i ∈ I.

Lemma 4.2. Let R be a Noetherian local ring with maximal ideal m and residue field k. Let M be a finitely generated R-module. Then M is free if and only if M is flat.

Proof. Suppose M is free, then M Rn for some n and since R ⊗ N N for all submodules N, so too is Rn ⊗ N N. Thus M is ﬂat.

Suppose M is ﬂat. Let x1,..., xn be a minimal generating system for M. By Nakayama’s lemma

n = dim(M/mM) = dim(k ⊗ M).

Let F be a free module of rank n, and let {a1,..., an} be a basis for F. Deﬁne φ : F → M by

φ(ai) = xi for i = 1,..., n. Then φ is surjective. Let N = ker φ. Since the minimal generating systems for F are just the bases, we know that dim(F/mF) = dim(k ⊗ F) = n. We have 0 → N → F → M → 0 exact, and since M is ﬂat by proposition 1.14 we have 0 → k ⊗ N → k ⊗ F →

k ⊗ M → 0 exact. Thus (k ⊗ F)/(k ⊗ N) k ⊗ M but then dim(k ⊗ M) = dim(k ⊗ F) implies that k ⊗ N = 0 and so N = 0 by corollary 1.3. Thus M F.

54 Proposition 4.4. Let R be a Noetherian ring, and let M be a ﬁnitely generated R-module. The following are equivalent

(1) M is ﬂat

(2) Mp is free for all p ∈ Spec(R)

(3) Mm is free for all m ∈ Max(R).

Proof. Suppose M is ﬂat. We know that ﬂatness is a local property by Proposition 3.8. Thus Mp

is a ﬂat Rp-module for all p ∈ Spec(R). Since Rp is always is a local ring, lemma 4.2 shows that

Mp is free for all p ∈ Spec(R).

Suppose Mm is free for all m ∈ Max(R). Then Mm is ﬂat Rm-module for all m ∈ Max(R), so that

M is a ﬂat R-module.

We should also note that the Noetherian property respects localization.

Proposition 4.5. If S is a multiplicatively closed subset of a ring R, then if M is a Noetherian R-module, S −1 M is a Noetherian S −1R-module.

Proof. Suppose N ≤ S −1 M. Then if f is the canonical homomorphism of localization, we have

−1 c c f (N) = N a submodule of M. Let x1,..., xn generate N . By a similar argument to that given

c ce in the proof of proposition 3.10, we must have f (N ) = N = N = (x1/1,..., xn/1) so that every submodule of S −1 M is ﬁnitely generated.

Note that the ﬁeld of fractions of Z, Q, is Noetherian as a Q-module. However, there is no way to generate Q with a ﬁnite number of elements as a Z-module. Thus for a Noetherian R-module M we need not have S −1 M a Noetherian R-module.

55 4.2 Algebraic Varieties; The Nullstellensatz

Let An(L) denote n-dimensional affine space over a field L. Let K ⊂ L be a subfield.

Deﬁnition 4.2. Let S = { fk}k∈I be a set of polynomials in K[X1,..., Xn]. Then

n n Z( fk)k∈I = {x ∈ A (L): f (x) = 0 for all f ∈ S } ⊂ A (L)

is an aﬃne algebraic K-variety. We will often shorten this to Z.

n For a subset Z0 ⊂ A (L), deﬁne the set I(Z0) = { f ∈ K[x1,..., xn]: f (x) = 0 for all x ∈ Z0}. I(Z0)

is called the ideal of Z0.

Given an ideal I0 ⊂ K[x1,..., xn] deﬁne the zero set of I0 by

n Z(I0) = {x ∈ A (L): f (x) = 0 for all f ∈ I0}

n Proposition 4.6. (1) For any Z0 ⊂ A (L), I(Z0) = r(I(Z0)).

(2) For any variety Z, Z(I(Z)) = Z.

(3) For any two varieties Z1, Z2, we have Z1 ⊂ Z2 if and only if I(Z1) ⊃ I(Z2). Moreover, Z1 (

Z2 if and only if I(Z1) ) I(Z2)

(4) For any two varieties Z1, Z2, we have I(Z1 ∪Z2) = I(Z1)∩ I(Z2) and Z1 ∪Z2 = Z(I(Z1)· I(Z2)).

(5) For any family {Zλ}λ∈Λ of varieties

\ [ Zλ = Z I(Zλ) λ∈Λ λ∈Λ

Proof.

n n (1) Suppose f ∈ I(Z0), then ( f (x)) = 0 = 0 for all x ∈ Z0. Suppose there is an n > 0 such that

n ( f (x)) = 0 for all x ∈ Z0. Then clearly f (x) = 0 for all x ∈ Z0.

56 (2) If x ∈ Z, then by deﬁnition f (x) = 0 for all f ∈ I(Z), so that x ∈ Z(I(Z)). If x ∈ Z(I(Z)) =

Z(I(Z( fk)k∈I)) then since { fk ∈ I} ⊂ I(Z) we have x ∈ Z(( fk)k∈I.

(3) Suppose Z1 ⊂ Z2, then if f ∈ I(Z2) we have f (x) = 0 for all x ∈ Z1 so that f ∈ I(Z1). Suppose

I(Z1) ⊃ I(Z2), then if x ∈ Z1, we have f (x) = 0 for all f ∈ I(Z2). Thus x ∈ Z(I(Z2)) = Z2.

We have Z1 ( Z2 if and only if there is an x ∈ Z(I(Z2)) = Z2 such that x < Z(I(Z1)) = Z1 which is true if and only if f (x) , 0 for some f ∈ I(Z1), which is true if and only if I(Z1) ) I(Z2).

(4) We have f ∈ I(Z1 ∪ Z2) if and only if f (x) = 0 for all x ∈ Z1 and x ∈ Z2 if and only if

f ∈ I(Z1) ∩ I(Z2). Since f g(x) = 0 if and only if f (x) = 0 or g(x) = 0 we have

Z( fk · gl)(k,l)∈I×J = ∩k∈I,l∈JZ( fk · gl) = ∩k∈I,l∈J Z( fk) ∪ Z(g j) = ∩k∈I Z( fk) ∪ ∩l∈J Z(g j) = Z1 ∪ Z2

(5) We have x ∈ Z( ∪λ∈ΛI(Zλ)) if an only if, for all λ ∈ Λ, f (x) = 0 for all f ∈ I(Zλ) if and only if

x ∈ ∩λ∈ΛZλ.

The previous proposition very naturally suggests a topology in which the varieties correspond to closed sets.

Deﬁnition 4.3. Let S = { fk}k∈I be a set of polynomials in K[X1,..., Xn]. Then the topology on An(L) whose closed sets are of the form

Z(S ) = {x ∈ An(L): f (x) = 0 for all f ∈ S } is just the Zariski topology for An(L). As a corollary to Hilbert’s Basis Theorem we have the following.

Corollary 4.6. Every decreasing chain

Z1 ⊃ Z2 ⊃ · · · ⊃ Zn ⊃ ...

57 of subsets of An(L) terminates.

Proof. Since K[x1,..., xn] is Noetherian,

I(Z1) ⊂ I(Z2) ⊂ · · · ⊂ I(Zn) ⊂ ...

terminates, so that there is a k such that I(Zn) = I(Zk) for all n ≥ k. Thus Zk = Zn for all n ≥ k.

We call a variety Z irreducible if whenever Z = Z1 ∪ Z2 for varieties Z1, Z2, then either Z = Z1 or

Z = Z2.

Proposition 4.7. A variety Z is irreducible if and only if I(Z) is a prime ideal.

Proof. Suppose Z is irreducible. Suppose f1, f2 are such that f1 f2 ∈ I(Z). Let H1 = Z( f1) and

H2 = Z( f2), then since f1 f2 ∈ I(Z), for all x ∈ Z we have f1(x) = 0 or f2(x) = 0 so that Z =

(Z ∩ H1) ∪ (Z ∩ H2). Hence Z = Z ∩ H1 or Z = Z ∩ H2. Since Z ⊂ H1 or Z ⊂ H2 we must have either f1 ∈ I(Z) or f2 ∈ I(Z) so that I(Z) is prime.

Suppose I(Z) is prime. Suppose Z1 and Z2 are varieties, each not equal to Z, such that

Z = Z1 ∪ Z2. Since I(Z) = I(Z1) ∩ I(Z2) and I(Z) , I(Z1), I(Z2) there are f1 ∈ I(Z1) \ I(Z), f2 ∈

I(Z2) \ I(Z) such that f1 f2 ∈ I(Z1) ∩ I(Z2) = I(Z1 ∪ Z2) = I(Z), which is impossible since I(Z) is prime.

Similarly a general topological space X is said to be irreducible if it satisﬁes one of the following equivalent conditions

1) If Z1 and Z2 are closed sets such that X = Z1 ∪ Z2 then X = Z1 or X = Z2.

2) Every nonempty open set is dense

3) Every pair of nonempty open sets U1 and U2 are such that U1 ∩ U2 , ∅.

58 Proof. By taking complements, we have (1) if and only if for every open set U1 and U2 such that

U1 ∩ U2 = ∅ either U1 = ∅ or U2 = ∅.

Suppose (2) is satisﬁed, then (1) is also. Let U be a nonempty, open subset of X. If U , X, then since X \ U ∪ U = X we have X \ U = X which implies that U = ∅, a contradiction. Suppose every nonempty open subset is dense. Since dense subsets intersect every nonempty

open set, (2) follows.

A subset X0 of a topological space X is irreducible if it is irreducible under the subspace topology.

0 0 Then by the above X irreducible if and only if for all open subsets U1, U2 such that U1 ∩X , ∅ ,

0 0 U2 ∩ X we have U1 ∩ U2 ∩ X , ∅.

0 0 Suppose X is irreducible. With U1, U2 nonempty open subsets of X such that U1 ∩ X , ∅ ,

0 0 0 0 U2 ∩ X implies that U1 ∩ U2 ∩ X , ∅, so we must have U1 ∩ U2 ∩ X , ∅. Thus X is irreducible. An irreducible component of a topolgical space X is deﬁned to be a maximal irreducible subset of X. By the above, the irreducible components are always closed. By a standard argument using Zorn’s lemma, every irreducible subset is contained in an irreducible component, and thus every topological space is covered by its irreducible components.

A topological space X such that every descending chain Z1 ⊃ Z2 ⊃ · · · ⊃ Zn ⊃ ... of closed subsets terminates is said to be Noetherian. Thus, An(L) and every variety Z with the subspace topology is Noetherian.

Lemma 4.3. Let R ⊂ S ⊂ T = R[x1,..., xn] be rings. Suppose R is Noetherian, and T is ﬁnitely generated as an S -module. Then S is ﬁnitely generated as a ring over R.

ik Proof. Let w1,..., wm generate T as an S -module. For each i, k = 1,..., m there are al ∈ S such that m X ik wiwk = al wl. l=1

0 ik ik Let S = R[{al }i,k,l=1,...,m] be the R-algebra generated by the al . Then since T contains the xi we Pm have equations xi = j=1 ai jw j with ai j ∈ S , and since every element of T is a polynomial in

59 the xi with coeﬃcients in R, given f ∈ T we can subsitute the equation for xi and then use the

0 equations for the wiwk to recover f from a linear combination of the w1,..., wm over S . Thus T

0 0 0 0 lies in S w1 + ··· + S wm and so T is a finitely generated S -module. Since R is Noetherian, S is Noetherian by the basis theorem. Moreover, since T is a finitely generated S 0-module, it is a Noetherian S 0-module. Since S 0 ⊂ S ⊂ T, we must have S a finitely generated S 0-module. Hence

S is ﬁnitely generated as a ring over R.

