Extensions of modules and uniform bounds of Artin-Rees type

by Janet Striuli

Submitted to the Department of Mathematics

and the Faculty of the Graduate School

of the University of Kansas in partial fulfillment of the requirements for the degree of Doctor of Philosophy

\overline{(Chair)}

Date Submitted:

Abstract

EXTENSIONS OF MODULES AND UNIFORM BOUNDS OF ARTIN-REES TYPE

Janet Striuli The University of Kansas Advisor: Craig Huneke August, 2005

This dissertation is about some aspects of commutative algebra. Two main themes are developed: extensions of modules and variations of the classical Artin-Rees Lemma.

In the first chapter we first explain some basic results on the Yoneda’s correspondence, we present Miyata’s theorem and we prove a generalization of it. We give many applications of this generalization and of its partial converse, particularly to Cohen-Macaulay rings of finite Cohen-Macaulay type. We introduce the notion of sparse modules, we study their properties and we show that Ext_{R}^{1}(M, N) is a sparse module if M and N are maximal Cohen-Macaulay modules over a ring of finite Cohen-Macaulay type.

The second and the third chapters are devoted to the study of variations of the classical Artin-Rees Lemma. First we give an easy proof that in one-dimensional excellent rings the strong uniform version of the Artin-Rees Lemma holds. We also give some examples to show that such property does not hold in two dimensional rings.

Finally we study a uniform Artin-Rees property for syzygies. We relate the uniform Artin-Rees property with uniform annihilators for a certain family of Tors. We show that in a one-dimensional Noetherian local ring any finitely generated R-module has the Artin-Rees property for syzygies.

to Mirco

iv Acknowledgments

My heartfelt thanks go to my advisor, Professor Craig Huneke, for teaching me commutative algebra, for his most valuable suggestions and comments, for his support over these years, for his dedication to students, for his enormous patience, for his enthusiasm for math, for the wonderful lectures he gives and, for all the nice conversations we had. For me he will always be an inspiring example throughout my professional life.

I would like to thank Daniel Katz, Satyagopal Mandal, Bangere Purnaprajna, Christian Schoeneich, who together with Craig Huneke are the members of my thesis committee.

I am grateful to Liam O’Carroll, Estela Gavosto and, once again, Craig Huneke and Daniel Katz for the support they gave me and the time they spent in helping me to apply for a postdoctoral position. I thank the Math Department for the support I received in these years.

All the people I met in the algebra community heve been very kind, I thank them for the valuable suggestions, for all the math I learnt, for the possibility of presenting my results.

Special thanks go to Yongwei Yao, for all his answers to questions about commutative algebra, to Giulio Caviglia, for his sharp mathematical thinking and for his support, to Manoj Kummini, for endless enthusiasm over math and everything and, for his highly professional skills with L^{A}TEX. I would like to remember Enrico, Catalin, Jooyong, Neil, Ananth, Glenn, Elisa, Laura, Diego, Erica, Lynn, Melissa, Amy, Jila, Eric, Bill, Ken, Garrin and Morgan, and all the other friends; I consider myself very lucky to have met them.

This thesis is dedicated to Mirco. I just cannot find the right words to express all my gratitude and my appreciation. It has been such a pleasure to share not only the joyful moments but also the difficult ones.

Contents

Abstract ii

Acknowledgments v

Introduction ix

1 Short exact sequences 1 1.1 The Yoneda correspondence . . . . . . . . . . . . . . . . . . . 2 1.2 Miyata’s Theorem and its extension . . . . . . . . . . . . . . . 13 1.3 More results on short exact sequences . . . . . . . . . . . . . . 23

1.3.1 Minimal generators in Ext_{R}^{1}(M, N) . . . . . . . . . . . 25 1.3.2 Additive sequences . . . . . . . . . . . . . . . . . . . . 27 1.3.3 Structure of Ext_{R}^{1}(M, N) . . . . . . . . . . . . . . . . . 32 1.3.4 Annihilators of short exact sequences . . . . . . . . . . 36

1.4 Sparse modules . . . . . . . . . . . . . . . . . . . . . . . . . . 37 1.5 Application to rings of finite Cohen-Macaulay type . . . . . . 45 1.6 Other applications . . . . . . . . . . . . . . . . . . . . . . . . 51 1.6.1 Efficient systems of parameters . . . . . . . . . . . . . 51 1.6.2 Splitting of syzygies . . . . . . . . . . . . . . . . . . . . 52

2 Artin Rees Numbers 58 2.1 One dimensional rings . . . . . . . . . . . . . . . . . . . . . . 59 2.2 Two dimensional rings . . . . . . . . . . . . . . . . . . . . . . 69

3 Artin-Rees bounds on syzygies 73

3.1 Strong uniform Artin-Rees bounds for syzygies and the residue

field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.2 Uniform Artin-Rees bounds for syzygies . . . . . . . . . . . . 76

3.2.1 Uniform Artin-Rees bounds and short exact sequences 76 3.2.2 Uniform Artin-Rees property for syzygies and annihi-

lators of the Tors modules . . . . . . . . . . . . . . . . 81

3.3 Uniform Artin-Rees property for syzygies for modules over

Cohen-Macaulay rings . . . . . . . . . . . . . . . . . . . . . . 85 3.4 Uniform Artin-Rees property for syzygies in rings of dimension

one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4 Conclusion 91

References 93 Introduction

In this dissertation we study the following topics: the Yoneda correspondence between short exact sequences and elements of the cohomology group Ext_{R}^{1}(−, −) and uniform versions of the classical Artin-Rees Lemma.

Chapter one is devoted to the study of the Yoneda correspondence and to some applications of the developed techniques. Let R be a Noetherian ring and M and N be finitely generated R-modules. There is a one to one correspondence between elements of Ext_{R}^{1}(M, N) and short exact sequences

α : 0 → N → X_{α} → M → 0,

modulo a certain equivalence relation. Under this correspondence the zero element is given by a split exact sequence. It is a theorem of Miyata [16] that a short exact sequence is split exact if and only if X_{α} ≅ N ⊕ M. We extend Miyata’s theorem to the following

Theorem 0.1. Let R be a Noetherian ring and I ⊂ R an ideal. Let α and β be two short exact sequences in Ext_{R}^{1}(M, N), with isomorphic middle modules. If α ∈ I Ext_{R}^{1}(M, N) then β ⊗ R/I is a split exact sequence.

We study some more properties of short exact sequences in Ext_{R}^{1}(M, N), for example when the minimal number of generators of the modules at the side sums up to the minimal number of generator of the module in the middle; in some situations, we can also say when a short exact sequence is a minimal generator of Ext_{R}^{1}(M, N).

Theorem 0.1 gives some interesting consequences on the structure of the module Ext_{R}^{1}(M, N) which are particularly helpful in the study of rings of finite Cohen-Macaulay type. A Cohen-Macaulay ring is said to be of finite Cohen-Macaulay type if there are a finite number of isomorphism classes of maximal Cohen-Macaulay modules. Rings of finite Cohen-Macaulay type have been widely studied, see [24]. In particular it is know that if (R, m) is a local ring of finite Cohen-Macaulay type and M and N are maximal Cohen-Macaulay modules then Ext_{R}^{1}(M, N) is a module of finite length, see [2] and [12] in this generality. We introduce the notion of sparse modules and derive some properties. We show that sparse modules are Artinian and that Ext_{R}^{1}(M, N) are sparse modules if M and N are maximal Cohen-Macaulay modules over a ring of finite Cohen-Macaulay type, giving a different proof from the one in [12]. Huneke-Leuschke, in the same paper, give a bound on the power of the maximal ideal that annihilates Ext_{R}^{1}(M, N). We are able to improve that bound and to give some information on the structure of the module of Ext_{R}^{1}(M, N). Specifically, we have the following theorem, see Proposition 1.59 and Proposition 1.60.

Theorem 0.2. Let (R, m, k) a Cohen-Macaulay local ring with infinite residue field. Let M and N be maximal Cohen-Macaulay R-modules and let h be the number of isomorphism classes of modules that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N). Then m^{h−1} Ext_{R}^{1}(M, N) = 0. Suppose that N = M_{1}, where M_{1} is the first syzygy in a free resolution of M, and suppose that the residue field k is algebraically closed. If

m^{h−2} Ext_{R}^{1}(M, N) ≠ 0

then Ext_{R}^{1}(M, N) = ⊕_{i=1}^{n}C_{i}, where each C_{i} is a cyclic R-module and the length λ(m^{j}C_{i}/m^{j+1}C_{i}) has value 0 or 1 for every integer j.

In the second chapter we study a uniform version of the Artin-Rees Lemma. Let R be a Noetherian ring and N ⊂ M be finitely generated Rmodules. If I ⊂ R is an ideal of R, the classical lemma of Artin-Rees states that there exists an integer k such that I^{n}M ∩ N = I^{n−k}(I^{k}M ∩ N), for n > k. For a proof see [15]. In particular a weaker version of the lemma says that there exists a k such that I^{n}M ∩ N ⊂ I^{n−k}N for every n > k. Notice that the integer k depends, a priori, on the ideal I and the two modules N and M. The uniform version (the strong and the weak one) of the Artin-Rees Lemma asks whether we can find the integer k that works for all the ideals of a certain family. This kind of questions were first raised by Eisenbud-Hochster in [8], where the family of ideals they considered is the family of maximal ideals. Duncan-O’Carroll [6] gave a positive answer for the strong uniform Artin-Rees Lemma for the family of maximal ideals in excellent rings. Huneke [11] showed that in many rings a weak version of the uniform Artin-Rees Lemma holds for the family of all ideals. In general the strong uniform Artin-Rees Lemma does not hold. Wang [22] gave a counterexample in a three dimensional ring, and we give two counterexamples in rings of dimension two (see Examples 2.15 and 2.16). Planas-Vilanova [19] showed that in a one-dimensional excellent ring the strong uniform version of the Artin-Rees Lemma holds for the family of all ideals. We give a much simpler proof of this fact (see Corollary 2.14).

In the third chapter we study a uniform version of the Artin-Rees Lemma for syzygies in a free resolution. Let (R, m) be a local Noetherian ring and M a finitely generated R-module. Let 𝔽 = {F_{i}} be a free resolution of M. Denote by M_{i} ⊂ F_{i} the i-th syzygy. We study if, given an ideal I ⊂ R, there exists an integer h such that

M_{i} ∩ I^{n}F_{i} ⊂ I^{n−h}M_{i},

for all n > h and for all i > 0. If such an integer h exists we say that M has the uniform Artin-Rees property for syzygies with respect to an ideal I. This question was first raised by Eisenbud-Huneke [9], who proved that such h exists when the module M has finite projective dimension on the punctured spectrum. We show that having the uniform Artin-Rees property for syzygies well behaves in short exact sequence. In particular if any of the two modules appearing in a short exact sequence have the property, then also the third one has it. We relate the uniform Artin-Rees property for syzygies with respect to an ideal I with uniform annihilators of the family of modules T_{i,n} = Tor_{i}^{R}(M, R/I^{n}). In Cohen-Macaulay rings any finitely generated Rmodule have the uniform Artin-Rees property for syzygies with respect to m-primary ideals. The proof of this fact was shown to me by Professor Katz. We show that in a one dimensional ring any finitely generated R-module has the uniform Artin-Rees property with respect to the maximal ideal m. In analogy to the classical version of the Artin-Rees Lemma, we can ask whether given an ideal I there exists an integer h such that

M_{i} ∩ I^{n}F_{i} = I(I^{n−1}F_{i} ∩ M_{i}),

for every n > h and for every i > 0. If such an integer h exists we say that the module M has the strong Artin-Rees property for syzygies. We show the following

Theorem 0.3. Let (R, m, k) be a local Noetherian ring. Then k has the strong uniform Artin-Rees property for syzygies with respect to the maximal ideal m.

Chapter 1

Short exact sequences

In this chapter we will study closely the correspondence between short exact sequences and elements of Ext_{R}^{1}(−, −). All rings are assumed to be commutative with identity and Noetherian; moreover all modules are finitely generated.

In the first section we go through the Yoneda correspondence, setting up the notation. A more extended description of such correspondence can be found in [14].

Under this correspondence, Miyata’s Theorem, see [16], gives a complete characterization of splitting exact sequences. In the second section we present an extension of this theorem.

We later study more properties of short exact sequences. In particular we give some situations when we are able to say if a short exact sequence is a minimal generator for Ext_{R}^{1}(M, N); we describe y Ext_{R}^{1}(M, N) as union of particular subsets of Ext_{R}^{1}(M, N), for y a non-zerodivisor on the ring R and on the module M; we also introduce the notion of sparse module and we derive some properties of such modules. The notion of sparse module rises from the behaviour of Ext_{R}^{1}(M, N), where R is a ring of finite Cohen-Macaulay type and M and N are maximal Cohen-Macaulay modules. We give some applications of such a situation: for example we give a new proof of the fact that Ext_{R}^{1}(M, N) is a module of finite length if M and N are maximal Cohen-Macaulay over a ring of finite Cohen-Macaulay type, improving the bound given in [12] on the power of the maximal ideal that kills the Ext module. Finally, we use the developed techniques to give a different proof of a theorem of O’Carroll and Popescu, [18].

1.1 The Yoneda correspondence

We describe the Yoneda correspondence between short exact sequences and elements of the group Ext_{R}^{1}. Let R be a Noetherian ring and M and N two finitely generated R-modules.

Definition 1.1. An extension of M by N is a short exact sequence

\xymatrix{α : 0 \ar[r] & N \ar[r] & X_{α} \ar[r] & M \ar[r] & 0.}  (1.1.1)

On the set of all extensions of M by N there is an equivalence relation such that the quotient set is naturally isomorphic to Ext_{R}^{1}(M, N).

Given two extensions of M by N, α as in the Definition 1.1 and

β : 0 → N → X_{β} → M → 0,  (1.1.2)

we say α ∼ β if there exists an R-homomorphism h making the following diagram commute

[figure: 0; N; X_{α}; h; M; 0; 0; N; X_{β}; M; 0.]

By the Snake Lemma, h is an isomorphism and therefore the relation ∼ is an equivalence relation among extensions.

Denote by ext(M, N) the set of all extensions of M by N modulo the equivalent relation ∼. In the following part of the section, we will go through the correspondence between ext(M, N) and Ext_{R}^{1}(M, N). Before doing so, it is useful to recall the construction of pushout. Discussion 1.2. Let α̃ be the following exact sequence:

\xymatrix{α̃ : M_{2} \ar[r]^{f} & M_{1} \ar[r]^{g} & M \ar[r] & 0,}

where M_{1}, M_{2} are finitely generated R−modules.

Suppose h : M_{2} → N is an R−homomorphism, define P as the cokernel of the map:

\xymatrix{M_{2} \ar[r]^{(h,−f)} & N ⊕ M_{1} \ar[r] & P \ar[r] & 0.}

Consider the following diagram:

[figure: α̃ :; M_{2}; f; h; M_{1}; j; g; M; 0; hα̃ :; N; i; P; π; M; 0,]

where the maps are defined as follows:

i(n) = \overline{(n, 0)} for any n ∈ N,
j(x) = \overline{(0, x)} for any x ∈ M_{1},
π(\overline{(n, 0)}) = g(x).

Then the diagram commutes and the bottom row is an exact sequence. Literally, the pushout of the diagram is the module P; we will abuse notation and call the exact sequence hα̃ the push out of α̃ via the homomorphism h.

Notice that if f is an injective map then so is i.

The pushout 1.2 has the following universal property:

Lemma 1.3. Suppose the following commutative diagram is given:

\xymatrix{μ : & M_{2} \ar[r]^{f_{2}} \ar[d]^{ϕ_{2}} & M_{1} \ar[d]^{ϕ_{1}} \ar[r]^{f_{1}} & M \ar[d]^{ϕ} \ar[r] & 0 \\ ν : & N_{2} \ar[r]^{g_{2}} & N_{1} \ar[r]^{g_{1}} & N \ar[r] & 0.}

Then the diagram factors through the pushout of μ via ϕ_{2}. More precisely there exists the following commutative diagrams:

[figure: μ :; M_{2}; f_{2}; ϕ_{2}; M_{1}; j; f_{1}; M; 0; ϕ_{2}μ :; N_{2}; i; P; π; ψ; M; ϕ; 0; ν :; N_{2}; g_{2}; N_{1}; g_{1}; N; 0,]

where ψj = ϕ_{1}.

Proof. Define ψ : P → N_{1} such that ψ(\overline{(n, m)}) = g_{2}(n) + ϕ_{1}(m). It is a well-defined map since:

ψ(0) = g_{2}(ϕ_{2}(m)) + ϕ_{1}(−f_{2}(m)) = 0,

for any m ∈ M_{1}. Moreover for any m ∈ M_{2} we have

ψj(m) = ϕ_{1}(m).

Immediately we have the following

Corollary 1.4. Let μ be an exact sequence M_{2} → M_{1} → M → 0 and let ϕ : M_{2} → N be an R-homorphism. If the pushout ϕμ is an extension of M by N then it is unique in ext(M, N).

Lemma 1.5. Let α̃ an exact sequence as in the previous discussion, and let β_{1} and β_{2} two extensions of M by N, obtained respectively by pushing out α̃ via h_{1} and h_{2}. Then β_{1} = β_{2} in ext(M, N) if and only if there exists a map ψ : M_{1} → N such that ψf = h_{1} − h_{2}.

Proof. Let

[figure: α̃ :; M_{2}; f; h_{1}; M_{1}; j_{1}; g; M; 0,; β_{1} :; 0; N; i_{1}; P_{1}; π_{1}; M; 0,]

and

[figure: α̃ :; M_{2}; f; h_{2}; M_{1}; j_{2}; g; M; 0,; β_{2} :; 0; N; i_{2}; P_{2}; π_{2}; M; 0,]

obtained as in the discussion 1.2.

Suppose there is a map ψ : M_{1} → N such that ψf = h_{1} − h_{2}. Define the map ϕ : P_{1} → P_{2} such that ϕ(\overline{(n, x)}) = \overline{(n + ψ(x), x)}, for any n ∈ N and any x ∈ M_{1}. Notice that it is a well-defined map since

ϕ(0) = ϕ(\overline{(h_{1}(n), −f(n))}) = \overline{(h_{1}(n) − ψf(n), −f(n))} = \overline{(h_{2}(n), −f(n))} = 0.

The map ϕ is the one that satisfies the definition of β_{1} = β_{2}.

On the other hand suppose that β_{1} = β_{2}. Then, by definition, we have the following diagram:

[figure: 0; N_{1}; i_{1}; P_{1}; ϕ; π_{1}; M; 0; 0; N_{1}; i_{2}; P_{2}; π_{2}; M; 0.]

For any x ∈ M_{1}, define

\overline{ψ}(x) = ϕj_{1}(x) − j_{2}(x) ∈ P_{2}.

Since π_{2}(\overline{ψ}(x)) = 0, there exists a map ψ : M_{1} → N such that i_{2}ψ = \overline{ψ}. For any n ∈ M_{2}, we have:

i_{2}ψf(n) = \overline{ψ}f(n) = ϕj_{1}(f(n)) − j_{2}(f(n)) = ϕi_{1}h_{1}(n) − i_{2}h_{2}(n)
= i_{2}h_{1}(n) − i_{2}h_{2}(n) = i_{2}(h_{1} − h_{2})(n).

By assumption i_{2} is an injective map and therefore ψf = h_{1} − h_{2}.

Let the following sequence be part of a projective resolution 𝔽 of M:

\xymatrix{… \ar[r] & P_{2} \ar[r]^{d_{2}} & P_{1} \ar[r]^{d_{1}} & P_{0} \ar[r]^{d_{0}} & M \ar[r] & 0.}

Recall that Ext_{R}^{1}(M, N) is the first cohomology of the complex Hom_{R}(𝔽, N),

\xymatrix{0 \ar[r] & Hom_{R}(M, N) \ar[r]^{d_{0}^{∗}} & Hom_{R}(P_{0}, N) \ar[r]^{d_{1}^{∗}} & … \\ … \ar[r] & Hom_{R}(P_{1}, N) \ar[r]^{d_{2}^{∗}} & Hom_{R}(P_{2}, N) \ar[r] & …,}

i.e.

Ext_{R}^{1}(M, N) = \frac{ker d_{2}^{∗}}{Image d_{1}^{∗}} .

An element in Ext_{R}^{1}(M, N) is given by an equivalence class ξ + Image d_{1}^{∗}, where ξ ∈ ker d_{2}^{∗}.

Define a map Ψ : Ext_{R}^{1}(M, N) → ext(M, N) such that Ψ(ξ + Image d_{1}^{∗}) is the short exact sequence γ obtained as a pushout of P_{1} → P_{0} → M → 0 via ξ, as described in discussion 1.2:

[figure: …; P_{2}; d_{2}; P_{1}; ξ; d_{1}; P_{0}; j; d_{0}; M; 0,; γ :; 0; N; i; P; M; 0.]

