HILBERT FUNCTIONS AND GRADED FREE RESOLUTIONS

A Dissertation

Presented to the Faculty of the Graduate School

of Cornell University

in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy

by

Christopher A. Francisco August 2004 \begin{matrix}c \\ ◯ \end{matrix} 2004 Christopher A. Francisco

ALL RIGHTS RESERVED HILBERT FUNCTIONS AND GRADED FREE RESOLUTIONS

Christopher A. Francisco, Ph.D. Cornell University 2004

This thesis discusses the connection between Hilbert functions and graded free resolutions. One can compute the Hilbert function of a module from its graded free resolution. However, there may be modules with the same Hilbert function but different graded Betti numbers. We investigate what graded free resolutions actually occur for a given Hilbert function. We provide a quick review of background material in the first chapter.

In Chapter 2, we consider the simplest case of this question: For what Hilbert functions is there only one possible set of graded Betti numbers? We characterize these Hilbert functions for ideals in k[x, y]. In more variables, the question is more difficult, and we give examples of some families of Hilbert functions for which there is a unique set of attainable graded Betti numbers.

The third chapter focuses on a stable ideals. Fix a Hilbert function H. We ask if there exists a stable ideal I with Hilbert function H such that the graded Betti numbers of any other stable ideal J with the same Hilbert function are at least as large as those of I. We answer the question affirmatively for ideals in k[x_{1}, x_{2}, x_{3}] but provide a counterexample in four or more variables.

In the fourth and fifth chapters, we investigate the Lex-Plus-Powers Conjecture on graded Betti numbers and the EGH Conjecture on Hilbert functions. We devote Chapter 4 to introducing versions of the conjectures and the relationships among them. In Chapter 5, we prove that the graded Betti numbers of a lex-plus-powers almost complete intersection are an upper bound for the graded Betti numbers of any ideal with the same Hilbert function and regular sequence in the same degrees. In addition, we give some related results on Hilbert functions.

We conclude in Chapter 6 by investigating a conjecture of Huneke and Srinivasan on the multiplicity of an ideal. They conjecture bounds for the multiplicity in terms of the graded free resolution. We prove some special cases of the conjecture for ideals with generators in high degrees relative to their regularity, survey some computational work, and discuss some reductions of the conjecture.

BIOGRAPHICAL SKETCH

Chris Francisco is proud to have been born a Jayhawk in Lawrence, Kansas, on July 26, 1977. He is the only child of Deborah and Ronald Francisco.

Chris spent his entire childhood in Lawrence, attending the Lawrence Public Schools, where he had some terrific mathematics instructors. After taking Ray Wilbur’s second- and third-semester calculus classes, Chris took some math courses at the University of Kansas his senior year in high school, including Phil Montgomery’s abstract algebra class. That course made him want to study algebra.

Chris graduated from Lawrence High School in 1995 and entered the University of Illinois at Urbana-Champaign in August 1995. Through the efforts of Sonia Carringer in the Campus Honors Program at Illinois, Chris met Graham Evans, an algebraist who agreed to mentor Chris starting early in his freshman year. Graham introduced Chris to commutative algebra, computer algebra systems, and the mathematical community in general, providing many opportunities for him.

At Illinois, Chris had a number of classes from wonderful professors who furthered his interest in algebra and mathematical research, including Nigel Boston, Bruce Reznick, Michio Suzuki, Dan Grayson, and Joseph Rotman. Chris did his first research by using computer algebra systems to work on problems in computational group theory and on elliptic curves that Nigel Boston suggested.

In the summer of 1998, Graham Evans made it possible for Chris to attend a workshop in commutative algebra and algebraic geometry at MSRI, and since then, Chris has been primarily interested in commutative algebra. After graduating from Illinois with a B.S. in mathematics and a B.A. in economics, Chris came to Cornell University for graduate school in August 1999 on an NSF Graduate Research Fellowship and began working with Mike Stillman.

August 1999 was also great for personal reasons; Chris met his wife, Cynthia, another entering math graduate student, that month. They got engaged a year later and were married in Lawrence on July 7, 2001.

In the fall of 2002, Chris passed his A-exam and received his M.S. in mathematics from Cornell. He continued his research in commutative algebra for the next few years, completing his Ph.D. thesis in the summer of 2004.

Next year, Chris and Cynthia will move to Columbia, Missouri, where Chris will begin a three-year postdoctoral position at the University of Missouri. He will still always cheer for the Jayhawks.

To my wife, Cynthia, and to my parents, Deborah and Ronald,

for their love and support.

ACKNOWLEDGEMENTS

I was fortunate to have help and support from many people in writing this thesis, and it is a pleasure to be able to thank some of them. First, thanks to my advisor, Mike Stillman, for all his work with me the last five years. I appreciate his willingness to mentor me early in my time at Cornell and through all the ups and downs of graduate school. Mike was a patient teacher from whom I learned a lot, and his insights were extremely helpful during the research process. Listening to his perspective on the work I was doing often guided me in the right direction and raised interesting questions to consider. Mike’s suggestions greatly improved my thesis, and he helped give me the confidence to do research on my own.

A number of other faculty have been particularly helpful to me at Cornell. I enjoyed all of Lou Billera’s courses in combinatorics, and he always knew what papers to recommend that I read. His help in the job search was also invaluable. Ken Brown was kind enough to agree to step in when I needed a third committee member before my A-exam. It is a pleasure to talk with him, and he always has insightful comments. Though she was not officially on my committee, Irena Peeva has done an incredible amount for me. I thank her for her wonderful courses on resolutions, for spending time to talk with me about my research, for inviting me to talk at an AMS meeting and to organize a session with her, and for her tireless efforts in my job search. I also thank Tom Rishel and Maria Terrell for their help in my development as a teacher at Cornell.

It is not an exaggeration to say that I owe much of my mathematical career to Graham Evans at the University of Illinois. He was a wonderful mentor for me, helping me learn mathematics and become a part of the mathematical community. His influence is evident throughout my thesis. Graham and his wife, Kay, helped make my college experience as enjoyable as it was, and I thank them for being surrogate parents to me. I wish them the best of luck in retirement.

I had the pleasure of meeting a number of people in commutative algebra the last few years, leading to many great discussions about questions on which I was working. At the risk of leaving someone out, I wish to thank Hara Charalambous, Brian Coolen, Susan Cooper, Vesko Gasharov, Leah Gold, Tony Geramita, Craig Huneke, Matt Rodriguez, Sindi Sabourin, Hal Schenck, Hema Srinivasan, and Adam Van Tuyl. Ben Richert deserves special mention for enjoyable conversations about what resolutions can occur for a given Hilbert function; one of these days we will actually write a paper together. Thanks also to the students in the commutative algebra and algebraic geometry group during my time at Cornell, including Achilleas, Leah, Jeff, Mauricio, and Steve. Additionally, thanks to Sam Hsiao for our mathematical conversations and help in my job search.

My friends at Cornell helped make graduate school much more enjoyable. Thanks to everyone who entered in 1999; we had a great group of people without whom core courses would have been even more miserable than they were. Special thanks to Aaron, Melanie, and Spencer for their friendship the last five years.

I saved my family for last. My wife, Cynthia, has been a wonderful source of encouragement and help throughout the last five years, and meeting her was by far the best part of graduate school. My parents have always done whatever they could to help me achieve what I wanted, and I am grateful to them for their efforts and love. Thanks also to my grandparents for their love and support.

Finally, I gratefully acknowledge that this material is based upon work supported under a National Science Foundation Graduate Research Fellowship and a Cornell Graduate School Fellowship.

TABLE OF CONTENTS

0 Introduction 1

1 Background material 7 1.1 Hilbert functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2 Free resolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.3 The Eliahou-Kervaire resolution . . . . . . . . . . . . . . . . . . . . 14 1.4 Connections between Hilbert functions and free resolutions . . . . . 16

2 Hilbert functions with a unique resolution 23 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.2 First examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.3 Hilbert functions over k[x, y] . . . . . . . . . . . . . . . . . . . . . . 27 2.4 Families of examples with “ghost terms” . . . . . . . . . . . . . . . 30

3 Minimal graded Betti numbers and stable ideals 36 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.2 Definitions and some previous work . . . . . . . . . . . . . . . . . . 38 3.3 Some lemmas and formulas . . . . . . . . . . . . . . . . . . . . . . . 40 3.4 Combinatorial formulas for the Betti numbers . . . . . . . . . . . . 42 3.5 The algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.6 Incomparability in more variables . . . . . . . . . . . . . . . . . . . 52

4 Conjectures about Lex-plus-powers ideals 55 4.1 Lex-plus-powers ideals . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.2 Conjectures on Hilbert functions . . . . . . . . . . . . . . . . . . . . 57 4.3 The Lex-Plus-Powers Conjecture . . . . . . . . . . . . . . . . . . . 62 4.4 Relating the EGH Conjecture to the LPP Conjecture . . . . . . . . 65

5 Almost complete intersections and the LPP Conjecture 68 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 5.2 Resolutions of monomial almost complete intersections . . . . . . . 70 5.3 The nonmonomial almost complete intersection case . . . . . . . . . 75 5.4 Complete intersections . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.5 Upper bounds on some Hilbert functions . . . . . . . . . . . . . . . 92 5.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

6 Bounding the multiplicity of ideals 100 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 6.2 Relation to resolutions with a given Hilbert function . . . . . . . . . 104 6.3 Some computational work on the upper bound . . . . . . . . . . . . 107 6.4 Truncation and a reduction in the Cohen-Macaulay case . . . . . . 119

Bibliography 126

Chapter 0

Introduction

One of the central objects of study in commutative algebra and algebraic geometry is a homogeneous ideal in k[x_{1}, …, x_{n}]. This thesis investigates two of the most important invariants of these modules: Hilbert functions and graded free resolutions. The Hilbert function, which gives the vector space dimension of a module in each degree, yields useful algebraic and geometric information, including the dimension and degree of a variety. The graded free resolution is a finer invariant; in particular, one can easily compute the Hilbert function from the resolution. Moreover, the resolution can provide additional geometric information. For example, the coordinate ring of any set of seven points in linearly general position in ℙ^{3} has the same Hilbert function. However, the graded Betti numbers of the coordinate rings tell whether the points lie on a curve of degree three [9].

Understanding the graded free resolution of a module can give substantial insight into the module’s structure. Though computer algebra systems have made it possible to find the graded free resolution of any particular module (within memory and time constraints), structure theorems for resolutions are known only in a small number of special cases.

One way to get more information about the structure of free resolutions is fix a Hilbert function and ask what graded free resolutions can occur for a module with that Hilbert function. In the first chapter, we describe a partial order on the graded free resolutions with a given Hilbert function, ordering the resolutions by inclusion of the multisets of shifts in the resolution. The partial order provides a systematic way of studying the graded Betti numbers that occur for a given Hilbert function. Additionally, incomparability in the partial order can also give geometric information as in the following example.

Example 0.1 We consider sets of nine points in ℙ^{2} with the same Hilbert function (1, 3, 6, 8, 9, 9, …), drawing from an example in [20]. Let 𝕏 be the set of points formed by the union of a complete intersection of two quadrics and a complete intersection of a line and a quintic. Let 𝕐 be the set of points formed by the union of a complete intersection of two lines and a complete intersection of a quadric and a quartic. Using methods from [20], one can easily show that the coordinate rings have the following resolutions:

0 ⟶ R(−5) ⊕ R(−6) ⟶ R(−3)^{2} ⊕ R(−5) ⟶ R ⟶ R/I_{𝕏} ⟶ 0
0 ⟶ R(−4) ⊕ R(−6) ⟶ R(−3)^{2} ⊕ R(−4) ⟶ R ⟶ R/I_{𝕐} ⟶ 0

These resolutions are incomparable in the partial order since there are copies of R(−5) but not R(−4) in the resolution of R/I_{𝕏} (and vice-versa in the resolution of R/I_{𝕐}). The geometric statement that corresponds to this algebraic fact is that we cannot deform 𝕏 into 𝕐 without passing through another special configuration of points. There are two choices for the special configuration: We can deform to a set of points ℤ that is a complete intersection of two cubics and has the following resolution:

0 ⟶ R(−6) ⟶ R(−3)^{2} ⟶ R ⟶ R/I_{ℤ} ⟶ 0

In the partial order, this resolution lies below those of both R/I_{𝕏} and R/I_{𝕐}. Alternatively, we can deform to a set of points 𝕎 with the resolution below:

0 → R(−4)⊕R(−5)⊕R(−6) ⟶ R(−3)^{2} ⊕R(−4)⊕R(−5) ⟶ R ⟶ R/I_{𝕎} → 0 𝕎 is the union of three complete intersections: two lines, a line and a cubic, and a line and a quintic. Note that the resolution of R/I_{𝕎} lies above the resolutions of the other three coordinate rings in the partial order.

We devote much of the rest of the thesis to studying the poset of graded Betti numbers for a given Hilbert function. We consider a number of related questions about the minimal and maximal elements of particular subposets. Though we shall frequently work with Artinian ideals, results about the graded Betti numbers of Artinian ideals are generally applicable to Cohen-Macaulay modules as well, so this is not a substantial restriction. Given a Cohen-Macaulay module R/I, we can find an Artinian module R′/I′ with the same graded Betti numbers.