Lemma 4.4. Let S = K(Z1,..., Zt) be the quotient field of K[x1,..., xn] over a field K, with t > 0. Then S is not finitely generated as a ring over K.

fi(Z1,...,Zt) Proof. Suppose x1,..., xm generate S as a ring over K, with xi = for fi, gi ∈ K[Z1,..., Zt] gi(Z1,...,Zt)

for each i = 1,..., m. Every gi can be written as a ﬁnite product of irreducible polynomials,

unique up to associates. Since we have S = K[x1,..., xm], every element of S can be expressed

as the quotient of two polynomials in K[Z1,..., Zt] whose denominator consists solely of those ir-

reducible polynomials which divide a gi. Since there are inﬁnitely many irreducible polynomials,

1 there is an irreducible polynomial p prime to every gi, whence p is not in S , a contradiction.

Proposition 4.8.

If A/K is an extension of ﬁelds and A arises from K through ring adjunction of ﬁnitely many elements, then A/K is algebraic.

Proof. We have A = K[Z1,..., Zs] for some s ∈ N. If A/K is transcendental and {Z1,..., Zt}

(t > 0) is a transcendence basis, we have K ⊂ S = K(Z1,..., Zt) ⊂ A, and {Z1,..., Zt} are

algebraically independent over K while each of Zt+1,..., Zs is algebraic over S . Thus A is a finite algebraic extension of S and so finitely generated as an S -module. By lemma 4.3, S is finitely

generated as a ring over K, which is impossible by lemma 4.4.

Theorem 4.2. Hilbert’s Nullstellensatz

If L is algebraically closed and I , K[x1,..., xn], then Z(I) , ∅.

60 Proof. We have I contained in some maximal ideal m of K[x1,..., xn]. We have A = K[x1,..., xn]/m a ﬁnitely generated K-algebra, so by proposition 4.8 A/K is algebraic and so there is a homomorphism φ : A → L since L is algebraically closed. Let ξi denote the image of xi in A, then

n (φ(ξ1), . . . , φ(ξn)) ∈ L is a zero of m and hence I.

Corollary 4.7. Let L/K be a field extension, with L algebraically closed. Then Z 7→ I(Z) defines a bijection from the set of affine K-varieties onto the set of ideals I0 of K[x1,..., xn] with I0 = r(I0). In fact

r(I0) = I(Z(I0)).

Proof. By proposition 4.6, we have injectivity. It suﬃces to show surjectivity by proving the for- mula. For an ideal I0 of K[x1,..., xn], r(I0) ⊂ I(Z(I0)). Let F , 0 ∈ I(Z(I0)).

In the polynomial ring K[x1,..., xn, t] form the ideal J = (I0, (F · t) − 1). Suppose (x1,..., xn, t) is a zero of J, then (x1,..., xn) a zero of I0 so F(x1,..., xn) · t − 1 = −1. However, (x1,..., xn, t) is a zero of F · t − 1, a contradiction. Since J therefore has no zeros, J = K[x1,..., xn, t] by the nullstellensatz. Thus Xs 1 = RkFk + S (Ft − 1) k=1 for some Rk, S ∈ K[x1,..., xn, t] and Fk ∈ I0. Let φ : K[x1,..., xn, t] → K(x1,..., xn) be the

1 homomorphism such that φ ﬁxes each xi and φ(t) = F . Then φ(S (Ft − 1)) = φ(S )(Fφ(t) − 1) = 0 and Xs 1 = φ(Rk)Fk k=1

Ak where each φ(Rk) can be written as ρ for some Ak ∈ K[x1,..., xn] and ρk ∈ N. If ρ = maxk=1,...,s ρk Fk we have s ρ X 0 F = AkFk k=1

0 ρ for some Ak ∈ K[x1,..., xn]. Thus F ∈ (F1,..., Fs) ⊂ I0, so F ∈ r(I0).

Deﬁnition 4.4. For a K-variety Z ⊂ An(L) we deﬁne the coordinate ring of Z to be K[Z] =

61 K[x1,..., xn]/I(Z). The elements φ ∈ K[Z] are functions φ : Z → L. Remarks:

Suppose φ = F + I(Z) for some F ∈ K[x1,..., xn]. Let ξi be the image of xi in K[Z], then the ξi are called the coordinate functions, which assign to every point (t1,..., tn) ∈ Z the i−th coordinate ti. If W is a subvariety of Z there is a notion of a vanishing ideal IZ(W) = I(W)/I(Z). Moreover,

W is an irreducible subvariety if and only if IZ(W) is prime in K[Z].

The maximal ideals of K[Z] = K[x1,..., xn]/I(Z) are in one to one correspondence with the maximal ideals m of K[x1,..., xn] such that m ⊃ I(Z). By the Nullstellensatz, if K is algebraically

n closed then m has a zero in x = (t1,..., tn) ∈ K . The polynomials not contained in m do not have

x as a zero, so that (x1 − t1,..., xn − tn) ⊂ m. However (x1 − t1,..., xn − tn) is the kernel of a surjective homomorphism from K[x1,..., xn] to K, and hence is maximal.

Thus m = (x1 − t1,..., xn − tn) = I(x). Since I(x) ⊃ I(Z) we must have x ∈ Z. Thus points of a variety correspond to maximal ideals in the coordinate ring.

4.3 Dimension of a ring

For a topological space X we deﬁne the dimension of X to be the supremum of all n ∈ Z such that there is a chain

Z0 ( Z1 ( Z2 ( ··· ( Zn of distinct irreducible, closed subsets of X. In a sense, this definition of dimension has the Zariski topology (or perhaps some sufficiently similar topology) already built in. For example, in any Hausdorff space, which the An(L) under the Zariski topology is not, the irreducible sets are singleton, so that in any Hausdorff space such as Rn or Cn the dimension is zero, rather than n. On the other hand, under the Zariski topology,

n by the Nullstellensatz C is homeomorphic to Max(R) where R = C[x1,..., xn], and in this case

62 will have dimension n. One can then deﬁne the (Krull) dimension of a ring R as

dim(Spec(R)) = dim R

so that dim R is the length of a maximal chain of prime ideals in R. To be more speciﬁc the Krull dimension is the supremum of the length of all chains of prime ideals

p0 ( p1 ( ··· ( pn where by “length” we mean the number n. Deﬁne the height of a prime ideal p, to be denoted h(p), as the supremum of the length of all chains

p0 ( p1 ( ··· ( pn

with p = pn. For a prime ideal p ∈ Spec(R), since the prime ideals of the localized ring Rp are in a one to one correspondence with the prime ideals of R contained in p we have ht(p) = dim(Rp). For a K-variety Z the chains of irreducible, closed subsets W in Z correspond bijectively to prime ideals IZ(W) in K[Z], so that dim Z = dim K[Z].

Proposition 4.9. For a ring R,

1) Max(R) is Noetherian if and only if J(R) is.

2) dim(Max(R)) = dim(J(R)) ≤ dim(Spec(R))

Proof. For a closed subset A of Max(R), let A denote its closure in J(R). For B a closed subset of J(R), let B∗ = B ∩ Max(R). We claim (A)∗ = A and B∗ = B. This will then give a one to one correspondence between closed subsets of Max(R) and closed subsets of J(R) which preserves inclusions and irreducibility. The validity of this claim will establish both parts of the proposition.

63 Since A is closed in Max(R) we have (A)∗ = A ∩ Max(R) = A by deﬁnition. T Now for B ⊂ J(R) we have I(B) = p∈B p so that B = {p ∈ J(R): p ⊃ I(B)}. Since every p ∈ B

T ∗ ∗ ∗ is an intersection of maximal ideals, we have I(B) = m∈B∗ m = I(B ), so that B = Z(I(B )) = Z(I(B)) = B.

Proposition 4.10. Every closed irreducible subset of J(R) has a unique generic point.

Proof. Let B be a closed irreducible subset of J(R). We want to show that there is a unique p ∈ J(R) such that p = B. We have I(B) prime, and Z(q) = {p ∈ J(R): p ⊃ q} contains I(B) if and only if I(B) ⊃ q. Thus the smallest such closed set is Z(I(B)) = B = {p ∈ J(R): p ⊃ I(B)} = I(B). If p is another generic point, then B = p = Z(I(p)) yet since I(p) = p we have Z(I(p)) = Z(p) and thus T I(B) = I(Z(p)). Hence I(B) = I(Z(p)) = q⊃p q = p, so p = I(B).

64 CHAPTER 5

THE FORSTER-SWAN THEOREM

Given a commutative ring R of krull dimension n and R-module M, all of whose localizations Mp for p ∈ Spec(R) can be generated by at least r elements, Forster showed that M can be generated by r + n elements. Serre suggested that, by replacing n with the dimension of Max(R), one could show that M is generated by r + d elements. Making use of the properties of the J-spectrum, i.e. Propostions 4.9 and 4.10, Swan was able demonstrate the validity of this conjecture. Let µ(M) denote the number of elements of a minimal generating system for a ﬁnitely generated

R-module M. For every prime ideal p ∈ Spec(R) we let µp(M) denote the number of elements in a minimal generating system for the Rp-module Mp. Moreover, we deﬁne the Kaplansky ideal I(M, r), for each integer r ≥ 0, by

X I(M, r) = Ann(M/hm1,..., mri).

{m1,...,mr}⊂M

Lemma 5.1. Let R be a ring and let M be an R-module.

(a) I(M, 0) = Ann(M)

(b) I(M, r) ⊂ I(M, r + 1)

(d) For any multiplicatively closed subset S of R, we have I(S −1 M, r) = S −1I(M, r) as ideals of S −1R.

65 (e) For p ∈ Spec(R) we have µp(M) ≥ r + 1 if and only if p ⊃ I(M, r).

Proof.

(a) By deﬁnition, I(M, 0) = Ann(M/h0i) = Ann(M).

(b) We have

X X I(M, r + 1) = Ann(M/hm1,..., mr+1i) ⊃ Ann(M/hm1,..., mr, 0i)

{m1,...,mr+1}⊂M {m1,...,mr,0}⊂M

= I(M, r)

(d) Since

−1 X −1 X −1 S I(M, r) = S Ann(M/hm1,..., mri) = AnnRS (S (M/hm1,..., mri)) {m1,...,mr}⊂M {m1,...,mr}⊂M

X −1 −1 X −1 = AnnRS (S M/S hm1,..., mri) = AnnRS (S M/hm1/1,..., mr/1i) −1 {m1,...,mr}⊂M {m1/1,...,mr/1}⊂S M

= I(S −1 M, r) the result follows.