Remark 1.6. Notice that

(1) γ is a short exact sequence. Indeed if i(n) = 0 for some n ∈ N then, by the definition of P,

(n, 0) = (ξ(p), −d_{1}(p)) for some p ∈ P_{1}.

Since ker(d_{1}) = Image(d_{2}), there exists a q ∈ P_{2} such that d_{2}(q) = p. But ξd_{2} = 0, since ξ ∈ ker(d_{2}^{∗}); therefore n = ξ(p) = ξd_{2}(q) = 0.

(2) The definition of γ does not depend on the representative of ξ+Image d_{1}^{∗}, by Lemma 1.5.

To define a map Φ : ext(M, N) → Ext_{R}^{1}(M, N), consider γ ∈ ext(M, N) and a projective resolution of M. Lift the identity map of M as in the following diagram:

[figure: …; P_{2}; ξ_{2}; d_{2}; P_{1}; ξ; d_{1}; P_{0}; ξ_{0}; d_{0}; M; 0,; γ :; 0; N; i; P; M; 0.]

and define Φ(γ) = ξ + Image d_{1}^{∗} ∈ Ext_{R}^{1}(M, N).

Remark 1.7. Notice that

(1) ξ ∈ ker(d_{2}^{∗}), since the squares are commutative.

(2) Φ is a well defined map by Lemma 1.5 and since any two lifting are homotopic.

The maps Φ and Ψ are inverse to each other. Easily we can see that ΦΨ(ξ + Image(d_{1}^{∗})) = ξ + Image(d_{1}^{∗}), since (…, 0, …, 0, ξ, j) is a lifting of id_{M}. On the other hand ΨΦ(γ) = γ by the uniqueness of the pushout, see Corollary 1.4. Remark 1.8. Often in the following we will construct γ from

\xymatrix{σ : 0 \ar[r] & K \ar[r] & P_{0} \ar[r]^{d_{0}} & M \ar[r] & 0.}

where K = ker d_{0}. Indeed if ξ ∈ ker(d_{2}^{∗}), Image d_{2} ⊂ ker ξ and therefore ξ induces a map

[figure: \overline{ξ} :; P_{1}/ Image d_{2}; P_{0}; P_{1}/ ker d_{1}; K]

Then it is easy to see that

Ψ(ξ + Image d_{1}^{∗}) = \overline{ξ}σ,

where \overline{ξ}σ is the pushout of σ via \overline{ξ}.

The map Φ induces a structure of R-module on ext(M, N).

Remark 1.9. The zero element in ext(M, N) corresponds to the pushout of σ via the 0 map, and by Lemma 1.5 via any map from K to N that can be extended to P_{0}. A representative extension for the zero element is the exact sequence 0 → N → N ⊕ M → M → 0 as the following diagram is commutative:

[figure: 0; K; 0; P_{0}; (0,π); π; M_{1}; 0; 0; N; N ⊕ M; M; 0.]

By the Snake Lemma, any representative of the zero element has the middle module isomorphic to M ⊕ N and it is a split exact sequence. On the other hand any short exact sequence with the middle module isomorphic to M ⊕N is a representative of the zero element. The proof of this fact is a theorem by Miyata [16] which we will discuss in the next section.

We will finish the section by describing the first part of the long exact sequence induced by applying the functor Hom to an extension of M by N. Before doing so we need to introduce the pullback of a short exact sequence. Discussion 1.10. Suppose

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0}

is a given extension of M by N. Let ϕ : M_{1} → M an R-homorphism. Denote by Q the kernel of the following map:

\xymatrix{0 \ar[r] & Q \ar[r] & X_{α} ⊕ M_{1} \ar[r]^{g−ϕ} & M \ar[r] & 0.}

Then the pullback of α via ϕ, which we will denote by αϕ, is the exact top row, as in the following diagram:

[figure: αϕ :; 0; N; i; Q; p; π; M_{1}; ϕ; 0; α :; 0; N; f; X_{α}; g; M; 0,]

where the maps are

i(n) = (f(n), 0), for any n ∈ N;
π(x, m) = g(x), for any (n, x) ∈ Q;
p(n, x) = x, for any , (n, x) ∈ Q.

The pullback of an extension has the following universal property, for the proof of which we refer the reader to [14].

Proposition 1.11. Given the following commutative diagram:

\xymatrix{α : 0 \ar[r] & N \ar[d]^{ϕ_{1}} \ar[r]^{f_{1}} & X_{α} \ar[d]^{ϕ_{2}} \ar[r]^{g_{1}} & M \ar[d]^{ϕ_{3}} \ar[r] & 0 \\ β : 0 \ar[r] & N \ar[r]^{f} & X_{β} \ar[r]^{g} & M \ar[r] & 0}

it factors through the pullback of β via ϕ_{3}, more precisely there exists a map ψ such that the ψp = ϕ_{2} and the following diagram is commutative:

[figure: α : 0; N; ϕ_{1}; f_{1}; X_{α}; ψ; g_{1}; M; 0; βϕ_{3} : 0; N; i; Q; p; π; M; ϕ_{3}; 0; β : 0; N; f; X_{β}; g; M; 0.]

Lemma 1.12. Given an extension α of M by N as in Discussion 1.10 and given a map ϕ : M_{1} → M, αϕ is a split exact sequence if and only if there exists a map \overline{ϕ} : M_{1} → X_{α} such that g\overline{ϕ} = ϕ.

In the following we will drop the notation ext(M, N). We go throught the proof of the following proposition for later reference.

Proposition 1.13. Let α be an extension of M by N

\xymatrix{0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Let A be a finitely generated R-module. There exists a long exact sequence:

\xymatrix{0 \ar[r] & Hom_{R}(A, N) \ar[r] & Hom_{R}(A, X_{α}) \ar[r] & Hom_{R}(A, M) \ar[r]^{d} & … \\ … \ar[r]^{d} & Ext_{R}^{1}(A, N) \ar[r]^{f^{∗}} & Ext_{R}^{1}(A, X_{α}) \ar[r]^{g^{∗}} & Ext_{R}^{1}(A, M),}

where

d(ϕ) = αϕ,
f^{∗}(β) = fβ,
g^{∗}(β) = gβ.

Proof. Exactness in Hom(A, M) is given by Lemma 1.12. Notice that by Corollary 1.4, g(f(α)) = (gf)(α) = 0. To prove exactness in Ext_{R}^{1}(A, X_{α}) we need to show that for any β ∈ Ext_{R}^{1}(A, X_{α}) such that gβ = 0 there exists a γ ∈ Ext_{R}^{1}(A, N) such that fγ = β.

Consider the following diagram:

[figure: N; f; β :; 0; X_{α}; g; h; L; p; j; A; 0; gβ :; 0; M; i; P; π; A; 0]

By Lemma 1.5, the splitting of the sequence gβ implies the existence of a map g′ : L → M such that g′h = g. Notice that g′hf = gf = 0, therefore Image fh ⊂ ker g′ =: L′. We claim that the sequence:

\xymatrix{γ : & 0 \ar[r] & N \ar[r]^{fh} & L′ \ar[r]^{p|_{L′}} & A \ar[r] & 0}  (1.1.3)

is a short exact sequence and fγ = β. To prove the first claim we need to check first that the restriction of p to L′ is still surjective. For it, let a ∈ A and l ∈ L such that p(l) = a. Let x ∈ X_{α} such that g(x) = g′(l). Then l −h(x) ∈ L′ and it maps to a under p. To conclude the first claim we are left to prove exactness in L′. Since ph = 0 we have Image(hf) ⊂ ker(p|_{L′}). For the other inclusion, take an l ∈ L′ such that p(l) = 0; since α is exact, there exists an x ∈ X_{α} such that f(x) = l. Moreover, g(x) = g′h(x) = g′(l) = 0 and therefore there exists an n ∈ N such that f(n) = x, proving exactness in L′.

For the second claim notice that the following diagram commutes, and by the Lemma 1.4, we have that fγ = β.

[figure: γ :; 0; N; f; hf; L′; p|_{L′}; A; 0; β :; 0; X_{α}; h; L; p; A; 0]

Using a dual proof and the universal property of the pullback one can prove the following:

Proposition 1.14. Let α be an extension of M by N

\xymatrix{0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Let A be a finitely generated R-module. There exists a long exact sequence:

\xymatrix{0 \ar[r] & Hom_{R}(M, A) \ar[r] & Hom_{R}(X_{α}, A) \ar[r] & Hom_{R}(N, A) \ar[r]^{δ} & … \\ … \ar[r]^{δ} & Ext_{R}^{1}(M, A) \ar[r]^{g^{∗}} & Ext_{R}^{1}(X_{α}, A) \ar[r]^{f^{∗}} & Ext_{R}^{1}(M, A),}

where

d(ϕ) = ϕα,
f^{∗}(β) = βf,
g^{∗}(β) = βg.

For the proof see [14], page 73-74.

1.2 Miyata’s Theorem and its extension

In this section we present an extension of Miyata’s Theorem, which characterizes when short exact sequences are split exact. In the following we refer to short exact sequences as elements of Ext, keeping in mind the correspondence of the previous section.

A very important tool used in this section is the following theorem. For the proof, in this context, we refer to [25]. For other proofs see [1], [3], [4], [10]. Theorem 1.15. Let (R, m) be a Noetherian local ring, and let M and N two finitely generated R-modules. Then M ≅ N if and only if

λ_{R}(Hom_{R}(M, L)) = λ_{R}(Hom_{R}(N, L)),

for any R-module L such that λ_{R}(L) < ∞ if and only if

λ_{R}(M ⊗ L) = λ_{R}(N ⊗ L)

for any R-module L such that λ_{R}(L) < ∞. Theorem 1.16 (Miyata). Let R a Noetherian ring and let

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0,}

be an element in Ext_{R}^{1}(M, N). Then α represents the zero element if and only if X_{α} ≅ M ⊕ N.

The proof provided here is due to Huneke-Katz. We first state and prove separately a Lemma for easy reference. Lemma 1.17. Let R be a Noetherian ring. Let α be the following short exact sequence:

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Denote by C the image of the connecting homomorphism δ (see the definition in Proposition 1.14), obtained by applying the functor Hom_{R}( , N) to α:

\xymatrix{0 \ar[r] & Hom_{R}(M, N) \ar[r] & Hom_{R}(X_{α}, N) \ar[r] & Hom_{R}(N, N) \ar[r]^{δ} & … \\ … \ar[r]^{δ} & Ext_{R}^{1}(M, N) \ar[r]^{g^{∗}} & Ext_{R}^{1}(X_{α}, N) \ar[r]^{f^{∗}} & Ext_{R}^{1}(N, N).}

Then α is a split exact sequence if and only if C = 0.

Proof. Suppose C = 0, then since α = δ(id_{N}) we have that α = 0 and hence α is a split exact sequence, see Remark 1.9.

On the other hand suppose that α is a split exact sequence, hence there exists an R-homomorphism f′ : X_{α} → N, such that f′f = id_{N}. To prove the lemma, we need to show that the map f^{∗} : Hom(X_{α}, N) → Hom(N, N) is a surjective map. But for any l ∈ Hom(N, N) we have l = f^{∗}(lf′).

Now the proof of Theorem 1.16.

Huneke-Katz. One direction of the Theorem has been already discussed in Remark 1.9: any representative of the zero element in Ext_{R}^{1}(M, N) has middle module isomorphic to M ⊕ N. For the other direction, by Lemma 1.17, it is enough to prove that C = 0. By way of contradiction, assume C ≠ 0 and let P ∈ Spec(R) be a minimal prime over Ann(C). By localizing at P, we may assume (R, m) local, and C is finite length. Since X_{α} ≅ M ⊕ N, we have the following exact sequence:

\xymatrix{0 \ar[r] & Ext_{R}^{1}(M, N)/C \ar[r] & Ext_{R}^{1}(M, N) ⊕ Ext_{R}^{1}(N, N) \ar[r] & Ext_{R}^{1}(N, N),}

and therefore

[figure: 0; H_{m}^{0}(\frac{Ext_{R}^{1}(M,N)}{C}); H_{m}^{0}(Ext_{R}^{1}(M, N)) ⊕ H_{m}^{0}(Ext_{R}^{1}(N, N)); …; H_{m}^{0}(Ext_{R}^{1}(N, N)).]

Since H_{m}^{0}(Ext_{R}^{1}(M, N)/C) = H_{m}^{0}(Ext_{R}^{1}(M, N))/C, by counting lengths we obtain C = 0, which is a contradiction.

Locally, the following is an equivalent statement: Theorem 1.18. Let (R, m) be a local Noetherian ring and let

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0,}

be an element in Ext_{R}^{1}(M, N). Then α represents the zero element if and only if there exists a surjection

\xymatrix{X_{α} \ar@{->>}[r] & M ⊕ N.}

Proof. The only thing to show is that in this case the surjection implies the existence of an isomorphism. Indeed, for any R-module of finite length L, we have the following inequalities:

λ(X_{α} ⊗ L) ≤ λ(M ⊗ L) + λ(N ⊗ L), by tensoring α by L,
≤ λ(X_{α} ⊗ L), by tensoring the surjection π by L.

Therefore λ(X_{α} ⊗ L) = λ((M ⊕ N) ⊗ L), for any R-module of finite length. Theorem 1.15 implies X_{α} ≅ M ⊕ N.

The dual statement of Theorem 1.18, where we replace the surjection by an injection, does not hold. For example if x, y are a regular sequence on (R, m), then we have the short exact sequence:

\xymatrix{α : & 0 \ar[r] & R \ar[r] & R^{2} \ar[r] & (x, y) \ar[r] & 0.}

By looking at the number of minimal generators of each module we conclude that α cannot be a split exact sequence. On the other hand there exists the injection 0 → R ⊕ (x, y) → R^{2}.

We can further relax the assumption on α. In particular we have the following Lemma 1.19. Let (R, m) be a Noetherian ring and let α̃ the following exact sequence of finitely generated R-modules:

\xymatrix{α̃ : & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Suppose there exists a surjection

\xymatrix{π : & X_{α} \ar@{->>}[r] & M ⊕ N.}

Then α̃ is a split short exact sequence.

Proof. By Theorem 1.18, it is enough to show that the map f is an injection. Suppose by way of contradiction that there exists a non zero element x ∈ N such that f(x) = 0. By the Krull Intersection Theorem there exists an n ∈ ℕ such that x ∉ m^{n}N. For such a choice of n, tensor the exact sequence α̃ by R/m^{n} to obtain:

\xymatrix{0 \ar[r] & C \ar[r] & N/m^{n}N \ar[r]^{f̄} & X_{α}/m^{n}X_{α} \ar[r]^{ḡ} & M/m^{n}M \ar[r] & 0,}

where C ≠ 0 since x mod m^{n}N ∈ C. Notice that by tensoring the surjection we still have a surjection:

\xymatrix{π̄ : & X_{α}/m^{n}X_{α} \ar@{->>}[r] & M/m^{n}M ⊕ N/m^{n}N.}

By counting the lengths of the modules we obtain

λ(X_{α}/m^{n}X_{α}) + λ(C) = λ(N/m^{n}N) + λ(M/m^{n}M)
≤ λ(X_{α}/m^{n}X_{α}), from the surjection π̄.

Hence C = 0, which is a contradiction.

We are ready to state and prove the main Theorem of the section.

Theorem 1.20. Let R be a Noetherian ring and let I ⊂ R be an ideal. Let M and N be finitely generated R-modules. Let α : 0 → N → X_{α} → M → 0 be a short exact sequence and denote by α⊗R/I the complex (not necessarily exact) 0 → N/IN → X_{α}/IX_{α} → M/IM → 0. If α ∈ I Ext_{R}^{1}(M, N) then α ⊗ R/I is a split exact sequence. Proof. We first show that it is enough to prove a local statement. Assume that the local version of the theorem holds, and let

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0}

be a short exact sequence in I Ext_{R}^{1}(M, N). By Corollary 1.4, α/1 ∈ Ext_{R_{m}}^{1}(M_{m}, N_{m}) is given by the sequence

\xymatrix{0 \ar[r] & N_{m} \ar[r]^{f/1} & (X_{α})_{m} \ar[r]^{g/1} & M_{m} \ar[r] & 0,}

for any maximal ideal m. Let C be the kernel of f ⊗ id_{R/I}, then Supp(C) ⊂ V (I). Let m ∈ V (I). α ∈ I Ext_{R}^{1}(M, N) implies α/1 ∈ IR_{m} Ext_{R_{m}}^{1}(M_{m}, N_{m}) and therefore α/1 ⊗ R_{m}/IR_{m} is a split exact sequence. In particular

C_{m} = ker(f ⊗ id)_{m} = ker(f/1 ⊗ id_{R_{m}/IR_{m}}) = 0,

proving that α⊗R/I is a short exact sequence. We need to show that α⊗R/I is actually a split exact sequence. By Lemma 1.17 it is enough to prove that Image(δ) = 0, where δ is the connecting homomorphism in the following sequence (for its definition see Proposition 1.14):

\xymatrix{Hom_{R}(\frac{X_{α}}{IX_{α}} , \frac{N}{IN}) \ar[r] & Hom_{R}(\frac{N}{IN} , \frac{N}{IN}) \ar[r]^{δ} & Ext_{R}^{1}(\frac{M}{IM} , \frac{N}{IN}).}

Call C = Image(δ), again Supp(C) ⊂ V (I). Let m any maximal ideal in V (I), then C_{m} = Image(\overline{δ}), where \overline{δ} is defined in Proposition 1.14:

\xymatrix{Hom_{R_{m}}(N_{m}/IN_{m}, N_{m}/IN_{m}) \ar[r]^{\overline{δ}} & Ext_{R_{m}}^{1}(M_{m}/IM_{m}, N_{m}/IN_{m}).}

But, by Lemma 1.17, Image(\overline{δ}) = 0 since α/1 ⊗ R_{m}/IR_{m} is a split exact sequence. Therefore C_{m} = 0 for any m ∈ V (I), which implies that C = 0.

We can assume (R, m) is a local Noetherian ring. By Lemma 1.19, it is enough to show that X_{α}/IX_{α} ≅ M/IM ⊕ N/IN. Since α ∈ I Ext_{R}^{1}(M, N), following the construction of the map Φ as in the previous section, one can choose a representative short exact sequence which is the pushout of the presentation of M

\xymatrix{σ : P_{1} \ar[r]^{d_{1}} & P_{0} \ar[r]^{d_{0}} & M \ar[r] & 0}

via a map ξ ∈ I Hom(P_{1}, N), more precisely via a map ξ ∈ I ker(d_{2}^{∗}). We have the following diagram

\xymatrix{β : P_{1} \ar[r]^{(d_{1},−ξ)} & P_{0} ⊕ N \ar[r] & X_{α} \ar[r] & 0.}

Let Ω be a finitely generated module of finite length such that IΩ = 0. Tensor both the sequences β and σ with Ω and denote by Image_{1} and by Image_{2} the images d_{1} ⊗ id(P_{1} ⊗ Ω) and (d_{1} ⊗ id, −ξ ⊗ id)(P_{1} ⊗ Ω). Since Image(ξ) ⊂ IN, it follows that Image_{1} = Image_{2} ⊂ P_{0} ⊗ Ω.

If λ_{R}(M) denotes the length of an R-module M, we have:

λ_{R}(X_{α} ⊗ Ω) = λ_{R}(P_{0} ⊗ Ω) + λ_{R}(N ⊗ Ω) − λ_{R}(Image_{1})
= λ_{R}(M ⊗ Ω) + λ_{R}(Image_{2}) + λ_{R}(N ⊗ Ω) − λ_{R}(Image_{1})
= λ_{R}(M ⊗ Ω) + λ_{R}(N ⊗ Ω).

Notice that if Y is any R-module, which is also an R/I-module, then we have that λ_{R}(Y) = λ_{R/I}(Y), hence the equality above is just:

λ_{R/I}(M/IM ⊗ Ω) + λ_{R/I}(N/IN ⊗ Ω) = λ_{R/I}(X_{α}/IX_{α} ⊗ Ω).

In this equality above we can choose any R/I module of finite length Ω, therefore by Theorem 1.15 we have:

M/IM ⊕ N/IN ≃ X_{α}/IX_{α}.

An immediate corollary is the following

Theorem 1.21. Let R be a Noetherian ring and I ⊂ R an ideal. Let α and β be two short exact sequences in Ext_{R}^{1}(M, N), with isomorphic middle modules. If α ∈ I Ext_{R}^{1}(M, N) then β ⊗ R/I is a split exact sequence.

Notice that this is an extension of Miyata’s Theorem. For, if I is the zero ideal then α is a split exact sequence and the middle module is isomorphic to M ⊕ N.