In Chapter 1, we introduce Hilbert functions, free resolutions, and related notions. The first three sections are standard background material that one can find in many texts on commutative algebra. In the fourth section, we present a number of tools that are helpful in determining whether a set of numbers can actually occur as the graded Betti numbers of a module with a particular Hilbert function. Here, we give the partial order on the graded Betti numbers of modules with the same Hilbert function. Work of Bigatti, Hulett, and Pardue proves that the poset that comes from this partial order always has a unique maximal element, but an example of Charalambous and Evans shows that there may be several minimal elements. Finally, we describe some recent work of Evans and Richert and of Peeva that yields tools that we will use in the rest of the thesis.

Our second chapter addresses the case in which we have the simplest poset: a Hilbert function for which there is only one possible set of graded Betti numbers. A recent theorem of Peeva describes what configurations of graded Betti numbers are candidates for resolutions of modules with a given Hilbert function. This result yields a number of easy examples of Hilbert functions admitting only one set of graded Betti numbers. In the third section, we characterize the Hilbert functions for modules over k[x, y] for which there is a unique set of graded Betti numbers, using a result of Charalambous and Evans about resolutions over a polynomial ring in two variables. Unfortunately, such a characterization is much harder in more variables. We give some families of examples of less obvious Hilbert functions admitting only one set of graded Betti numbers in the final section, using a tool of Evans and Richert that rules out certain cancellations in Betti diagrams.

Because there is not always a uniquely minimal element in the partial order for a given Hilbert function, we have to restrict to a smaller class of ideals to try to obtain a poset in which there is a single smallest element. In the third chapter, we focus on stable ideals. Deery asked in [7] whether for each Hilbert function H, there exists a stable ideal I such that if J is another stable ideal with the same Hilbert function, the graded Betti numbers of J are at least as large as those of I. In other words, if we restrict to stable ideals with a given Hilbert function, is there a uniquely minimal set of graded Betti numbers in the partial order? We answer the question affirmatively in three variables, presenting an algorithm to construct the ideal. In the final section, we give a counterexample for any polynomial ring in at least four variables.

In Chapters 4 and 5, we consider Artinian ideals with a regular sequence in prescribed degrees. The knowledge that an ideal contains a regular sequence in certain degrees provides helpful algebraic and geometric information. We restrict to these ideals and use this structure to bound their graded Betti numbers. The motivation for this work comes from [10]. There, Eisenbud, Green, and Harris conjecture a bound for the growth of a Hilbert function of an Artinian ideal with regular sequence in prescribed degrees. The ideals that provide the bound in the EGH Conjecture are monomial ideals called lex-plus-powers ideals. They are the analogues of Artinian lexicographic ideals in the case in which we require that our ideals contain a regular sequence in given degrees. This conjecture, which is related to Cayley-Bacharach theory in algebraic geometry, inspired the Lex-Plus-Powers (LPP) Conjecture, the analogue of the Bigatti-Hulett-Pardue theorem on graded Betti numbers. It asserts that if we have a lex-plus-powers ideal L and another ideal I with the same Hilbert function and regular sequence in the same degrees, the graded Betti numbers of I are at most those of L. In Chapter 4, we discuss a number of versions of the EGH conjecture and introduce the LPP Conjecture. We also prove that the LPP Conjecture implies the EGH Conjecture.

We devote Chapter 5 to the special case of when the LPP ideal is an almost complete intersection. Our main result is the LPP Conjecture in the case when L is an almost complete intersection. In the process, we find an explicit comparison map between two Koszul complexes of the same length that helps detect nonminimality in a mapping cone resolution. We also prove some upper bounds on the Hilbert functions of almost complete intersections. The LPP Conjecture is known in few cases, and it is a good illustration of how difficult it is to prove theorems about resolutions without significant restrictions on the modules one is considering.

The final chapter of this thesis addresses a conjecture of Huneke and Srinivasan on the multiplicity of a Cohen-Macaulay ideal in a polynomial ring. We study the bottom of the poset of graded Betti numbers for a given Hilbert function, presenting a new way of viewing the problem as a question about the minimal elements in the poset. Our approach shows that if the conjecture is true for all the minimal elements in the poset for a fixed Hilbert function, it is true for all modules with that Hilbert function. In the third section, we discuss some computational work using the computer algebra system Macaulay 2 [23]. The results from this work show that the conjecture of Huneke and Srinivasan, if true, implies that a number of sets of potential graded Betti numbers cannot occur for modules with a given Hilbert function; the conjecture gives nontrivial knowledge about the central question of the thesis. In the final section, we focus on the upper bound of the conjecture in the Cohen-Macaulay case, proving the bound for ideals with generators in high degrees relative to their regularity. We also provide a reduction of the upper bound in the Cohen-Macaulay case to Artinian ideals with their generators all in a single degree.

Background material

In this chapter, we provide a quick introduction to Hilbert functions and graded free resolutions and give some results we will use in the rest of the thesis. There are many good references for the material on Hilbert functions; two excellent resources are Chapter 4 in the book of Bruns and Herzog [3] and Valla’s survey article [41]. Eisenbud’s book [8] is probably the best place to learn about free resolutions.

Throughout this thesis, let R = k[x_{1}, …, x_{n}] be the polynomial ring in n variables over a field k. We will usually use S to denote particular polynomial rings, such as k[x, y]. In addition, all rings in the thesis will be commutative with identity.

1.1 Hilbert functions

We begin by introducing graded rings.

Definition 1.1 A ring A is a graded ring if it has an abelian group decomposition

A = A_{0} ⊕ A_{1} ⊕ A_{2} ⊕ ⋯

and satisfies A_{i}A_{j} ⊂ A_{i+j} for all i, j ≥ 0. We call the elements of A_{i} homogeneous elements of degree i, and an ideal I in A is called a homogeneous ideal if it can be generated by homogeneous elements of A.

The ring R = k[x_{1}, …, x_{n}] graded by degree is the example we shall use most often. In this case, R_{d} consists of the homogeneous polynomials of R of degree d, and homogeneous ideals are those that can be generated by polynomials each of which has all its terms of the same degree. All ideals we discuss in this thesis will be homogeneous.

We are interested in modules over graded rings that respect the grading of the underlying ring.

Definition 1.2 Let M be a module over a graded ring A. We say that M is a graded module over A if

M = ⨁_{i∈ℤ}M_{i}

as abelian groups and A_{i}M_{j} ⊂ M_{i+j} for all i and j.

For example, let I be a homogeneous ideal in R. Then R/I is a graded module over R with (R/I)_{d} = R_{d}/I_{d}.

We can now define the Hilbert function.

Definition 1.3 Let M be a finitely generated graded module over the ring R, graded by degree. The Hilbert function of M is the function

H(M, d) = dim_{k} M_{d}.

It is sometimes useful to express this information as a formal power series

∑_{d} H(M, d)t^{d}

called the Hilbert series. Moreover, as we will discuss in Section 4, when M is an R-module, the Hilbert series of M is a rational function

\frac{f_{M}(t)}{(1 − t)^{n}} .

The numerator of this rational function comes from the graded Betti numbers of M, quantities introduced in the next section. In addition, Hilbert showed that for large d, the Hilbert function H(M, d) agrees with a polynomial called the Hilbert polynomial of M. Note that the Hilbert function of R itself is H(R, d) = \binom{n−1+d}{d}.

We shall mostly consider modules of the form R/I; in this case, H(R/I, d) = dim_{k} R_{d} − dim_{k} I_{d}. For monomial ideals I, this number is easy to compute: H(R/I, d) is the number of monomials in R_{d} but not in I_{d}. For nonmonomial ideals, the following result of Macaulay reduces the problem to the case of monomial ideals. Let > be a monomial order on R, and let in_{>}(I) be the set of leading terms of the elements of a homogeneous ideal I in R.

Theorem 1.4 The set of monomials not in in_{>}(I) forms a basis for R/I. Moreover, R/in_{>}(I) has the same Hilbert function as R/I.

Example 1.5 Let I = (x_{1}^{2}, x_{1}x_{2}, x_{2}^{3}, x_{3}^{3}) ⊂ S = k[x_{1}, x_{2}, x_{3}]. To find the Hilbert function of S/I, we count the number of monomials of S in each degree not in I. The only monomial of degree zero in S is 1, and it is not in I, so H(S/I, 0) = 1. None of x_{1}, x_{2}, and x_{3} is in I, so H(S/I, 1) = 3. In degree 2, there are four monomials of S not in I, x_{1}x_{3}, x_{2}^{2}, x_{2}x_{3}, and x_{3}^{2}; thus H(S/I, 2) = 4. It is also easy to see that H(S/I, 3) = 3 and H(S/I, 4) = 1. All monomials of S of degree greater than or equal to five are in I, and therefore H(S/I, d) = 0 for d ≥ 5. We shall abbreviate this data by writing H(S/I) = (1, 3, 4, 3, 1).

In studying Hilbert functions, certain monomial ideals called lexicographic ideals prove particularly useful. We recall their definition.

Definition 1.6 An ideal L is called a lexicographic ideal if in each degree d, L is generated by the first dim_{k} L_{d} monomials in descending lexicographic order.

For example, in k[a, b, c], (a^{2}, ab, ac^{2}, b^{4}) is a lexicographic ideal. Macaulay showed that given a module R/I, there exists a lexicographic ideal L ⊂ R such that R/L has the same Hilbert function.

Consequently, we can determine what numerical functions occur as the Hilbert function of a module R/I for some I by restricting our attention to lexicographic ideals. Macaulay characterizes the valid Hilbert functions for polynomial rings modulo homogeneous ideals in [33], and one can express this characterization using sums of binomial coefficients. It is not hard to show that, if a and d are positive integers, a has a unique decomposition as

a = \binom{b_{d}}{d} + \binom{b_{d−1}}{d − 1} + ⋯ + \binom{b_{2}}{2} + \binom{b_{1}}{1},

where b_{d} > b_{d−1} > ⋯ > b_{1} ≥ 0. Green [24] calls this the d-th Macaulay representation of a. We can then define a new object a^{<d>} to be

a^{<d>} = \binom{b_{d} + 1}{d + 1} + \binom{b_{d−1} + 1}{d} + ⋯ + \binom{b_{2} + 1}{3} + \binom{b_{1} + 1}{2}.

With this notation, we have the following theorem from [33]. Theorem 1.7 (Macaulay) Let H be a function from ℕ to ℕ. Then H is the Hilbert function of a module R/I if and only if H(0) = 1 and for all d ≥ 1,

H(d + 1) ≤ (H(d))^{<d>}.

Moreover, this bound is attained by quotients by lexicographic ideals with no generators in degree d + 1.

Example 1.8 We show that (1, 3, 6, 11) is not the Hilbert function of a quotient of k[x_{1}, …, x_{n}] by some homogeneous ideal. Consider the growth from degree two to degree three. The second Macaulay representation of six is \binom{4}{2}. Thus 6^{<2>} = \binom{5}{3} = 10 < 11, and therefore the growth from 6 to 11 is impossible.

1.2 Free resolutions

The other invariant of a module in which we are most interested is its free resolution.

Definition 1.9 Let A be a ring, and let M be an A-module. A free resolution of M over A is a complex

⋯ ⟶ F_{n} ⟶^{d_{n}} ⋯ ⟶ F_{1} ⟶^{d_{1}} F_{0} ⟶ 0,

where each F_{i} is a free A-module, im d_{i+1} = ker d_{i} for all i ≥ 1, and coker d_{1} = M.

For clarity, we will often put M in the complex above immediately to the right of F_{0} and require that the complex be exact everywhere. If F is a free resolution of M, and the modules F_{i} in F are zero for all i > r but F_{r} ≠ 0, then we say that F has length r. Additionally, if A is local or graded with maximal ideal (in the graded case, maximal homogeneous ideal) 𝔪, and we insist that the image of d_{i} be contained in 𝔪F_{i−1} for each i ≥ 1, then we have a minimal free resolution of M, and we can always construct a minimal resolution from a nonminimal one. If we write the d_{i} as matrices, the condition on the image of the d_{i} is equivalent to requiring that the matrix representing each d_{i} has all of its entries in 𝔪.

If A and M are graded, then we can always construct a minimal graded free resolution of M over A. In this case, we insist that that the maps d_{i} be homogeneous of degree zero; this means that the maps take homogeneous elements of degree t to homogeneous elements of degree t. We also require that each F_{i} be a graded A-module.

In order to write down a graded free resolution, we need a bit of notation. Let M be a graded module. We define M(d), the d-th shift (or twist) of M, to be the module M but graded such that M(d)_{t} = M_{d+t}. Thus M(−d)_{d} = M_{0}. Also, multiplying by a form of degree d is a map of degree zero from M(−d) to M.