(e) Note that for a ring S and a S -module N, Ann(N) = S if and only if N = 0. Suppose that

µp(M) ≥ r + 1, then µp(M) > r and so

X I(Mp, r) = AnnRp (Mp/hm1/1,..., mr/1i) , Rp

{m1/1,...,mr/1}⊂Mp

since Mp has no generating sets with at most r generators. Hence, I(Mp, r) is contained in the

66 maximal ideal pRp. Thus I(Mp, r) = I(M, r)p ⊂ pRp ( Rp which implies that I(M, r) ∩ R \ p = ∅ and so I(M, r) ⊂ p.

If µp(M) ≤ r then I(Mp, r) = Rp so that, from the proof of proposition 3.10, we must have

I(M, r) ∩ R \ p = ∅ and thus p 1 I(M, r).

For a local ring R, Nakayama’s lemma tells us that given a ﬁnitely generated R-module M, if m ∈

M is such that m < mM, then we have that the image of m in M/mM is an element of a basis for the R/m-vector space M/mM. We try to extend this notion to non-local rings with the following deﬁnition

Deﬁnition 5.1. An element m ∈ M is called basic at p ∈ Spec(R) if m/1 < pMp.

Recall that the support of M is deﬁned to be the set Supp(M) of prime ideals p such that Mp , 0.

Now by Nakayama’s lemma µp(M) = dimRp/pRp (Mp/pMp). We note that

µp(M/hxi) = dimRp/pRp Mp/hx/1i pMp + hx/1i /hx/1i = dimRp/pRp Mp/ pMp + hx/1i .

Moreover, we have dimRp/pRp Mp/ pMp + hx/1i = dimRp/pRp (Mp/pMp) − 1 = µp(M) − 1 if and only if x/1 < pMp. Thus, x is basic at p if and only if µp(M/hxi) = µp(M) − 1. It then follows that

x is basic at p ∈ Supp(M) if and only if x/1 is in a minimal generating system for the Rp-module

Mp. In what follows, take X = J(R) to be the J-spectrum of R. For all m ∈ M let X(m) denote the set of p ∈ X such that m is basic at p. As before, for an ideal I of R, let V(I) = {p ∈ X : p ⊃ I}. Note that by Nakayama’s lemma, if m is basic at a prime p it lies in a minimal generating system for

Mp. Then we must have m basic at p ∈ V(I) if and only if its image m in M/IM is basic at p.

Lemma 5.2. Let X = J(R) be Noetherian and suppose that d = dim X < ∞. Let M be a ﬁnitely generated R-module. Then for any ideal I of R and any m ∈ M, X(m) ∩ V(I) has only ﬁnitely many minimal elements.

67 Proof. Let m be the image of m in the R/I-module M/IM. Since V(I) is in a bijective correspondence with J(R/I), we have X(m) ∩ V(I) = X(m). We can therefore assume without loss that I = (0). By proposition 4.10, the irreducible components of X have a unique generic point. If

{p1,..., pt} are the generic points corresponding to each irreducible component, then X(m) =

St i=1 X(mi) where mi is the image of m in the R/pi-module M/pi M. Since R/pi is an integral domain (for each i = 1,..., t), without loss we may assume R is itself an integral domain. Now we proceed by induction on d. For d = 0, R is an Artinian integral domain, and thus R is a ﬁeld, so that J(R) consists of a single point. Let d > 0, if m is basic at (0), then (0) is the only minimal element in X(m). If m is not basic at

(0) there is an f , 0 ∈ R such that f m = 0. If m is the image of m in M/ f M then X(m) = X(m) and since dim(J(R/ f )) < d, we are done.

‘

Lemma 5.3. Let M be a ﬁnitely generated R-module, and let m ∈ M. Suppose that X = J(R) is

Noetherian and that dim X < ∞. Let um = max{µp(M) + dim(V(p)) : p ∈ X(m)}, then there are

only ﬁnitely many p ∈ X(m) with µp(M) + dim V(p) = um.

Proof. Suppose p ∈ X(m) is such that µp(M) + dim V(p) = um. Let µp(M) = r, then r > 0 and by lemma 5.1 we have p ⊃ (I(M, r − 1)), p 2 I(M, r). Suppose there is a prime ideal q ∈ X(m) with p % q ⊃ I(M, r − 1). Then µp(M) = r ≥ µq(M) ≥ r, so

um = µp + dim V(p) < µq(M) + dim V(q) ≤ um

since dim V(p) < dim V(q), but this is impossible. Thus p is minimal over I(M, r − 1), and since p ∈ X(m) ∩ V(I(M, r − 1)) by lemma 5.2 there are only ﬁnitely many p ∈ X(m) such that p is minimal over I(M, r − 1). Moreover, since there are only ﬁnitely many distinct ideals I(M, r), as

M is ﬁnitely generated, we are done.

68 Deﬁnition 5.2. A submodule U ⊂ M is called k-times basic (for k ∈ N) at p ∈ Spec(R) if

µp(M) − µp(M/U) ≥ k.

Now since µp(M) − µp(M/U) is the number of generators of a minimal system for Mp which are also in Up, U is k-times basic if and only if Mp has a minimal generating system with at least k elements in Up. Of particular importance, for all m ∈ M, U = (m) is 1-times basic at p if and only if m is basic at p.

Lemma 5.4. For {m1,..., mt} ⊂ M and {p1,..., pr} ⊂ Spec(R). Let (m1,..., mt) be ki-times basic, for some ki with 0 ≤ ki < t, at each of the pi (i = 1,..., r). Then there are elements a1,..., at−1 ∈

R such that (m1 + a1mt,..., mt−1 + at−1mt) is ki-times basic at pi for each i = 1,..., r.

Proof. For r = 1 we have no loss in generality by making the assumption that R is a local ring and m = p1. Now let U = (m1,..., mt), with images mi of mi in M/mM, we can form the exact sequence 0 → (U + mM)/mM → M/mM → M/(U + mM) → 0

so µ(M) − µ(M/U) = dimR/m(m1,..., mt) ≥ k1 where k1 < t. Now if dimR/m(m1,..., mt) ≥ k1 then by taking a1 = a2 = ··· = at−1 = 0 we have the desired result. Since dimR/m(m1,..., mt) ≥ k1 we have dimR/m(m1,..., mt−1) ≥ k1 − 1. If dimR/m(m1,..., mt−1) = k1 − 1 then mt is linearly independent of {m1,..., mt−1}, since including mt in the generating system necessarily increases

the dimension. If m1,..., mk1−1 are linearly independent then dimR/m(m1,..., mk−1) = k1−1 so that

dimR/m(m1,..., mt−1) ≥ k1 − 1, and we take ai = 0 for i = 1,..., k1 − 1, k1 + 1,..., t − 1 and ak1 = 1.

Let l = dimR/m(m1,..., mk1−1, mk1 + mt,..., mt−1). If dimR/m(m1,..., mt−1) = k1 − 1 then mt is

linearly independent over m1,..., mk1−1, mk1 ,..., mt−1 so that l ≥ k1. If dimR/m(m1,..., mt−1) ≥ k1 we still have l ≥ k1.

Thus (m1 + a1mt,..., mk1 + ak1 mt,..., mt−1 + at−1mt) is k1-times basic.

69 Now let r > 1 and suppose the result holds for prime ideals p1,..., pr−1. Reorder the prime ideals

0 0 so that pr is minimal in the set {p1,..., pr}. By the induction hypothesis there are a1,..., at−1 ∈ R 0 0 such that (m1 + a1mt,..., mt−1 + at−1mt) is ki-times basic at each pi (i = 1,..., r − 1). Now we have Tr−1 Tr−1 i=1 pi 1 pr so there is an a ∈ i=1 pi with a < pr. So

0 0 V = (m1 + a1mt,..., mt−1 + at−1mt, amt)

is kr-times basic at pr since a/1 is a unit in Rpr and thus Vpr ≥ (m1,..., mt)pr , but by hypothesis

(m1,..., mt)pr already contains at least kr elements of a minimal generating system for Mpr . Now

00 00 by the base case applied with V in place of {m1,..., mt} we get a1 ,..., at−1 ∈ R such that

0 00 0 00 U = (m1 + a1mt + a1 amt,..., mt−1 + at−1mt + a1 amt)

is kr-times basic at pr. Since a ∈ pi for each i < r, if

0 0 0 U = (m1 + a1mt,..., mt−1 + at−1mt)

0 0 then for each i < r we have µpi (M/U) = µpi (M/U ) as the images in Mpi /pi Mpi of each mk +akmt + 00 0 00 0 00 ak amt with k = 1,..., t − 1 are such that m1 + akmt + ak amt = m1 + akmt since ak amt ∈ pi. But 0 then we must have µpi (M) − µpi (M/U) = µpi (M) − µpi (M/U ) ≥ ki for all i < r. Thus U is ki-times

0 00 basic at pi for each i < r as well. Thus with ai = ai + ai a we have

(m1 + a1mt,..., mt−1 + at−1mt)

ki-times basic at each pi with i = 1,..., r.

Theorem 5.1. Let X = J(R) be Noetherian and of ﬁnite dimension.

Let M = (m1,..., mt), and suppose that µp(M) + dim V(p) < t for all p ∈ X(mt). Then there are

70 elements a1,..., at−1 ∈ R such that M = (m1 + a1mt,..., mt−1 + at−1mt).

Proof. If t = 1 there’s nothing to prove. If X(mt) = ∅ then m does not lie in a minimal generating

system for any prime ideal p ∈ X so that Mp = (m1/1,..., mt−1/1) for all p ∈ X. It follows that

M = (m1,..., mt−1). This is the desired result with a1 = a2 = ··· = at−1 = 0.

If X(mt) , ∅, then

u = max{µp(M) + dim V(p): p ∈ X(mt)} > 0

and by lemma 5.3 there are only ﬁnitely many p ∈ X(mt) with u = µp + dim V(p), by lemma 5.4 we can ﬁnd an a1 ∈ R such that m1 + a1mt is basic at each of these p.

0 0 Now let M = M/(m1 + a1mt). Then M is generated by the images of m2,..., mt, which we

0 0 0 denote by m2,..., mt respectively. We must have mt basic at a prime p only if mt is basic at p. If 0 p ∈ X(mt) is such that u = µp + dim V(p) then m1 + a1mt is basic at p so that µp(M ) < µp(M). It then follows that

0 0 0 u = max{µp(M ) + dim V(p): p ∈ X(mt )} < u < t and therefore u0 < t − 1.

0 0 0 We now proceed by induction on t. For t = 2 we have u < 1 so that X(mt ) = ∅ and thus M = (0), which implies that M = (m1 + a1mt). For t > 2, suppose that the result holds for t − 1. Since

0 0 0 0 0 0 u < t − 1 we must have elements a2,..., at−1 ∈ R such that M = (m2 + a2mt ,..., mt−1 + at−1mt ), but then M = (m1 + a1mt, m2 + a2mt,..., mt−1 + at−1mt).

Corollary (The Forster-Swan Theorem) Let X = J(R) be Noetherian of ﬁnite Krull dimension. Then for every ﬁnitely generated R-module M

µ(M) ≤ u = max{µp(M) + dim V(p): p ∈ X ∩ Supp(M)}.