Proof. Let

\xymatrix{α : 0 \ar[r] & N \ar[r] & X_{α} \ar[r] & M \ar[r] & 0,}

and

\xymatrix{β : 0 \ar[r] & N \ar[r] & X_{β} \ar[r] & M \ar[r] & 0,}

where X_{α} ≅ X_{β}. Since α ∈ I Ext_{R}^{1}(M, N), α ⊗ R/I is a split exact sequence, by Theorem 1.20. Therefore,

X_{β}/IX_{β} ≅ X_{α}/IX_{α}
≅ M/IM ⊕ N/IN.

By Lemma 1.19, the complex N/IN → X_{β}/IX_{β} → M/IM → 0 is a split exact sequence.

In general, given a short exact sequence α : 0 → N → X → M → 0, tensoring by R/I gives a long exact sequence with homology groups. As we have seen, if α ∈ I Ext_{R}^{1}(M, N) implies that the first three modules in this long exact sequence form a split short exact sequence. We can ask the following

Question 1.22. If α ∈ I Ext_{R}^{1}(M, N), does the long exact sequence

⋯ → Tor_{i}(R/I, N) → Tor_{i}(R/I, X_{α}) → Tor_{i}(R/I, M) → …

split in short split exact sequences

0 → Tor_{i}(R/I, N) → Tor_{i}(R/I, X_{α}) → Tor_{i}(R/I, M) → 0

for every i?

The converse of Theorem 1.20 does not hold in general, as the following example shows.

Example 1.23. Let R = k[|x^{2}, x^{3}|]. Every non-zero element α ∈ Ext_{R}^{1}(k, R) is a minimal generator and hence is not in m Ext_{R}^{1}(k, R).

Let α : 0 → R → P → k → 0, where P is the pushout of the following diagram:

[figure: 0; m; ψ; i; R; k; 0; 0; R; P; k; 0,]

where ψ is the R-homomorphism sending x^{2} to x^{3} and x^{3} to x^{4}. The short exact sequence α is not split exact because there is no map from R that extends ψ, hence α is not in m Ext_{R}^{1}(k, R). On the other hand, the minimal number of generators of P is 2 and hence P/mP ≃ k ⊕ k.

However, the following holds

Proposition 1.24. Let M and N be finitely generated R-modules, let y ∈ R be a non-zerodivisor on R, M, N and let α be the short exact sequence

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r] & M \ar[r] & 0,}

Suppose that X_{α}/yX_{α} ≃ M/yM ⊕ N/yN. Then α ∈ y Ext_{R}^{1}(M, N).

Proof. Since y is a non-zerodivisor on N we have the following exact sequences:

\xymatrix{0 \ar[r] & N \ar[r]^{y} & N \ar[r]^{π} & N/yN \ar[r] & 0,}

and, by applying the functor Hom_{R}(M, ),

\xymatrix{… \ar[r] & Ext_{R}^{1}(M, N) \ar[r]^{y} & Ext_{R}^{1}(M, N) \ar[r]^{π^{∗}} & Ext_{R}^{1}(M, N/yN).}

By exactness, to show that α ∈ y Ext_{R}^{1}(M, N), it is enough to show that π^{∗}(α) = 0.

Call ϕ : Ext_{R}^{1}(M, N/yN) → Ext_{R/yR}^{1}(M/yM, N/yN) the isomorphism that takes a short exact sequence β : 0 → N/yN → Y → M → 0 to the short exact sequence β ⊗ R/yR : 0 → N/yN → Y/yY → M/yM → 0, which is exact because y is a non-zerodivisor on M. Since π^{∗}(α) is

\xymatrix{0 \ar[r] & N/yN \ar[r] & X_{α}/yf(N) \ar[r] & M \ar[r] & 0,}

ϕπ^{∗}(α) is the short exact sequence:

\xymatrix{0 \ar[r] & N/yN \ar[r] & X_{α}/yX_{α} \ar[r] & M/yM \ar[r] & 0,}

which is split exact. Hence π^{∗}(α) = 0.

Question 1.25. Does Proposition 1.24 hold for a sequence regular on R, M and N, of length greater than one?

1.3 More results on short exact sequences

In this section we present some results about short exact sequences that will be used in further chapters. Assume that (R, m, k) is a local Noetherian ring, unless otherwise specified. Recall that, as general notation, we denote by X_{α} the module that appears in the middle of a short exact sequence α ∈ Ext_{R}^{1}(M, N). Let us start with the following

Definition 1.26. Let α : 0 → N → X_{α} → M → 0 be a short exact sequence. We say that α is additive if

μ(M) + μ(N) = μ(X_{α}).

The following Lemma gives a necessary and sufficient condition for a short exact sequence to be additive.

Lemma 1.27. Let ξ : 0 → M_{1} → F → M → 0 the initial part of a minimal free resolution of M and let α be the pushout of ξ via ϕ, where ϕ ∈ Hom_{R}(M_{1}, N), (see 1.8). Then, α is additive if and only if ϕ(M_{1}) ⊂ mN.

Proof. First of all notice that the map ϕ is not uniquely determined by the short exact sequence α; on the other hand, by Lemma 1.5, if ϕ_{1} and ϕ_{2} are two maps which give the same short exact sequence α, there exists a map f : F_{0} → N such that ϕ_{1} = ϕ_{2} + fi. Since M_{1} ⊂ mF, we have that

ϕ_{1}(M_{1}) ⊂ mN if and only if ϕ_{2}(M_{1}) ⊂ mN.

Denote by i the inclusion of M_{1} in F. By definition of the pushout, X_{α} is the cokernel of the map (i, −ϕ), as in the following short exact sequence:

\xymatrix{ν_{α} : 0 \ar[r] & M_{1} \ar[r]^{(i,−ϕ)} & F_{0} ⊕ N \ar[r]^{π} & X_{α} \ar[r] & 0.}

Tensor ν_{α} by R/m and denote the operation ⊗ R/m by ¯ . We have:

\overline{X_{α}} = \overline{F_{0}} ⊕ (\overline{N}/⟨\overline{ϕ}(\overline{l}) | \overline{l} ∈ \overline{M_{1}}⟩),

where \overline{ϕ} : \overline{M_{1}} ⟶ \overline{N} is the map induced by ϕ. If ϕ(M_{1}) ⊂ mN then the module ⟨\overline{ϕ}(\overline{l}) | \overline{l} ∈ \overline{L}⟩ = 0, hence:

μ(X_{α}) = μ(N) + μ(F_{0}) = μ(N) + μ(M).

On the other hand, suppose there exists an l ∈ M_{1} such that ϕ(l) ∉ mN, then ⟨\overline{ϕ}(\overline{l}) | \overline{l} ∈ \overline{L}⟩ ≠ 0 implies μ(\overline{N}/⟨\overline{ϕ}(\overline{l}) | \overline{l} ∈ \overline{L}⟩) < μ(\overline{N}). Therefore,

μ(X_{α}) < μ(F_{0}) + μ(N) = μ(M) + μ(N).

Corollary 1.28. In the previous lemma, assume N ≃ k^{n} in the short exact sequence α : 0 → N → X_{α} → M → 0. Then, α is split exact if and only if α is additive.

Proof. If α is split exact then α is additive.

On the other hand, keeping the notation of the above lemma, suppose that α is additive, then by the previous lemma we have that ϕ(M_{1}) ⊂ mN, where ϕ is as in the previous lemma. But N is a vector space and mN = 0. This says that the short exact sequence α is induced as a pushout by the 0 map from M_{1} to N. Therefore α is split exact.

1.3.1 Minimal generators in Ext_{R}^{1}(M, N) Before studying more in detail additive short exact sequences, we give some situations where we can describe minimal generators for Ext_{R}^{1}(M, N). First, as a corollary of Theorem 1.20, we have

Corollary 1.29. Let α be a short exact sequence in Ext_{R}^{1}(M, N) ≠ 0. and let M_{1} be the first syzygy of M in a minimal free resolution. If Ext_{R}^{1}(X_{α}, M_{1}) = 0 then α is a minimal generator of Ext_{R}^{1}(M, N).

Proof. By way of contradiction, assume α ∈ m Ext_{R}^{1}(M, N). By Theorem 1.20, α ⊗ R/m is a split exact sequence and therefore

μ(M) + μ(N) = μ(X_{α}).

On the other hand, if 0 → M_{1} → F → M → 0 is the beginning of a minimal free resolution for M, there exists an R-homomorphism ϕ : M_{1} → N such that X_{α} is the cokernel as in the following short exact sequence:

\xymatrix{β : 0 \ar[r] & M_{1} \ar[r]^{(i,−ϕ)} & F ⊕ N \ar[r] & X_{α} \ar[r] & 0,}

where i is the inclusion. Since Ext_{R}^{1}(X_{α}, M_{1}) = 0, β is a split exact sequence and X_{α} ⊕ M_{1} ≅ F ⊕ N. Therefore we have the following contradiction:

μ(M) + μ(N) = μ(X_{α}) ≤ μ(X_{α}) + μ(M_{1}) = μ(F) + μ(N) = μ(M) + μ(N).

It follows that μ(M_{1}) = 0 whixh implies that M is free and Ext_{R}^{1}(M, N) = 0, which is a contradiction.

The same conclusion holds if we have Ext_{R}^{1}(X_{α}, N) = 0 instead of the condition Ext_{R}^{1}(X_{α}, M_{1}) = 0. For this, we need some preparatory propositions.

For easy reference, recall the following: Lemma 1.30 (Schanuel). Suppose that

α : 0 → N_{α} → X_{α} → M → 0

β : 0 → N_{β} → X_{β} → M → 0

are two short exact sequences where X_{α} and X_{β} are two free modules. Then,

N_{β} ⊕ X_{α} ≅ X_{β} ⊕ N_{α}.

The previous lemma can be generalized in the following way: Lemma 1.31. Suppose that

α : 0 → N_{α} → X_{α} → M → 0

β : 0 → N_{β} → X_{β} → M → 0

are two short exact sequences such that Ext_{R}^{1}(X_{α}, N_{β}) = Ext_{R}^{1}(X_{β}, N_{α}) = 0. Then,

N_{β} ⊕ X_{α} ≅ X_{β} ⊕ N_{α}.

Proof. Apply the functor Hom( , N_{β}) to α to obtain

⋯ → Hom(N_{α}, N_{β}) → Ext_{R}^{1}(M, N_{β}) → Ext_{R}^{1}(X_{α}, N_{β}) = 0. The surjectivity of the last map implies that there exists an R-homorphism ϕ such that β is given by the following pushout

[figure: α : 0; N_{α}; ϕ; X_{α}; M; 0; β : 0; N_{β}; X_{β}; M; 0,]

where X_{β} is given as in the following short exact sequence:

ν : 0 → N_{α} → N_{β} ⊕ X_{α} → X_{β} → 0.

Since ν is an element of Ext_{R}^{1}(X_{β}, N_{α}) = 0, we have that

N_{β} ⊕ X_{α} ≅ X_{β} ⊕ N_{α}.

Proposition 1.32. Let M and N finitely generated R-modules and assume that Ext_{R}^{1}(M, N) ≠ 0. Let α be a non zero element in Ext_{R}^{1}(M, N). If Ext_{R}^{1}(X_{α}, N) = 0 then α is a minimal generator of Ext_{R}^{1}(M, N).

Proof. In this proof we keep the notation as in Corollary 1.29. By way of contradiction, assume α ∈ m Ext_{R}^{1}(M, N). By Theorem 1.20, α ⊗ R/m is a split exact sequence. Therefore, μ(X_{α}) = μ(M) + μ(N). On the other hand, by Lemma 1.31 N ⊕ F_{0} ≅ M_{1} ⊕ X_{α}, which implies the following equalities:

μ(X_{α}) + μ(M_{1}) = μ(N) + μ(F_{0}) = μ(N) + μ(M) = μ(X_{α}).

It follows that M_{1} = 0, which implies that M is a free module and hence Ext_{R}^{1}(M, N) = 0, contradicting the assumptions.

1.3.2 Additive sequences We now study additive sequences; we will prove first that they form a submodule of Ext_{R}^{1}(M, N).

Proposition 1.33. Let α and β be two short exact sequences in Ext_{R}^{1}(M, N). If α and β are additive then α + β is additive.

Proof. Assume that α and β are given respectively as pushouts of

\xymatrix{ξ : 0 \ar[r] & M_{1} \ar[r] & F \ar[r] & M \ar[r] & 0,}

via the homomorphism ϕ_{1}, ϕ_{2} ∈ Hom_{R}(M_{1}, N). Then α +β is the pushout of ξ via the homomorphisms ϕ_{1}+ϕ_{2}. Since ϕ_{i}(M_{1}) ⊂ mN, (ϕ_{1}+ϕ_{2})(M_{1}) ⊂ mN and hence, by Lemma 1.27, α + β is additive.

Proposition 1.34. Let α ∈ Ext_{R}^{1}(M, N) be an additive short exact sequence. Let N′ be a finitely generated R-module and let ϕ ∈ Hom_{R}(N, N′). If β ∈ Ext_{R}^{1}(M, N′) is the pushout of α via ϕ, then β is additive.

Similarly, let M′ be a finitely generated R-module and let ϕ ∈ Hom_{R}(M′, M). If β is the pullback of α via ϕ, then β is additive.

Proof. For the first part of the proposition, we have the following diagram

[figure: α : 0; N; ϕ; X_{α}; M; 0; β : 0; N′; X_{β}; M; 0,]

where X_{β} is the cokernel of the short exact sequence

\xymatrix{0 \ar[r] & N \ar[r] & N′ ⊕ X_{α} \ar[r] & X_{β} \ar[r] & 0.}

If β is not additive then we have the following contradiction:

μ(X_{β}) + μ(N) < μ(M) + μ(N′) + μ(N)
= μ(X_{α}) + μ(N′)
≤ μ(X_{β}) + μ(N).

The proof goes the same way for the second statment in the proposition. Indeed, we have the following diagram:

[figure: β : 0; N; X_{β}; M′; ϕ; 0; α : 0; N; X_{α}; M; 0,]

where X_{β} is the kernel of the short exact sequence

\xymatrix{0 \ar[r] & X_{β} \ar[r] & M′ ⊕ X_{α} \ar[r] & M \ar[r] & 0.}

If β is not additive then we have the following contradiction

μ(X_{β}) + μ(M) < μ(M) + μ(N) + μ(M′)
= μ(X_{α}) + μ(M′)
≤ μ(X_{β}) + μ(M).

Definition 1.35. Denote by < M, N >⊂ Ext_{R}^{1}(M, N) the set of all elements in Ext_{R}^{1}(M, N) which are additive.

Corollary 1.36. < M, N > is an R-module.

Proof. Closure under multiplication by an element of R is assured by Proposition 1.34; indeed given a short exact sequence α ∈ Ext_{R}^{1}(M, N) and an element r ∈ R, rα corresponds to the pushout of α via multiplication by r. Finally, closure under sum is assured by Proposition 1.33.

If we apply Proposition 1.13 to an additive short exact sequence, then the maps can be restricted to additive short exact sequences. In particular, Proposition 1.37. Let α ∈< M, N >. Write

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Let A be a finitely generated R-module. There exists a long exact sequence:

\xymatrix{0 \ar[r] & Hom_{R}(A, N) \ar[r] & Hom_{R}(A, X_{α}) \ar[r] & Hom_{R}(A, M) \ar[r]^{d} & … \\ … \ar[r]^{d} & < A, N > \ar[r]^{f^{∗}} & < A, X_{α} > \ar[r]^{g^{∗}} & < A, M >,}

where

d(ϕ) = αϕ,
f^{∗}(β) = fβ,
g^{∗}(β) = gβ.

Proof. The maps d and f^{∗} are well-defined by Proposition 1.34. By Proposition 1.13, we have that Image d ⊂ ker f^{∗} and Image f^{∗} ⊂ ker g^{∗}. To prove the proposition we need to check that ker g^{∗} ⊂ Image f^{∗}.

For, let β ∈< A, X_{α} > be such that g^{∗}(β) = 0. Recall that in Proposition 1.13 we constructed the short exact sequence γ (1.1.3) such that f^{∗}(γ) = β. In particular, we have the following situation:

\xymatrix{γ : 0 \ar[r] & N \ar[r] & L′ \ar[r] & A \ar[r] & 0, \\ β : 0 \ar[r] & X_{α} \ar[r] & L \ar[r] & A \ar[r] & 0,}

where L and L′ are related by the following short exact sequence:

\xymatrix{0 \ar[r] & L′ \ar[r] & L \ar[r] & M \ar[r] & 0.}

Assume that γ is not additive, then we have the following contradiction:

μ(N) + μ(A) = μ(X_{α}) − μ(M) + μ(A),  since α is additive,
= μ(L) − μ(M),  since β is additive
≤ μ(L′),  by the relation between L and L′.

Since the other inequality always holds, we have that γ has the right number of generators.

With a similar proof, the following Proposition holds (see Proposition 1.14):

Proposition 1.38. Let α be an extension of M by N

\xymatrix{0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r]^{g} & M \ar[r] & 0.}

Let A be a finitely generated R-module. There exists a long exact sequence:

\xymatrix{0 \ar[r] & Hom_{R}(M, A) \ar[r] & Hom_{R}(X_{α}, A) \ar[r] & Hom_{R}(N, A) \ar[r]^{d} & … \\ … \ar[r]^{d} & < M, A > \ar[r]^{g^{∗}} & < X_{α}, A > \ar[r]^{f^{∗}} & < M, A >,}

where

d(ϕ) = ϕα,
f^{∗}(β) = βf,
g^{∗}(β) = βg.

1.3.3 Structure of Ext_{R}^{1}(M, N) In this section we describe y Ext_{R}^{1}(M, N), where y is a non-zerodivisor on R, M and N as a disjoint union of subsets. This description will be particularly useful in studying rings of finite Cohen-Macaulay type. Recall that for any short exact sequence α, we denote by X_{α} the module which appears in the middle. Also, given an R-module X we denote by [X] the isomorphism class of X.

For any R-module X that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N), define the set:

E_{[X]} := {α ∈ Ext_{R}^{1}(M, N) such that X_{α} ≅ X}.

For any element y ∈ m define the set:

S_{y} := {[X] such that ∃α ∈ y Ext_{R}^{1}(M, N) and X_{α} ≅ X}.

with these definitions in mind, we are ready to state and prove the following:

Theorem 1.39. Let (R, m, k) be a Noetherian local ring with strictly positive depth. Assume that y is a non-zerodivisor on R, M and N. Then

y Ext_{R}^{1}(M, N) = ⋃_{X∈S_{y}} E_{[X]}.

Proof. The fact that y Ext_{R}^{1}(M, N) ⊂ ⋃_{X∈S_{y}} E_{[X]} is clear by the definition of the sets S_{y}. For the other inclusion, it is enough to show that if

E_{[X]} ∩ y Ext_{R}^{1}(M, N) ≠ ∅,

then E_{[X]} ⊂ y Ext_{R}^{1}(M, N). Let α ∈ E_{[X]} ∩y Ext_{R}^{1}(M, N) and β ∈ E_{[X]}. Since X_{α} ≅ X_{β} ≅ X we can apply Theorem 1.18 to conclude that β ⊗ R/(y) is a split exact sequence. Since y is a non-zerodivisor on M, N and R we can apply Proposition 1.24 to conclude that β ∈ y Ext_{R}^{1}(M, N).

Notice that the sets E_{[X]} are not submodules in general. Proposition 1.40. For every E_{[X]} ⊂ Ext_{R}^{1}(M, N), where X ≇ M ⊕N, there exists an integer n such that m^{n} Ext_{R}^{1}(M, N) ∩ E_{[X]} = ∅.

To prove the proposition we need to recall the following Theorem due to Guralnick, [10]. Theorem 1.41. Let (R, m) be a Noetherian local ring. Let M and N be finite R-modules. Then there exists a nonnegative integer h := h(M, N) such that for any n ∈ ℕ and for any R-linear map σ ∈ Hom_{R}(M/m^{n+h}M, N/m^{n+h}N), there exists an R-homomorphism τ ∈ Hom_{R}(M, N) such that σ and τ induce the same homorphism in Hom_{R}(M/m^{n}M, N/m^{n}N). Corollary 1.42. Let α ∈ Ext_{R}^{1}(M, N). There exists an integer h(X_{α}, N) such that if α ⊗ R/m^{h+1} is a split short exact sequence then α is a split exact sequence. Proof. Write

\xymatrix{α : 0 \ar[r] & N \ar[r]^{f} & X_{α} \ar[r] & M \ar[r] & 0.}

Let h(X_{α}, N) the integer given by Theorem 1.41. For every R-module X and Y, given a map l ∈ Hom_{R}(X, Y), denote by l_{n} the map

l ⊗_{R} id_{R/m^{n}} : X/m^{n}X → Y/m^{n}Y.