Since the polynomial ring R = k[x_{1}, …, x_{n}] is graded by degree, every graded module over R has a graded free resolution. Moreover, the Hilbert Syzygy Theorem says that for each R-module M, there exists a free resolution of M of length at most n. These observations show that for any homogeneous ideal I ⊂ R, we can write down a minimal graded free resolution of R/I of the form

0 ⟶ ⨁_{j} R(−j)^{β_{nj}} ⟶^{d_{n}} ⋯ ⟶ ⨁_{j} R(−j)^{β_{1j}} ⟶^{d_{1}} R ⟶ R/I ⟶ 0.

It is not hard to show that the minimal graded free resolution of R/I is unique up to change of basis. Thus we are justified in calling the β_{ij} in the resolution above the graded Betti numbers of R/I. The graded Betti numbers are also the dimension over k of the graded pieces of certain Tor modules:

β_{ij}^{R/I} = dim_{k} Tor_{i}(R/I, k)_{j}.

The total Betti numbers are simply the sum of the graded Betti numbers at each step in the resolution:

β_{i}^{R/I} = ∑_{j} β_{ij}^{R/I} = dim_{k} Tor_{i}(R/I, k).

Example 1.10 Let S = k[a, b], and let I = (a^{3}, ab^{2}, b^{4}). Here is the minimal graded free resolution of S/I:

0 ⟶ S(−5)^{2} \underrightarrow{(\begin{matrix}−b^{2} & 0 \\ a^{2} & −b^{2} \\ 0 & a \end{matrix})} S(−3)^{2} ⊕ S(−4) \underrightarrow{(a^{3} ab^{2} b^{4})} S ⟶ S/I ⟶ 0

Note that each of the columns of the leftmost matrix dot to zero with the 1 × 3 matrix of generators. We say that each of the columns of the 3 × 2 matrix is a syzygy on the generators of I. Each of these syzygies is a relation on the generators, and we call them first syzygies. If the resolution were longer, we would have minimal relations on these syzygies, which would be called second syzygies, and so on. The fact that each column (each syzygy) in the 3 × 2 matrix in our example dots to zero with the matrix of generators is a consequence of the resolution being a complex.

Frequently, we shall primarily be concerned with the graded Betti numbers of a module and not the maps in the resolution. The computer algebra system Macaulay 2 [23] has a convenient way of displaying this information in a compact form. The graded Betti number β_{i,i+j} is displayed in column i and row j, and the rows and columns are numbered starting from zero.

\begin{matrix}0: & β_{00} & β_{11} & β_{22} & … \\ 1: & β_{01} & β_{12} & β_{23} & … \\ 2: & β_{02} & β_{13} & β_{24} & … \\ 3: & β_{03} & β_{14} & β_{25} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}

For instance, in the resolution in Example 1.10, the nonzero graded Betti numbers are β_{00} = 1, β_{13} = 2, β_{14} = 1, and β_{25} = 2. The corresponding Macaulay 2 diagram, with the total Betti numbers in the row labeled “total,” is:

column # 0 1 2
S/I: total: 1 3 2
0:  1 . .
1:  …
2:  . 2 .
3:  . 1 2

Column i represents the module F_{i} in the resolution of S/I. A slight confusion arises when we resolve the module S/I instead of I. Column one corresponds to the generators of I, and columns i ≥ 2 represent (i − 1)-st syzygies, so column two shows the first syzygies. Note that all the graded Betti numbers for a fixed degree occur on one upward-sloping diagonal in the diagram. The Betti diagram is particularly nice because it is easy to see the degrees of syzygies at a glance; for example, in the diagram above, one can see that there are no linear syzygies on the generators in degree three because there is a period, symbolizing zero, next to the “2” representing the two generators of degree three. In addition, one can readily see the regularity of S/I in the Betti diagram; this is the greatest integer j such that there exists i with β_{i,i+j}^{S/I} ≠ 0. In the example above, the regularity of S/I is three; it is the label on the last row in the Betti diagram.

1.3 The Eliahou-Kervaire resolution

We have already seen that lexicographic ideals play a significant role in the study of Hilbert functions, and we shall describe their importance in the theory of graded free resolutions in the next section. In this section, we discuss how to compute the graded Betti numbers of these ideals and some generalizations using a result of Eliahou-Kervaire from [12].

Let u be a monomial in R = k[x_{1}, …, x_{n}], and let max(u) denote the largest index i such that x_{i} divides u. For example, max(x_{2}^{4}x_{3}x_{5}) = 5. We use this idea to define stable ideals.

Definition 1.11 An ideal I is said to be stable if I is a monomial ideal, and for each monomial m in I and all i < max(m), x_{i}m/x_{max(m)} is also in I.

All lexicographic ideals are stable, and there are many others as well. For example, the ideal (a^{2}, ab, b^{2}, bc) in k[a, b, c] is stable. Borel-fixed ideals in characteristic zero are also stable; in fact, they satisfy a stronger condition and are called strongly stable. See the discussion before Proposition 3.12 for more details. When the base field has characteristic zero, generic initial ideals are strongly stable; since generic initial ideals are an important object in commutative algebra and algebraic geometry, it is particularly interesting to study the resolutions of stable ideals.

Eliahou and Kervaire exploit the combinatorial structure of stable ideals to describe their minimal free resolutions in [12]. If I is a stable ideal, then the graded Betti numbers of R/I are given by the formula

β_{qj}^{R/L} = ∑_{deg(m)=j−q+1^{m∈G(I)}} \binom{max(m) − 1}{q − 1},

where G(I) is the (unique) minimal monomial generating set of I. Thus if x_{1}x_{5} is a minimal generator of a stable ideal L, then it yields one minimal generator, four first syzygies, six second syzygies, four third syzygies, and one fourth syzygy. Hence it is easy to find the graded Betti numbers of a stable ideal simply by looking at its minimal generating set. Also, the formula shows that if a stable ideal L has no minimal generators in degree d, then all the entries in row d − 1 of the Betti diagram of R/L are zero.

Example 1.12 Let L = (a^{2}, ab, ac^{2}, b^{3}, b^{2}c, bc^{4}, c^{6}). Then L is a lexicographic ideal in S = k[a, b, c], and the Hilbert function of S/L is (1, 3, 4, 2, 2, 1). We can resolve S/L with the Eliahou-Kervaire resolution:

S/L: total: 1 7 10 4
0: 1 …
1: . 2 1 .
2: . 3 5 2
3: ….
4: . 1 2 1
5: . 1 2 1

Note that since L has no generators in degree four, row three contains only zeros.

1.4 Connections between Hilbert functions and free resolutions

Most of this thesis is dedicated to exploring the relationship between the Hilbert function of an ideal and its graded free resolution. In this section, we collect some results about this connection and tools that we shall use in later chapters.

In his 1890 paper [29], Hilbert used the graded free resolution of a module to compute its Hilbert function. The key observation is that vector space dimension is additive on exact sequences, so we can look at the degree d portion of the resolution of a module M to find H(M, d). Suppose the following is a graded free resolution of M over R:

0 ⟶ ⨁_{j} R(−j)^{β_{nj}} ⟶^{d_{n}} ⋯ ⟶ ⨁_{j} R(−j)^{β_{1j}} ⟶^{d_{1}} R ⟶ 0

By additivity, we have

H(M, d) = ∑_{i=0}^{n} ∑_{j}(−1)^{i}β_{ij} dim_{k} R(−j)_{d} = ∑_{i=0}^{n} ∑_{j}(−1)^{i}β_{ij}\binom{n − 1 + d − j}{d − j}.

Consequently, we can write the Hilbert series of M as the rational function

\frac{1}{(1 − t)^{n}} ∑_{i=0}^{n} ∑_{j}(−1)^{i}β_{ij}t^{j}.

These formulas show how the graded Betti numbers of different modules with the same Hilbert function may vary. Suppose M has graded Betti numbers β_{ij}^{M}. Suppose another module N has the same graded Betti numbers except that, for some integers r, s, and j, β_{rj}^{N} = β_{rj}^{M} − 1 and β_{r+2s+1,j}^{N} = β_{r+2s+1,j}^{M} − 1. Then the alternating sums above are unchanged since the terms with i = r and i = r+2s+1 have opposite sign in the sums. Hence M and N have the same Hilbert function. We define this idea more formally later in the section.

Consequently, given a graded free resolution of a module, we know exactly what its Hilbert function is. What happens if we fix the Hilbert function and ask what graded free resolutions can occur for a module with that Hilbert function? This question has been the subject of considerable research in the last fifteen years. We wish to know the possible graded Betti numbers than can occur for a module with a given Hilbert function. Because one can compute the Hilbert function of a module from its graded free resolution, the alternating sum of the Betti numbers in each degree is fixed for a particular Hilbert function.

To study this problem further, it is useful to impose a partial order on the graded Betti numbers of modules with the same Hilbert function.

Definition 1.13 Suppose M and N are R-modules with the same Hilbert function. We say that β^{M} ≤ β^{N} if and only if β_{ij}^{M} ≤ β_{ij}^{N} for all i and j.

This is a strong condition because the inequality has to go the same way for each of the graded Betti numbers. Thus there may be many incomparable resolutions for a given Hilbert function.

Using this partial order, Bigatti [2] and Hulett [30] proved Theorem 1.14, the most important result known in this area, over a field of characteristic zero. Pardue [36] generalized the result to positive characteristic. Recall that for each module R/I, there is a lexicographic ideal L such that R/I and R/L have the same Hilbert function.

Theorem 1.14 Let I be a homogeneous ideal in R = k[x_{1}, …, x_{n}], and let L be the lexicographic ideal in R such that R/I and R/L have the same Hilbert function. Then β^{R/I} ≤ β^{R/L}.

Theorem 1.14 is especially powerful because it puts an upper bound on the possible resolutions with a given Hilbert function, making the search for all resolutions with a particular Hilbert function a bounded problem.

Let us combine Theorem 1.14 now with our observations about how the Hilbert function can stay the same even with changes in the graded Betti numbers. We need a definition first.

Definition 1.15 Let {β_{ij}} be a set of graded Betti numbers. Fix some integers r, s, and j, and replace β_{rj} by β_{rj} − 1 and β_{r+2s+1,j} by β_{r+2s+1,j} − 1. We call this a cancellation. If s = 0, we call this a consecutive cancellation.

Now suppose I is a homogeneous ideal in R, and let L be the lexicographic ideal in R with the same Hilbert function. By Theorem 1.14, β^{R/I} ≤ β^{R/L}. Since R/I and R/L have the same Hilbert function, we must be able to obtain the graded Betti numbers of R/I by making cancellations in the graded Betti numbers of R/L.

Example 1.16 In the ring S = k[a, b, c], the ideal (a^{2}, ab, ac^{2}, b^{3}, b^{2}c^{2}, bc^{3}, c^{5}) is the lexicographic ideal L such that S/L has Hilbert function (1, 3, 4, 3, 1). The ideals I = (a^{2}, b^{2}, c^{3}) and L have the same Hilbert function. Theorem 1.14 says that β^{S/I} ≤ β^{S/L}. One can obtain the graded Betti diagram of S/I by cancelling an even and an odd syzygy in the same degree from the resolution of S/L. The following are the Betti diagrams of S/L and S/I:

S/L: total: 1 7 10 4  S/I: total: 1 3 3 1
0: 1 …  0: 1 …
1: . 2 1 .  1: . 2 . .
2: . 2 3 1  2: . 1 1 .
3: . 2 4 2  3: . . 2 .
4: . 1 2 1  4: … 1

Note that in this case, we get the resolution of S/I by making all cancellations possible in the resolution of S/L.

Peeva [37] has recently proven a particularly strong result that describes more precisely what sets of numbers are candidates for being the graded Betti numbers of a module with a given Hilbert function.

Theorem 1.17 Let I be an ideal in R = k[x_{1}, …, x_{n}], and let L be the lexicographic ideal in R with the same Hilbert function. Then the graded Betti numbers of R/I are obtained from a sequence of consecutive cancellations in the graded Betti diagram of R/L.

In Example 1.16, we have a resolution of length three, so the fact that the cancellations have to be consecutive does not matter. However, in four or more variables, it makes a significant difference. We shall need the full strength of Theorem 1.17 in Chapter 2.

Theorem 1.14 tells us that the lexicographic ideal has uniquely maximal graded Betti numbers among all ideals with a given Hilbert function. At the bottom of the partial order, the situation is more complicated. Charalambous and Evans showed that there can be incomparable resolutions for a given Hilbert function that are both minimal in the partial order [4]. Moreover, Evans and Richert proved that even if there are both unique maximal and unique minimal resolutions for a given Hilbert function, not every potential resolution between those two need occur [14].

The technique Evans and Richert use in their proof is especially noteworthy. It represents one of the few general methods we have to show that a particular set of numbers below the graded Betti numbers of the lexicographic ideal in the partial order cannot be the graded Betti numbers of a module. The following theorem of Evans and Richert from [14] allows us, in a number of cases, to avoid ad hoc arguments that would otherwise be necessary to rule out specific configurations of Betti numbers.