Proof. Let {x1,..., xn} be a minimal generating system for M.

71 0 0 Suppose n > u, and let u = max{µp + dim V(p): p ∈ X(xn)}. Since u ≤ u < n, we must have by theorem 5.1 elements a1,..., an−1 ∈ R such that (x1 + a1 xn,..., xn−1 + an−1 xn), which contradicts the minimality of {x1,..., xn}. Thus µ(M) ≤ u.

As an example, the ring of integers OK of an algebraic number ﬁeld K is a Noetherian ring of dimension one, and it’s localization at every prime ideal is a principal ideal domain. Hence, the

Forster-Swan theorem tells us that any ideal of OK can be generated by 2 elements.

72 CHAPTER 6

PROJECTIVE MODULES, QUILLEN’S LOCAL-GLOBAL PRINCIPLE, THE EISENBUD-EVANS CONJECTURE, AND SERRE’S PROBLEM

In the famous text “Faisceaux algebriques´ coherents”´ (Coherent algebraic sheaves, 1955), Jean- Pierre Serre wrote:

r “Note that when V = K (in which case A = K[X1,..., Xr]), we do not know if there exist projective A-modules of finite type that are not free, or equivalently, if there exist algebraic vector bundles with base Kr that are not trivial.” In this work, Serre introduced the theory of sheaves into algebraic geometry, wherein a sheaf is essentially a system of local coefficients over a space. Sheaves were originally considered by Leray in the 1940s. The importance of sheaf theory is that it gives strong relations between the local and global properties of a space. The question which ultimately became known as Serre’s Problem arose from considerations of a correspondence between modules and sheaves. A vector bundle could roughly be thought of as a family of vector spaces parametrized by a smooth man- ifold, and by considering the algebraic structure of the “sections” of vector bundles over algebraic varieties, which are groups of certain continuous functions, one is led to a correspondence between projective modules and vector bundles. In this correspondence, the free modules correspond to the trivial bundles. Thus, one has the equivalent statement in terms of algebraic vector bundles. The motivation for this consideration is then more clear, in that the affine n-space over a field should behave like a contractible space in topology, which has only trivial bundles. After nearly twenty years of extraordinary effort on the part of the mathematical community, Quillen and Suslin independently proved Serre’s conjecture in January of 1976. A large body

73 of work preceded their solutions. In 1958 Seshadri proved the conjecture for the case of two variables. In 1963, Bass showed that non-ﬁnitely generated projective modules were free for

K[x1,..., xn]. In 1971, Ojanguren, Sharma, and Sridharan showed that for n ≥ 2 Serre’s conjecture does not hold in the non-commutative case. In 1974 Murthy-Towber proved the result for the case of K[x1, x2, x3], resolving confusion caused by a claimed counterexample in this case. Also in 1974, Roitman proved Serre’s conjecture, for an infinite field, in the case of projective modules of rank n. The methods used to demonstrate these special cases of the problem did not figure strongly into the proofs given by Quillen and Suslin, with the exception of Horrocks’ theorem, which was proved in 1964 and used to great effect by Quillen. Quillen’s proof ultimately rests on a remarkable local-global principle concerning the concept of of “extended modules”, which allowed for the reduction to the case of local rings with n = 1. Shortly after the original proofs, Vaserstein and Suslin presented elementary proofs using tech- niques involving unimodular rows. This established the study of unimodular rows as a topic of research in commutative algebra. Moreover, the attempt to prove Serre’s Conjecture provided a tremendous stimulus to the development of the field of algebraic K-theory. In 1971, Eisenbud and Evans used the notion of a basic element to extend results on K-theory and unimodular rows to the setting of finitely generated modules. In particular, the Eisenbud-Evans theorem extended Serre’s theorem on free summands of projective modules to the non-projective case. An immediate consequence of this result is the Forster-Swan theorem. Even better, Eisenbud-Evans conjectured an improvement on the Forster-Swan bound, which was proved by Sathaye and Mohan Kumar. Throughout, R is a non-zero ring.

Deﬁnition 6.1. For a ring R , 0, M an R-module, we say that M is projective if there is an R-

0 0 module M such that M ⊕ M is free. We say that M is locally free if Mp is a free Rp-module for all p ∈ Spec(R). Given a projective R-module M, it is clear from corollary 3.2 that for any multiplicatively closed

74 subset S ⊂ R we must have S −1 M a projective S −1R-module. Just as easily, direct summands of M are projective, and if N is projective then M ⊕ N is projective.

Deﬁnition 6.2. Let f g 0 −→ M0 −→ M −→ M00 → 0 be an exact sequence of R-modules and R-module homomorphisms. We say the sequence splits if there is an R-module homomorphism ζ : M00 → M such that g ◦ ζ = 1. If such a sequence splits we have an internal direct sum decomposition M = M0 ⊕ ζ(M00). On the other hand, if M = U ⊕ f (M0) = U ⊕ M0 for some U ≤ M, then since g ◦ f = 0 we have U M00

−1 under g|U . If ψ = g|U then g ◦ ψ = 1, and the sequence splits. Moreover if the sequence splits we have g ◦ ζ = 1 and the diagram

ζ M00 M

g◦ζ g M00

00 00 00 00 commutes. Thus if φ : HomR(M , M) → HomR(M , M ) is the natural map, taking h : M → M to g ◦ h, then for all R-linear ρ : M00 → M00 we have φ(ζ ◦ ρ) = ρ so that the natural map is

00 surjective. On the other hand, if φ, as above, is surjective we must have a ζ ∈ HomR(M , M) such that φ(ζ) = g ◦ ζ = 1.

00 00 00 Thus, the sequence splits if and only if the natural map HomR(M , M) → HomR(M , M ) is surjective.

Proposition 6.1. The following statements are equivalent

a) M is projective.

b) The functor HomR(M, ) is exact.

0 00 0 c) For any surjective R-linear map f : M → M , the natural map φ : HomR(M, M ) →

00 HomR(M, M ) given by φ(g) = f ◦ g is a surjective R-linear map.

75 d) Any exact sequence of R-modules of the form 0 → C → D → M → 0 splits.

Proof.

0 f g 00 (a)→ (b) We know that HomR(M, ) is left exact. Let 0 −→ N −→ N −→ N → 0 be exact. We

0 00 must have 0 → HomR(M, N ) → HomR(M, N) → HomR(M, N ) exact, so it suﬃces to show that

00 0 HomR(M, N) → HomR(M, N ), which maps φ to g ◦ φ, is surjective. Let F = M ⊕ M be a free

0 00 00 module, for some R-module M . Suppose φ ∈ HomR(M, N ), then extend φ to φ : F → N by

0 deﬁning φ to be zero on M and equal to φ on M. Let {e1,..., en} be a basis for F, and let φ(ei) =

yi. The yi completely determine φ, and since g is surjective there are xi ∈ N such that g(xi) = yi

0 0 0 0 for each i. Let φ : F → N be given by φ (ei) = xi, then g ◦ φ = φ. Hence, φ|M is the desired R-linear homomorphism.

(b)→ (c) This follows directly from the deﬁnition of an exact sequence.

(c)→ (d) Since HomR(M, D) → HomR(M, C) is surjective, there is a γ : M → D such that 1 = f ◦ γ. Alternatively, we have the commutative diagram

γ f D M 0

since f is surjective.

(d)→ (a) Let F be a free R-module, then there is an exact sequence 0 → K → F → M → 0 which

and this sequence splits. Thus F = M ⊕ K.

Now given two R-modules M,N, and a multiplicatively closed subset S ⊂ R. Recall that for an

R-linear map f : M → N by the universal property of localization there is a unique RS -linear map

−1 −1 fS : S M → S N given by fS (m/s) = f (m)/s.

−1 −1 The mapping φ : HomR(M, N) → HomS −1R(S M, S N) given by φ( f ) = fS is clearly R-linear.

76 Deﬁne

−1 −1 −1 h : S (HomR(M, N)) → HomS −1R(S M, S N)

−1 by h( f /t) = fS /t, which clearly deﬁnes an S R-linear mapping. We claim that if M is ﬁnitely generated, h is injective.

Let x1,..., xn generate M. For f ∈ HomR(M, N) and t ∈ S suppose that f /t ∈ ker h. Then

0 0 0 f (xk)/t = 0 so there is a t ∈ S such that t f (xk) = 0 for all k = 1,..., t. Hence t f = 0, and thus f /t = 0.

6.1 Finitely Presentable Modules

Lemma 6.1. Suppose that there exists an exact sequence of R-modules

0 → K → Rn → M → 0

where K is ﬁnitely generated. Then M is ﬁnitely generated. Moreover, given an R-module N and

−1 −1 −1 a multiplicatively closed subset S ⊂ R the map h : S (HomR(M, N)) → HomS −1R(S M, S N) is an isomorphism.

n Proof. Let im K = hx1,..., xti. Since the sequence is exact, M = R /hx1,..., xti and is ﬁnitely generated. Thus h is injective. It remains to show that h is surjective.

Suppose M is a free module. Let {x1,..., xr} be a basis for M. Lr → Deﬁne φ : HomR(M, N) k=1 N by φ( f ) = ( f (x1),..., f (xr)). Since every R-linear map

f : M → N is determined by the f (xi), this gives a well-deﬁned isomorphism. We have a commutative diagram

−1 h −1 −1 HomR(M, N) ⊗ S R HomS −1R(M ⊗ S R, N ⊗ S R)

ϕ ϕ

r −1 r −1 ⊕k=1N ⊗ S R, ⊕k=1(N ⊗ S R)

77 and since φ is an isomorphism, h is an isomorphism.

We have K Rt. In the general case, since Hom( , N) is left exact and S −1R is ﬂat, a commuta-

−1 −1 −1 tive diagram, with MS = S M, NS = S N,and RS = S R

n t 0 HomR(M, N) ⊗ RS HomR(R , N) ⊗ RS HomR(R , N) ⊗ RS

h ξ η ⊕n ⊕t 0 HomRS (MS , NS ) HomRS ( k=1RS , NS ) HomRS ( k=1RS , NS )

with both rows exact. Since Rn, Rt are both free modules, ξ and η are isomorphisms. It follows that h is an isomorphism

The exact sequence 0 → K → Rn → M → 0 in the lemma is an example of a presentation of a module.

Deﬁnition 6.3. Let {xi}i∈I be a generating system for M, with I potentially inﬁnite, then a presen- f tation of M belonging to the generating system is the exact sequence 0 → K → ⊕i∈IR −→ M → 0

where f maps the basis element ei of ⊕i∈IR to mi for each i ∈ I. K is called the module of relations. If there is an n ∈ N and an exact sequence 0 → K → Rn −→ M → 0 with K finitely generated, we say that M is finitely presentable. In particular, in a Noetherian ring R every finitely generated R-module is finitely presentable by Hilbert’s basis theorem. The finitely presented modules over a ring behave similarly to finite dimensional vector spaces over a field, giving a new way to view linear systems of equations. For a finitely presented mod-

ule M, if M = (m1,..., mn) and K = (v1,..., vm) and we write the vi as the rows of an m×n matrix

78 A, then M is uniquely determined by A, i.e.

    m1     m2 A   = 0. (6.1)  .   .      mn

m A n n Since M is isomorphic to the cokernel of the linear mapping R −→ R , i.e. M R /(v1,..., vm). We call A the relation matrix, and later we will give a description of when two matrices present isomorphic R-modules.