Assume that α ⊗ R/m^{h+1} is a split exact sequence, then there exists an R/m^{h+1}-homorphism

g′ : X_{α}/m^{h+1}X_{α} → N/m^{h+1}N

such that f_{h+1}g′ = id_{h+1}. By Theorem 1.41, we can lift g′ to an R-homorphism g ∈ Hom_{R}(X_{α}, N) in such a way that g and g′ induce the same homorphism g_{1} in Hom_{R}(X_{α}/mX_{α}, N/mN). Therefore we have the following chain of equalities:

(gf)_{1} = g_{1}f_{1}
= (g_{h+1} ⊗ id_{R/m})(f_{h+1} ⊗ id_{R/m})
= (g_{h+1}f_{h+1}) ⊗ id_{R/m}
= id_{h+1} ⊗ id_{R/m}
= id_{1}

This implies that gf(N) = N + mN and hence, by Nakayama’s Lemma, gf is a surjective map from N to itself. Since N is Noetherian, gf is an isomorphism and α is a split exact sequence.

Proof. of Proposition 1.40. Let X ≇ M ⊕N and let α ∈ E_{[X]} ⊂ Ext_{R}^{1}(M, N). Let h(X_{α}, N) be as in Theorem 1.41. We claim that for any n > h(X_{α}, N),

E_{[X]} ∩ m^{n} Ext_{R}^{1}(M, N) = ∅.

Assume, by way of contradiction, that the intersection is not empty and let β be an element of it. Since X_{α} ≅ X_{β} ≅ X, by Theorem 1.20, β ∈ m^{n} Ext_{R}^{1}(M, N) implies α ⊗ R/m^{n} is a split exact sequence. Since n ≥ h + 1, α ⊗ R/m^{h+1} is a split exact sequence and by the above corollary it follows that α is a split exact sequence. This is a contradiction, since

X_{α} ≅ X ≇ M ⊕ N.

It would be nice to know when the sum of two sequences in E_{[X]} is still in E_{[X]}. In general this is hard to know, even in the case when X is a free module. Before giving an example, we prove the following

Lemma 1.43. Let α ∈ E_{[F]} ∩ Ext_{R}^{1}(M, N), where F is a free R-module such that μ(F) = μ(M). For any ϕ ∈ Hom_{R}(N, N), ϕα ∈ E_{[F]} if and only if ϕ is an automorphism.

Proof. Write

[figure: α : 0; N; ϕ; f; F; M; 0; ϕα : 0; N; P; M; 0,]

where P is the cokernel of the following exact sequence:

\xymatrix{ν : 0 \ar[r] & N \ar[r]^{(ϕ,−f)} & N ⊕ F \ar[r] & P \ar[r] & 0.}

If ϕ is an automorphism, then by Snake Lemma, we have P ≅ F. If P is a free module the short exact sequence ν is split exact. Therefore there exist a map ψ = (ψ_{1}, ψ_{2}), such that ψ_{1}ϕ − ψ_{2}f = id_{N}. Since f(N) ⊂ mF, by Nakayama’s lemma ψ_{1}ϕ is a surjective map. Since N is Noetherian, ψ_{1}ϕ is an isomorphism. It follows that ϕ is an isomorphism.

Example 1.44. In this example we show that E_{F} ⊂ Ext_{R}^{1}(M, M_{1}) is not closed under addition modulo m Ext_{R}^{1}(M, N). Assume M_{1} is a syzygy in a minimal free resolution, α : 0 → M_{1} → F → M → 0. Assume that M_{1} = A ⊕ B and let β be the pushout of α via ϕ

( \begin{matrix}id_{A} & 0 \\ 0 & − id_{B} \end{matrix} ) .

Then α + β is the pushout of α via 1 + ϕ

( \begin{matrix}id_{A} & 0 \\ 0 & 0 \end{matrix} ) .

By Lemma 1.43, β ∈ E_{[F]} but α+β ∉ E_{[F]}. Moreover, since (1+ϕ)(N) ⊈ mN, α + β is not additive and therefore not in m Ext_{R}^{1}(M, N).

1.3.4 Annihilators of short exact sequences We end the section with a proposition about the annihilators of short exact sequences. We will use this proposition later in the chapter, in Proposition 1.60. Proposition 1.45. Let α and β be two short exact sequences in Ext_{R}^{1}(M, N). If Ext_{R}^{1}(X_{α}, N) = Ext_{R}^{1}(X_{β}, N) = 0 then Ann(α) = Ann(β). Proof. Apply the functor Hom_{R}( , N) to the short exact sequence

\xymatrix{α : 0 \ar[r] & N \ar[r] & X_{α} \ar[r] & M \ar[r] & 0,}

to obtain the long exact sequence

\xymatrix{… \ar[r] & Hom_{R}(X_{α}, N) \ar[r] & Hom_{R}(N, N) \ar[r] & Ext_{R}^{1}(M, N) \ar[r] & 0.}

Since Hom_{R}(N, N) surjects onto Ext_{R}^{1}(M, N) and β ∈ Ext_{R}^{1}(M, N), then there exists a ϕ ∈ Hom_{R}(N, N) such that β is the pushout of α via ϕ. Therefore Ann(α) ⊂ Ann(β). Since we can repeat the same argument replacing α by β, we have the required equality.

The following examples were shown to me by Giulio Caviglia and they show two short exact sequences with the same modules but with different annihilators.

Example 1.46. Let α and β the following sequences of ℤ-modules:

\xymatrix{α : 0 \ar[r] & ℤ^{2} ⊕ ℤ_{2} \ar[r]^{f_{α}} & ℤ^{2} ⊕ ℤ_{4} ⊕ ℤ_{2} \ar[r]^{g_{α}} & ℤ_{4} ⊕ ℤ_{2}^{2} \ar[r] & 0,}

where f_{α}(x, y, z) = (4x, y, 2z, 0) and g_{α}(x, y, z, t) = (x, z, t).

\xymatrix{β : 0 \ar[r] & ℤ^{2} ⊕ ℤ_{2} \ar[r]^{f_{β}} & ℤ^{2} ⊕ ℤ_{4} ⊕ ℤ_{2} \ar[r]^{g_{β}} & ℤ_{4} ⊕ ℤ_{2}^{2} \ar[r] & 0,}

where f_{β}(x, y, z) = (2x, 2y, 0, z) and g_{β}(x, y, z, t) = (z, x, y). Notice that β = β_{1} ⊕ β_{2} ⊕ β_{2}, where β_{1} is the split exact sequence

0 → ℤ_{2} → ℤ_{2} ⊕ ℤ_{4} → ℤ_{4} → 0

and β_{2} is the generator of Ext_{ℤ}^{1}(ℤ_{2}, ℤ). The annihilator of β is therefore 2ℤ. On the other hand α is the direct sum of the split exact sequence

0 → ℤ → ℤ ⊕ ℤ_{2} → ℤ_{2} → 0

and the generators of Ext_{ℤ}^{1}(ℤ_{4}, ℤ) and Ext_{ℤ}^{1}(ℤ_{2}, ℤ_{2}). Therefore Ann(α) = 4ℤ. The example shows also that Ann α and Ann β do not have the same integral closure (see exercise A3.29, page 656, [7]).

1.4 Sparse modules

In this section we introduce the notion of sparse modules and we study them. Sparse modules are introduced to better understand the structure of the modules Ext_{R}^{1}(M, N), where M and N are maximal Cohen-Macaulay modules over a ring of finite Cohen-Macaulay type (see the following section). Definition 1.47. Let M be a finitely generated module over a local Noetherian ring (R, m) of positive depth. M is said to be sparse if there is only a finite number of submodules xM, where x ∈ m is a non-zerodivisor on R.

Being sparse is closed under taking direct sums and quotients. On the other hand, submodules of sparse modules need not to be sparse, as the following example shows.

Example 1.48. Let R = k[|x, y|], with k infinite. Let M be the cokernel of the following map:

\xymatrix{R^{4} \ar[r]^{β} & R^{2}}

where β is given by the following matrix:

( \begin{matrix}0 & x & 0 & y^{2} \\ x & y & y^{2} & 0 \end{matrix} ) .

Let m_{1}, m_{2} be elements of M corresponding to (1, 0), (0, 1) ∈ R^{2}. Notice that m^{2}M = 0. A basis for mM is given by ym_{1} and ym_{2}. The multiplication by an element ax+by ∈ m\m^{2} takes u_{1}m_{1} +u_{2}m_{2} ∈ M\mM to v_{1}ym_{1} +v_{2}ym_{2}, where v_{1} and v_{2} are given as follows:

( \begin{matrix}v_{1} \\ v_{2} \end{matrix} ) = ( \begin{matrix}b & 0 \\ −a & b \end{matrix} ) ( \begin{matrix}u_{1} \\ u_{2} \end{matrix} ) .

It follows that the only submodules, which are not zero, of the form lM with l ∈ m are xM and (x + y)M. If N = Rm_{1}, mN is a two dimensional vector space and therefore it cannot be sparse.

Proposition 1.49. Let (R, m) be a local ring of positive depth. If M is a sparse module then M is an Artinian module.

Proof. We can choose a set of generators of the maximal ideal which are non-zero-divisors on the ring R, say x_{1}, …, x_{n}. For each i = 1, …, n, there exists h and s such that x_{i}^{h}M = x_{i}^{h+s}M, since M is sparse and powers of regular elements are regular. By Nakayama’s lemma, x_{i}^{h}M = 0. Therefore there exists a power of the maximal ideal that annihilates M. Proposition 1.50. Let (R, m, k) be a local ring of positive depth. Assume that the residue field is infinite. If M is a sparse module then there exists an element l ∈ m such that lM = mM. Proof. Let l_{1}M, …, l_{h}M the list of all possible modules of the form xM, where x ∈ m is a regular element on R. This list is finite since M is sparse. Let x_{1}, …, x_{n} be a list of minimal generators for the maximal ideal m. Choose a sequence of vectors v_{i} = (v_{i}^{1}, …, v_{i}^{n}) ∈ k^{n} such that v_{1}, …, v_{n} are linearly independent and for i > n, v_{i} is linearly independent with any subset of n − 1 vectors of v_{1}, …, v_{i−1}. After at most hn + 1 steps, there exists a j (1 ≤ j ≤ h) and n linearly independent vectors (which for simplicity we rename v_{1}, …, v_{n}) such that

l_{j}M = y_{s}M, for any s = 1, …, n,

where y_{s} = ∑_{i=1}^{n} v_{s}^{i}x_{i}. Since the vectors v_{i} are linearly independent, the y_{i} are a system of generators for m. We claim that mM = l_{j}M. For, if a ∈ m then a = ∑ r_{i}y_{i} and aM ⊂ y_{1}M + ⋯ + y_{h}M ⊂ l_{j}M so that mM ⊂ l_{j}M. Since the other inclusion always holds, we have equality. Corollary 1.51. Let (R, m, k) be a local Noetherian ring of positive depth and infinite residue field. Let M a finitely generated sparse R-module. Denote by μ the number of minimal generators. Then μ(m^{h}M) ≤ μ(M), for any h ≥ 0. Proof. By Proposition 1.50, for any integer k, we have that m^{k}M = l^{k}M, from which the proposition follows.

It would be very interesting to find a characterization of sparse modules. The following example shows what can happen in a very specific situation.

Example 1.52. Let (R, m, k) be a local ring with algebraically closed residue field k. Suppose that the maximal ideal is generated by three elements, l_{1}, l_{2}, l_{3} and that M is a R-module generated by two elements, m_{1}, m_{2}. Assume that M is sparse and that m^{2}M = 0 so that mM is a vector space. By Corollary 1.51, μ(mM) ≤ 2. Assume that μ(mM) = 2 and fix a basis v_{1}, v_{2}. We can describe the multiplication of M by an element of the maximal ideal, by a 2 × 2 matrix with entries homogeneous linear polynomials with coefficients in k. Indeed, for any element a in m, we can write a = xl_{1} + yl_{2} + zl_{3} + ξ, where ξ ∈ m^{2} and x, y, z ∈ k. Since m^{2}M = 0, for any m ∈ M we have am = (xl_{1} + yl_{2} + zl_{3})m. If m = b_{1}m_{1} + b_{2}m_{2}, then am = c_{1}v_{1} + c_{2}v_{2}, where

( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = ( \begin{matrix}α_{11}x + β_{11}y + γ_{11}z & α_{12}x + β_{12}y + γ_{12}z \\ α_{21}x + β_{21}y + γ_{21}z & α_{22}x + β_{22}y + γ_{22}z \end{matrix} ) ( \begin{matrix}b_{1} \\ b_{2} \end{matrix} ) ,

and the columns of the matrix represents am_{i} in terms of the v_{1}, v_{2}. Denote the 2 × 2 matrix by A(x, y, z). Studying when the module is sparse is equivalent to study when the matrix A(x, y, z) has a finite number image spaces, while x, y, z vary.

The entries of the matrix represents four lines in ℙ^{2}, therefore we have the following cases.

(1) There exists no point P (x_{0}, y_{0}, z_{0}) where three of the four lines intersect. We will show that in this case M cannot be sparse.

After a change of coordinates we can write the matrix A(x, y, z) as

( \begin{matrix}x & y \\ z & ax + by + cz \end{matrix} ) ,

where none of the a, b, c are zero. We are looking for conditions forcing this matrix to have a finite number of images. If the rank of the matrix is two then the image is the all vector space mM. Therefore, it is enough to study what happen when the rank of the matrix is one. The determinant being zero gives us the conic C : ax^{2} + bxy + cxz − yz = 0. Either the conic C is irreducible or not. In the first case a + bc ≠ 0. For any point P = (x_{0}, y_{0}, z_{0}) ≠ (0, 1, 0) the image of A(x_{0}, y_{0}, z_{0}) is generated by the vector (x_{0}, z_{0}). On the other hand we claim that the map that takes a point P (x_{0}, y_{0}, z_{0}) ≠ (0, 1, 0) of the conic to the vector (x_{0}, z_{0}) is injective. If the claim is true the points of the conic different from (0, 1, 0) are in one to one correspondence with one dimensional subspaces of k^{2}. Since the conic is irreducible and the field algebraically closed, there are infinitely many different possible images of the matrix A(x, y, z) and therefore the module cannot be sparse. To prove the claim, notice that for any point (x_{0}, y_{0}, z_{0}) ≠ (0, 1, 0), y_{0} is uniquely given by \frac{−ax_{0}^{2}−cx_{0}z_{0}}{bx_{0}−z_{0}} , where bx_{0} − z_{0} ≠ 0. For it, if x_{0} = 0 and bx_{0} − z_{0} = 0 then z_{0} = 0 which is a contradiction since our point is different from (0, 1, 0). If x_{0} ≠ 0 and bx_{0} − z_{0} = 0 then

x_{0}^{2}(a + bc) = x_{0}^{2}a + x_{0}^{2}bc = x_{0}^{2}a + x_{0}cz_{0} = y_{0}(bx_{0} − z_{0}) = 0

which leads to the contradiction a + bc = 0 since we are assuming the the conic C is irreducible.

Suppose that the conic is reducible in two lines

cx − y = 0 and − bx + z = 0.

The vectors (s, t) for any value of s, t are all possible images of the matrix A(P) for some P in the conic, more in particular in the first line.

(2) There exists a point P ∈ P^{2} where at least three of the four lines rapresented by the entries of the matrix, intersect. There are two subcases:

(a) The fourth line goes through P. We will show that in this case there are always a infinite number of image spaces. A change of coordinates the matrix A(x, y, z) to the form:

( \begin{matrix}x & y \\ x + y & ax + by + cz \end{matrix} ) ,

where c ≠ 0. Again, we need to study what happen when the rank of the matrix is one which means that we have to study the family of matrix obtained by specializing x, y, z with any point P ∈ C where C is given by the equation ax^{2}+bxy+cxz−xy−y^{2} = 0 and it is always irreducible. For any P (x_{0}, y_{0}, z_{0}) ∈ C but P ≠ (0, 0, 1) the image of the matrix A(P) is generated by the vector (x_{0}, x_{0} + y_{0}). The map the takes a point P ∈ C different from (0, 0, 1) to the vector generating the image, is a one to one map. For it, it is enough to notice that given x_{0} and y_{0} not both equal to zero, there exists a unique z_{0} such that (x_{0}, y_{0}, z_{0}) ∈ C: z_{0} = \frac{x_{0}y_{0}+y_{0}^{2}−ax_{0}^{2}−bxy}{cx_{0}}. Notice that the denominator is always different from zero since c is and

x_{0} = 0 would imply y_{0} = 0 which is a contradiction.

(b) The fourth lines does not go through P. We will show that the module M is sparse. After a change of coordinates, we can write the matrix A(x, y, z) as follows:

( \begin{matrix}x & y \\ x + y & ax + by \end{matrix} )

As previously, we reduce ourselves to study what happen to the images along the conic C, which in this case is given by the equation ax^{2} + (b − 1)xy − y^{2} = 0. It is easy to see that the number of images are finite. For example, if a = 0 and b ≠ 1, then C is the union of the two lines y = 0 and y = (b − 1)x. For any point P in the first line different from (0, 0, 1) the image of A(P) is generated by the vector (1, 1). For any point P in the second line different from (0, 0, 1) the image of A(P) is generated by the vector (1, b − 1). Notice that the two lines meet exactly in (0, 0, 1) in which the rank of the matrix A is zero.

Suppose that R is a standard graded ring over k. Assume k is infinite and R has depth at least one. Let M be a finitely generated graded R-module. Then, we can formulate the following

Definition 1.53. M is said to be homogeneously sparse if there are a finite number of submodules xM where x ∈ m is a homogeneous non-zerodivisor of R.

Question 1.54. In the case of graded modules over graded ring, being sparse implies being homogeneously sparse. Is the converse true?

Theorem 1.55. Let R be a standard graded ring over an algebraically closed field k and assume that R_{1} is generated by two elements. Let M be a standard graded module over R. Then, M is homogeneously sparse if and only if M is Artinian and there exists a linear form l ∈ m such that lM = mM, where m is the homogeneous maximal ideal.

Proof. The proof for the “only if” direction is the graded version of Proposition 1.49 and Proposition 1.50.

For the other direction, suppose that M is Artinian and that there exists an l ∈ m such that lM = mM. Since M is Artinian, it is enough to show that there are finitely many submodules of the form fM, where f is a homogeneous polynomial of degree less than d, for some d > 0. Since k is algebraically closed, any homogeneous polynomial in two variables can be factored into linear terms; hence it is enough to show that there are finitely many submodules of the form tM, where t is a linear form. Moreover, since M = ⊕M_{i}, it is enough to show that there are finitely many tM_{i}, where t is a linear form and M_{i} is the i-th degree component of the module M. It is enough to show that there are finitely many k vector spaces of the form tM_{0}, for t a linear form. Indeed, if we want to show that there are finitely many tM_{i} for i > 0, replace the module M by the module generated by M_{i}. Without loss of generality, let R_{1} be generated by l and s. Since lM = mM, μ(mM) ≤ μ(M). Suppose m_{1}, ..., m_{h} are minimal generators in degree zero, such that lm_{1}, ..., lm_{h} generate mM. Complete m_{1}, ..., m_{h} to a minimal system of generators of M, m_{1}, ..., m_{n} say. Let αl + βs be a general linear form.

Let A to be the matrix where the columns are the images in M_{1} of (αl+βs)m_{i} in terms of the lm_{i}, i = 1, …, h:

( \begin{matrix}α + βa_{11} & βa_{12} & … & βa_{1h} & … & βa_{1n} \\ βa_{21} & α + βa_{22} & … & βa_{2h} & … & βa_{1n} \\ … & … & … & … & … & … \\ βa_{h1} & … & … & α + βa_{hh} & … & βa_{hn} \end{matrix} ) .

If there are only finitely many (α, β) such that the first h−columns have a zero determinant, then the module M is sparse. Since the determinant of the first h-columns is a homogeneous polynomial of degree h in two variables, it is enough to show that it is not identically zero. But the values α = 1, β = 0 give a nonzero determinant.

1.5 Application to rings of finite Cohen-Macaulay type

In this section we apply the previous material to rings of finite Cohen-Macaulay type. These rings are widely studied; Yoshino’s monograph [24] is a comprehensive source for information about rings of finite Cohen-Macaulay type. We use the theory of sparse modules developed in the previous section to prove a theorem of Auslander [2], which states that if R is a complete local ring of finite Cohen-Macaulay type then R is isolated singularity. We are able to prove this fact without assuming that the ring is complete (see also the work of Huneke-Leuschke [12]). Specifically, we prove that the modules Ext_{R}^{1}(M, N) are sparse. We give a bound for the power of the maximal ideal that annihilates Ext_{R}^{1}(M, N) which improves the one given in [12].

In the following of the section, (R, m, k) is a Notherian local ring. Fix two finitely generated R-modules M and N. For any α ∈ Ext_{R}^{1}(M, N) denote X_{α} the module which appears in the middle of the short exact sequence α. Recall the following

Definition 1.56. A local Noetherian ring (R, m) is said to be of finite Cohen-Macaulay type if it is Cohen-Macaulay and if there exists a finite number of isomorphism classes of indecomposable maximal Cohen-Macaulay modules.