Theorem 1.18 Let R = k[x_{1}, …, x_{n}], and let M be a graded R-module. Let i ≥ 2 be a positive integer. If t is the smallest integer such that β_{it}^{M} ≠ 0, then the sum of the graded Betti numbers β_{i−1,j}^{M} for all j < t is at least i.

The theorem follows from a clever application of a result of Evans and Griffith [13] on the heights of order ideals of elements of i-th syzygy modules. Evans and Richert also give a corresponding result for the dual of M; see [14].

We give an example to show how this result can be used.

Example 1.19 In Chapter 6, we shall discuss a conjecture that relates the multiplicity of an ideal to the shifts in its graded free resolution. In investigating this conjecture, we need to determine whether certain Betti diagrams below that of a lexicographic ideal can occur. Let us consider the Hilbert function (1, 3, 6, 10, 15, 16, 15, 10) in the ring S = k[x_{1}, x_{2}, x_{3}]. The resolution of the lexicographic ideal corresponding to this Hilbert function is below:

total: 1 27 47 21
0: 1 …
1: ….
2: ….
3: ….
4: . 5 6 2
5: . 4 7 3
6: . 7 13 6
7: . 11 21 10

If we cancel all the Betti numbers corresponding to generators of the ideal of degree greater than five and then make all the other cancellations that may be possible, we obtain this Betti diagram:

total: 1 27 47 21
0: 1 …
1: ….
2: ….
3: ….
4: . 5 2 2
5: … 1
6: ….
7: . . 15 10

We wish to know if the Betti diagram with the cancellations can occur. The first generator in the third place in the resolution occurs in degree 4 + 3 = 7. The sum of the Betti numbers in column two (remember, the columns are numbered starting from zero) in rows zero through four in the Betti diagram is two. By Theorem 1.19, a necessary condition for such a resolution to exist is that this number must be at least three. Therefore the second Betti diagram does not occur for any graded module.

We state a final standard result in this section that provides a way to pass from considering Cohen-Macaulay modules to modules of finite length.

Lemma 1.20 Let I be an ideal of codimension c in R = k[x_{1}, …, x_{n}] such that R/I is Cohen-Macaulay. Then there exists an ideal I′ ⊂ R′ = k[x_{1}, …, x_{c}] such that R′/I′ is Artinian and has the same graded Betti numbers as R/I.

We shall use the lemma in the following way in Chapter 6: Suppose we want to show there is no Cohen-Macaulay module with a particular set of graded Betti numbers. Then it is enough to show that there is no Artinian module with those graded Betti numbers. Additionally, Lemma 1.20 means, for example, that results about the graded Betti numbers of Artinian modules in three variables hold more generally for Cohen-Macaulay codimension three modules. Thus considering only Artinian modules is not as much of a restriction as it might seem.

In the next five chapters, we use the tools from this chapter to investigate what resolutions do occur for modules with a given Hilbert function.

Hilbert functions with a unique

resolution

We begin our investigation of the possible resolutions for a given Hilbert function by considering the simplest case: For what Hilbert functions H is there only one possible set of graded Betti numbers that can occur in the resolution of a module with Hilbert function H?

2.1 Introduction

Let R = k[x_{1}, …, x_{n}]. If H is the Hilbert function of a module R/I for some homogeneous ideal I in R, we know that there exists a lexicographic ideal L in R such that the Hilbert functions of R/I and R/L are the same. Theorem 1.14 shows that the graded Betti numbers of R/I are at most those of R/L, but it is possible that there are many different sets of graded Betti numbers for modules that have Hilbert function H. The partial order on resolutions of modules with the same Hilbert function can be complicated with many incomparable elements.

In this chapter, we consider the simplest poset that can arise: the case in which there is only one set of graded Betti numbers possible for a given Hilbert function. We ask the following question:

Question 2.1 Let R = k[x_{1}, …, x_{n}]. What Hilbert functions H have the property that all modules with Hilbert function H have the same graded Betti numbers?

There are a number of easy examples of Hilbert functions for which this occurs, and we discuss these in the next section, relying on some recent work of Peeva [37]. In Section 3, we characterize the Hilbert functions of homogeneous ideals in k[x, y] that have this property. Finally, in Section 4, we construct a number of infinite families of Hilbert functions that admit only one set of graded Betti numbers. These families are different from the examples in [37] in that there are potential cancellations in the resolution of the lexicographic ideal, but Theorem 1.18 forbids them.

2.2 First examples

We seek Hilbert functions for which all modules with that Hilbert function have the same set of graded Betti numbers. Recall that for each valid Hilbert function, there is a quotient by a lexicographic ideal with that Hilbert function. In addition, every resolution for a module with a given Hilbert function is obtained through a sequence of consecutive cancellations of even and odd syzygies of the same degree in the resolution of the lexicographic ideal with that Hilbert function. Thus we seek lexicographic ideals L with resolution such that no configuration of graded Betti numbers below the resolution of L occur for any module with the same Hilbert function.

There are a few easy examples of this. (See also [38].) Let 𝔪 = (x_{1}, …, x_{n}) ⊂ R. Then R/𝔪^{d} has a d-linear resolution for all d ≥ 1; that is, for i ≥ 1, the i-th free module in the resolution, if nonzero, is a direct sum of copies of R(−d − i + 1). For example, if S = k[x_{1}, x_{2}, x_{3}], the resolution of S/𝔪^{4} is

0 → S(−6)^{10} → S(−5)^{24} → S(−4)^{15} → S → S/𝔪^{4} → 0,

represented by this Betti diagram:

S/𝔪^{4}: total: 1 15 24 10
0: 1 …
1: ….
2: ….
3: . 15 24 10

Obviously, no cancellation is possible in this Betti diagram, and therefore, since H(S/𝔪^{4}) = (1, 3, 6, 10), there is only one possible resolution for a module with Hilbert function (1, 3, 6, 10).

This is not such an interesting case, however, since R/𝔪^{4} is the only ideal in the polynomial ring in three variables corresponding to that Hilbert function. For an example with more content, consider the Hilbert function H = (1, 2, 3, 2, 2) in the ring S = k[x, y]. Let L be the lexicographic ideal (x^{3}, x^{2}y, xy^{4}, y^{5}); then S/L has Hilbert function H. Let I be the ideal

I = ((x − y)^{3}, (x − y)^{2}(x + y), (x − y)(x + y)^{4}, (x + y)^{5}).

Then I ≠ L, S/I also has Hilbert function H, and both modules have the following resolution:

S/I and S/L: total: 1 4 3
0: 1 . .
1: …
2: . 2 1
3: …
4: . 2 2

Note that no cancellation is possible in this Betti diagram, and therefore this is the only possible graded Betti diagram for a module with Hilbert function H.

Peeva gives the following easy corollary to Theorem 1.17 that generalizes this idea [37].

Corollary 2.2 Suppose a lexicographic ideal L does not have minimal monomial generators in consecutive degrees. If I is a homogeneous ideal with the same Hilbert function as L, then I and L have the same graded Betti numbers.

Example 2.3 Let L be the lexicographic ideal in k[x_{1}, …, x_{4}] corresponding to the Hilbert function (1, 4, 6, 10, 15). Then the quotient has the Betti diagram below:

total: 1 25 62 54 16
0: 1 ….
1: . 4 6 4 1
2: . . . . .
3: . . . . .
4: . 21 56 40 15

Because L does not have minimal generators in consecutive degrees, Corollary 2.2 shows that all ideals in k[x_{1}, …, x_{4}] corresponding to Hilbert function (1, 4, 6, 10, 15) have the same graded Betti numbers. Note that the only possibility for cancellation is the third syzygy (in column four in the resolution) of degree five with a generator of degree five, but Theorem 1.17 forbids this cancellation since it is not a consecutive cancellation.

One can actually prove a slightly stronger statement. It is no problem for L to have a single generator in the lowest degree d in which L is nonzero and then have multiple generators in degree d+1. This follows from the location of syzygies in the Eliahou-Kervaire resolution; the generator in degree d does not yield a first syzygy of degree d + 1. Example 2.4 Let L be the lexicographic ideal in S = k[x, y, z] such that S/L has Hilbert function (1, 3, 5, 4, 5). The graded Betti diagram of S/L is the following:

S/L: total: 1 10 16 7
0: 1 …
1: . 1 . .
2: . 3 5 2
3: ….
4: . 6 11 5

L has a single minimal generator in degree two and three in degree three, but since there is no first syzygy of degree three, no cancellation may occur.

Additionally, we can rephrase Corollary 2.2 in terms of the Hilbert function while also incorporating the notion discussed in the example above. Let H be a Hilbert function for R modulo a homogeneous ideal. Suppose d is the first integer for which H(d) < \binom{n−1+d}{d}; that is, d is the first degree in which we have a minimal generator. Then H admits a unique set of graded Betti numbers if there is no i such that

h_{i+1} < h_{i}^{<i>} and h_{i+2} < h_{i+1}^{<i+1>},

except possibly for

h_{d} = \binom{n − 1 + d}{d} − 1 and h_{d+1} < h_{d}^{<d>}.

In English, the condition says that the lexicographic ideal corresponding to H never has two minimal generators in consecutive degrees except possibly for one in the lowest degree d and additional generators in degree d + 1.

2.3 Hilbert functions over k[x, y]

In two variables, we can characterize all Hilbert functions that admit only one set of graded Betti numbers. We mention that this characterization is probably part of the folklore; we think it was at least known to A. Geramita before we discovered it.

We first need a result of Charalambous and Evans [4].

Theorem 2.5 Let L be a lexicographic ideal in S = k[x, y]. Let D be any graded Betti diagram that can be obtained from cancelling any number of even and odd syzygies of the same degree from the graded Betti diagram of S/L. Then there exists a monomial ideal I in S with graded Betti diagram D.

Charalambous and Evans give an algorithm for constructing the monomial ideal I from the desired Betti diagram D. Their theorem is particularly significant because it means that there are no “gaps” in the partial order; that is, any graded Betti diagram under that of the lexicographic ideal really does occur. This shows that to answer Question 2.1 for k[x, y], we need to find lexicographic ideals in two variables that do not have a generator and syzygy in the same degree. The Hilbert functions of these ideals will be exactly those for which there is only one possible set of graded Betti numbers.

What are the Hilbert functions of modules S/I in S = k[x, y]? The Hilbert function of S is (1, 2, 3, 4, …); that is, H(S, d) = d+1. Suppose the first generator of an ideal I ⊂ S is in degree t. Then h_{t} := H(S/I, t) ≤ t. By Macaulay’s Theorem,

H(S/I, t + 1) ≤ h_{t}^{<t>} ≤ (\binom{t}{t} + ⋯ + \binom{1}{1})^{<t>} = \binom{t + 1}{t + 1} + ⋯ \binom{2}{2} = t.

Therefore after the Hilbert function stops increasing, it will never increase again. Macaulay’s Theorem shows that if a lexicographic ideal L ⊂ S has a generator in a degree less than d but no generators in degree d, then H(S/L, d) = H(S/L, d − 1).

We combine these computations with the Eliahou-Kervaire resolution to prove a lemma about the resolution of ideals in k[x, y].

Lemma 2.6 Let S = k[x, y], and let L be a lexicographic ideal in S. Let H = (1, h_{1}, h_{2}, …) be the Hilbert function of S/L. Then H has two consecutive drops (h_{i} > h_{i+1} > h_{i+2} for some i) if and only if L has minimal generators in degrees i + 1 and i + 2 and a minimal first syzygy in degree i + 2.

Proof: First assume that there exists an i such that h_{i} > h_{i+1} > h_{i+2}. By the computations above, L has minimal generators in degrees i + 1 and i + 2. If x^{i+1} is a minimal generator of L, then L must have an additional generator of degree i + 1, namely x^{i}y, or else h_{i+1} = h_{i}. If x^{i+1} is not a minimal generator of L, then clearly L has a minimal generator in degree i + 1 involving y, for L can have only one pure power of x as a minimal generator. Therefore, by the Eliahou-Kervaire resolution, L has a minimal first syzygy of degree i + 2.

Now assume instead that L has minimal generators in degrees i + 1 and i + 2 and a minimal first syzygy in degree i + 2. By the Eliahou-Kervaire resolution, the only way L can have a minimal first syzygy of degree i + 2 is for L to have a minimal generator of degree i + 1 involving y. Therefore either both x^{i+1} and x^{i}y are minimal generators of L, in which case h_{i} = i + 1 > i ≥ h_{i+1}, or i + 1 is not the lowest degree in which L has a minimal generator, so by our computations before the lemma, h_{i} > h_{i+1}. In addition, because L has a minimal generator of degree i + 2, and there are minimal generators in a degree lower than i + 2, h_{i+1} > h_{i+2}, and we are done. □

We give an example to illustrate Lemma 2.6.

Example 2.7 Let S = k[x, y], and let I = (x^{3}, x^{2}y^{3}, xy^{4}, y^{6}). Then S/I has the graded Betti diagram:

S/I: total: 1 4 3
0: 1 . .
1: …
2: . 1 .
3: …
4: . 2 2
5: . 1 1

The Hilbert function corresponding to this ideal is (1, 2, 3, 3, 3, 1, 0), and the two consecutive drops at the end indicate the presence of minimal generators in degrees five and six and the minimal first syzygy of degree six.