Proposition 6.2. Suppose that the sequence of R-modules, with P ﬁnitely presentable,

0 −→ M −→ N −→ P → 0 is exact. Then the sequence splits if and only if

0 −→ Mp −→ Np −→ Pp → 0 splits for all p ∈ Spec(R). Moreover, the sequence splits if and only if the corresponding sequences split for all m ∈ Max(R).

Proof. Since P is ﬁnitely presentable, we may assume that M is ﬁnitely generated. The sequence splits if and only if φ : HomR(P, N) → HomR(P, P) is surjective.

Now φ is surjective if and only if φp : HomR(P, N)p → HomR(P, P)p is surjective for all p ∈ Spec(R). By the lemma 6.1 , φp is surjective if and only if φp : HomRp (Pp, Np) → HomRp (Pp, Pp).

Yet, φp is surjective if and only if 0 −→ Mp −→ Np −→ Pp → 0 splits. The same argument holds for m ∈ Max(R) in place of p ∈ Spec(R).

79 Corollary 6.1. Let M be a finitely presentable R-module, U ⊂ M a finitely generated submodule. Then M/U is also finitely presentable. U is a direct summand if and only if Um is a direct summand of Mm for all m ∈ Max(R).

Proof. Since M is ﬁnitely presentable, there is an exact sequence

f g 0 −→ K −→ Rn −→ M → 0 with K a ﬁnitely generated R-module. Let φ : M → M/U be the canonical map, and let g0 = φ ◦ g.

By exactness, U g−1(U)/K. Since ker g0 = g−1(U) and U and K are finitely generated, ker g0 is g0 finitely generated. We have the exact sequence 0 → ker g0 → Rn −→ M/U → 0, so M/U is finitely presentable. Now 0 → U → M → M/U → 0 is exact, and U is a direct summand if and only if this sequence splits. Yet, since M/U is finitely presentable, by proposition 6.2 the sequence splits if and only if

0 → Um → Mm → (M/U)m → 0 splits for all m ∈ Max(R). And since 0 → Um → Mm →

(M/U)m → 0 splits for all m ∈ Max(R) if and only if Um is a direct summand for all m ∈ Max(R), the second assertion follows.

We have even more local-global principles.

Proposition 6.3. Let M be an R-module. If P, Q are submodules of M, then P = Q if and only if

Pm = Qm for all m ∈ Max(R).

Proof. The ( =⇒ ) direction is clear.

By corollary 3.2, ((P + Q)/Q)m = (Pm + Qm)/Qm and ((P + Q)/P)m = (Pm + Qm)/Pm. Now if

Pm = Qm for all m ∈ Max(R) then (Pm + Qm)/Qm = (0) = (Pm + Qm)/Pm so by proposition 3.6

(P + Q)/Q = (0) = (P + Q)/P and thus P = Q.

In particular, if f, g ∈ R are such that X f ∪ Xg ∈ Spec(R) and M f is ﬁnitely generated as an

R f -module while Mg is ﬁnitely generated as an Rg-module, then M is ﬁnitely generated as an R-

80 module. For if the images of x1,..., xm ∈ M generate M f and y1 ..., yn ∈ M generate Mg then

let N = (x1,..., xm, y1,..., yn). If m ∈ Max(R) then m ∈ X f or m ∈ Xg so Mm = Nm for

all m ∈ Max(R). Hence, M = (x1,..., xm, y1,..., yn). We have an analogous result for ﬁnitely presented modules.

Proposition 6.4. For f, g ∈ R with X f ∪ Xg = Spec(R) let M f be ﬁnitely presentable as an

R f -module and Mg be ﬁnitely presentable as an Rg-module. Then M is a ﬁnitely presentable R- module.

˜ n α˜ ˜ Proof. We have an exact sequence of R f -modules 0 → K → R f −→ M f → 0 with K ﬁnitely n generated. Let {e1/1,..., en/1} be a basis for R f , and let φ : M → M f be the canonical map. n For each k = 1,..., n there are mk ∈ M and sk ∈ R \{ f }n≥1 such thatα ˜(ek/1) = mk/sk. Then

s = s1 s2 ··· sk is such that sα˜(ek/1) ∈ im φ for each k = 1,..., n, i.e. there are xk ∈ M such that

n sα˜(ek) = xk/1. Thusα ˜ induces the R-linear map α : R → M with α(ek) = xk. w α In other words, there is an exact sequence of R-modules 0 → K −→ F −→ M where F is a ﬁnite

w f α f rank free module and the induced sequence of R f -modules 0 → K f −−→ F f −−→ M f → 0 is exact 0 0 0 w 0 α with K f ﬁnitely generated. Let 0 → K −→ F −→ M be the corresponding sequence for Mg. Let

0 0 0 0 α ⊕ −α : F ⊕ F → M be the map acting like α on F and −α on F . Since Spec(R) = X f ∪ Xg we

0 0 have (α ⊕ −α )m surjective for all m ∈ Max(R), so by proposition 3.7 α ⊕ α is surjective and

α⊕α0 0 → ker α ⊕ −α0 → F ⊕ F0 −−−→ M → 0

is exact. Let U = ker α ⊕ −α0. It then suﬃces to show that U is ﬁnitely generated.

0 0 Since α f is surjective, given bases B and B for F f and F f respectively, we can deﬁne a map φ 0 0 which satisﬁes (α f ◦ φ)(x) = α f (x) on the basis elements x ∈ B . By extending linearly, we 0 0 0 get a homomorphism φ : F f → F f with α f ◦ φ = α f . Let β : F f ⊕ F f → F f be given by 0 0 0 β(x, y) = x − φ(y). Then (α f ◦ β)(x, y) = α f (x) − α f (y) = (α f ⊕ −α f )(x, y) for all (x, y) ∈ F f ⊕ F f .

0 −1 −1 Thus U f = ker α f ⊕ −α f = β (ker α f ) = β (im w f ). Then x ∈ U f if and only if β(x) = w f (y)

81 0 0 for some y ∈ K f , so β induces β : U f → K f with β (x) = y, which is well-deﬁned since w f is injective. Moreover, ker β = ker β0. Let G = ker β = ker β0, then we have a commutative diagram

0 0

G G

0 0 U f F f ⊕ F f M f 0

β0 β

0 K f F f M f 0

0 0

with exact rows and columns. Since F f is a free R f -module, hence projective, by proposition 6.1

0 the middle column splits, and so G is a direct summand of, and hence an image of, F f ⊕ F f .

By proposition 1.8, G is finitely generated. Since K f is finitely generated with K f U f /G, we must have U f finitely generated. By a similar argument, Ug is finitely generated, and hence U is finitely generated.

We now want to return to our problem of studying when two matrices present isomorphic R- modules. Thus, our approach is to compare presentations of isomorphic modules. Let, for j = 1, 2

g j f j 0 → K j −→ F j −→ M j → 0

be an exact sequence of R-modules, where the F j are free.

Proposition 6.5. For sequences of R-modules as above:

1) If there is an isomorphism i : M1 → M2, then there exists an f ∈ Aut(F1 ⊕ F2) = { f ∈

82 End(F1 ⊕ F2): f is an isomorphism } such that the diagram

f1⊕0 F1 ⊕ F2 M1

f i

0⊕ f2 F1 ⊕ F2 M2

commutes. We have f (F1 ⊕ F2) = F1 ⊕ K2 we identify K j with g j(K j) ⊂ F j.

2) If there is an f ∈ Aut(F1 ⊕ F2) with f (K1 ⊕ F2) = F1 ⊕ K2, then there is an isomorphism

i : M1 → M2 such that the diagram above is commutative.

Proof.

1) Since the F j are free, given an isomorphism i : M1 → M2, by deﬁning on their respective

(canonical) bases and extending linearly, we can obtain R-linear mappings h1 : F1 → F2 and

0 00 h2 : F2 → F1 satisfying i ◦ f1 = f2 ◦ h1 and f2 = i ◦ f1 ◦ h2. Now, deﬁne f , f : F1 ⊕ F2 → F1 ⊕ F2

0 00 0 00 by f (x, y) = (x, y − h1(x)) and f (x, y) = (x − h2(y), y). Clearly, f and f are bijective, and thus

0 00 0−1 00 f , f ∈ Aut(F1 ⊕ F2). Let f = f ◦ f . Since

00 (i f1 ⊕ f2)( f (x, y)) = i f1(x) + f2(y) − i f1h2(x) = i f1(x) = (i ◦ f1 ⊕ 0)(x, y) and

0 (i f1 ⊕ f2)( f (x, y)) = i f1(x) + f2(y) − f2h1(x) = f2(x) = (0 ⊕ f2)(x, y)

00 0 we have (0 ⊕ f2) ◦ f = i ◦ ( f1 ⊕ 0) as f = f ◦ f . Thus, the diagram commutes.

Now, by exactness ker f1 ⊕ 0 = K1 ⊕ F2 and ker 0 ⊕ f2 = F1 ⊕ K2. Since, (0 ⊕ f2) f (ker f1 ⊕ 0) =

(i ◦ ( f1 ⊕ 0))(ker f1 ⊕ 0) = i(0) = 0 we must have f (K1 ⊕ F2) = ker 0 ⊕ f2 = F1 ⊕ K2.

0 00 2) Let g = g1 ⊕ 1F2 : K1 ⊕ F2 → F1 ⊕ F2 and g = 1F1 ⊕ g2 : F1 ⊕ K2 → F1 ⊕ F2, then

0 00 M1 = Coker g while M2 = Coker g . Since F1 ⊕ F2 is free, we can ﬁnd a linear map i : M1 → M2

00 0 0 00 with g ◦ α = i ◦ g . Since f (K1 ⊕ F2) F1 ⊕ K2, and im g = K1 ⊕ F2, im g = F1 ⊕ K2 we have

83 a commutative diagram with exact rows

0 0 g 0 0 im g F1 ⊕ F2 Coker(g ) 0

f f i

00 00 g 00 0 im g F1 ⊕ F2 Coker(g ) 0

and since f is an isomorphism, so too is i.

Corollary 6.2. Let there be given two exact sequences, for j = 1, 2,

g f 0 j j F j −→ F j −→ M j → 0

0 with free R-modules F j, F j. Then M1 M2 if and only if there is an f ∈ Aut(F1 ⊕ F2) and 0 0 g ∈ Aut(F1 ⊕ F2 ⊕ F1 ⊕ F2) such that the diagram

(g ⊕1 ,0) 0 0 1 F2 F1 ⊕ F2 ⊕ F1 ⊕ F2 F1 ⊕ F2

g f (0,1 ⊕g ) 0 0 F1 2 F1 ⊕ F2 ⊕ F1 ⊕ F2 F1 ⊕ F2

⊕ 0 0 commutes. Here, (g1 1F2 , 0) acts like g1 on F1, 1 on F2, and 0 on F1 and F2. Similarly for

(0, 1F1 ⊕ g2).