Proposition 1.57. Let (R, m) be a ring of finite Cohen-Macaulay type and let M and N be finitely generated maximal Cohen-Macaulay R-modules. Then Ext_{R}^{1}(M, N) is a sparse R-module.

Proof. If R is of finite Cohen-Macaulay type then there is a finite number of modules that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N). Therefore there is a finite number of possible sets E_{[X]} and hence, by Proposition 1.39, there is a finite number of submodules of the type y Ext_{R}^{1}(M, N), where y is a non-zerodivisor on R (and on M and N, since they are maximal Cohen-Macaulay).

Corollary 1.58. Let (R, m) a ring of finite Cohen-Macaulay type. Then it has a isolated singularity.

Proof. It is enough to show that Ext_{R}^{1}(M, N) are modules of finite length for any maximal Cohen-Macaulay modules M and N. For suppose this is true and set d = dim(R). For every prime ideal p ≠ m, consider a free resolution of R/p. At the d-th step we get a short exact sequence, which is an element of Ext_{R}^{1}(Ω_{d}(R/p), Ω_{d+1}(R/p))

0 → Ω_{R}^{d+1}(R/p) → F_{d} → Ω_{R}^{d}(R/p) → 0.

Both the modules at the side are maximal Cohen-Macaulay therefore after localizing at p this is a split exact sequence. This implies that both Ω_{R}^{d+1}(R/p)_{p} and Ω_{R}^{d}(R/p)_{p} are free R_{p}-modules. Therefore, the residue field of R_{p} has a finite free resolution, and so R_{p} is a regular local ring. By Proposition 1.57, Ext_{R}^{1}(M, N) is a sparse module and hence Artinian, by Proposition 1.49.

The fact that rings of finite Cohen-Macaulay type have isolated singularities was proved first by Auslander [2] in the case when R is complete. In this generality it is a theorem due to Huneke-Leuschke [12]. Notice that another proof of this fact follows from Proposition 1.40. Indeed, assume there is a finite number of isomorphic classes of modules X that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N), [X_{1}], …, [X_{t}], [X_{t+1}], where X_{t+1} ≅ M ⊕ N; as in Proposition 1.40, take

n = max{h(X_{i}, N) + 1 | i = 1, …, t}.

Then E_{[X_{i}]} ∩ m^{n} Ext_{R}^{1}(M, N) = ∅, for i = 1, …, t. Since

Ext_{R}^{1}(M, N) = ∪_{i=1}^{n}E_{[X_{i}]} ∪ E_{[M⊕N]},

we have

m^{n} Ext_{R}^{1}(M, N) = (m^{n} Ext_{R}^{1}(M, N)) ∩ (∪_{i=1}^{n}E_{[X_{i}]} ∪ E_{[M⊕N]}) = E_{[M⊕N]} = 0.

In their paper Huneke and Leuschke gave a bound on the power of the maximal ideal that kills the R-module Ext_{R}^{1}(M, N). In particular, if h is the number of modules that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N) then m^{h} Ext_{R}^{1}(M, N) = 0. We are able to improve the bound.

Proposition 1.59. Let (R, m, k) a Cohen-Macaulay local ring with infinite residue field. Let M and N be maximal Cohen-Macaulay R-modules and let h be the number of isomorphism classes of modules that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N). Then m^{h−1} Ext_{R}^{1}(M, N) = 0.

Proof. Assume by way of contradiction that m^{h−1} Ext_{R}^{1}(M, N) ≠ 0 and let

X_{1}, …, X_{h−1},
X_{h} ≅ M ⊕ N

be a list of representatives for all the isomorphism classes of modules that can fit in the middle of a short exact sequence in Ext_{R}^{1}(M, N). To simplify notation, denote by E_{i} the set E_{[X_{i}]}. Define two sets of positive integers, S_{1} := {0, …, h − 1} and S_{2} := {1, …, h − 1}. Define a map ϕ from S_{1} to the power set of S_{2}, in such a way that r ∈ ϕ(i) if and only if there exists a minimal generator for m^{i} Ext_{R}^{1}(M, N) which is in E_{r}. We claim that the sets ϕ(i) are disjoint subsets of S_{2}. If the claim holds, we have the desired contradiction since the cardinality of S_{2} is strictly smaller than the cardinality of S_{1}. Before proving the claim, recall that, by Proposition 1.50, there exists a non-zerodivisor l ∈ m such that m Ext_{R}^{1}(M, N) = l Ext_{R}^{1}(M, N). To prove the claim, assume by way of contradiction that there exist i < j such that r ∈ ϕ(i)∩ϕ(j). Pick α and β in E_{r} to be minimal generators for l^{j} Ext_{R}^{1}(M, N) and l^{i} Ext_{R}^{1}(M, N) respectively. Since X_{α} ≅ X_{β} ≅ X_{r}, we can apply Theorem 1.18 to conclude that β ⊗ R/(l^{j}) is a split exact sequence and Proposition 1.24 to prove that β ∈ l^{j} Ext_{R}^{1}(M, N) = m^{j} Ext^{1}(M, N) ⊂ ml^{i} Ext_{R}^{1}(M, N), contradicting the fact that β is a minimal generator for l^{i} Ext_{R}^{1}(M, N).

Using the decomposition of the Ext modules in subset E_{i} we can give more information on the structure of the Ext.

Proposition 1.60. In the same setup of the previous proposition, assume that N = M_{1} where M_{1} is the first syzygy in a free resolution of M. Assume that the residue field k is algebraically closed and that

m^{h−2} Ext_{R}^{1}(M, N) ≠ 0.

Then Ext_{R}^{1}(M, N) is the direct sum of cyclic R-modules, whose Hilbert functions have as only values 0 and 1.

Proof. In this proof we will use the notation of the proof of the above proposition. In particular, S_{1} := {0, …, h−2} and S_{2} := {1, …, h−1} are two sets of positive integers. In the above proof we showed that the sets ϕ(i), where ϕ is the map defined above between S_{1} and the power set of S_{2}, are disjoint for any i ∈ S_{2}. Therefore, since m^{h−2} Ext_{R}^{1}(M, M_{1}) ≠ 0, we have that the cardinality of ϕ(i) is one, for every i ∈ S_{1}. In particular, all the short exact sequences which are minimal generators for the R-modules m^{i} Ext_{R}^{1}(M, M_{1}) have isomorphic modules in the middle. By Proposition 1.32, the initial part of the minimal free resolution of M, α : 0 → M_{1} → F → M → 0, is a minimal generator for the R-module Ext_{R}^{1}(M, M_{1}) and hence any other minimal generator of the R-module Ext_{R}^{1}(M, M_{1}) belongs to E_{1}, where we define E_{1} to be the set of all short exact sequences having a free module isomorphic to F in the middle. By Proposition 1.50, there exists an element l ∈ m\m^{2}, which is a non-zero divisor on R, such that

l^{h−2} Ext_{R}^{1}(M, M_{1}) = m^{h−2} Ext_{R}^{1}(M, M_{1}) ≠ 0;

hence there exists a β ∈ E_{1} such that l^{j}β ≠ 0, for j = 1, …, h − 2. By Proposition 1.45, l^{j}γ ≠ 0, for every γ ∈ E_{1} and j = 1, …, h − 2.

Let α_{1}, …, α_{m} be a minimal set of generators for Ext_{R}^{1}(M, M_{1}). We first claim that for any j ∈ S_{2}, l^{j}α_{1}, …, l^{j}α_{m} are minimal generators for m^{j} Ext_{R}^{1}(M, M_{1}). Assume there exists a linear combination ∑_{i=1}^{m} λ_{i}l^{j}α_{i} = 0, where λ_{i} ∈ k, and not all equal to zero, hence l^{j}(∑_{i=1}^{m} λ_{i}α_{i}) = 0. Since ∑_{i=1}^{m} λ_{i}α_{i} ∈ E_{1}, applying Proposition 1.45 we get that l^{j}γ = 0 for every γ ∈ E_{1}, contradicting the last statement of the above paragraph.

The last step is to prove that we can complete l to a minimal system of generators for m, say l, l_{1}, …, l_{n}, such that l_{i} Ext_{R}^{1}(M, N) = 0, for every i = 1, …, n. To prove this, we claim that if l, l_{1}, …, l_{n} is a system of minimal generators for m, then there exists a λ_{i} ∈ k such that (l_{i} − λ_{i}l)γ = 0 for every γ ∈ E_{1} and for every i = 1, …, n. To prove the claim, notice that the multiplication by l_{i}:

[figure: \frac{Ext_{R}^{1}(M,M_{1})}{m Ext_{R}^{1}(M,M_{1})}; l_{i}; \frac{m Ext_{R}^{1}(M,M_{1})}{m^{2} Ext_{R}^{1}(M,M_{1})},]

is a k-linear map between k^{m} and itself. Since k is algebraically closed, there exists an eigenvalue λ_{i} and an eigenvector γ_{i}. we can write l_{i}γ_{i} = λ_{i}lγ_{i}. This means that (l_{i} − λ_{i}l)γ_{i} = 0, since γ_{i} ∈ E_{1}, by Proposition 1.45, we have (l_{i} − λ_{i}l)γ = 0 for every γ ∈ E_{1}. By replacing l_{i} with l_{i} − λ_{i}l, we have the claim.

This shows that we can write Ext_{R}^{1}(M, M_{1}) = ⊕_{i=1}^{m}Rα_{i}, and each Rα_{i} has the property of the Hilbert function to have possible as only values 0 and 1.

Remark 1.61. Notice that the same conclusion of the above proposition holds if all the minimal generators for Ext_{R}^{1}(M, N) have a free module in the middle.

1.6 Other applications

1.6.1 Efficient systems of parameters Recall the following definitions:

Definition 1.62. Let R be a T algebra and let μ the multiplication map from R ⊗_{T} R → R. The Noetherian differ is N_{T}^{R} = μ(Ann_{R⊗_{T} R}(ker(μ))).

Definition 1.63. A system of parameters x_{1}, …, x_{n} is an efficient system of parameters if for any i = 1, …, n there is a regular subring T_{i} of R over which R is finite and such that x_{i} belongs to the Noetherian different N_{T_{i}}^{R}.

The proofs above can be used to give an improvement of the following proposition (Proposition 6.17, [24]):

Proposition 1.64. Let x = {x_{1}, …, x_{n}} be an efficient system of parameters. Let α : 0 → N → X → M → 0 be a short exact sequence of maximal Cohen-Macaulay modules. Denote by x^{2} the ideal generated by x_{1}^{2}, …, x_{n}^{2} and let \overline{α} be the short exact sequence 0 → N/x^{2}N → X/x^{2}X → M/x^{2}M → 0. If \overline{α} is split exact then α is split exact.

We can substitute x^{2} simply by x and prove the following Proposition:

Proposition 1.65. Let x = {x_{1}, …, x_{n}} be an efficient system of parameters. Let α : 0 → N → X → M → 0 be a short exact sequence of maximal Cohen-Macaulay modules and let \overline{α} be 0 → N/xN → X/xX → M/xM → 0. If \overline{α} is split exact then α is split exact.

Proof. Denote by x_{i} = {x_{1}, …, x_{i}} and by α_{i} the following short exact sequence:

\xymatrix{0 \ar[r] & N/x_{i}N \ar[r] & X/x_{i}X \ar[r] & M/x_{i}M \ar[r] & 0.}

We will show by descending induction on i that α_{i} is split exact. The case i = 0 will give the thesis of the improved Proposition 1.65. The assumption say that α_{n} is a split exact sequence, which is the case i = n. Suppose that the case i = k is true. Then, by Proposition 1.24,

α_{k−1} ∈ x_{k} Ext_{R/x_{k−1}R}^{1}(M/x_{k−1}M, N/x_{k−1}N) ≃ x_{k} Ext_{R}^{1}(M, N/x_{k−1}N).

By definition of efficient systems of parameters, there exists a regular subring S_{k} such that the extension S_{k} ⊂ R is finite and x_{k} is in N_{S_{k}}^{R}, the Noetherian different. By a result proved by Wang ([21], Proposition 5.9), we have that

N_{S_{k}}^{R} Ann(Ext_{S_{k}}^{1}(M, N/x_{k−1}N)) ⊂ Ann(Ext_{R}^{1}(M, N/x_{k−1}N)).

But M is a free S_{k}−module so that Ann(Ext_{S_{k}}^{1}(M, N/x_{k−1}N)) is the unit ideal of S_{k} and hence x_{k} ∈ Ann(Ext_{R}^{1}(M, N/x_{k−1}N)), which implies that α_{k−1} is a split exact sequence.

1.6.2 Splitting of syzygies In this section (R, m) is a local Noetherian ring, and M is a finitely generated R-module. We denote by Ω_{R}^{i}(M) the i-th syzygy of M in a minimal free resolution over the ring R. In this section we give a simpler proof of the following theorem Theorem 1.66. Let (R, m) be a local Noetherian ring and let z = (z_{1}, …, z_{n}) be a regular sequence on the ring and on a finitely generated R-module M. Then, z Ext_{R}^{i}(M, ) = 0, for i ≥ 1, if and only if

Ω_{R}^{j}(M/(z_{1}, …, z_{n})M) ≅ ⨁_{i=0}^{j} Ω_{R}^{i}(M)^{\binom{j}{i}},

for every j = 1, …, n.

This theorem was first proved by O’Carroll and Popescu in [18], Theorem 2.1.

Let

\xymatrix{ξ : 0 \ar[r] & Ω_{R}^{1}(M) \ar[r] & F_{0} \ar[r]^{π} & M \ar[r] & 0}  (1.6.1)

be part of a minimal free resolution of M. If z is a regular element on R and M, then Ω_{R}^{1}(M/zM) = Ω_{R}^{1}(M) + zF_{0}. Remark 1.67. Let z be a regular element on the ring R and on a module M, then there exists the following short exact sequence:

\xymatrix{γ(M, z) : 0 \ar[r] & Ω_{R}^{1}(M) \ar[r] & Ω_{R}^{1}(M/zM) \ar[r]^{f} & M \ar[r] & 0,}

where f(a + zb) = π(b), for a ∈ Ω_{R}^{1}(M) and b ∈ F_{0}. Notice that the map is well defined, in fact if a + zb = a_{1} + zb_{1}, for a, a_{1} ∈ Ω_{R}^{1}(M) and for b, b_{1} ∈ F_{0}, then 0 = π(a) − π(a_{1}) = π(a − a_{1}) = zπ(b_{1} − b) ∈ M. Since z is a nonzerodivisor on M, we have that π(b_{1} − b) = 0, showing that the map f is well-defined. Moreover for every element a+zb in ker f we have that π(b) = 0 and hence b and a + zb ∈ Ω_{R}^{1}(M).

Lemma 1.68. With the same notation as in Remark 1.67, the following holds

γ(M, z) ∈ z Ext_{R}^{1}(M, Ω_{R}^{1}(M)),

and

γ(M, z) = 0 if and only if z Ext_{R}^{1}(M, ) = 0.

Proof. It is enough to show that γ(M, z) = zξ, where ξ is as in 1.6.1. The following diagram shows that γ(M, z) is the pushout of ξ and therefore proves the claim:

[figure: ξ : 0; Ω_{R}^{1}(M); z; F_{0}; z; M; 0; γ(M, z) : 0; Ω_{R}^{1}(M); Ω_{R}^{1}(M) + zF_{0}; f; M; 0.]

Remark 1.69. Notice that z Ext_{R}^{1}(M, ) = 0 implies z Ext_{R}^{i}(M, ) = 0 for every i ≥ 2. Indeed, by the previous lemma, we have that M ⊕ Ω_{R}^{1}(M) = Ω_{R}^{1}(M/zM) and hence

0 = z Ext_{R}^{i}(M/zM, ) = z Ext_{R}^{i−1}(Ω_{R}^{1}(M/zM), )
= z Ext_{R}^{i−1}(M, ) ⊕ z Ext_{R}^{i−1}(Ω_{R}^{1}(M), ),

so that each summand at the right hand side is zero. Remark 1.70. Lemma 1.68 and Remark 1.69 give the proof of Theorem 1.66 for n = 1. Indeed, M ⊕ Ω_{R}^{1}(M) ≅ Ω_{R}^{1}(M/zM) if and only if γ(M, z) = 0 (by Miyata’s Theorem), iff and only if z Ext_{R}^{1}(M, ) = 0 (by Lemma 1.68), if and only if z Ext_{R}^{i}(M, ) = 0 (by Remark 1.69).

Remark 1.71. The above proof shows that

γ(M, z) ∈ z Ext_{R}^{1}(M, Ω_{R}^{1}(M)) ⊂ m Ext_{R}^{1}(M, Ω_{R}^{1}(M)).

Therefore by 1.20 μ(M) + μ(Ω_{R}^{1}(M)) = μ(Ω_{R}^{1}(M/zM)).

Before proving the theorem we need another lemma. If z is a non-zerodivisor, we denote M/zM by \overline{M}.

Lemma 1.72. Let z be a non-zerodivisor on the ring and on a finitely generated R-module M, then Ω_{R}^{i}(Ω1 ; \overline{R}(\overline{M})) = Ω_{R}^{i+1}(\overline{M}).

Proof. Since z is a non-zero divisor on M, we have zF_{0} ∩ Ω_{R}^{1}(M) = zΩ_{R}^{1}(M). Therefore there exists the following short exact sequence:

\xymatrix{0 \ar[r] & zF_{0} \ar[r] & Ω_{R}^{1}(M) + zF_{0} \ar[r] & Ω_{R}^{1}(M)/zΩ_{R}^{1}(M) \ar[r] & 0.}

Notice that zF_{0} is a free module and hence, using the Horseshoe Lemma we have:

Ω_{R}^{i}(Ω_{R}^{1}(M + zF_{0})) ≅ Ω_{R}^{i}(Ω_{R}^{1}(M)/zΩ_{R}^{1}(M)),

for every i > 0. In particular we have:

Ω_{R}^{i+1}(\overline{M}) ≅ Ω_{R}^{i}(Ω_{R}^{1}(\overline{M}))
≅ Ω_{R}^{i}(Ω_{R}^{1}(M) + zF_{0})
≅ Ω^{i}(Ω_{R}^{1}(M)/zΩ_{R}^{1}(M))
≅ Ω^{i}(Ω1 ; \overline{R}(\overline{M})).

We are ready to give a proof of Theorem 1.66 Proof. One direction of the theorem is easy to prove. Indeed, if

Ω_{R}^{n}(M/(zM)) ≅ ⨁_{i=0}^{n} Ω_{R}^{i}(M)^{\binom{n}{i}}

then

(z) ⊂ Ann_{R}(Ext_{R}^{i+n}(M/zM, )
= Ann_{R}(Ext_{R}^{i}(Ω_{R}^{n}(M/zM), ))
= Ann_{R}(⊕_{j=0}^{j=n} Ext_{R}^{i}(Ω_{R}^{j}(M), ))
⊂ Ann_{R}(Ext_{R}^{i}(M, )), for every i ≥ 1.

We will prove the other direction of the statement by induction on n. The case n = 1 is given by Remark 1.70. Assume that the conclusion holds for i ≤ n. Denote by \overline{M} = M/(z_{1} … z_{n})M and consider the sequence γ(Ω_{R}^{n}(\overline{M}), z_{n+1}), where z_{n+1} is a non-zero divisor on \overline{M} and therefore on Ω_{R}^{n}(\overline{M}). Assume that (z_{1}, …, z_{n+1}) Ext_{R}^{i}(M, ) = 0, in particular we have

(z_{1}, …, z_{n}) Ext_{R}^{i}(M, ) = 0

and , by induction,

Ω_{R}^{n}(\overline{M}) ≅ ⨁_{i=0}^{n} Ω_{R}^{i}(M)^{\binom{n}{i}}.  (1.6.2)

By Lemma 1.68,

z_{n+1} Ext_{R}^{1}(Ω_{R}^{n}(\overline{M}), ) = 0 if and only if γ(Ω_{R}^{n}(\overline{M}), z_{n+1}) = 0.

Moreover, γ(Ω_{R}^{n}(\overline{M}), z_{n+1}) = 0 if and only if

Ω_{R}^{1}(\frac{Ω_{R}^{n}(\overline{M})}{z_{n+1}Ω_{R}^{n}(\overline{M})}) ≅ Ω_{R}^{n}(\overline{M}) ⊕ Ω_{R}^{n+1}(\overline{M})

For the left hand side, we have

Ω_{R}^{1}(\frac{Ω_{R}^{n}(\overline{M})}{z_{n+1}Ω_{R}^{n}(\overline{M})}) ≅ Ω_{R}^{1}(Ω_{R/z_{n+1}R}^{n}(\frac{M}{(z_{1}, … z_{n}, z_{n+1})M}))
≅ Ω_{R}^{n+1}(\frac{M}{(z_{1}, …, z_{n+1})M}),

where the last congruence is true by Lemma 1.72. Putting this together with equation 1.6.2 we obtain

Ω_{R}^{n+1}(M/(z_{1}, …, z_{n+1})M) ≅ Ω_{R}^{n}(\overline{M}) ⊕ Ω_{R}^{n+1}(\overline{M})
≅ Ω_{R}^{n}(\overline{M}) ⊕ Ω_{R}^{1}(Ω_{R}^{n}(\overline{M}))
≅ ⊕_{j=0}^{n}Ω_{R}^{j}(M)^{\binom{n}{j}} ⊕ Ω_{R}^{1}(⊕_{j=0}^{n}Ω_{R}^{j}(M)^{\binom{n}{j}})
≅ ⊕_{j=0}^{n+1}Ω_{R}^{j}(M)^{\binom{n+1}{j}}.