It is now easy to prove the following theorem.

Theorem 2.8 Let S = k[x, y], and let H = (1, h_{1}, h_{2}, …) be a Hilbert function for a module of the form S/I. Then all modules of the form S/I with Hilbert function H have the same set of graded Betti numbers if and only if there are never two consecutive drops in H (that is, if there is no i such that h_{i} > h_{i+1} > h_{i+2}). Proof: Suppose first that there exists an i such that h_{i} > h_{i+1} > h_{i+2}. Let L be the lexicographic ideal such that S/L has Hilbert function H. Lemma 2.6 implies that L has minimal generators in degrees i+1 and i+2 and a minimal first syzygy of degree i + 2. By Theorem 2.5, the generator and syzygy of degree i + 2 can be cancelled to yield a Betti diagram for a module S/I with the same Hilbert function as S/L, and this Betti diagram will obviously be below that of the lexicographic ideal in the partial order.

For the other implication, we must show that if there is more than one possible set of graded Betti numbers for H, then there are two consecutive drops in H. Since there are multiple possible Betti diagrams corresponding to H, there must be some cancellation possible in the resolution of the lexicographic ideal L. Therefore, for some i, L must have minimal generators in degrees i + 1 and i + 2 and a minimal first syzygy in degree i + 2. Hence by Lemma 2.6, H has two consecutive drops. □

2.4 Families of examples with “ghost terms”

In this section, we shall discuss some infinite families of Hilbert functions for which there is a unique set of graded Betti numbers. In each case, the resolution of the lexicographic ideal appears to allow some cancellation. However, we shall use Theorem 1.18 to show that no cancellation may occur; the terms that look like they may cancel but cannot are often called ghost terms.

To find examples of Hilbert functions for which there is a unique set of graded Betti numbers, we want to find lexicographic ideals whose resolutions do not allow any cancellations. Our proof in each of the cases below is to resolve the lexicographic ideal with a particular Hilbert function, using the Eliahou-Kervaire resolution, and then to use Theorem 1.18 to show that none of the potential cancellations can occur. Proposition 2.9 Let R = k[x_{1}, …, x_{n}] with n ≥ 3. Let H be the Hilbert function

(1, n, \binom{n + 1}{2} − 2, \binom{n + 1}{3} + \binom{n − 1}{2} − 1, \binom{n + 2}{4} + \binom{n − 1}{3} + \binom{n − 2}{2}) .

Then there is a unique set of graded Betti numbers corresponding to H. Proof: We begin by analyzing the Hilbert function itself. Let L be the lexicographic ideal such that R/L has Hilbert function H. Since H(1) = n, L has no linear generators. However, H(2) is two fewer than the maximum possible, and therefore L has two minimal generators of degree two, namely x_{1}^{2} and x_{1}x_{2}. To find the number of minimal generators L has in degree three, we need to compare H(3) to H(2)^{<2>}, the value of H if there were no minimal generators in degree three. We have

\binom{n + 1}{2} − 2 = \binom{n}{2} + \binom{n − 2}{1},

so

H(2)^{<2>} = \binom{n + 1}{3} + \binom{n − 1}{2}.

This is one more than H(3), and hence L has one minimal generator in degree three, which must be x_{1}x_{3}^{2}.

Next, we investigate how H grows from degree three to degree four. It is easy to compute that

H(3) = \binom{n + 1}{3} + \binom{n − 2}{2} + \binom{n − 3}{1},

and hence

H(3)^{<3>} = \binom{n + 2}{4} + \binom{n − 1}{3} + \binom{n − 2}{2} = H(4).

Consequently, L has no minimal generators in degree four. Since H(5) = 0, if 𝔪 = (x_{1}, …, x_{n}), L = (x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}^{2}) + 𝔪^{5}. The graded Betti diagram of R/L has the following shape, where * indicates possibly nonzero entries.

0: 1 . . . . . . .
1: . 2 1 . . . . .
2: . 1 2 1 ….
3: . . . . . . . .
4: . * * * * …

It follows from Theorem 1.17 that the only possible cancellation in this Betti diagram would occur in degree three, producing the Betti diagram below:

0: 1 . . . . . . .
1: . 2 . . . . . .
2: . . 2 1 ….
3: . . . . . . . .
4: . * * * * …

By Theorem 1.18, the two first syzygies of degree at most four are not enough to allow the second syzygy of degree five. Therefore this Betti diagram cannot occur. Hence the only Betti diagram that can occur for a module R/I with Hilbert function H is the Betti diagram of the lexicographic ideal L. □

Our next example uses three generators in degree two instead of the two in Proposition 2.9. This provides for two possible cancellations, so there are more cases to check.

Proposition 2.10 Let R = k[x_{1}, …, x_{n}] with n ≥ 4. Let H be the Hilbert function

(1, n, \binom{n + 1}{2} − 3, \binom{n + 1}{3} + \binom{n − 2}{2} − 1, \binom{n + 2}{4} + \binom{n − 2}{3} + \binom{n − 3}{2}) .

Then there is a unique set of graded Betti numbers corresponding to H.

Proof: Let L be the lexicographic ideal in R such that R/L has Hilbert function H. Then L has three minimal generators of degree two, and they must be x_{1}^{2}, x_{1}x_{2}, and x_{1}x_{3}. One computes easily that

H(2) = \binom{n + 1}{2} − 3 = \binom{n}{2} + \binom{n − 3}{1}.

Therefore

H(2)^{<2>} = \binom{n + 1}{3} + \binom{n − 2}{2},

and hence L has a single minimal generator, x_{1}x_{4}^{2}, in degree three. Because

H(3) = \binom{n + 1}{3} + \binom{n − 2}{2} − 1 = \binom{n + 1}{3} + \binom{n − 3}{2} + \binom{n − 4}{1},

we have

H(3)^{<3>} = \binom{n + 2}{4} + \binom{n − 2}{3} + \binom{n − 3}{2} = H(4),

and hence L has no minimal generators in degree four. Thus

L = (x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}, x_{1}x_{4}^{2}) + 𝔪^{5},

where 𝔪 = (x_{1}, …, x_{n}). The Betti diagram of R/L is below, where * again indicates a possibly nonzero entry.

0: 1 . . . . . . . .
1: . 3 3 1 . . . . .
2: . 1 3 3 1 ….
3: . . . . . . . . .
4: . * * * * * …

There are two possible cancellations, in degree three and in degree four. If we make only the cancellation in degree three, we get the following diagram:

0: 1 . . . . . . . .
1: . 3 2 1 . . . . .
2: . . 3 3 1 ….
3: . . . . . . . . .
4: . * * * * * …

But then we have only two first syzygies of degree at most three and a second syzygy of degree four, which Theorem 1.18 does not allow. If we make only the cancellation in degree four instead, we have the diagram below:

0: 1 . . . . . . . .
1: . 3 3 . . . . . .
2: . 1 2 3 1 ….
3: . . . . . . . . .
4: . * * * * * …

Then we have three second syzygies of degree at most five, but we have a third syzygy of degree six, which Theorem 1.18 forbids. Finally, if we make both cancellations, we have

0: 1 . . . . . . . .
1: . 3 2 . . . . . .
2: . . 2 3 1 ….
3: . . . . . . . . .
4: . * * * * * …,

which gives the same problem with the second syzygies of degree at most five and the third syzygy of degree six. Therefore there is only one possible Betti diagram corresponding to H. □

The logical next step is to try four minimal generators in degree two with one in degree three in n ≥ 5 variables. However, this yields the Betti diagram

0: 1 . . . . . . . . .
1: . 4 6 4 1 . . . . .
2: . 1 4 6 4 1 ….
3: . . . . . . . . . .
4: . * * * * * * …,

and one can cancel the generator in degree three with a first syzygy of the same degree without violating Theorem 1.18.

Propositions 2.9 and 2.10 are by no means exhaustive; one can use Theorem 1.18 to find more examples of Hilbert functions admitting only one set of graded Betti numbers. In particular, there are several easy examples if one allows linear generators. Most of these have the same flavor, so we give only one here.

It is not hard to check that, in n ≥ 3 variables, the Hilbert function

(1, n − 1, \binom{n}{2} − 1, \binom{n}{3} + \binom{n − 1}{2} − 1, \binom{n + 1}{4} + \binom{n − 1}{3} + \binom{n − 2}{2})

permits only one set of graded Betti numbers. (Note that this is not the same as our two previous examples; for example, when n = 4, the Hilbert function in Proposition 2.9 is (1, 4, 8, 12, 17), and Proposition 2.10 yields Hilbert function (1, 4, 7, 10, 15), while with n = 5, this example gives Hilbert function (1, 4, 9, 15, 22).) The corresponding lexicographic ideal is L = (x_{1}, x_{2}^{2}, x_{2}x_{3}^{2}) + 𝔪^{5}, where 𝔪 = (x_{1}, …, x_{n}). This yields the Betti diagram

0: 1 1 . . . . . .
1: . 1 1 . . . . .
2: . 1 2 1 ….
3: . . . . . . . .
4: . * * * * …,

and Theorem 1.18 shows that cancellation in degree three cannot occur because of the second syzygy in degree five.

In this chapter, we have given a number of examples of Hilbert functions for which there is a unique set of graded Betti numbers possible. It is much more difficult, however, to give necessary conditions. A solution to Question 2.1 that characterizes Hilbert functions admitting a unique set of graded Betti numbers would represent great progress in the investigation of what resolutions can occur for a given Hilbert function. The result of Peeva, Corollary 2.2, gives a sufficient condition, and the families in the examples above yield additional examples that are not immediately obvious from the resolution of the lexicographic ideal. It seems likely, however, that more powerful tools for determining what sets of graded Betti numbers can occur will be necessary for a complete solution to Question 2.1.

Minimal graded Betti numbers and

stable ideals*

In Chapter 1, we saw that the resolution of the lexicographic ideal is uniquely maximal in the partial order on the graded Betti numbers of modules with a given Hilbert function, but there may not be a unique minimal element in the partial order. In this chapter, we investigate the graded Betti numbers of stable ideals. Instead of considering all ideals with a fixed Hilbert function, we restrict our attention only to stable ideals with a given Hilbert function. We determine over which polynomial rings it is always possible to find a uniquely minimal resolution in this poset.

3.1 Introduction

In [4], Charalambous and Evans characterize all the possible graded resolutions of Artinian codimension two ideals. They start with the resolution of a lexicographic ideal with a given Hilbert function, and then they prove that all resolutions obtained from cancelling syzygies of the same degree from that resolution do occur for some module with the same Hilbert function. One would like to be able to find a similar characterization in higher dimensions. However, Charalambous and Evans show that the situation is more complicated as soon as we move to three variables. For all n ≥ 3, they provide a Hilbert function in k[x_{1}, …, x_{n}] for which

*Portions of this chapter are reprinted from “Minimal graded Betti numbers and stable ideals,” by Christopher A. Francisco, which appeared in Communications in Algebra 31 (2003), no. 10, 4971-4987 [15]. there are incomparably minimal graded Betti numbers. Therefore there does not have to be a unique minimal element in the partial order on graded Betti numbers of modules with the same Hilbert function.

This is one reason that determining the sets of graded Betti numbers that do occur for some module is much harder in three or more variables. Deery proposes in [7] restricting to the class of stable ideals to overcome the incomparability problem that exists when considering all ideals in k[x_{1}, x_{2}, x_{3}]. Stable ideals are of particular interest because they occur as generic initial ideals, and they have proven tremendously useful in reduction arguments in the areas of Hilbert functions and graded Betti numbers recently. Our main result in this chapter is the following theorem.

Theorem 3.10 Let S = k[x_{1}, x_{2}, x_{3}], and let H be a Hilbert function for S modulo a homogeneous ideal. Then there exists a stable ideal I such that H(S/I) = H, and given any stable ideal J ⊂ S with H(S/J) = H, β^{S/I} ≤ β^{S/J}.

We prove Theorem 3.10 by giving an algorithm to produce I. Moreover, the ideal I is actually strongly stable, a more restrictive condition than stability. We prove this in Theorem 3.12. Finally, Deery conjectured that a slightly weaker statement than Theorem 3.10 would be true in more variables, but in the final section, we give a family of counterexamples.

I thank the referee of the published version of this chapter that appeared in the Communications in Algebra for his or her many helpful suggestions.

3.2 Definitions and some previous work

For the rest of the chapter, let S = k[x_{1}, x_{2}, x_{3}], where k is a field. In this section, we introduce some terminology from Deery’s paper [7] and discuss some particular types of stable ideals.

Recall from Chapter 1 that for a monomial m, we define max(m) to be the largest index i such that x_{i} divides m. For example, max(x_{1}^{2}) = 1, and max(x_{1}x_{2}^{3}x_{4}) = 4. We used this notion in Chapter 1 to define stable ideals.

Definition 1.11 An ideal I is said to be stable if I is a monomial ideal, and for each monomial m in I and all i < max(m), x_{i}m/x_{max(m)} is also in I. (We define a stable set of monomials analogously.)