Proof. Since M1 Coker (g1 ⊕ 1F2 , 0), and M2 Coker(0, 1F1 ⊕ g2), if such a diagram is given,

M1 M2.

Suppose M1 M2. Let K j = ker f j, then by proposition 6.5 part 1), there is an f ∈ Aut(F1 ⊕ F2) and an isomorphism f 0 such that

0 g1⊕1 F1 ⊕ F2 K1 ⊕ F2 F1 ⊕ F2

f 0 f

0 1⊕g2 F1 ⊕ F2 F1 ⊕ K2 F1 ⊕ F2

84 commutes. Applying proposition 6.6 part 1) to f 0 now gives the desired result.

We will denote by Mn×m(R) the R-module consisting of all n × m matrices whose coeﬃcients are in R. Let Gln(R) denote the group of invertible n × n matrices in R, i.e. the matrices whose

0 determinant is a unit of R. Every ring homomorphism f : R → R induces a mapping Mn×m(R) →

0 0 Mn×m(R ) and a group homomorphism Gln(R) → Gln(R ) by sending the coeﬃcients ai j to f (ai j).

Deﬁnition 6.4. We say two matrices A1, A2 ∈ Mn×m(R) are equivalent if there is an A ∈ Gln(R) and a B ∈ Glm(R) such that

−1 A1 = A · A2 · B

and we denote this equivalence by A1 ∼ A2.

Let there be given two exact sequences,

g f 0 j j F j −→ F j −→ M j → 0

0 n j 0 n j 0 where F j = R and F j = R , with n j, n j ∈ N, for each j = 1, 2. 0 0 With respect to the canonical basis, g1 is given by an n1×n1 matrix B1 and g2 is given by an n2×n2 matrix B2. With (g1 ⊕ 1F2 , 0) and (0, 1F1 ⊕ g2) deﬁned as in corollary 6.2, these maps correspond to n × m matrices of the form

    B 0   0 0   1        B˜ =   and B˜ =   1  0 In2  2 In1 0           0 0   0 B2

0 0 where n = n1 + n2 + n1 + n2 and m = n1 + n2. For f and g deﬁned as in corollary 6.2, let g and f be represented, with respect to the canonical basis, by matrices G ∈ Glm(R) and F ∈ Gln(R) respectively. Then, formulated in these terms, corollary 6.2 says that M1 M2 if and only if

−1 B˜1 = G · B˜2 · F . Thus, B1 and B2 present the same R-module if and only if B˜1 ∼ B˜2.

85 6.2 Quillen’s Local-Global Principle

We will now turn our attention to matrices with coeﬃcients in the polynomial ring R[x]. Given

A ∈ Mn×m(R[x]) and some element y ∈ T of an R-algebra T, let A(y) denote the image of A in

Mn×m(T) under the homomorphism X 7→ y. For example, A(0) ∈ Mn×m(R) is the matrix obtained from A by replacing x with 0 in each coeﬃcient of A.

Deﬁnition 6.5. We call matrices A1, A2 ∈ Mn×m(R[x]) locally equivalent for some m ∈ Max(R) if the images of A1, A2 in Mn×m(Rm[x]) are equivalent.

We can establish a local-global principle for the equivalence of matrices over R[x] due to Vaser- stein, originally contained in a letter to Bass. For more information, see Lam [10].

Lemma 6.2. Let there be given matrices A1 ∈ Mn×m(R[x]), A2 ∈ Mm×r(R[x]) and A3 ∈ Mn×r(R[x]).

Let S be a multiplicatively closed subset S ⊂ R. Let Ai (i = 1, 2, 3) be the matrix corresponding

−1 to Ai under the canonical map R[x] → S R[x]. Suppose that A1 · A2 = A3 and A1(0) · A2(0) =

A3(0). Then there is an s ∈ S such that A1(sx) · A2(sx) = A3(sx).

Proof. Since A1(0) · A2(0) = A3(0) all the coeﬃcients of A1A2 − A3 have a constant term of

0. Moreover, A1A2 − A3 is mapped to zero under the canonical homomorphism Mn×r(R[x]) →

−1 Mn×r((S R)[x]). Hence, there is an s ∈ S which lies in the annihilator of all the coeﬃcients of

the matrix A1A2 − A3, and since the constant terms are all zero this gives

A1(sx) · A2(sx) − A3(sx) = 0.

Theorem 6.1. A ∈ Mn×m(R[x]) is equivalent to A(0) if and only if A is locally equivalent to A(0) for all m ∈ Max(R).

Proof. Suppose A and A(0) are locally equivalent for all m ∈ Max(R).

86 Let I be the set of all a ∈ R with the property that for all f, g ∈ R[x] with f − g ∈ aR[x] we must have A( f ) ∼ A(g) (in the “global” sense).

We claim that I is an ideal of R. Let a1, a2 ∈ R, then if f, g ∈ R[x] are such that f − g = (r1a2 +

r2a2)φ for some φ ∈ R[x] and r1, r2 ∈ R we must have ( f − r1a1φ) − g = r2a2φ ∈ a2R[x] and thus

A(g) ∼ A( f − r1a1φ) ∼ A( f ) by the deﬁning property of I. Thus a1 + a2 ∈ I. It is obvious that ra ∈ I for all r ∈ R and a ∈ I. Thus, it suﬃces to show that 1 ∈ I for then A( f ) ∼ A(g) for all f, g ∈ R[x], and thus A(x) ∼ A(0). We use the familiar method of adding an extra variable. With another indeterminate y, we get

A(x + y) = C(x + y)A(0)D(x + y) = C(x + y)C(x)−1A(x)D(x)−1D(x + y).

Let C∗ = C(x + y)C(x)−1 and D∗ = D(x)−1D(x + y). Now we can write C∗ as

∗ m C = C0(x) + C1(x)y + ··· + Cm(x)y

∗ ∗ −1 ∗ −1 for some Ci ∈ Mn×n(Rm[x]) where C0 is the unit matrix. Similarly for D , (C ) , and (D ) . Now by collecting denominators of all the coeﬃcents of the matrix, we can obtain an a0 ∈ R \ m

0 −1 such that C(x + a y)C(x) is the image of a matrix in Mn×n(R[x, y]) under the canonical homomorphism. Moreover, a0 can be chosen such that the substitution y 7→ a0y leaves D∗, (C∗)−1, and

∗ −1 (D ) the images of matrices with coeﬃcients in R[x, y]. Let Γ(x, y) ∈ Mn×n(R[x, y]) be such that

0 −1 0 Γ(x, 0) is the unit matrix and Γ(x, y) = C(x + a y)C(x) and let Γ (x, y) ∈ Mn×n(R[x, y]) be such that Γ0(x, 0) is the unit matrix and Γ0(x, y) = C(x)C(x + a0y)−1. Our previous comments guarantee the existence of such matrices. Then

Γ(x, 0) · Γ0(x, 0) = 1

87 and Γ(x, y) · Γ0(x, y) = 1.

Thus, by lemma 6.2 Γ(x, y)·Γ0(x, y) = 1 and so we may assume C(x+a0y)C(x)−1 is the image of an invertible matrix Γ(x, y) with Γ(x, 0) the unit matrix. Similarly we may assume D(x)−1D(x + a0y) is the image of an invertible matrix ∆(x, y) with ∆(x, 0) the unit matrix.

Now, over Rm[x, y] we have

A(x + a0y) = C(x + a0y)c(x)−1A(x)D(x)−1D(x + a0y) and over R[x] A(x) = Γ(x, 0)A(x)∆(x, 0).

Hence, by lemma 6.2 there is an a00 ∈ R \ m such that, for a = a0a00, we have

A(x + ay) = Γ(x, a00y)A(x)∆(x, a00y) over R[x, y]. If f, g, φ ∈ R[x] are given with f − g = aφ then

A( f ) = A(g + aφ) = Γ(g, a00φ)A(g)∆(g, a00φ) so that A( f ) ∼ A(g). Hence, a ∈ I, but since a0, a00 < m we must have a < m. Since m was arbitrary, we must have I = R.

The implication in the forward direction is obvious.

For an R-module N, let N[x] denote the the set of all polynomials in x with coeﬃcients in N. Here, two polynomials are added term-wise, and the R-module structure of N naturally induces

P j P j an R[x]-module structure on N. Deﬁne φ : N[x] → R[x] ⊗R N by j n j x 7→ j(x ⊗ n j). If

88 P i P j f (x) = i ai x ∈ R[x] and j n j x = g(x), then

X X k X X k X X i j X i X j φ( f (x)·g(x)) = φ(ain j x ) = (x ⊗ain j) = (x x ⊗ain j) = (ai x )⊗ (n j x ) k i+ j=k k i+ j=k i j i j

= f (x)φ(g(x)).

P j P j Hence, φ is R-linear. Deﬁne ψ : R[x] × N → N[x] by ψ( j a j x , n) = j a jnx , so that ψ is clearly

P j R-bilinear. Thus, there is an induced linear map Ψ : R[x] ⊗ N → N[x] with Ψ( j a j x ⊗ n) =

P j j a jnx . Clearly, Ψ ◦ φ = 1 and φ ◦ Ψ = 1, so that N[X] R[x] ⊗ N. We call N[x] the extension module of N to R[x].

Deﬁnition 6.6. An R[x]-module M is called extended (from R) if there is an R-module N such that M N[x] as R[x]-modules. M is called locally extended for an m ∈ Max(R) if the Rm[x]- module Mm is extended from Rm.

If M N[x] then M/xM N as R-modules. For we have, making use of the fact that N[x]

N ⊗R R[x] and proposition 1.14,

M/xM (N⊗RR[x])⊗R[x]R[x]/xR[x] N⊗R(R[x]⊗R[x]R[x]/xR[x]) N⊗RR[x]/xR[x] N⊗RR N.

The following local-global principle for extended modules is a key step in Quillen’s proof resolving Serre’s problem.

Theorem 6.2. Quillen’s Local-Global Principle A ﬁnitely presentable R[x]-module M is extended if and only if it is locally extended for all m ∈ Max(R).

Proof. Since M is ﬁnitely presentable there is an exact sequence of R[x] modules

g f R[x]m −→1 R[x]n −→1 M → 0

89 and so taking quotients we get an exact sequence

g f Rm −→1 Rn −→1 M/xM → 0.

Let B ∈ Mn×m(R[x]) be the matrix of g1 with respect to the canonical basis, so that the matrix

B(0) corresponds to g1. The same can be done for f1. Let N M/(x)M, then we have an exact sequence g [x] f [x] Rm[x] −−−→1 Rn[x] −−−→1 N[x] → 0

m n where g1[x] maps x to x and maps polynomials in R [x] into R [x] by applying g1 to the coeﬃ-

m m n cients in R . Similarly for f1[x]. Thus, if g2 : R[x] → R[x] is the map described by B(0), and

similarly for f2, we have an exact sequence of R[x]-modules

g f R[x]m −→2 R[x]n −→2 N[x] → 0.