On the other hand, again by the induction hypothesis,

z_{n+1} Ext_{R}^{1}(Ω_{R}^{n}(\overline{M}), ) = 0

if and only if z_{n+1} Ext_{R}^{1}(Ω_{R}^{j}(M), ) ≅ z_{n+1} Ext_{R}^{j+1}(M, ) = 0, for j = 0, …, n.

Chapter 2

Artin Rees Numbers

We say that a Noetherian ring R has the uniform Artin-Rees property with respect to the set of ideals W if for any given two finitely generated R-modules N ⊆ M there exists an integer h, depending only on M and N, such that for any I ∈ W and for any n ≥ h we have I^{n}M ∩ N ⊆ I^{n−h}N. Such h is said to be a uniform Artin-Rees number, and we denote the minimum of such h by ar(N ⊂ M, W).

We say that a Noetherian ring R has the strong uniform Artin-Rees property with respect to a set of ideals W if for any given two modules N ⊆ M there exists an integer s, depending on N and M, such that for any I ∈ W and for any n ≥ s we have I^{n}M ∩ N = I(I^{n−1}M ∩ N). We call such a s a strong uniform Artin-Rees number and, more precisely, we denote it as AR(N ⊆ M, W) or AR(N ⊆ M) whenever the set W is the set of all ideals. In this last case we say that the ring R has the strong uniform Artin-Rees property.

The question of whether an excellent ring has the strong uniform Artin-Rees property with respect to the family of maximal ideals, was raised by Eisenbud-Hochster in [8] and positively answered by Duncan-O’Carroll in [6]. O’Carroll [17] proved that the family of principal ideals has the strong uniform Artin-Rees property. Huneke [11] proved that many Noetherian rings have the uniform Artin-Rees property for the family of all ideals in the ring. Planas-Vilanova [19] proved that one dimensional excellent rings have the strong uniform Artin-Rees property; here we give a simple proof of this fact. We also give some examples in two-dimensional rings. A three-dimensional ring fails in general to have a strong uniform Artin-Rees number as it is shown in [22].

2.1 One dimensional rings

In this section we prove that excellent rings of dimension one have the strong uniform Artin-Rees property. Our main reduction is to show that it is enough to find a uniform bound for the family of all m-primary ideals, as in the case of the uniform Artin-Rees property. To do this we first need a lemma.

Lemma 2.1. Let (R, m) be a Noetherian local ring and let N_{1}, N_{2} ⊆ M be finitely generated R-modules. Then there exists h_{0} = h_{0}((N_{1} + N_{2}) ⊆ M) such that for any h ≥ h_{0} we have:

N_{1} ∩ (N_{2} + m^{h}M) ⊆ (N_{1} ∩ N_{2}) + m^{h−h_{0}}N_{1}.

Proof. Let h_{0} be chosen such that for any h ≥ h_{0} we have

m^{h}M ∩ (N_{1} + N_{2}) = m^{h−h_{0}}(m^{h_{0}}M ∩ (N_{1} + N_{2})) ⊂ m^{h−h_{0}}(N_{1} + N_{2}).

Then the following holds for h > h_{0}:

N_{1} ∩ (N_{2} + m^{h}M) = N_{1} ∩ (N_{2} + (m^{h}M ∩ (N_{1} + N_{2}))
⊆ N_{1} ∩ (N_{2} + m^{h−h_{0}}(m^{h_{0}}M ∩ (N_{1} + N_{2})))
⊆ N_{1} ∩ (N_{2} + m^{h−h_{0}}(N_{1} + N_{2}))
⊆ N_{1} ∩ N_{2} + m^{h−h_{0}}N_{1}.

Remark 2.2. Notice that if h_{0} is an integer that satisfies Lemma 2.1, any bigger integer does also. Proposition 2.3. Let R be a Noetherian ring and N ⊆ M two finitely generated R-modules. Suppose that there exists a strong uniform Artin-Rees number s(N ⊆ M, W) where W = {I ⊆ R | nilrad(I) = m is a maximal ideal}. Then there exists a strong uniform Artin-Rees number s(N ⊆ M), with respect to the family of all ideals of R. Proof. We actually prove that the Artin-Rees number s = s(N ⊆ M, W) works for all ideals. Suppose by contradiction that there exists I ⊂ R and n ≥ s such that I^{n−s}(I^{s}M ∩N) ≠ I^{n}M ∩N; since this inequality is preserved after localizing at some maximal ideal m, we may assume (R, m) local. On the other hand, for all h >> 0 and for such a fixed n and s, we have:

I^{n}M ∩ N ⊆ (I + m^{h})^{n}M ∩ N
⊆ (I + m^{h})^{n−s}((I + m^{h})^{s}M ∩ N), by the definition of s,
⊆ I^{n−s}((I + m^{h})^{s}M ∩ N) + m^{h}M, by expanding the powers,
⊆ I^{n−s}((I^{s} + m^{h})M ∩ N) + m^{h}M
= I^{n−s}((I^{s}M + m^{h}M) ∩ N) + m^{h}M.

Let h_{0} be an integer depending on (I^{s}M + N) ⊆ M that satisfies Lemma 2.1 with N_{1} = N, N_{2} = I^{s}M. By Remark 2.2, we may assume h_{0} ≥ n − s. Applying Lemma 2.1, we have (I^{s}M + m^{h}M) ∩ N ⊆ (I^{s}M ∩ N) + m^{h−h_{0}}M, for any h > h_{0}. Therefore, the following holds:

I^{n−s}((I^{s}M + m^{h}M) ∩ N) + m^{h}M ⊆ I^{n−s}(I^{s}M ∩ N + m^{h−h_{0}}M) + m^{h}M,
⊆ I^{n−s}(I^{s}M ∩ N) + m^{h−h_{0}+n−s}M
+m^{h}M,
⊆ I^{n−s}(I^{s}M ∩ N) + m^{h−h_{0}+n−s}M.

Putting together the right and the left end of the chain of inclusions, we have

I^{n}M ∩ N ⊆ I^{n−s}(I^{s}M ∩ N) + m^{h−h_{0}+n−s}M,

for any h > h_{0}. By taking the intersection of the right side of the inclusion over all h > h_{0}, we can conclude I^{n}M ∩ N ⊆ I^{n−s}(I^{s}M ∩ N). Since the reverse inclusion always holds, we conclude I^{n−s}(I^{s}M ∩ N) = I^{n}M ∩ N, contradicting the assumption.

The following are standard reductions.

Proposition 2.4. Let R → S be a faithfully flat extension. If the strong uniform Artin-Rees property holds for S, then it holds for R.

Proof. Let N ⊆ M be two finitely generated R-modules, let I be an ideal of R and let s be a strong uniform Artin-Rees number for N ⊗_{R} S ⊂ M ⊗_{R} S.

Then for any n ≥ s and for any ideal I of R, we have:

(I^{n}M ∩ N) ⊗_{R} S = (I^{n}M ⊗_{R} S) ∩ (N ⊗_{R} S)
= ((IS)^{n}M ⊗_{R} S) ∩ (N ⊗_{R} S)
= (IS)^{n−s}(((IS)^{s}M ⊗_{R} S) ∩ (N ⊗_{R} S))
= I^{n−s}(I^{s}M ∩ N) ⊗_{R} S.

Since S is faithfully flat, it follows that I^{n}M ∩ N = I^{n−s}(I^{s}M ∩ N).

Remark 2.5. Note that if dim R = 0 then R has the strong uniform Artin-Rees property. In fact there are finitely many maximal ideals. Let h be an integer such that m^{h}R_{m}, for all maximal ideals in R. Since for every primary ideal, h is a strong uniform Artin-Rees number, then by Proposition 2.3 h is a strong uniform Artin-Rees number for all the ideals of R. Remark 2.6. Let (R, m) be a local Noetherian ring. Then R ⟶ R[x]_{mR[x]} is a faithfully flat extension and R[x]_{mR[x]} has an infinite residue field.

Proposition 2.7. Suppose (R, m) is a one-dimensional Noetherian local ring with infinite residue field. Then there exists an integer h, depending only on the ring, such that for any m-primary ideal I there exists y ∈ I such that I^{n} = yI^{n−1}, for any n ≥ h.

Proof. First suppose that R is Cohen-Macaulay and let e be the multiplicity of the ring. By [20], Theorem 1.1, Chapter 3, for every m-primary ideal, we have that μ(I) ≤ e, where μ(I) denotes the minimal number of generators of I. Therefore, we have μ(I^{e}) ≤ e < e + 1. Hence, by [20], Theorem 2.3, Chapter 2, there exists y ∈ I such that I^{e} = yI^{e−1}, so that for any n ≥ e = h we have I^{n} = yI^{n−1}.

Next suppose depth(R) = 0, and let 0 = q_{1}∩q_{2} ⋯∩q_{s+1} be a minimal primary decomposition of 0 where q_{s+1} is m-primary and set J = q_{1} ∩q_{2} ⋯∩q_{s}. Then R/J is Cohen-Macaulay and there exists a h_{0} such that m^{h_{0}}J = 0. Let e_{1} be the multiplicity of R/J then, by the above case, there exists a y ∈ I such that for any n ≥ e_{1} we have I^{n} ⊆ yI^{n−1} + J. By Huneke [11] Theorem 4.12, there exists a h_{1}, depending just on R and J such that for any n ≥ h_{1} and for any ideal I ⊂ R we have

I^{n} ∩ J ⊂ I^{n−1}J.

Hence, for any n ≥ h = max{e_{1}, h_{0} + 1, h_{1}}:

I^{n} ⊆ yI^{n−1} + I^{n} ∩ J ⊂ yI^{n−1} + I^{n−1}J ⊂ yI^{n−1} + m^{n−1} ∩ J = yI^{n−1}.

Proposition 2.8. Let (R, m) be a one-dimensional Noetherian local ring. Then R has the strong uniform Artin-Rees property.

Proof. Without loss of generality, by Proposition 2.4 and Remark 2.6, we may assume that R has infinite residue field. Let t be the integer as in Proposition 2.7 and let I be an m-primary ideal. If M/N is Cohen-Macaulay, then we can choose y ∈ I such that y is a non-zerodivisor in M/N, so that for n ≥ t,

I^{n}M ∩ N = yI^{n−1}M ∩ N,
⊆ y(I^{n−1}M ∩ N), since y is a non-zerodivisor on M/N,
⊆ I(I^{n−1}M ∩ N), since y ∈ I,

proving that t is a strong Artin-Rees number. Now suppose that M/N is not Cohen-Macaulay and let M′/N = H_{m}^{0}(M/N).

We have that M/M′ is Cohen-Macaulay, so there exists a uniform strong Artin-Rees number s for M′ ⊂ M, and there exists l such that m^{l}M′ ⊆ N. Let h = max{l, s} and let t = h + l. For any n ≥ t and for any I ⊆ R we have:

I^{n}M ∩ N = I^{n}M ∩ M′ ∩ N,  since N ⊂ M′,
= I^{n−h}(I^{h}M ∩ M′) ∩ N,  since M/M′ is Cohen-Macaulay,
⊆ I^{n−h}(I^{h}M ∩ M′),  since n − h > l and I^{l}M′ ⊂ N,
= I^{n−h−l}I^{l}(I^{h}M ∩ M′).
= I^{n−t}(I^{l}(I^{h}M ∩ M′) ∩ N),  since I^{l}M′ ⊂ N,
⊆ I^{n−t}(I^{h+l}M ∩ M′ ∩ N),
= I^{n−t}(I^{t}M ∩ N),

proving that t is a strong Artin-Rees number.

Let I = (f_{1}, …, f_{n}) be an ideal in R minimally generated by the polynomials f_{i}. Map the polynomial ring, with the standard grading, R[x_{1}, …, x_{n}], onto the Rees algebra R[It] by sending x_{i} to f_{i}t. Let L be the kernel of this map. Then L is an homogeneous ideal and the rt(I) is defined to be the minimum integer s such that the ideal L can be genenerated by elements of degree less than or equal to s. This number does not depend on the choice of the generators of the ideal I; see for example [19]. The following lemma had been basically proved by Wang in [22].

Lemma 2.9. Let (R, m) be a Noetherian local ring and J be an ideal of R; denote R̄ = R/J. Let I = (x_{1}, …, x_{m}) be an ideal of R and suppose that rt(IR̄) ≤ h, for some h > 0. Then for any n ≥ h,

I^{n} ∩ J = I^{n−h}(I^{h} ∩ J).

Proof. Let n ≥ h and let x ∈ I^{n} ∩ J. Then there exists F (X_{1}, …, X_{m}) ∈ R[X_{1}, …, X_{m}], a homogeneous polynomial of degree n, such that F (x_{1}, …, x_{m}) = x. Going modulo J, F̄ is a relation on the x̄_{i}’s, so by hypothesis there are polynomials G_{i} of degree h, and H_{i}, of degree n − h, such that F̄ = ∑ Ḡ_{i}H̄_{i} in R̄[X_{1}, …, X_{m}] and Ḡ_{i} are relations on the x̄_{i}. Therefore F = ∑ G_{i}H_{i} + K for some K ∈ R[X_{1}, …, X_{m}] of degree n and coefficients in J. Since:

K(x_{1}, …, x_{m}) ∈ JI^{n} ⊂ I^{n−h}(I^{h} ∩ J),
G_{i}(x_{1}, …, x_{m}) ∈ I^{h} ∩ J, and
H_{i}(x_{1}, …, x_{m}) ∈ I^{n−h},

x = F (x_{1}, …, x_{m}) = ∑ G_{i}(x_{1}, …, x_{m})H_{i}(x_{1}, …, x_{m}) ∈ I^{n−h}(I^{h} ∩ J).

Remark 2.10. See [19], Lemma 6.3. If (R, m) is a one dimensional, Cohen-Macaulay local ring and I is an m-primary ideal, then rt(I) ≤ e, where e is the multiplicity of R.

Lemma 2.11. Let (R, m) a Noetherian local ring. Suppose J is an ideal of R such that dim(R/J) ≤ 1. Then there exists an integer k such that for any n > k and for any m-primary ideal I, I^{n} ∩ J = I(I^{n−1} ∩ J).

Proof. If dim(R/J) = 0 then there exists a power of the maximal ideal m^{h} ⊂ J. Therefore for n ≥ h + 1 and for any ideal I we have the following:

I^{n} ∩ J = I^{n} = II^{n−1} = I(I^{n−1} ∩ J).

Suppose that dim(R/J) = 1. If R/J is Cohen-Macaulay then the conclusion holds by the previous remark and by Lemma 2.9.

Suppose R/J has dimension one and it is not Cohen-Macaulay. Let J ⊂ J′ such that R/J′ is Cohen-Macaulay and let l such that m^{l}J′ ⊂ J, so that for every ideal I ⊂ R we have I^{l}J′ ⊂ J. By the Cohen-Macaulay case there exists an Artin-Rees number s = s(J′ ⊂ R). Let h = max{s, l} and let k = h + l. Then, with an argument we already used, for any n > k we have:

I^{n} ∩ J = I^{n} ∩ J′ ∩ J
= I^{n−h}(I^{h} ∩ J′) ∩ J
= I^{n−h}(I^{h} ∩ J′)
= I^{n−h−l}(I^{h+l} ∩ J′)
= I^{n−k}(I^{k} ∩ J′ ∩ J)
= I^{n−k}(I^{k} ∩ J).

This concludes the proof of the theorem.

Using techniques from [6], we can prove the following more general proposition.

Proposition 2.12. Suppose N ⊆ M are finitely generated modules over a Noetherian ring. Let J = Ann(M/N). Suppose there exist a k(J ⊂ R), such that I^{n} ∩ J = I^{n−k}(I^{k} ∩ J), for any ideal I ⊂ R and for any n > k. Assume that there is a strong uniform Artin-Rees number s(N/JM ⊂ M/JM). Then I^{n}M ∩ N = I^{n−s}(I^{s}M ∩ N), for any ideal I of R and for any n > k.

Proof. Let ϕ : R^{m} → M, a surjection of a free module onto M. Denote by K = ker(ϕ) and by L = ϕ^{−1}(N), the pre-image of the submodule N ⊂ M. Then, as shown in [6], it is enough to show that there exists a k such that for any n > k and for any ideal I ⊂ R, we have I^{n}R^{m} ∩ L = I^{n−k}(I^{k}R^{m} ∩ L). Indeed,

ϕ^{−1}(I^{n}M ∩ N) = ϕ^{−1}(I^{n}M) ∩ ϕ^{−1}(N)
= (I^{n}R^{m} + K) ∩ L
= K + (I^{n}R^{m} ∩ L)
= K + I^{n−k}(I^{k}R^{m} ∩ L);

by applying ϕ across the equality between the first and the last term we obtain the claim. Therefore without loss of generality we may assume M is a free module. By hypothesis, for any n > s and for any ideal I, we have

I^{n}M ∩ N ⊂ I^{n−s}(I^{s}M ∩ N) + JM.

Therefore

I^{n}M ∩ N ⊂ I^{n−s}(I^{s}M ∩ N) + JM ∩ I^{n}M = I^{n−s}(I^{s}M ∩ N) + (I^{n} ∩ J)M,

where the last equality holds since M is a free module. If n > k, we have I^{n} ∩ J = I^{n−s}(I^{s} ∩ J). Hence,

I^{n}M ∩ N = I^{n−s}(I^{s}M ∩ N) + I^{n−s}(I^{s} ∩ J)M
= I^{n−s}(I^{s}M ∩ N) + I^{n−s}(I^{s}M ∩ JM)
⊂ I^{n−s}(I^{s}M ∩ N), since JM ⊆ N.

Corollary 2.13. Let (R, m) a local Noetherian ring. Given two finitely generated R-modules N ⊂ M such that dim(M/N) ≤ 1, there exists an Artin-Rees number s(N ⊂ M). In other words, there exist an integer s such that for any n > s and for any ideal I ⊂ R we have I^{n}M ∩ N = I(I^{n−1}M ∩ N).

Proof. Apply Lemma 2.12 and Lemma 2.11.

The proof of the following corollary can be found in [19]. We give it here for easy reference.

Corollary 2.14. Let R be an excellent ring. Let N ⊂ M be finitely generated R-modules such that dim(M/N) ≤ 1. Then there exists a strong uniform Artin-Rees number s(N ⊂ M).

Proof. Let J = Ann(M/N). Notice that s(N ⊂ M) = s(N ⊂ M, W), where W is the family of maximal ideals in V (J). Let {P_{1}, …, P_{n}} = min(R/J) define the following subset of V (J):

A_{1} = Ass(R/J)
A_{2} = Ass(M/N)
Σ_{1} = ∪_{j>1,j<n,i_{1}<...,<i_{j}}V (P_{i_{1}} + ⋯ + P_{i_{j}})
Σ_{1} = ∅ if n = 1
Σ_{2} = ∪_{i=1}^{n} Sing(A/P_{i}).

All the sets above are finite sets of maximal ideals of R/J. For each such maximal ideal m there exists a strong Artin-Rees number for N_{m} ⊂ M_{m}. Let s be the maximum of such s_{m}. Denote R/J by \overline{R}. Let m ∉ S := A_{1} ∪ A_{2} ∪ Σ_{1} ∪ Σ_{2}, then (M/N)_{m} is a Cohen-Macaulay module over (\overline{R})_{m}, which is a Cohen-Macaulay local ring with a unique minimal prime P such that (\overline{R}/P)_{m} is a DVR with uniform parameter t. Notice that P is the image in \overline{R} of P_{i} for some i. Since (M/N)_{m} is a Cohen-Macaulay module over a Cohen-Macaulay ring it is enough to show that there a uniform bound for e(\overline{R})_{m} for any m ∉ S, see the proofs of Proposition 2.10, Proposition 2.11 and Proposition 2.12. Let

n = max_{i}{∑_{j=0}^{k} μ(P_{i}^{j}) | i = 1, …, n},

where k be a minimum integer such that the k-th power of the nilradical of \overline{R} is zero. Then

e(\overline{R}_{m}) = μ(m^{i}\overline{R}_{m}), for i >> 0,
= μ((P + t)^{i}\overline{R}_{m}), since (\overline{R}/P \overline{R})_{m} is a DVR,
= ∑_{j=0}^{k} μ(P^{j}\overline{R}_{m}),
= ∑_{j=0}^{k} μ(P^{j}) ≤ n, by the definition of n.