To check whether a monomial ideal is stable, it is enough to check the condition on minimal generators, and an ideal is stable if and only if the set of monomials in each degree is stable. Recall from Chapter 1 that the graded Betti numbers of stable ideals are given by

β_{qj}^{R/I} = ∑_{deg(m)=j−q+1^{m∈G(I)}} \binom{max(m) − 1}{q − 1},  (3.1)

where G(I) is the minimal monomial generating set of a stable ideal I ⊂ R = k[x_{1}, …, x_{n}]. We seek to minimize this quantity.

To do this, we shall consider some special stable ideals. In [7], Deery defines the notions of reverse-lex ideals and weakly reverse-lex ideals (Deery calls them “almost reverse-lex” ideals) to prove some minimality properties. Recall that if x_{1}^{a_{1}} ⋯ x_{n}^{a_{n}} and x_{1}^{b_{1}} ⋯ x_{n}^{b_{n}} are monomials of the same degree, then x_{1}^{a_{1}} ⋯ x_{n}^{a_{n}} > x_{1}^{b_{1}} ⋯ x_{n}^{b_{n}} in the reverse-lex order if and only if the last nonzero entry of (a_{1} − b_{1}, …, a_{n} − b_{n}) is negative.

Definition 3.1 Call a monomial ideal I a reverse-lex ideal if in each degree, I is generated by an initial reverse-lex segment.

For example, the ideal (x_{1}^{3}, x_{1}^{2}x_{2}, x_{1}x_{2}^{2}, x_{2}^{3}) ⊂ k[x_{1}, x_{2}, x_{3}] is a reverse-lex ideal. Of course, if one replaces “reverse-lex” with “lexicographic,” one obtains the definition of a lexicographic ideal. In [7], Deery shows that if I is a reverse-lex ideal in R = k[x_{1}, …, x_{n}], and J is a stable ideal with the same Hilbert function, then the graded Betti numbers in the resolution of R/I are at most those of R/J. Marinari and Ramella have a similar result in [34]. Unfortunately, Hilbert functions admitting a reverse-lex ideal are rather uncommon, and it is beneficial to consider a generalization introduced in [7].

Definition 3.2 Let I be a monomial ideal, and for each degree d, let m_{d} be the minimal monomial generator of I in degree d that is smallest with respect to reverse-lex order. We say that I is a weakly reverse-lex ideal if all monomials of degree d greater than m_{d} in the reverse-lex order are in I.

Obviously any reverse-lex ideal is a weakly reverse-lex ideal, and it is easy to see that weakly reverse-lex ideals are stable. The ideal (x_{1}) + (x_{1}, x_{2}, x_{3})^{3} is not a reverse-lex ideal since in degree 2, x_{2}^{2} is not in the ideal, while x_{1}x_{3} is. However, it is a weakly reverse-lex ideal.

In [7], Deery modifies combinatorial methods that Bigatti uses in [2] to show that in three variables, the weakly reverse-lex ideal has uniquely minimal graded Betti numbers among all stable ideals with the same Hilbert function. This is a useful generalization since weakly reverse-lex ideals can attain many more Hilbert functions than reverse-lex ideals. Nevertheless, there are still many Hilbert functions for which there is no quotient corresponding to a weakly reverse-lex ideal in S. Consider, for example, the Hilbert function (1, 3, 3, 4). We need to put three monomials into the ideal in degree 2, and they must be x_{1}^{2}, x_{1}x_{2}, and x_{2}^{2}. But this puts seven monomials into the ideal in degree three, and there can be at most six. Thus there is no weakly reverse-lex ideal such that the quotient can have Hilbert function (1, 3, 3, 4).

Theorem 3.10 guarantees that we can always find a stable ideal I ⊂ S with uniquely minimal graded Betti numbers among all stable ideals with the same Hilbert function. Our algorithm in Section 5 to produce this I returns a weakly reverse-lex ideal if there exists one corresponding to the given Hilbert function. If not, the algorithm returns an ideal that is as close to a weakly reverse-lex ideal as possible. We make this notion precise in Section 5.

3.3 Some lemmas and formulas

In this section, we gather some results and formulas about stable ideals that we shall need in the proof of Theorem 3.10. Many of the techniques in the next two sections come from Bigatti [2] and Deery [7]. Following their notation, if T is a finite set of monomials in k[x_{1}, …, x_{n}], let m_{i}(T) be the number of monomials u in T such that max(u) = i. Similarly, let m_{≤i}(T) be the number of monomials u in T such that max(u) ≤ i. Finally, let X_{n}T denote the set of monomials {x_{i}u|1 ≤ i ≤ n and u ∈ T }. We have the following proposition from [2] and [7]:

Proposition 3.3 Let T be a stable set of monomials of degree d. Then: (i) X_{n}T is stable. (ii) m_{i}(X_{n}T) = m_{≤i}(T). (iii) |X_{n}T | = ∑_{i=1}^{n} m_{≤i}(T).

Given a stable ideal, this result is extremely useful in helping determine how many monomials are in the ideal in a high degree when one knows what the ideal looks like in low degrees. We are particularly interested in the case n = 3. Suppose a stable ideal I ⊂ S is generated in degree d and lower, and suppose we form a new stable ideal J = (I, m) by adding to I a minimal monomial generator m of degree d. If max(m) = 2, then Proposition 3.3 tells us that, comparing J to I, J has one more monomial in degree d, two more in degree d + 1, and, in general, s + 1 more monomials in degree d + s. If instead max(m) = 3, then J has exactly one more monomial than I in each degree d and higher.

In the proof of Theorem 3.10, we shall need some information about the growth of a Hilbert function for a quotient of S. Macaulay’s characterization of Hilbert functions in Theorem 1.7 says that H = (h_{0}, h_{1}, …) is a valid Hilbert function if and only if h_{0} = 1 and for all d ≥ 1, h_{d+1} ≤ h_{d}^{<d>}. We are interested in Hilbert functions for quotients in three variables, so h_{d} ≤ \binom{d+2}{d} = \binom{d+2}{2} for each d. In order to determine whether we have a valid Hilbert function, the next lemma is helpful.

Lemma 3.4 Let a ≥ 1 be an integer, and let l be an integer such that 1 ≤ l ≤ \binom{a+2}{a}. If m_{l} is the unique integer such that 1 + ⋯ + m_{l} < l ≤ 1 + ⋯ + (m_{l} + 1) (equivalently, \binom{m_{l}+1}{2} < l ≤ \binom{m_{l}+2}{2}), then [\binom{a+2}{a} − l]^{<a>} = [\binom{a+2}{a} − l] + (a − m_{l}).

Proof: First suppose that \binom{a+2}{a} − l ≥ a + 1. Since l ≥ 1, the a-th Macaulay expansion of \binom{a+2}{a} − l begins with \binom{a+1}{a}, and each additional term is of the form \binom{y+1}{y}, \binom{y}{y}, or zero. When passing to [\binom{a+2}{a} − l]^{<a>}, it is clear that the values of the latter two types of terms are unchanged, and sending \binom{y+1}{y} to \binom{y+2}{y+1} adds one to the previous sum. To compute [\binom{a+2}{a} − l]^{<a>}, we work from the Macaulay expansion

\binom{a + 2}{a} − l = \binom{a + 1}{a} + \binom{a}{a − 1} + ⋯ + \binom{z + 1}{z} + \binom{z − 1}{z − 1} + ⋯ + \binom{b}{b}.

Clearly [\binom{a+2}{a} − l]^{<a>} = [\binom{a+2}{a} − l] + (a − z + 1). Therefore it suffices to show that z = m_{l} + 1. Note that

(a + 1) + a + ⋯ + (z + 1) + z > \binom{a + 2}{2} − l ≥ (a + 1) + a + ⋯ + (z + 1).

Multiplying by -1 and adding \binom{a+2}{2}, we obtain

1 + ⋯ + (z − 1) < l ≤ 1 + ⋯ + z.

Consequently, m_{l} = z − 1 as desired.

If instead \binom{a+2}{a} − l ≤ a, all the terms in the Macaulay expansion are 1, and so the “upper a” operation should not add anything. Note that 1 + ⋯ + a < l ≤ 1 + ⋯ + (a + 1). Hence m_{l} = a, and this agrees with our formula. □

3.4 Combinatorial formulas for the Betti numbers

We wish to find a condition under which the graded Betti numbers of a given stable ideal will be minimal among the Betti numbers of all stable ideals with a fixed Hilbert function. It proves helpful to relate Formula 3.1 from Section 2 for the graded Betti numbers to the m_{≤i} defined in the previous section. To do this, we again borrow some definitions and results from Bigatti and Deery. First, we compare m_{≤i} for two stable sets, taking a result from [7]. Lemma 3.5 Let T and U be stable sets of the same degree such that m_{≤i}(T) ≤ m_{≤i}(U) for all i. Then m_{≤i}(X_{n}T) ≤ m_{≤i}(X_{n}U).

Next, we make two definitions from [7] that will allow us to draw conclusions about graded Betti numbers and total Betti numbers (the sum of the graded Betti numbers for a fixed syzygy). For a finite set of monomials T, we define

b_{qj}(T) = ∑_{deg(m)=j−q+1^{m∈T}} \binom{max(m) − 1}{q − 1} and b_{q}(T) = ∑_{j=0}^{∞} b_{qj}(T).

The first step in relating the m_{≤i} to the Betti numbers is the following proposition of Bigatti from [2] and [7]:

Proposition 3.6 If T is a stable set of monomials of all the same degree in S = k[x_{1}, x_{2}, x_{3}], then

b_{q}(T) = \binom{2}{q − 1}|T | − ∑_{i=1}^{2} [m_{≤i}(T)\binom{i − 1}{q − 2}].

We need two more results before we can prove our main lemma. The first is Deery’s adaptation in [7] of a formula from [2], and its proof follows from the previous results.

Lemma 3.7 Let T and U be stable sets of monomials of the same degree such that |T | = |U| and m_{≤i}(T) ≥ m_{≤i}(U) for each i. Then (i) b_{q}(T) ≤ b_{q}(U) and (ii) b_{q}(X_{n}T) ≥ b_{q}(X_{n}U).

Now let M(I_{d}) be the set of monomials in I of degree d. Our final tool is the following from [2] and [7] (note that the subscripts on I are incorrect in [7]):

Lemma 3.8 If I is a stable ideal, then for j ≥ q, β_{qj}(S/I) = b_{q}(M(I_{j−q+1}))− b_{q}(X_{3}M(I_{j−q})).

Finally, we may state the lemma that we need to prove the minimality of the graded Betti numbers in the following section. The proof is based on Deery’s proof of Theorem 3.10 in [7].

Lemma 3.9 Let I ⊂ S be a stable ideal such that if J ⊂ S is any stable ideal with the same Hilbert function, then m_{≤i}(I_{d}) ≥ m_{≤i}(J_{d}) for all i and d. Then for all q and j, β_{qj}(S/I) ≤ β_{qj}(S/J).

Proof: By Lemma 3.7, b_{q}(M(I_{d})) ≤ b_{q}(M(J_{d})) and b_{q}(X_{3}M(I_{d})) ≥ b_{q}(X_{3}M(J_{d})) for all d. Now if j < q, clearly β_{qj}(S/I) = β_{qj}(S/J) = 0. Suppose j ≥ q. By the previous lemma,

β_{qj}(S/I) = b_{q}(M(I_{j−q+1})) − b_{q}(X_{3}M(I_{j−q}))
≤ b_{q}(M(J_{j−q+1})) − b_{q}(X_{3}M(J_{j−q}))
= β_{qj}(S/J).
□

Since we are working in three variables, if two stable ideals I and J have the same Hilbert function, m_{≤1}(I_{d}) = m_{≤1}(J_{d}) and m_{≤3}(I_{d}) = m_{≤3}(J_{d}) in each degree d. Hence in the previous lemma, it suffices to note that I and J are stable ideals with the same Hilbert function such that m_{≤2}(I_{d}) ≥ m_{≤2}(J_{d}). That is, we seek a stable ideal I that has as many monomials in each degree that do not involve x_{3} as possible.

3.5 The algorithm

In this section we present our main result, an algorithm to obtain an ideal in S with uniquely minimal graded Betti numbers among all stable ideals with a given Hilbert function. The algorithm will give us the following theorem. Theorem 3.10 Let S = k[x_{1}, x_{2}, x_{3}], and let H be a Hilbert function for S modulo a homogeneous ideal. Then there exists a stable ideal I such that H(S/I) = H, and given any stable ideal J ⊂ S with H(S/J) = H, β^{S/I} ≤ β^{S/J}.

Note that the algorithm for producing the ideal I in Theorem 3.10 assumes that our Hilbert function has finite length (that is, that the Hilbert function is eventually zero). We describe how to remove that hypothesis in Lemma 3.11.

Input: A finite length Hilbert function H for S = k[x_{1}, x_{2}, x_{3}] modulo some

homogeneous ideal.

Output: A stable ideal I in S such that S/I has Hilbert function H and uniquely

minimal graded Betti numbers among all quotients of S with Hilbert function H

that correspond to stable ideals.