Now by corollary 6.2 we have M N[x] if and only if the (2m) × 2(n + m)-matrices

    B 0 0 0        0   A = 0 I  and A = I 0   n  n      0 0 0 B(0)

By performing elementary row and column operations, which will respectively correspond to multiplication on the left and right by an invertible matrix, it’s easy to see that A0 ∼ A(0). Hence, by theorem 6.1, M N[x] if and only if A and A(0) are locally equivalent for every m ∈ Max(R).

Since R[x]m Rm[x] and localization at m preserves the above exact sequences, we must have M extended if and only if it is locally extended for all m ∈ Max(R).

90 6.3 Projective Modules and Serre’s Problem

We easily have the following as a consequence of proposition 6.1

Proposition 6.6. Let M be a finitely generated projective R-module. Then there is a finitely generated R-module M0 such that M ⊕ M0 is free. In particular, M is finitely presentable.

Proof. Let 0 → M0 → F → M → 0 be an exact sequence with F a finite rank free R-module. By proposition 6.1, the sequence splits, and so F M ⊕ M0. Since M0 is an image of F, it is finitely generated. It follows that M is finitely presentable.

In what follows, we will need to know under what conditions a projective module is free. We will ﬁrst show this problem is trivial over local rings, i.e. every projective module is free.

Proposition 6.7. Let R be a local ring with maximal ideal m. If M is a ﬁnitely presentable R- module, then the following are equivalent.

1) M is free.

f g 2) There is an exact sequence of R-modules 0 → K −→ P −→ M → 0, where P is projective and the map K/mK → P/mP induced by f is injective.

Proof. If M is free, let P = M and K = (0). f g Suppose there is an exact sequence 0 → K −→ P −→ M → 0 where P is projective and the

f0 induced map K/mK → P/mP is an injection. If µ(M) = r there is an exact sequence 0 → K0 −→

g0 F0 −→ M → 0 where F0 is a free R-module of rank r. Since P is projective and g0 is surjective, by proposition 6.2(c) there is an R-linear h : P → F0 with g = g0 ◦ h. Under the induced maps,

91 where the map induced by h for example takes x + mP to h(x) + mF0, the diagram

P/mP

M/mM

F0/mF0

commutes. Since F0/mF0 and M/mM are both R/m-vector spaces of dimension r, the map in-

duced by g0 is bijective. From the commutativity of the diagram, the map induced by h is neces-

sarily surjective, so that F0 = h(P) + mF0. By Nakayama’s lemma, F0 = h(P), i.e. h is surjective. Now if we identify K with it’s image under f , we have x ∈ K if and only if g(x) = 0. But then

g(x) = g0(h(x)) = 0 so h(x) ∈ im f0, i.e. there is a y ∈ K0 such that h(x) = f0(y). Thus h induces

0 0 0 an h : K → K0 where h (x) = y, where y is as above. It’s clear that ker h = ker h . Let F1 = ker h. We have a commutative diagram with exact rows and columns

0 0

F1 F1

f g 0 K P M 0

h0 h

f0 g0 0 K0 F0 M 0

0 0

Because M is ﬁnitely presentable, we have an exact sequence 0 → K1 → F → M → 0 where

F is a ﬁnite rank free module and K1 is ﬁnitely generated. The argument above applies to this sequence, so we can substitute it for 0 → K → P → M → 0. The middle column still splits, so

we have F1 ﬁnitely generated. Then K0 K1/F1 is ﬁnitely generated.

92 Consider the original diagram modulo m.

F1/mF1 F1/mF1

0 K/mK P/mP M/mM 0

K0/mK0 F0/mF0 M/mM 0

0 0

The middle row is exact by hypothesis, and because F0 is free the sequence 0 → F1 → P →

F0 → 0 splits. Thus the ﬁnitely generated module F1 is a direct summand of P, and so as an

R/m-vector space F1/mF1 a direct summand of P/mP. Thus the middle column is exact. Thus

we have a commutative diagram with exact rows and columns. Moreover, K0/mK0 → F0/mF0 is injective.

Since F0/mF0 → M/mM is an isomorphism, by exactness we must have K0/mK0 = (0) Since K0

is ﬁnitely generated, Nakayama’s lemma gives K0 = (0).

Corollary 6.3. A ﬁnitely generated module over a local ring is projective if and only if it is free.

Proof. If M is projective take P to be M and K = (0) in the exact sequence 0 → K → P → M →

0 of proposition 6.7(b).

Corollary 6.4. For a ﬁnitely generated module M over an arbitrary ring R the following statements are equivalent

1) M is projective.

2) M is ﬁnitely presentable and locally free.

93 Proof. If M is projective, then by proposition 6.6 it is finitely presentable. Since Rm is a local ring for all m ∈ Max(R), it is locally free by corollary 6.3. If M is finitely presentable and locally free, there is an exact sequence 0 → K → F → M → 0 with F a finite rank free R-module and K a finitely generated R-module. The sequence splits locally for every m ∈ Max(R), so by proposition 6.2 it splits globally. Thus M is a direct summand of the free module F, and so M is projective.

A key step in the proof of the Serre Conjecture will be the following monic inversion principle. Horrocks established this result for local rings, while the work of Quillen and Suslin extended the result to arbitrary ground rings.

Theorem 6.3. Horrocks’ Theorem Let R be a local ring with maximal ideal m, M a ﬁnitely

generated projective R[x]-module. Suppose there is a monic polynomial f ∈ R[x] such that M f is

a free R[x] f -module. Then M is a free R[x]-module.

Proof. a) M/F is ﬁnitely presentable.

n Since { f }n≥0 has no zero-divisors, we can view M as a submodule of M f and multiply any basis

for M f by a common denominator. Choose a basis E for the free R[x] f -module M f which consists of elements of M, we can do so after clearing denominators. Let F be the submodule of M spanned by F. Now, let P = M/F, so that P f = M f /F f = 0 and thus there is an n ∈ N such that f n annihilates P. Moreover

P (M/F)/( f n(M/F)) (M/ f n M) ((F + f n M)/ f n M)

and M/ f n M is a finitely generated projective module over S = R[x]/( f n). If deg( f ) = t, then B = {1,..., xt−1} generates S as an R-module, i.e. S is a free R-module of finite rank. Thus, M/ f n M is a free R-module of finite rank. Since (F + f n M)/ f n M is finitely generated as an R-module, and

94 we have an exact sequence 0 → (F + f n M)/ f n M → M/ f n M → P → 0 we must have P ﬁnitely presentable. b) M is the direct sum of P and F. ¯ If f is the image of f in (R/m)[x], then clearly (F/mF) f¯ = (M/mM) f¯ and the canonical map F/mF → M/mM is injective. Since R[x] is a free R-module, M is also a projective R-module. Now 0 → F → M → P → 0 is exact, so by proposition 6.7 P is a free R-module and so

M P ⊕ F. c) Determine the relation matrix for M.

Let p1,..., ps ∈ M be such that their images in P form a basis for P as a free R-module, and let

{ps+1,..., pt} be an R[x]-basis for F. Since every element of M can be written as a sum of ele-

ments of P and F, for each k = 1,..., s there exist αki ∈ R and bk j ∈ R[x] such that

Xs Xt −xpk = αki pi + bk j p j. i=1 j=s+1

Ps P Whenever we are given a relation i=1 ai pi + b j p j = 0 with ai, b j ∈ R[x] we can use the equations above to obtain αi ∈ R and b˜j ∈ R[x] with

Xs Xt αi pi + b˜j p j = 0. i=1 j=s+1

But M P ⊕ F so that αi = b˜j = 0.

Ps Pt From the equations 0 = i=1 αki pi + xpk + j=s+1 bk j p j, if A is the s × s matrix whose coeﬃcients

95 are the αki and B is the s × t − s matrix whose coeﬃcients are the bk j, then we must have

  p   1   p   2 (A + xIs | B)   = 0  .   .       pt 

Hence, the R[x]-relation matrix of M relative to the generating system {p1,..., pt} is given by

(A + xIs | B). d) Express the relation matrix for M with coeﬃcients in R.

We can use the division algorithm in polynomial rings to obtain B0 ∈ Ms×(t−s)(R) and B˜ ∈ Ms×(t−s)(R[x])

with B = B0 + (A + xI)B˜. Thus, the relations can be written in the form

      p  p  p   1  s+1  s+1       p  p  p   2 ˜  s+2  s+2 (A + xIs) ·   + B   + B0   = 0.  .   .   .   .   .   .              ps  pt   pt 

And so by replacing the pi with suitable linear combinations of the p j such that

    p  p   1  s+1     p  p   2 ˜  s+2   + B   ∈ Ms×1(R)  .   .   .   .          ps  pt 

we can assume without loss that the matrix B only has coeﬃcients in R. e) Enlarge F at the expense of P until no more of P remains.

We claim that the ideal in R[x] generated by the s × s minors of (A + xIs | B) is all of R[x]. By

96 proposition 6.3, it is enough to show this locally for all m ∈ Max(R). By corollary 6.4, Mm is

a free R[x]m-module, and since (Mm) f = (Fm) f the rank of Mm is the same as the rank of Fm,

t namely t− s. Since Mm is free the exact sequence 0 → K → R[x]m → Mm → 0, with K the R[x]m-

submodule spanned by the rows of the relation matrix, splits. Thus, K is a free R[x]m-module of

t rank s. It follows that the rows of the relation matrix can be extended to a basis for R[x]m, and so

at least one s × s minor must be a unit in R[x]m.

Thus if g = det(A + xIs) and I is the ideal in R generated by the coefficients of B, then over R[x] we have (g, I) = (1) = R[x]. Since g is a monic polynomial the ring T = R[x]/(g) is a free R- module. Since T = T · I, we have I = R. Since R is a local ring, this implies that B contains a coefficient which is a unit in R. By elementary row and column operations, bring the relation matrix into the form    0      0 A0 + xI | B0   s−1 .  .      0      0 0 ··· 0 1 where A0, B0 have coefficients in R. Since we have only made elementary operations, by the argu-

0 0 ment above the ideal in R[x] generated by the (s − 1) × (s − 1) minors of (A + xIs−1 | B ) is all of

R[x]. Repeating this process, we can bring the relation matrix into the form (0 | Is). This implies that M is a free R[x]-module of rank t − s.

To extend Horrocks’ theorem to arbitrary ground rings, we will make some use of the construc-

tion called the ﬁber product to “glue” modules. The idea is that if Spec(R) = X f ∪ Xg we can

obtain global information from the local information on X f and Xg so long as this information is

compatible at the boundary X f g.

Given R-modules M1, M2,N and homomorphisms αi : Mi → N (i = 1, 2), then the “ﬁber product’

consists of the R-module P = M1ΠN M2 with homomorphisms βi : P → Mi (i = 1, 2), such that

97 the diagram β1 M1ΠN M2 M1

β2 α1

α2 M2 N

commutes. All we will need is the following proposition.

Proposition 6.8. Let f, g ∈ R. Let M1 be an R f -module, M2 an Rg-module, and suppose there is an isomorphism of R f g-modules α :(M1)g → (M2) f .