2.2 Two dimensional rings

The following example was inspired by the example in [22], it shows that the uniform Artin-Rees property does not hold for two dimensional rings.

Example 2.15. Let R = k[x, y, z]/(z^{2}). Consider the following family of ideals:

I_{n} =< x^{n}, y^{n}, x^{n−1}y + z >,

for any n ∈ ℕ. Let J the ideal generated by z. We claim that I_{n}(I_{n}^{n−1} ∩ J) ≠ I_{n}^{n} ∩ J, for any n. In particular we will show that

x^{(n−1)^{2}}y^{n−1}z ∈ I_{n}^{n} ∩ J but x^{(n−1)^{2}}y^{n−1}z ∉ I_{n}(I_{n}^{n−1} ∩ J).

Denote x^{(n−1)^{2}}y^{n−1}z by ξ. Notice that I_{n} is a homogeneous ideal if we assign degree one to x and y and degree n to z. With such assigment, the degree of ξ is (n−1)^{2}+n−1+n = n^{2}. Since x^{(n−1)^{2}}y^{n−1}z = (x^{n−1}y+z)^{n}−(x^{n})^{n−1}y^{n} ∈ I_{n}^{n} the first claim holds. Suppose that x^{(n−1)^{2}}y^{n−1}z ∈ I_{n}(I_{n}^{n−1}∩J), this remains true going modulo the elements x^{(n−1)^{2}+1}, y^{n}. Notice that I_{n}^{n−1} mod (x^{(n−1)^{2}+1}, y^{n})R is generated by

< x^{n(n−1−i)}(x^{n−1}y + z)^{i} | i = 0, 1, …, n − 1 > .

Moreover,

x^{n(n−1−i)}(x^{n−1}y + z)^{i} = x^{n(n−1−i)}(x^{(n−1)i}y^{i} + x^{(n−1)(i−1)}y^{i−1}z)
= x^{n^{2}−n−i}y^{i} + x^{n^{2}−2n−i+1}y^{i−1}z.

But n^{2}−n−i ≥ (n−1)^{2}+1 for i ≤ n−2. Therefore I_{n−1}^{n} mod (x^{(n−1)^{2}+1}, y^{n}) is generated by

< x^{(n−1)^{2}}y^{n−1} + x^{(n−1)(n−2)}y^{(n−2)}z, x^{n^{2}−2n−i+1}y^{i−1}z | i = 1, …, n − 2 > .

Denote by

f = x^{(n−1)^{2}}y^{n−1} + x^{(n−1)(n−2)}y^{(n−2)}z

and by

g_{i} = x^{n^{2}−2n−i+1}y^{i−1}z.

Notice that ξ ∉ I_{n}G, where G is the ideal generated by the g_{i}’s. Indeed, the degree of x in the g_{i} is n^{2}−2n−i+1 and (n^{2}−2n−i+1)+(n−1) > n^{2}−2n+1 for i ≤ n − 2. Let hf + ∑ h_{i}g_{i} be a homogeneous element of I_{n}^{n−1} ∩ J that appear in the expression on ξ as element of I_{n}(I_{n}^{n−1} ∩ J). By the previous observation, we can assume h ≠ 0. Let m be a homogeneous monomial of h. If z does not divide m, then

mf = m′(x, y)x^{(n−1)(n−2)+1}y^{(n−2)}z

or

mf = m′(x, y)x^{(n−1)(n−2)}y^{(n−2)+1}z;

if z does divide m then mf = m′(x, y)x^{(n−1)^{2}}y^{n−1}z, with m′ possibly a unit. By a degree counting we can see that deg(hf) ≥ n^{2} − n + 1. Therefore, for any element i ∈ I_{n−1} we have deg(ihf) > n^{2} = deg(ξ). This shows a contradiction.

The following example shows that the Artin-Rees property fails in a two dimensional ring, even if the ring is reduced. Notice that, by Corollary 2.13, R cannot be a two dimensional domain.

Example 2.16. Let R = k[x, y, z]/(xz). Consider the following family of ideals:

I_{n} = (x^{n}, y^{n}, x^{n−1}y + z^{n}),

for any n ∈ ℕ. Let J = (z). Again, we claim that for any positive integer n, I_{n}(I_{n}^{n−1} ∩ J) ≠ I_{n}^{n} ∩ J. More in particular we claim that

z^{n^{2}} ∈ I_{n}^{n} ∩ J but z^{n^{2}} ∉ I_{n}(I_{n}^{n−1} ∩ J).

Indeed, z^{n^{2}} = (x^{n−1}y + z^{n})^{n} − (x^{n})^{n−1}y^{n} ∈ I_{n}^{n} and trivially z^{n^{2}} ∈ J. On the other hand I_{n}^{n−1} is generated by:

< x^{n(n−1)}, x^{(n−1)^{2}}y^{n−1} + z^{n(n−1)}, y^{n}L, x^{(n−1)^{2}+i}y^{n−1−i} | i = 1, … n − 1 >,

for some ideal L in R. Notice that if z^{n^{2}} ∈ I_{n}(I_{n}^{n−1} ∩ J) then this holds going modulo y^{n}. Moreover, if a homogeneous element of I_{n}^{n−1} ∩ (z)

f(x, y)x^{n(n−1)} + g(x, y, z)(x^{(n−1)^{2}}y^{n−1} + z^{n(n−1)}) + ∑ h_{i}(x, y)x^{(n−1)^{2}+i}y^{n−1−i}

is in J then f, g, h_{i} must be in the maximal homogeneous ideal (they cannot be units since n^{2}−n, (n−1)^{2}+i > (n−1)^{2}). Writing g(x, y, z) = g′′(x, y)+zg′. Since the expression above has to belong to the ideal generated by z, we see that

f(x, y)x^{n(n−1)} + g′′(x, y)x^{(n−1)^{2}}y^{n−1} + ∑ h_{i}(x, y)x^{(n−1)^{2}+i}y^{n−1−i} = 0.

But if this is the case, since xz = 0 in R, we have

fx^{n(n−1)} + g(x^{(n−1)^{2}}y^{n−1} + z^{n(n−1)}) + ∑ h_{i}x^{(n−1)^{2}+i} = zg′z^{n(n−1)}.

But zg′z^{n(n−1)} is an homogeneous element of degree at least n^{2} − n + 1 and the multiplication by any element in I_{n} increases the degree by n. Therefore any element in I_{n}(I_{n}^{n} ∩ J) has degree at least n^{2} + 1 while z^{n^{2}} has degree strictly smaller.

Chapter 3

Artin-Rees bounds on syzygies

In this chapter we study a uniform property for syzygies.

Definition 3.1. Uniform Artin-Rees for syzygies. Let (R, m, k) be a local Noetherian ring and let M be a finitely generated R-module. Let 𝔽 = {F_{i}} be a free resolution of M. Denote by M_{i} ⊂ F_{i−1} the i-th syzygy. We say that M has the uniform Artin-Rees property for syzygies with respect to an ideal I if there exists an integer h such that

M_{i} ∩ I^{n}F_{i−1} ⊂ I^{n−h}M_{i},  (3.0.1)

for all n > h and for all i > 0. We denote the minimum integer h that satisfies 3.0.1 by ar_{syz}(I, M). If 3.0.1 holds for every ideal I ⊂ R then we say that M has the uniform Artin-Rees property for syzygies.

We can make a stronger definition:

Definition 3.2. Strong uniform Artin-Rees for syzygies. In the same setting of the above, we say that M has the strong uniform Artin-Rees property for syzygies with respect to an ideal I if there an integer h such that

M_{i} ∩ I^{n}F_{i−1} = I(M_{i} ∩ I^{n−1}F_{i−1}),  (3.0.2)

for all n > h and for all i > 0. We denote the minimum integer h that satisfies 3.0.2 by AR_{syz}(I, M). If 3.0.2 holds for every ideal I ⊂ R then we say that M has the strong uniform Artin-Rees property for syzygies.

Eisenbud and Huneke [9] showed that there M has the uniform Artin-Rees property for syzygies when the module M has finite projective dimension on the punctured spectrum. Throughout this chapter (R, m, k) will be a Noetherian local ring and we will denote by M_{i} the i-th syzygy of a minimal free resolution of a finitely generated R-module M while Ω_{i}(k) is the i-th syzygy of the residue field k. In this chapter we show that the residue field has the strong uniform Artin-Rees property for syzygies with respect to the maximal ideal m. We also study uniform Artin-Rees property for syzygies and how the property behaves in short exact sequence. We show that the property with respect to an ideal J ⊂ R relates to having a uniform annihilator for the family of modules T_{i,n} = Tor_{i}^{R}(M, R/J^{n}). By an argument due to Katz, we can show that over Cohen-Macaulay rings, any R module has the uniform Artin-Rees property with respect to a m-primary ideal.

It is easy to see that a module has the (strong) uniform Artin-Rees property for syzygies if and only if one of its syzygies has it.

3.1 Strong uniform Artin-Rees bounds for syzygies and the residue field

The residue field k has the strong uniform Artin-Rees property for syzygies. This fact is a consequence of the content of Corollary 3.16, in [13].

Theorem 3.3 (Levin). Let (R, m, k) be a local Noetherian ring and let 𝔽 = {F_{i}, d_{i}} be a minimal free resolution of k. Denote by Ω_{i}(k) = ker d_{i−1}. Then there exists an integer r_{0} such that for n ≥ n_{0} and for all i ≥ 1:

Tor_{i}^{R}(k, R/m^{n}) ≅ \frac{m^{n−1}Ω_{i}(k)}{m^{n}Ω_{i}(k)}.

Remark 3.4. If M is a finitely generated R-module we have

\frac{M_{i} ∩ m^{n}F_{i−1}}{m^{n}M_{i}} = Tor_{i}^{R}(M, R/m^{n}).

Indeed, from the short exact sequence 0 → M_{i} → F_{i−1} → M_{i−1} → 0, by tensoring with R/m^{n} we have

0 → Tor_{1}^{R}(M_{i−1}, R/m^{n}) → M_{i}/m^{n}M_{i} → F_{i−1}/m^{n}F_{i−1} → M_{i−1}/m^{n}M_{i−1} → 0.

From this last exact sequence we can see that the modules

Tor_{1}^{R}(M_{i−1}, R/m^{n}) = Tor_{i}^{R}(M, R/m^{n})

and \frac{M_{i}∩m^{n}F_{i−1}}{m^{n}M_{i}} are the same module being the kernel of the middle map.

Corollary 3.5. Let (R, m, k) be a local Noetherian ring. Then k has the strong uniform Artin-Rees property for syzygies.

Proof. Let n_{0} as in the previous theorem, then for any n ≥ n_{0}, we have:

\frac{Ω_{i}(k) ∩ m^{n}F_{i−1}}{m^{n}Ω_{i}(k)} = Tor_{i}^{R}(k, R/m^{n}) ≅ \frac{m^{n−1}Ω_{i}(k)}{m^{n}Ω_{i}(k)},

where the first equality holds by Remark 3.4 and the second isomorphism is given by the previous Theorem. In particular the two modules

\frac{m^{n−1}Ω_{i}(k)}{m^{n}Ω_{i}(k)} ⊆ \frac{Ω_{i}(k) ∩ m^{n}F_{i−1}}{m^{n}Ω_{i}(k)}

have the same length and therefore they are equal. We have the following chain of inclusions,

Ω_{i}(k) ∩ m^{n}F_{i−1} = m^{n−1}Ω_{i}(k)
= m(m^{n−2}Ω_{i}(k))
⊂ m(Ω_{i}(k) ∩ m^{n−1}F_{i−1}).

3.2 Uniform Artin-Rees bounds for syzygies

3.2.1 Uniform Artin-Rees bounds and short exact sequences Recall from the previous chapter that ar(J, N ⊂ M) is the minimal integer such that for any n ≥ ar(J, N ⊆ M) we have J^{n}M ∩ N ⊂ J^{n−h}N. The proof of the following proposition is in [11].

Proposition 3.6. Let M_{0} ⊆ M_{1} ⊆ ⋯ ⊆ M_{n} be finitely generated R-modules over a Noetherian ring and let J to be an ideal of R. Then

(1) ar(J, M_{0} ⊂ M_{n}) ≤ ∑_{i=1}^{n} ar(J, M_{i−1} ⊂ M_{i});

(2) If furthermore M_{i}/M_{i−1} ≅ R/I_{i} are cyclic modules for i = 1, …, n, then ar(J, M_{0} ⊂ M_{n}) ≤ ∑_{i=1}^{n} ar(J, I_{i} ⊂ R).

The following definition and remark are due to Yongwei Yao, see [23].

Definition 3.7. For any ideal J ⊂ R and any finitely generated R-module, define

ar(J, M) = max{ar(J, M′ ⊂ M′′) | M′′/M′ ≅ M}.

Notice that the maximum in the definition is well defined by Proposition 3.6(2). Remark 3.8. (1) Let α : 0 → N → X → M → 0 be a short exact se-

quence. Then ar(J, X) ≤ ar(J, N) + ar(J, M). Indeed, let X_{1} ⊂ X_{2} two finitely generated R-modules such that X_{2}/X_{1} ≅ X and let L_{1} a finitely generated R-module such that X_{1} ⊂ L_{1} ⊂ X_{2} and L_{1}/X_{1} ≅ L. Then

X_{2}/L_{1} ≅ (X_{2}/X_{1})/(L_{1}/X_{1}) ≅ X/L ≅ M.

By definition and by Proposition 3.6(1), we have

ar(J, X_{1} ⊂ X_{2}) ≤ ar(J, X_{1} ⊂ L_{1})+ar(J, L_{1} ⊂ X_{2}) ≤ ar(J, L)+ar(J, M).

By taking the maximum over finitely generated R-moduls X_{1} ⊂ X_{2} such that X_{1}/X_{2} ≅ X on the left side we get the desired result.

(2) ar(J, M) = ar(J, N ⊂ F) where F is a free R-module. Indeed, let M_{1} ⊂ M_{2} two finitely generated R-modules such that M_{2}/M_{1} ≅ M. It is enough to show that for every n > ar(J, N ⊂ F) = k, m^{n}M_{2} ∩ M_{1} ⊂ m^{n−k}M_{1}. Since F is a free module we have the following commutative diagram:

[figure: 0; N; F; h; f; M; 0; 0; M_{1}; M_{2}; g; M; 0.]

Notice that h(F) + M_{1} = M_{2} and h(F) ∩ M_{1} = h(N). Therefore, for every n > k

J^{n}M_{2} ∩ M_{1} = J^{n}(h(F) + M_{1}) ∩ M_{1}
= J^{n}M_{1} + J^{n}h(F) ∩ M_{1}
⊆ J^{n}M_{1} + h(J^{n}F) ∩ h(N)
⊆ J^{n}M_{1} + J^{n−k}h(N)
⊆ J^{n−k}M_{1}.

A priori, the property of having a uniform Artin-Rees bound for syzygies depends on the free resolution chosen. It turns out that it is enough to look at one particular free resolution. Remark 3.9. Let M be a finitely generated R-module, and let G and F be a free and a minimal free resolution of M. Denote by M′_{i} and M_{i} the ith syzygies respectively of G and F. Then there exists an isomorphism of complex ϕ_{i−1} : G_{i−1} → F_{i−1} ⊕ H_{i}, where H = {H_{i}} is the trivial complex, see [7] page 495. Therefore,

J^{n}G_{i−1} ∩ M′_{i} ⊂ J^{n−h}M′_{i} if and only if
ϕ_{i−1}(J^{n}G_{i−1} ∩ M′_{i}) ⊂ ϕ_{i−1}(J^{n−h}M′_{i}) if and only if
J^{n}ϕ_{i}(G_{i−1}) ∩ ϕ_{i}(M′_{i}) ⊂ J^{n−h}ϕ_{i}(M′_{i}) if and only if
J^{n}(F_{i−1} ⊕ H_{i}) ∩ (M_{i} ⊕ H_{i}) ⊂ J^{n−h}(M_{i} ⊕ H_{i}) if and only if
J^{n}F_{i−1} ∩ M_{i} ⊂ J^{n−h}M_{i}.

Proposition 3.10. Let α : 0 → N → X → M → 0 be a short exact sequence. Assume that N and M have the uniform Artin-Rees property with respect to an ideal J ⊂ R. Then X has the uniform Artin-Rees property on syzygies with respect to J and

ar_{syz}(J, X) ≤ ar_{syz}(J, M) + ar_{syz}(J, N).

Proof. Let F = {F_{i}} and G = {G_{i}} be minimal free resolutions of M and N. Then X has a free resolution, not necessarily minimal, where the i-th free module is the direct sum of F_{i} and G_{i}. By Remark 3.8(2), we have that

ar(J, M_{i}) = ar(J, M_{i+1} ⊂ F_{i+1}) ≤ ar_{syz}(M)

and

ar(J, N_{i}) = ar(J, N_{i+1} ⊂ F_{i+1}) ≤ ar_{syz}(J, N),

for any i > 0. Therefore,

ar(J, X_{i+1} ⊂ F_{i+1} ⊕ G_{i+1}) = ar(J, X_{i}),  by Remark 3.8(2),
≤ ar(J, M_{i})
+ ar(J, N_{i}),  by Remark 3.8(1),
= ar(J, M_{i+1} ⊂ F_{i+1})
+ ar(J, N_{i+1} ⊂ G_{i+1}),  by Remark 3.8(2),
≤ ar_{syz}(J, N) + ar_{syz}(J, M), by hypothesis.

and hence

ar_{syz}(J, X) = sup_{i∈ℕ}{ar(J, X_{i} ⊆ F_{i} ⊕ G_{i})} ≤ ar_{syz}(J, N) + ar_{syz}(J, M).

More is true for short exact sequences.

Proposition 3.11. Let α : 0 → N → X → M → 0 be a short exact sequence. If two modules have the uniform Artin-Rees property for syzygies with respect to an ideal J then also the third does. Proof. If M and N have the uniform Artin-Rees property for syzygies, then so does X, by Proposition 3.10. Assume N and X have the uniform Artin-Rees property for syzygies with respect to an ideal J. Let F a free module surjecting on X. We have the following diagram:

[figure: 0; N; X; M; 0; 0; 0; F; F; 0; L; P]

and hence, by Snake Lemma, we have the following short exact sequence:

\xymatrix{0 \ar[r] & L \ar[r] & P \ar[r] & N \ar[r] & 0.}

By hypothesis, L and N have the uniform Artin-Rees property on syzygies with respect to J. By Proposition 3.10, P and hence M have the uniform Artin-Rees property on syzygies with respect to J.

Assume that M and X have the uniform Artin-Rees property on syzygies with respect to J. Consider the following diagram:

\xymatrix{0 \ar[r] & N \ar[r] & X \ar[r] & M \ar[r] & 0 \\ 0 \ar[r] & F \ar[r] \ar@{->>}[u] & H \ar@{->>}[u] \ar[r] & G \ar@{->>}[u] \ar[r] & 0, \\ 0 \ar[r] & N′ \ar[r] \ar@{^{(}->}[u] & X′ \ar@{^{(}->}[u] \ar[r] & M′ \ar@{^{(}->}[u] \ar[r] & 0.}

where each row and column is exact, and where F, G and H are free modules. From the above diagram we have the following:

[figure: 0; N; X; M; 0; 0; N; H/N′; G; 0,; 0; 0; X′/N′; M′; 0.]

Since M has the uniform property, X′/N′ ≅ M′ has it. By Proposition 3.10, H/N′ has the uniform property with respect to J. Since the middle row is split exact, N has the uniform property with respect to J.

It seems hard to prove the uniform Artin-Rees property for syzygies by using a prime filtration.

3.2.2 Uniform Artin-Rees property for syzygies and annihilators of the

Tors modules Fix an integer i; the classical Artin-Rees Lemma, applied to the i-th syzygy, gives uniform annihilators for the family of modules T_{n} = Tor_{i}^{R}(M, R/J^{n}).

Proposition 3.12. Let (R, m, k) be a local Noetherian ring. Let M be a finitely generated R-module, J an ideal of R and let i be a fixed integer. Then there exists an integer h such that for every n > 0

J^{h} Tor_{i}^{R}(M, R/J^{n}) = 0.

Moreover there exists an integer k such that

Ann(Tor_{i}^{R}(M, R/J^{k+i})) = Ann(Tor_{i}^{R}(M, R/J^{k})),

for every i > 0.

Proof. By the usual Artin-Rees Lemma, there exists a strong Artin-Rees number AR(J, M_{i} ⊂ F_{i−1}) = h. For any n > h, we have J^{n}F_{i−1} ∩ M_{i} = J(J^{n−1}F_{i−1} ∩ M_{i}). We claim that for any n > h the following inclusion holds

Ann(Tor_{i}(M, R/J^{n})) ⊂ Ann(Tor_{i}(M, R/J^{n+1}))

To prove the claim, let c ∈ Ann(Tor_{i}(M, R/J^{n})) then, by 3.4

c(J^{n}F_{i−1} ∩ M_{i})) ⊂ J^{n}M_{i}.