Initialization: p:=first positive degree in which H is zero; I:=(x_{1}, x_{2}, x_{3})^{p}; H̄:=Hilbert function of S/I; Δ:=H̄ − H;

begin while Δ ≠ {0, …, 0} do

t:=first degree in which Δ is not zero;

M:=list of all monomials of degree t in descending reverse-lex order; while entry t of Δ is not zero do

m:=first monomial in M not in I; H_{0}:=Hilbert function of S/(I, m); if (max(m)=2 and H_{0} − H is not weakly increasing)

then M := M \ all monomials of degree t not involving x_{3}; else if (max(m)=1 or 2) or (max(m)=3 and (I, m) is stable)

then I := (I, m); H̄:=Hilbert function of S/I; M := M \ {m}; Δ := H̄ − H;

return I; end

Our algorithm works by adding monomials into the ideal being constructed in

reverse-lex order as much as possible. The list Δ indicates how many monomials

we still need to add into the ideal in each degree to attain the Hilbert function H.

As long as the Hilbert function allows it, we put in generators only involving x_{1}

and x_{2}. At some point, it is possible that adding in such a generator will make the

Hilbert function too small in a higher degree; by Proposition 3.3, this occurs when Δ is no longer weakly increasing. Then we must resort to adding generators with nonzero powers of x_{3} while being careful to maintain the stability of the ideal.

The main work in the proof is showing that if we cannot add a generator involving only x_{1} and x_{2}, then there exists a monomial involving x_{3} that we can add that keeps the ideal stable. We prove this by supposing that no such monomial exists and enumerating the monomials that must already be in the ideal. This allows us to compute a formula for the Hilbert function in each degree. Consequently, we can use the bound on the growth of the Hilbert function from Lemma 3.4 to show that the Hilbert function grows faster than is allowed, so our input was not a valid Hilbert function.

Proof of Theorem 3.10: Assume first that the Hilbert function H is eventually zero. For each H, there is a quotient corresponding to a lexicographic ideal attaining that Hilbert function. Thus Proposition 3.3 shows that Δ, the difference vector of the Hilbert function of S/I at a point in the algorithm and H, must initially be weakly increasing. The first generator is forced, and it must be a power of x_{1}. If one needs to add a second generator, it must be x_{1} to one fewer power times a power of x_{2} to ensure stability. Clearly Δ stays weakly increasing since the ideal is currently a lexicographic ideal.

Suppose now that we have added w > 1 generators into our ideal, maintaining a stable ideal with weakly increasing Δ and putting in the monomials in the order dictated by the algorithm. If we need not add in more monomials (that is, if Δ is the zero vector), we are completely finished, for the ideal has the desired Hilbert function. Otherwise, we need to put in a new monomial in, say, degree t. If not all x_{1}^{t−d}x_{2}^{d} are in I, we attempt to add the first one (in reverse-lex order) that is not already in. If Δ remains weakly increasing, then we are done since the ideal clearly remains stable. If not, then we must add in a monomial with nonzero power of x_{3}. Proposition 3.3 shows that in this case, there is some s ≥ 0 such that Δ_{t+s} = Δ_{t+s+1} = y for some y > 0. We wish to show that we can find a monomial m of degree t involving x_{3} to add to the ideal that keeps the ideal stable; if so, Δ will remain weakly increasing by the formulas in Proposition 3.3.

We claim that we can always do this. Suppose not, and suppose that x_{1}^{t}, x_{1}^{t−1}x_{2}, …, and x_{1}^{t−d}x_{2}^{d} are in I. Under our assumption, there does not exist a monomial m of degree t such that (I, m) is stable and the Hilbert function is not too small. Hence we may conclude that x_{1}^{t−1}x_{3}, x_{1}^{t−2}x_{2}x_{3}, …, x_{1}^{t−d}x_{2}^{d−1}x_{3}, x_{1}^{t−2}x_{3}^{2}, …, x_{1}^{t−d}x_{2}^{d−2}x_{3}^{2}, …, x_{1}^{t−d}x_{3}^{d} are all in I. Therefore in degree t, there is one monomial involving only x_{1}; there are d monomials u with max(u) = 2, and there are \binom{d+1}{2}+r monomials u with max(u) = 3, where r ≥ 0. By (ii) of Proposition 3.3, it is easy to compute that there are d + s monomials involving x_{2} but not x_{3} in degree t + s, and there are d + s + 1 in degree t + s + 1. Similarly, we can count the monomials in each degree in I that have nonzero powers of x_{3}. There are \binom{s+1}{2} +sd+ \binom{d+1}{2} +r in degree t + s and \binom{s+2}{2} +(s + 1)d +\binom{d+1}{2} +r in degree t + s + 1.

Thus in I,

# of monomials in degree t + s = \binom{s + 2}{2} + (s + 1)d + \binom{d + 1}{2} + r,

and

# of monomials in degree t + s + 1 = \binom{s + 3}{2} + (s + 2)d + \binom{d + 1}{2} + r.

Consequently, we obtain formulas for the Hilbert function in degrees t + s and t + s + 1:

h_{t+s} = \binom{t + s + 2}{2} − [\binom{s + 2}{2} + (s + 1)d + \binom{d + 1}{2} + r + y] ,

and

h_{t+s+1} = \binom{t + s + 3}{2} − [\binom{s + 3}{2} + (s + 2)d + \binom{d + 1}{2} + r + y] .

As a result, we have h_{t+s+1} − h_{t+s} = t + s + 2 − (s + 2) − d = t − d. Thus h_{t+s} + t − d = h_{t+s+1} ≤ h_{t+s}^{<t+s>}. We wish to find h_{t+s}^{<t+s>}. Let l be the number of monomials in the desired final ideal in degree t + s. By Lemma 3.4,

h_{t+s}^{<t+s>} = h_{t+s} + t + s − m_{l}, \binom{m_{l} + 1}{2} < l ≤ \binom{m_{l} + 2}{2}.

Hence h_{t+s} + t − d ≤ h_{t+s} +t + s − m_{l}. Therefore s + d ≥ m_{l}.

Thus if we cannot find a monomial with nonzero power of x_{3} to add that keeps the ideal stable, then s + d ≥ m_{l}. We have

l = \binom{s + 2}{2} + (s + 1)d + \binom{d + 1}{2} + r + y
= \binom{s + d + 2}{2} + r + y
≤ \binom{m_{l} + 2}{2}
≤ \binom{s + d + 2}{2}.

This is a contradiction because r + y > 0. Hence by induction, we can always find the monomial we need to keep the ideal stable and to ensure that Δ stays weakly increasing.

The algorithm will clearly terminate because the Hilbert function is eventually zero. The minimality of the graded Betti numbers is immediate since the algorithm maximizes the number of monomials in each degree not involving x_{3} subject to the constraints of the Hilbert function and keeping the ideal stable.

The nonfinite length case follows from Lemma 3.11 and finishes the proof. □

Lemma 3.11 Suppose that for all finite length Hilbert functions H for modules of the form S = k[x_{1}, x_{2}, x_{3}] modulo a homogeneous ideal, the conclusion of Theorem 3.10 is true. Then the conclusion of Theorem 3.10 also holds for H that are not eventually zero in high degree.

Proof: Consider a Hilbert function H that is never zero in positive degree. The idea is to truncate the Hilbert function at an appropriate place so that we may work with something of finite length. Since the lexicographic ideal has the most generators in each degree of any ideal with a given Hilbert function, we first compute the highest degree d of a minimal generator for the lexicographic ideal L whose quotient has Hilbert function H. Writing H = (1, h_{1}, h_{2}, …, h_{d}, h_{d+1}, …), let H′ = (1, h_{1}, h_{2}, …, h_{d}, h_{d+1}). Then H′ has finite length, so we can use H′ as the input in the finite length algorithm, obtaining output I′. Now the Hilbert function of S/I′ is H′, so it agrees with H through one degree beyond the degree of the maximal generator of L. We then remove all the generators of I′ of degree higher than d, forming a new ideal I. Both L and I have generators of degree no higher than d, and the Hilbert functions of S/L and S/I agree through degree d + 1. By the Gotzmann Persistence Theorem (see [22]), since S/L has Hilbert function H, S/I must also, and its graded Betti numbers obviously have the minimality property we want. □

The ideal we obtain from our algorithm satisfies a more restrictive condition than stability. Recall that we call a monomial ideal strongly stable if whenever a monomial m ∈ I, then x_{i}m/x_{j} ∈ I for each i < j.

Theorem 3.12 The above algorithm produces a strongly stable ideal.

Proof: It suffices to check the strongly stable property on minimal generators. Suppose x_{1}^{a}x_{2}^{b}x_{3}^{c} is a minimal generator of I. We need to show that x_{1}^{a+1}x_{2}^{b−1}x_{3}^{c} ∈ I, for the other monomials to check are automatically in I by stability. We proceed by induction on c. If c = 0, x_{1}^{a+1}x_{2}^{b−1} ∈ I because it comes before x_{1}^{a}x_{2}^{b} in reverselex order, so we would have chosen it first. Now suppose that c ≥ 1 and that the strongly stable property holds for all monomials with power of x_{3} at most c − 1. Stability implies that both x_{1}^{a+1}x_{2}^{b}x_{3}^{c−1} and x_{1}^{a}x_{2}^{b+1}x_{3}^{c−1} are in I. Note that x_{1}^{a+1}x_{2}^{b−1}x_{3}^{c} >_{rlex} x_{1}^{a}x_{2}^{b}x_{3}^{c}, so if x_{1}^{a+1}x_{2}^{b−1}x_{3}^{c} is not in I, we must not have chosen it in our algorithm because its inclusion in I would have ruined stability. Thus either x_{1}^{a+2}x_{2}^{b−1}x_{3}^{c−1} or x_{1}^{a+1}x_{2}^{b}x_{3}^{c−1} is not in I. But by the induction hypothesis, x_{1}^{a+1}x_{2}^{b}x_{3}^{c−1} ∈ I implies that x_{1}^{a+2}x_{2}^{b−1}x_{3}^{c−1} ∈ I, and similarly, x_{1}^{a}x_{2}^{b+1}x_{3}^{c−1} ∈ I implies that x_{1}^{a+1}x_{2}^{b}x_{3}^{c−1} ∈ I. This is a contradiction, so x_{1}^{a+1}x_{2}^{b−1}x_{3}^{c} ∈ I after all. □

Example 3.13 For an example of how the algorithm works, we consider the Hilbert function H = (1, 3, 3, 4, 4). The first positive degree in which H is zero is p = 5. We set I = (x_{1}, x_{2}, x_{3})^{5}, and then H̄ = (1, 3, 6, 10, 15), and Δ = (0, 0, 3, 6, 11). To simplify notation, let 𝔪 = (x_{1}, x_{2}, x_{3}).

As we enter the first loop, we see that t, the first degree in which Δ is nonzero, is two, and M = (x_{1}^{2}, x_{1}x_{2}, x_{2}^{2}, x_{1}x_{3}, x_{2}x_{3}, x_{3}^{2}). The first monomial in M and not in I is x_{1}^{2}, and H_{0}, the Hilbert function of S/(I, x_{1}^{2}), is (1, 3, 5, 7, 9). We reset I = (x_{1}^{2}) + 𝔪^{5}, H̄ = (1, 3, 5, 7, 9), and M = (x_{1}x_{2}, x_{2}^{2}, x_{1}x_{3}, x_{2}x_{3}, x_{3}^{2}).

In the next trip through the loop, we have Δ = (0, 0, 2, 3, 5). We add the monomial x_{1}x_{2} to the ideal, and we obtain I = (x_{1}^{2}, x_{1}x_{2}) + 𝔪^{5}, H̄ = (1, 3, 4, 5, 6), and M = (x_{2}^{2}, x_{1}x_{3}, x_{2}x_{3}, x_{3}^{2}).

Now we have Δ = (0, 0, 1, 1, 2), and the situation is more complicated. The next monomial to try to add to the generating set of I is x_{2}^{2}. However, the Hilbert function of S/((x_{1}^{2}, x_{1}x_{2}, x_{2}^{2}) + 𝔪^{5}) is H_{0} = (1, 3, 3, 3, 3). Note that max(x_{2}^{2}) = 2, and H_{0} − H = (0, 0, 0, −1, −1), which is not weakly increasing. (This shows that there is no weakly reverse-lex ideal in S that corresponds to this Hilbert function.) Therefore we set M = (x_{1}x_{3}, x_{2}x_{3}, x_{3}^{2}), eliminating all the monomials that do not involve x_{3}, and we start the loop again by setting m = x_{1}x_{3}.

At this point, (I, m) = (x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}) + 𝔪^{5}. In the “else” portion of the loop, since max(m) = 3, we need to check that (I, m) is stable, which it is. Thus we let our new I be (x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}) + 𝔪^{5}, and S/I has Hilbert function H̄ = (1, 3, 3, 4, 5). That makes Δ = (0, 0, 0, 0, 1).