Let N = (M2) f . Let α1 be the composition of the canonical map M → (M1)g with α. Let α2 :

M2 → (M2) f be the canonical map.

If P = M1ΠN M2 is the ﬁber product formed with respect to α1,α2 then the canonical mappings ∼ ∼ βi : P → Mi (i = 1, 2) induce isomorphisms P f −→ M1, Pg −→ M2 (as R f and Rg modules, respectively). We say that P arises through gluing M1 and M2 over X f g with respect to α (where as usual X = Spec(R)).

For the details of the proof of proposition 6.8 , consult [9, pgs. 88 − 91]. We now give the extension to arbitrary ground rings due to Quillen and Suslin.

Theorem 6.4. If M is a ﬁnitely generated projective R[x]-module, f ∈ R[x] is a monic polynomial such that M f is a free R[x] f -module, then M is a free R[x]-module.

Proof. a) M is extended.

For m ∈ Max(R), Mm is a ﬁnitely generated projective Rm[x]-module, for which (Mm) f is a free

(Rm) f [x]-module. Thus, by Horrocks’ theorem, Mm is a free Rm[x]-module. Hence, Mm is extended. By Quillen’s local-global principle, M is globally extended.

Thus, there is an R-module N with N[x] M. Moreover, N M/xM and N[x]/(x − 1)N[x] N so it suﬃces to show that M/(x − 1)M is a free R-module.

98 b) Extension of M “into projective space”.

−1 n −1 −1 Consider the ring R[x ], then if S = {x }n≥0 we have S R[x] = R[x]x R[x, x ], the ring of

−n −1 −1 −1 −1 Laurent polynomials. If T = {x }n≥0 then T R[x ] = R[x ]x−1 R[x]x via the map x 7→ 1/x.

−1 Thus, the open set Xx ⊂ Spec(R[x]) is identiﬁed with the open set Xx−1 ⊂ Spec(R[x ]). The space which arises from this identiﬁcation is called the “projective line” over R.

n n−1 −n −(n−1) −1 Let f = x + an−1 x + ··· + a0 and g = a0 x a1 x + ··· + a1 x + 1 for some ai ∈ R.

−n −n −1 −1 −1 Now g = x f and x is a unit of R[x, x ]. We must then have R[x, x ] f R[x, x ]g, and so

−1 −1 (Mx) f (Mx)g is a free R[x, x ]g-module, as M f is a free R[x] f -module. Now (x , g) = (1) in

−1 −1 R[x ] so that Spec(R[x ]) = Xx−1 ∪ Xg.

0 φ M Mx

0 (Mx)g Mg

−1 −1 By gluing Mx to the free R[x ]g-module (Mx)g over the boundary Xgx−1 , we can get an R[x ]-

0 0 −1 module M for which Mg is a free R[x ]g-module of the same rank as (Mx)g. 0 → 0 → Moreover, by proposition 6.8, φ : M Mx is such that the induced map Mx−1 Mx is an 0 −1 0 0 0 isomorphism, so Mx−1 Mx as R[x, x ]-modules. Since Mx−1 and Mg are free, we must have M ﬁnitely presentable by proposition 6.4. c) M is free.

0 −1 −1 Given m ∈ Max(R), we have (Mm)x−1 (Mm)x a free module over the ring Rm[x, x ]. As x is

−1 0 −1 monic in R[x ] by Horrocks’ theorem Mm is a free Rm[x ]-module, and thus extended over Rm. Because M0 is ﬁnitely presentable and locally extended, Quillen’s local global principle guaran-

tees that M0 is extended. Thus, M0 N0[x−1] for some R-module N0 with N0 M0/x−1 M0 M0/(x−1 − 1)M0.

0 −1 −1 0 0 Since Mg is a free R[x ]g-module with g − 1 ∈ (x ) and the natural map M → Mg induces

0 −1 0 0 −1 0 0 −1 0 an isomorphism Mg/x Mg M /x M , we must have M /(x − 1)M a free R-module. Yet,

99 − − 0 −1 − 0 0 − 0 − M/(x 1)M Mx/(x 1)Mx Mx−1 /(x 1)Mx−1 M /(x 1)M so that M/(x 1)M is a free R-module, and we’re done.

We are now in a position to prove Serre’s conjecture

Theorem 6.5. Serre’s Conjecture If K is a principal ideal domain, then all ﬁnitely generated

projective K[x1,..., xn]-modules are free

Proof. For n = 0 the assertion holds since submodules of a ﬁnite rank free module over a principal ideal domain are free. Let n > 0, and suppose the statement has been shown for polynomial rings in n − 1 variables.

Let M be a ﬁnitely generated projective module over K[x1,..., xn] and let S be the multiplica-

−1 tively closed subset consisting of all the monic polynomials in K[x1]. Now, S M is a projective

−1 −1 module over S K[x1,..., xn] = (S K[x1])[x2,..., xn]

−1 It then suﬃces to show that S K[x1] is a ring in which every ideal is principal, for then by the

−1 induction hypothesis Ms is a free S K[x1,..., xn]-module and thus there is an f ∈ S with M f

a free K[x1,..., xn] f -module. From this fact, theorem 6.4 shows that M is a free K[x1,..., xn] module.

−1 Since K[x1] is a unique factorization domain, its localization R = S K[x1] is a unique factoriza-

tion domain. The dimension of K[x1] is one. By proposition 3.10 the prime ideals of R are in one

to one correspondence with the prime ideals p of K[x1] with p ∩ S = ∅. Hence, the longest possi-

−1 ble chain of prime ideals in R is (0) ⊂ S p where p ∈ Spec(K[x1]). In other words dim(R) ≤ 1.

−1 Let p ∈ K be prime, then p = (p) ∈ Spec(K[x1]) with and p ∩ S = ∅, so that (0) ( S p is a chain of prime ideals in R and dim R ≥ 1. Hence R is a unique factorization domain of dimension one, and is thus a principal ideal domain.

6.4 The Eisenbud-Evans Conjectures

We will use the following lemma.

100 Lemma 6.3. Let X be the spectrum or J-spectrum of R, and suppose that X is Noetherian. For

elements m1,..., mt ∈ M and all p ∈ X suppose that

µp(M) − µp(M/hm1,..., mti) ≥ min t, dim V(p) + 1.

Then there are only ﬁnitely many p ∈ X with both dim V(p)+1 < t and µp(M)−µp(M/hm1,..., mti) = dim V(p) + 1.

Proof. Let U = (m1,..., mt). Let p ∈ X with dim V(p)+1 < t and µp(M)−µp(M/U) = dim V(p)+1 be given. Let r = µp(M/U), then it’s enough to show that p is minimal in the set Ar(M/U) = {q ∈

X : µq(M/U) ≥ r}. For if this were the case, since an easy consequence of lemma 5.1(e) that

Ar(M/U) is a closed set in X, and thus has only ﬁnitely minimal elements since X is Noetherian.

Then, since there are only ﬁnitely many distinct Ar(M/U) we would have the desired result.

Suppose there exists a q ∈ Ar(M/U) with q ( p. Then dim V(q) > dim V(p) and since µp(M/U) = r ≤ µq(M/U) by lemma 5.1(e) we have µq(M/U) = r and also µq(M) ≤ Mp(M). Since t > dim V(p) + 1 = µp(M) − µp(M/U) ≥ µq(M) − µq(M/U) ≥ min{t, dim V(q + 1} > dim V(p) + 1 we obtain a contradiction. Thus p is minimal in Ar(M/U).

We can now establish the Eisenbud-Evans theorem for basic elements. The theorem tells us that if a submodule U of M is dim V(p) + 1 times basic in M at p for every prime ideal p of R, then U has a basic element.

Theorem 6.6. Eisenbud-Evans theorem Let X be the spectrum or the J-spectrum of a ring R.

Suppose X is Noetherian, and let M be a ﬁnitely generated R-module. Suppose that m1,..., mt ∈ M are such that for all p ∈ X we have

µp(M) − µp(M/hm1,..., mti) ≥ min{t, dim V(p) + 1}

then there are a2,..., at ∈ R such that m1 + a2m2 + ··· + atmt is basic for all p ∈ X.

101 (s) (s) Proof. For each s ∈ N satisfying 1 ≤ s ≤ t we will construct elements m1 ,..., mt ∈ M of the form t (s) X mi = mi + ai jm j j=s+1 with ai j ∈ R, for each i = 1,..., s, such that for all p ∈ X

(s) (s) µp(M) − µp(M/hm1 ,..., mt i) ≥ min{s, dim V(p) + 1}.

Thus, if we can carry out this construction for s = 1 we have an m = m1 + a2m2 + ··· + atmt with

µp(M) − µp(M/(m)) ≥ 1 for all p ∈ X, which renders m basic for all p ∈ X, and so the s = 1 case proves the theorem.

To establish the s = t case, we can simply take the given elements m1,..., mt. Suppose that for

(s) (s) some s with 1 < s ≤ t the desired elements m1 ,..., mt have already been obtained. By lemma (s) (s) 6.3 the set of all p ∈ X with dim V(p) + 1 < s and µp(M) − µp(M/hm1 ,..., ms i) = dim V(p) + 1 (s) s is ﬁnite. Let p1,..., pr be all of these primes. By lemma 5.4, since (m1 ,..., ms) is dim V(pi) + 1 (s) (s) (s) (s) times basic for each i = 1,..., r, there exist ai ∈ R such that (m1 + a1ms ,..., ms−1 + as−1ms ) is (s−1) s (s) dim V(pi) + 1 times basic for each i. Let mi = mi + aims ( i = 1,..., s − 1). We must have

− h (s−1) (s−1)i ≥ p µpi (M) µpi (M/ m1 ,..., ms−1 ) dim V( i) + 1 for each i = 1,..., r.

Now for p ∈ X \{p1,..., pr} we have

(s−1) (s−1) (s) (s) µp(M) − µp(M/hm1 ,..., ms−1 i) ≥ µp(M) − µp(M/hm1 ,..., mt i) − 1 ≥ min{s − 1, dim V(p) + 1}

(s−1) (s−1) so that the elements m1 ,..., ms−1 are as desired. Since we have established the s = t case, we iterate this process to obtain the s = 1 case, which establishes the theorem.

102 Most of the central theorems concerning modules over Noetherian rings, namely Bass’s cancella- tion theorem, Serre’s splitting theorem, and the Forster-Swan theorem can be obtained as a consequence of the Eisenbud-Evans theorem. In the case of a Noetherian ring R, Eisenbud and Evans also conjectured a bound on the number of generators for a ﬁnitely generated module over the polynomial ring R[x] which improved on the Forster-Swan bound.

Theorem 6.7. Eisenbud-Evans bound Let S be a Noetherian ring, and let R = S [x]. If M is a ﬁnitely generated R-module, then

µ(M) ≤ Max{µp(M) + dim R/p : p ∈ Spec(R), dim R/p < dim R} and this bound is the best possible.

This result was first shown by Sathaye for integral domains which are finitely generated as alge- bras over an infinite ground field, while Mohan Kumar resolved the general case. A proof, using the calculus of unimodular rows and the generalized dimension function of Plumstead can be found in [8,pg. 239].

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