Therefore,

c(J^{n+1}F_{i−1} ∩ M_{i}) ⊂ cJ(J^{n}F_{i−1} ∩ M_{i}) ⊂ J^{n+1}M_{i},

proving that c ∈ Ann(Tor_{i}^{R}(M, R/J^{n+1})). Because the ring R is Noetherian, these annihilators are eventually stable. Assume that h is the integer such that Ann(Tor_{i}^{R}(M, R/J^{h+i})) ⊂ Ann(Tor_{i}^{R}(M, R/J^{h})), for every i > 0. Then J^{h} ⊂ Ann(Tor_{i}^{R}(M, R/J^{n})), for every n > 0.

In particular the annihilator of the R-module

L_{i} = ⊕_{n} Tor_{i}^{R}(M, R/m^{n})

is m-primary. If the uniform Artin-Rees Lemma for syzygies holds, we have a uniform annihilator for the family of modules T_{i,n} = Tor_{i}^{R}(M, R/J^{n}). In fact, the following Lemma holds.

Lemma 3.13. Let (R, m) be local ring. If M has the uniform Artin-Rees property for syzygies then there exists an integer h such that for any i > 0 and for any n > 0, J^{h} ⊂ Ann(Tor_{i}^{R}(M, R/J^{n})).

Proof. Notice that J^{h} Tor_{i}^{R}(M, R/J^{n}) = 0 for every i > 0 and for every n ≤ h. Let 𝔽_{•} = {F_{i}} be a minimal free resolution of M. If M has the uniform Artin-Rees property for syzygies, then there exists an integer h such that J^{n}F_{i−1} ∩ M_{i} ⊂ J^{n−h}M_{i}, for every n > h and for every i > 0. By Remark 3.4 we have

Tor_{i}^{R}(M, R/J^{n}) = (J^{n}F_{i−1} ∩ M_{i})/J^{n}M_{i} ⊂ J^{n−h}M_{i}/J^{n}M_{i},

for every n > h and for every i > 0, showing that

J^{h} Tor_{i}^{R}(M, R/J^{n}) = 0.

The existence of an element which uniformly annihilates the family of modules T_{i,n} = Tor_{i}^{R}(M, R/m^{n}) is sometimes equivalent to the uniform Artin-Rees property for syzygies with respect to the maximal ideal m.

Proposition 3.14. Let (R, m) be a local Noetherian ring and let M be a finitely generated maximal Cohen-Macaulay R-module. Assume that there exists an element c ∈ R which is regular on the module M and on R. Then M has the uniform Artin-Rees property on syzygies with respect to the maximal ideal m if and only if there exists a non-zerodivisor c such that

c Tor_{i}^{R}(M, R/m^{n}) = 0

for every n ≫ 0 and for every i > 0, and M/cM satisfies the uniform Artin-Rees property for syzygies with respect to the maximal ideal m.

Proof. By Lemma 3.13, if M has the Artin-Rees property on syzygies with respect too the maximal ideal m, then there exists an integer h such that m^{h} ⊂ Ann(Tor_{i}^{R}(M, R/m^{n})), for any i > 0 and for any n > h. Since R is Cohen-Macaulay of positive dimension, there exists a non-zero divisor as in the proposition. For the other direction, let (𝔽, δ) be a minimal free resolution of M and notice that, since M is maximal Cohen-Macaulay and c a non-zerodivisor on R, 𝔽 ⊗ R/(c) is a minimal free resolution for M/cM. Denote by the operation of going modulo c. By assumption, \overline{M} has the uniform Artin-Rees property on syzygies with respect to the maximal ideal m, therefore there exists a h such that

m^{n}\overline{F}_{i} ∩ \overline{M}_{i} ⊂ m^{n−h}\overline{M}_{i},  (3.2.1)

for every i > 0 and for every n > h. Let u ∈ m^{n}F_{i−1} ∩ M_{i}. By taking a larger h, we can assume that 3.2.1 holds and c ∈ ⋂_{i>0,n>h} Ann Tor_{i}^{R}(M, R/m^{n}). For such values of n, since

Tor_{i}^{R}(M, R/m^{n}) = (m^{n}F_{i−1} ∩ M_{i})/m^{n}M_{i},

we have that cu ∈ m^{n}M_{i}. This means that there exist y ∈ F_{i+1}, u_{j} ∈ F_{i+1} and m_{j} ∈ m^{n}, such that

cu = cδ(y) = ∑ m_{j}δ(u_{j}).

This implies that cy − ∑ m_{j}u_{j} ∈ M_{i+1} and therefore

\overline{∑ m_{j}u_{j}} ∈ \overline{M}_{i+1} ∩ m^{n}\overline{F}_{i+1} ⊂ m^{n−h}\overline{M}_{i+1}.

This shows that ∑ m_{j}u_{j} ∈ m^{n−h}F_{i+1} +cF_{i+1} so that we can write ∑ m_{j}u_{j} = z +cy′. Then, cy′ ∈ m^{n−h}F_{i+1} ∩cF_{i+1}. By the usual Artin-Rees Lemma, there exists an l such that (c) ∩ m^{n} ⊂ m^{n−l}(c), for any n > l. Therefore,

cy′ ∈ cm^{n−h−l}F_{i+1}, for every n > h + l, which implies that y′ ∈ m^{n−h−l}F_{i+1} being c a non-zerodivisor. If u′ = δ(y′) then u′ ∈ m^{n−h−l}M_{i}. We claim that u ∈ m^{n−h−l}M_{i}, proving the uniform Artin-Rees property for syzygies with respect to the maximal ideal m, with bound h + l. For,

cu = δ(cy) = δ(∑ m_{j}u_{j}) = δ(z + cy′) = cδ(y′) = cu′

But c is a non zero divisor and therefore u = u′ ∈ m^{n−l−h}M_{i}, as desired.

3.3 Uniform Artin-Rees property for syzygies for modules over Cohen-Macaulay rings

We show that any module over a Cohen-Macaulay ring has the uniform Artin-Rees property for syzygies with respect to the maximal ideal m. Recall the following

Definition 3.15. Let R be a Noetherian ring and I ⊂ R an ideal. J ⊂ I is a reduction of I if there exists an integer h such that for any n > h we have I^{n} = J^{n−h}I^{h}. The integer h is call a reduction number.

For easy reference we recall here the following theorem due to Rees

Theorem 3.16. Let R be a ring and M an R-module. Let a_{1}, …, a_{n} be a sequence regular on M and I = (a_{1}, …, a_{n}). Let x_{1}, …, x_{n} be indeterminates over R. If g ∈ M[x_{1}, …, x_{n}] is homogeneous of degree d and g(a_{1}, …, a_{d}) ∈ I^{d+1}M then g has the coefficients in IM.

The proof can be found in [5], page 6.

Lemma 3.17. Let (R, m) be a Noetherian local ring. Let K be a submodule of a free module F and let M = F/K. Let J = (a_{1}, …, a_{l}) be an ideal generated by a sequence which is regular on M. Then J^{n}F ∩ K = J^{n}K for any n > 0. Proof. Let m ∈ J^{n}F ∩ K then there exists a homogeneous polynomial g of degree n in l variable and with coefficients in F such that g(a_{1}, …, a_{l}) = m. By going modulo K, we have a homogeneous polynomial \overline{g} of degree n in M[x_{1}, …, x_{n}] such that \overline{g}(a_{1}, …, a_{l}) = 0. We want to prove that \overline{g} is the zero polynomial. By applying Theorem 3.16, since \overline{g}(a_{1}, …, a_{l}) = 0 ∈ I^{n+1}M, we have that the coefficients of \overline{g} are in IM. Therefore there exists a homogeneous polynomial g_{1} of degree n + 1 in M[x_{1}, …, x_{n}] such that g_{1}(a_{1}, …, a_{l}) = \overline{g}(a_{1}, …, a_{l}) = 0. By repeating this argument we can see that the coefficients of \overline{g} are in I^{n}M for every n and therefore, by Krull Intersection Theorem, they are zero.

Theorem 3.18. Let (R, m, k) be a local Noetherian ring of dimension d, let M be a finitely generated R-module which is maximal Cohen-Macaulay and let I be a m-primary ideal. Let F be a free module and ϕ : F → M a surjection. Assume that the residue field is infinite. Then there exists an integer h, not depending on the module M, such that for every n > h we have

I^{n}F ∩ K ⊂ I^{n−h}K.

Assume further that R is Cohen-Macaulay then any module (not necessarly maximal Cohen-Macaulay) M has the uniform Artin-Rees property for syzygies with respect to any m-primary ideal.

Proof. Let J = (x_{1}, …, x_{d}) ⊂ I be a minimal reduction such that x_{1}, … x_{d} is a regular sequence on M. Let h be the reduction number. Let K = ker(ϕ). For every i > 0 and for every n > h we have

I^{n}F ∩ K = J^{n−h}I^{h}F ∩ K  (3.3.1)
⊂ J^{n−h}F ∩ K  (3.3.2)
= J^{n−h}K ⊂ J^{n−h}K,  (3.3.3)

where the last equality is given by the previous lemma with d = l, which we can apply since F/K ≅ M is a maximal Cohen-Macaulay module. Assume now that R is Cohen-Macaulay and let M be any finitely generated R-module. By replacing M with its d-th syzygy we can assume that M is maximal Cohen-Macaulay. Recall that M_{i} is the i-th syzygy of M. Notice that we can use equation 3.3.1 replacing K by M_{i} and F by F_{i−1}.

Lemma 3.13 gives the following

Corollary 3.19. Let (R, m, k) be a Cohen-Macaulay local ring with infinite residue field and let M be a maximal Cohen-Macaulay module. Given a mprimary ideal, there exists an integer h such that for any i > 0 and for any n > 0

J^{h} ⊂ Ann(Tor_{i}^{R}(M, R/J^{n}))

.

3.4 Uniform Artin-Rees property for syzygies in rings of dimension one

By applying Theorem 3.18 with d = 1 we obtain Proposition 3.20. Let (R, m, k) be a local Noetherian ring of dimension one with infinite residue field. There exists an integer h such that for every finitely generated R-module M of depth 1 and for every n > h

m^{h} Tor_{1}^{R}(M, R/m^{n}) = 0.

Note that in the previous corollary the integer h does not depend on the module.

Proposition 3.21. Let (R, m, k) be a local Noetherian ring of dimension one with infinite residue field. Let M be any finitely generated R-module which is a first syzygy. There exists an integer l, not depending on the module, such that for every n > 0

m^{l} Tor_{1}^{R}(M, R/m^{n}) = 0.

Proof. Without loss of generality we may assume that depth(M) = 0 and M is not a finite length module. Consider the following short exact sequence

0 → H_{m}^{0}(M) → M → M/H_{m}^{0}(M) → 0.

After tensoring by R/m^{n} we obtain the following exact sequence

Tor_{1}^{R}(R/m^{n}, H_{m}^{0}(M)) → Tor_{1}^{R}(R/m^{n}, M) → Tor_{1}^{R}(R/m^{n}, M/H_{m}^{0}(M)).

Chose a positive integer h_{1} such that m^{h_{1}}H_{m}^{0}(R) = 0. Since M is a first syzygy, we have H_{m}^{0}(M) ⊂ H_{m}^{0}(F) where F is some free module. Therefore m^{h_{1}}H_{m}^{0}(M) = 0 and m^{h_{1}} Tor_{1}^{R}(R/m^{n}, H_{m}^{0}(M)) = 0, for every n > 0. Since M/H_{m}^{0}(M) has positive depth, by the previous proposition there exists an integer h_{2}, not depending on the module M/H_{m}^{0}(M), such that m^{h_{2}} Tor_{1}^{R}(M/H_{m}^{0}(M), R/m^{n}) = 0, for every n > h_{2}. Let l = h_{1}h_{2}. For every n > h_{2} and therefore for every n > l, we have m^{l} Tor_{1}^{R}(M, R/m^{n}) = 0. On the other hand, m^{l} ⊂ Ann(Tor_{1}^{R}(M, R/m^{n})) for every n ≤ l.

Corollary 3.22. Let (R, m, k) be a local Noetherian ring of dimension one with infinite residue field. Let M be any finitely generated R-module. There exists an integer l, depending on the module, such that for every n > 0 and for every i > 0

m^{l} Tor_{i}^{R}(M, R/m^{n}) = 0.

Proof. By Proposition 3.21 there exists an integer q such that for every n > 0 and for every i > 1 we have

m^{q} Tor_{i}^{R}(M, R/m^{n}) = m^{q} Tor_{1}^{R}(M_{i−1}, R/m^{n}) = 0.

By taking i = 1 in Remark 3.12, there exists an integer p such that

m^{p} Tor_{1}^{R}(M, R/m^{n}) = 0,

for all integer n > 0. Let l be the maximum between p and q.

Recall the following definition

Definition 3.23. Let (R, m) be a local Noetherian ring. an element x ∈ m is said to be superficial if there exists an integer c such that

(m^{n} : x) ∩ m^{c} = m^{n−1},

for all n > c.

Superficial elements always exist if the residue field k is infinite, see [20] page 7. If x is superficial for R then for every free module F we have

(0 :_{F} x) ∩ m^{c}F ⊂ (m^{n} :_{F} x) ∩ m^{c}F = m^{n−1}F,

for all n > c. Therefore (0 :_{F} x) ∩ m^{c}F = 0 by Krull Intersection Theorem. Finally we have the following

Theorem 3.24. Let (R, m, k) be a Noetherian local ring of dimension one with infinite residue field. Then any finitely generated R-module M has the uniform Artin-Rees property for syzygies with respect to the maximal ideal.

Proof. Let x ∈ m be an element which is a reduction of m and a superficial element for R. Since x is superficial we have (0 :_{F_{i−1}} x) ∩ m^{c}F_{i−1} = 0, for every n > c. Let h_{1} be an integer as in Corollary 3.22. Let h_{2} be a reduction number of (x) ⊂ m. We may assume that both h_{1} and h_{2} are bigger then c. Let h = max{h_{1}, h_{2}}. Then for every i > 0 we have

x^{h}(m^{n}F_{i−1} ∩ M_{i}) ⊂ m^{n}M_{i} = x^{n−h}m^{h}M_{i}.

Let u ∈ m^{n}F_{i−1} ∩ M_{i} then x^{h}u = x^{n−h}v, where v ∈ m^{h}M_{i}. Then for every n > 2h we have x^{h}(u−x^{n−2h}v) = 0. Since h > c we that u−x^{n−2h}v ∈ m^{c}F_{i−1}. Therefore, since x is superficial by applying the equation (0 : x) ∩ m^{c} = 0 we have u = x^{n−2h}v ∈ m^{n−2h}M_{i}.

Chapter 4 Conclusion In this thesis we study problems in commutative algebra. The modules we study are finitely generated R-modules over Noetherian rings. In the first chapter of this thesis we study closely the correspondence between short exact sequences and elements of the Ext module. We prove an extension of Miyata’s theorem. On one hand the theorem it is interesting on its own, on the other hand the many applications of this theorem and of its partial converse, show that this extension is actually a nice tool for dealing with short exact sequences. We can give simpler proofs of previous results, we can improve bounds, we can descrive minimal generators of the Ext module and we can give, in certain cases, new information about the structure of the Ext modules.

In the second and third chapter we study uniform versions of the classical Artin-Rees Lemma. In the second chapter we study the strong uniform Artin-Rees property, where an unique integer makes the classical Artin-Rees Lemma working for an infinite family of ideals. We see that the strong uniform Artin-Rees property holds for one dimensional excellent rings. We give examples of two dimensional rings in which the property fails. In the third chapter we study the uniform property for syzygies. The ring we work with is always a local Noetherian ring. In this case the ideal is fixed and we search for an integer that makes the weaker version of the Artin-Rees Lemma hold for all the syzygies inside the free modules of a free resolution. We investigate also the strong uniform Artin-Rees property for syzygies. We show that the residue field k has the strong uniform Artin-Rees property for syzygies with respect to the maximal ideal and that any module over a local Noetherian ring of dimension one has the uniform Artin-Rees property for syzygies with respect to any m-primary ideal. The fact that the uniform Artin-Rees Lemma for syzygies with respect to any m-primary ideals holds in Cohen-Macaulay rings of any dimensions gives an indication that the the property might hold in other higher dimensional rings. On the other hand it would be interesting to know if any finitely generated R-module of finite length has the strong uniform Artin-Rees property for syzygies with respect to the maximal ideal.

References

[1] M. Auslander, Representation theory of finite-dimensional algebras, Algebraists’ homage: papers in ring theory and related topics (New Haven, Conn., 1981), Contemp. Math., vol. 13, Amer. Math. Soc., Providence, R.I., 1982, pp. 27–39. MR 84b:16031

[2] Maurice Auslander, Isolated singularities and existence of almost split sequences, Representation theory, II (Ottawa, Ont., 1984), Lecture Notes in Math., vol. 1178, Springer, Berlin, 1986, pp. 194–242. MR 87j:13029

[3] Maurice Auslander and Idun Reiten, Modules determined by their composition factors, Illinois J. Math. 29 (1985), no. 2, 280–301. MR 86i:16032

[4] Klaus Bongartz, A generalization of a theorem of M. Auslander, Bull. London Math. Soc. 21 (1989), no. 3, 255–256. MR 90b:16031

[5] Winfried Bruns and Jürgen Herzog, Cohen-Macaulay rings, Cambridge Studies in Advanced Mathematics, vol. 39, Cambridge University Press, Cambridge, 1993. MR MR1251956 (95h:13020)

[6] A. J. Duncan and L. O’Carroll, A full uniform Artin-Rees theorem, J. Reine Angew. Math. 394 (1989), 203–207. MR MR977443 (90c:13011)

[7] David Eisenbud, Commutative algebra, Graduate Texts in Mathematics, vol. 150, Springer-Verlag, New York, 1995, With a view toward algebraic geometry. MR 97a:13001

[8] David Eisenbud and Melvin Hochster, A Nullstellensatz with nilpotents and Zariski’s main lemma on holomorphic functions, J. Algebra 58 (1979), no. 1, 157–161. MR MR535850 (80g:14002)

[9] David Eisenbud and Craig Huneke, A finiteness property of infinite resolutions, Journal of Algebra (to appear).

[10] Robert M. Guralnick, Lifting homomorphisms of modules, Illinois J. Math. 29 (1985), no. 1, 153–156. MR 86g:13006

[11] Craig Huneke, Uniform bounds in Noetherian rings, Invent. Math. 107 (1992), no. 1, 203–223. MR 93b:13027

[12] Craig Huneke and Graham J. Leuschke, Two theorems about maximal Cohen-Macaulay modules, Math. Ann. 324 (2002), no. 2, 391–404. MR 2003j:13011

[13] Gerson Levin, Local rings and Golod homomorphisms, J. Algebra 37 (1975), no. 2, 266–289. MR 55 #2878

[14] Saunders Mac Lane, Homology, Classics in Mathematics, Springer-Verlag, Berlin, 1995, Reprint of the 1975 edition. MR 96d:18001 [15] Hideyuki Matsumura, Commutative algebra, second ed., Mathematics

Lecture Note Series, vol. 56, Benjamin/Cummings Publishing Co., Inc., Reading, Mass., 1980. MR MR575344 (82i:13003)

[16] Takehiko Miyata, Note on direct summands of modules, J. Math. Kyoto Univ. 7 (1967), 65–69. MR 35 #5434

[17] L. O’Carroll, A note on Artin-Rees numbers, Bull. London Math. Soc. 23 (1991), no. 3, 209–212. MR 92i:13001a

[18] Liam O’Carroll and Dorin Popescu, Free resolutions for deformations of maximal Cohen-Macaulay modules, Comm. Algebra 28 (2000), no. 11, 5329–5352. MR MR1785504 (2001i:13015)

[19] Francesc Planas-Vilanova, The strong uniform Artin-Rees property in codimension one, J. Reine Angew. Math. 527 (2000), 185–201. MR 2001g:13051

[20] Judith D. Sally, Numbers of generators of ideals in local rings, Marcel Dekker Inc., New York, 1978. MR MR0485852 (58 #5654)

[21] Hsin-Ju Wang, On the Fitting ideals in free resolutions, Michigan Math. J. 41 (1994), no. 3, 587–608. MR 96b:13013

[22] , Some uniform properties of 2-dimensional local rings, J. Algebra 188 (1997), no. 1, 1–15. MR 98e:13006

[23] Yongwei Yao, Linear growth property of Tor and Ext, preliminary version.

[24] Yuji Yoshino, Cohen-Macaulay modules over Cohen-Macaulay rings, London Mathematical Society Lecture Note Series, vol. 146, Cambridge University Press, Cambridge, 1990. MR 92b:13016

[25] , On degenerations of Cohen-Macaulay modules, J. Algebra 248 (2002), no. 1, 272–290. MR 2002m:16016