We have to add one final generator in degree four. Our M is now the list of all monomials of degree four in descending reverse-lex order. The first monomial in M that is not in I is x_{2}^{4}. Since max(x_{2}^{4}) = 2, we need only check to see if the difference between H_{0}, the Hilbert function of S/(I, x_{2}^{4}), and H is weakly increasing. But H_{0} = H, so we are done.

Thus the output of the algorithm is

I = (x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}, x_{2}^{4}, x_{2}^{3}x_{3}^{2}, x_{2}^{2}x_{3}^{3}, x_{2}x_{3}^{4}, x_{3}^{5}).

Note that I is actually a lexicographic ideal in S. Consequently, there is a unique set of graded Betti numbers possible for any stable ideal in S corresponding to Hilbert function H. The Betti diagram of S/I is the following:

S/I: total: 1 8 12 5
0: 1 …
1: . 3 3 1
2: ….
3: . 1 1 .
4: . 4 8 4

We conclude that the potential cancellation of a generator and first syzygy in degree five in the diagram above cannot occur for a stable ideal. (This is also clear from the structure of the Eliahou-Kervaire resolution.)

These results mean that one must go beyond the class of stable ideals in three variables to find the incomparability behavior that Charalambous and Evans exhibit with monomial ideals in three variables that are not stable. We consider stable ideals in more variables next, where the combinatorics is harder to unravel.

3.6 Incomparability in more variables

Our motivation in higher dimensions is the following conjecture of Deery from [7].

Conjecture 3.14 Let H be a Hilbert function such that there exists a weakly reverse-lex ideal I ⊂ R = k[x_{1}, …, x_{n}] with H(R/I) = H. Then β^{R/I} ≤ β^{R/J} for all stable J with H(R/J) = H.

Initially, one might try to be even more ambitious, hoping to find a algorithm similar to that of Section 5 for Hilbert functions in R = k[x_{1}, …, x_{n}], where n ≥ 4. It is quickly apparent that naively trying the analogous procedure fails. Consider, for example, the Hilbert function (1, 4, 6, 9) in k[x_{1}, x_{2}, x_{3}, x_{4}]. Putting in the first three monomials of degree two in reverse-lex order keeps the difference of the Hilbert functions vector weakly increasing, but then we can add only one more monomial in degrees two and three. No choice of degree two monomial will satisfy this condition. We should expect to have to make minor adjustments to account for the change in combinatorics from three to four variables; in particular, the requirement for Δ to be weakly increasing is a consequence of the behavior of Hilbert functions of ideals in k[x_{1}, x_{2}, x_{3}]. The larger problem is that one can no longer simply worry about the number of monomials not involving the last variable.

In lieu of such an algorithm, we would at least like to find some minimality result that applies to Hilbert functions for which there is no quotient corresponding to a reverse-lex ideal. In presenting Conjecture 3.14, Deery notes that one cannot use the same methods of proof in four or more variables as in three variables by exhibiting an example for which the techniques of Section 4 fail. We show that Deery’s example is actually a counterexample to his Conjecture 3.14 (modifying it to use finite length modules). This also proves the impossibility of an algorithm like the one in Section 5 in more than three variables.

Theorem 3.15 Let R = k[x_{1}, …, x_{n}], where n ≥ 4. There exists a Hilbert function H, a weakly reverse-lex ideal I, and another stable ideal J such that R/I and R/J have Hilbert function H, and β^{R/I} and β^{R/J} are incomparable. In addition, there are incomparably minimal sets of graded Betti numbers among stable ideals corresponding to the Hilbert function H.

Proof: Let 𝔪 = (x_{1}, …, x_{n}) ⊂ R. Define ideals

I = (x_{1}^{4}, x_{1}^{3}x_{2}, x_{1}^{2}x_{2}^{2}, x_{1}x_{2}^{3}, x_{2}^{6}) + 𝔪^{8}

and

J = (x_{1}^{4}, x_{1}^{3}x_{2}, x_{1}^{3}x_{3}, x_{1}^{3}x_{4}, x_{1}^{2}x_{2}^{3}, x_{1}x_{2}^{4}, x_{2}^{5}) + 𝔪^{8}.

It is easy to check that I and J have the same Hilbert function H; when n = 4, for example, R/I and R/J both have Hilbert function (1, 4, 10, 20, 31, 43, 55, 67). Moreover, I is a weakly reverse-lex ideal, and J is stable. (In fact, both ideals are strongly stable.) The graded free resolutions of R/I and R/J are as follows, where an asterisk represents a possibly nonzero entry:

R/I: 0: 1 . . . . . . .  R/J: 0: 1 . . . . . . .
1: . . . . . . . .  1: . . . . . . . .
2: . . . . . . . .  2: . . . . . . . .
3: . 4 3 . . . . .  3: . 4 6 4 1 …
4: . . . . . . . .  4: . 3 3 . . . . .
5: . 1 1 . . . . .  5: . . . . . . . .
6: . . . . . . . .  6: . . . . . . . .
7: . * * * * …  7: . * * * * …

Obviously these sets of graded Betti numbers are incomparable. In addition, there is no cancellation possible in the Betti diagram for R/I, so β^{R/I} is minimal in the partial order. Therefore there are incomparably minimal sets of graded Betti numbers among stable ideals with Hilbert function H. □

Conjectures about Lex-plus-powers

ideals*

In this chapter, we introduce lex-plus-powers ideals, which are Artinian monomial ideals that play a significant role in several important conjectures in commutative algebra and algebraic geometry. We discuss some of these conjectures and investigate the relationships among them. Throughout the chapter, let R = k[x_{1}, …, x_{n}].

4.1 Lex-plus-powers ideals

We have seen the important role lexicographic ideals play in the study of Hilbert functions and graded free resolutions. In order to get more information about the growth of Hilbert functions and the resolutions that occur, we impose additional constraints on the ideals we consider. For example, instead of looking at all ideals with a given Hilbert function, we can restrict to ideals with a particular Hilbert function that also contain a regular sequence in prescribed degrees. In that case, there may not be a lexicographic ideal that satisfies those restrictions. Lex-plus-powers ideals are the analogues of Artinian lexicographic ideals in this new setting. Definition 4.1 Let a_{1} ≤ ⋯ ≤ a_{n} be positive integers. We call L an (a_{1}, …, a_{n}) −lex-plus-powers (LPP) ideal if:

(1) L is minimally generated by x_{1}^{a_{1}}, …, x_{n}^{a_{n}} and monomials m_{1}, …, m_{l}, and

(2) If r is a monomial, deg r = deg m_{i}, and r >_{lex} m_{i}, then r ∈ L.

*Portions of this chapter are reprinted from “Almost complete intersections and the Lex-Plus-Powers Conjecture,” by Christopher A. Francisco, which appeared in the Journal of Algebra 276 (2004), no. 2, 737-760 [16].

Example 4.2 We check that L = (x_{1}^{2}, x_{2}^{3}, x_{3}^{3}, x_{1}x_{2}^{2}, x_{1}x_{2}x_{3}) is a (2, 3, 3)−LPP ideal. It contains appropriate powers of the variables, and we need only check the second condition for the other two generators. Since x_{1}^{3}, x_{1}^{2}x_{2}, and x_{1}^{2}x_{3} are all in L, L is an LPP ideal.

In general, one builds an LPP ideal by first forming the regular sequence of maximal length. Then, in order to get the desired Hilbert function, one adds more generators in descending lexicographic order in each degree.

For the rest of this section, we develop some properties of LPP ideals that correspond to useful characteristics of lexicographic ideals. We obtain tools to assist in our analysis of the relationship among conjectures on LPP ideals.

First, we find a basis in each degree for a quotient by an LPP ideal.

Proposition 4.3 Let L be an (a_{1}, …, a_{n})−LPP ideal in R = k[x_{1}, …, x_{n}]. Then the smallest H(R/L, d) monomials in lexicographic order in degree d not divisible by any x_{i}^{a_{i}} form a basis for (R/L)_{d}.

Proof: Suppose that x^{e} = x_{1}^{e_{1}} ⋯ x_{n}^{e_{n}} ∈ L is not divisible by any x_{i}^{a_{i}}. Let x^{d} = x_{1}^{d_{1}} ⋯ x_{n}^{d_{n}} be a monomial not divisible by any x_{i}^{a_{i}} such that deg x^{e} = deg x^{d} and x^{d} >_{lex} x^{e}. We show that x^{d} ∈ L. If x^{e} is a minimal generator of L, then x^{d} ∈ L by the definition of an LPP ideal. If not, then there exists a minimal generator s of L with s = x_{1}^{e_{1}−g_{1}} ⋯ x_{n}^{e_{n}−g_{n}}, each g_{i} ≥ 0, and some g_{i} > 0. Let m be the monomial of degree deg s equal to

m = x_{1}^{d_{1}} ⋯ x_{i−1}^{d_{i−1}}x_{i}^{r_{i}},

where i is chosen such that

d_{1} + ⋯ + d_{i−1} < deg s ≤ d_{1} + ⋯ + d_{i}

and such that deg m = deg s.

We show that m ≥_{lex} s. Since deg m = deg s, the definition of an LPP ideal would then imply that m ∈ L. Since m divides x^{d}, this proves that x^{d} ∈ L.

First, suppose that d_{1} = e_{1} − g_{1}, …, d_{i−1} = e_{i−1} − g_{i−1}. Then r_{i} ≥ e_{i} − g_{i} since deg m = deg s, and m ≥_{lex} s. Otherwise, for some j < i, we have d_{1} = e_{1} − g_{1}, …, d_{j−1} = e_{j−1} − g_{j−1}, and d_{j} ≠ e_{j} − g_{j}. If d_{j} < e_{j} − g_{j}, then d_{1} ≤ e_{1}, …, d_{j−1} ≤ e_{j−1}, and d_{j} < e_{j} − g_{j} ≤ e_{j}. But this contradicts the fact that x^{d} >_{lex} x^{e}. Therefore d_{j} > e_{j} − g_{j}, and m ≥_{lex} s. □

Corollary 4.4 Suppose that L_{1} and L_{2} are two (a_{1}, …, a_{n})−LPP ideals with H(R/L_{1}, d) = H(R/L_{2}, d), and suppose that all the minimal generators of L_{1} and L_{2} that are not pure powers have degree at most d. Then (L_{1})_{d+1} = (L_{2})_{d+1}.

Proof: We have (L_{1})_{d} = (L_{2})_{d}, and the only new generators that occur for either ideal in degree d + 1 are possibly some x_{i}^{d+1}. Therefore (L_{1})_{d+1} = (L_{2})_{d+1}. □

In other words, specifying the a_{i} and the value of the Hilbert function in degree d fully describes the monomials in an (a_{1}, …, a_{n})−LPP ideal in degree d. Consequently, this information determines the Hilbert function in degree d+1, assuming any minimal generators in degree d + 1 are pure powers.

4.2 Conjectures on Hilbert functions

In this section, we introduce several versions of conjectures on the growth of Hilbert functions of Artinian ideals. Most of the conjectures are due to Eisenbud, Green, and Harris, and they appear (sometimes implicitly) in [10]. Eisenbud, Green, and Harris were interested in these questions because they are closely related to problems in Cayley-Bacharach theory in algebraic geometry [10, 11].

The first conjecture is the easiest to state because the combinatorics is simple.

Conjecture 4.5 (EGH Conjecture for quadrics) Let R = k[x_{1}, …, x_{n}], and let I be an ideal containing a regular sequence of length n in degree two. Suppose

H(R/I, d) = \binom{b_{d}}{d} + \binom{b_{d−1}}{d − 1} + ⋯ + \binom{b_{1}}{1}

is the d-th Macaulay representation of H(R/I, d). Then

H(R/I, d + 1) ≤ \binom{b_{d}}{d + 1} + \binom{b_{d−1}}{d} + ⋯ + \binom{b_{1}}{2}.

The inequality is reminiscent of the Kruskal-Katona Theorem from combinatorics; the shift in the binomial coefficients is the same. Note that the conjectured bound in Conjecture 4.5 is much stronger than what Macaulay’s Theorem yields since we only increment the bottom number in each binomial coefficient.

A (2, …, 2)−LPP ideal attains the bound of Conjecture 4.5, and so the bound is sharp if it is true. Eisenbud, Green, and Harris argue that the case in which I contains the squares of all the variables follows from the Kruskal-Katona Theorem, as does the case in which the quadrics are a sufficiently general regular sequence.

There are two reasons to restrict to quadrics in Conjecture 4.5. First, this is the case that implies the geometric Cayley-Bacharach conjecture in which Eisenbud, Green, and Harris were interested. Second, it is much more difficult to write down explicit formulas if we allow the regular sequence to be in degrees higher than two. Nevertheless, using the notion of LPP ideals, it is still possible to formulate the analogous conjecture. We begin with a general version and then modify it to something a bit less cumbersome.

Conjecture 4.6 (EGH Conjecture) Fix a nonnegative integer d. Let I be a homogeneous ideal containing a maximal length regular sequence in degrees a_{1} ≤ ⋯ ≤ a_{n}. Suppose we can form a monomial ideal L with minimal generators x_{1}^{a_{1}}, …, x_{n}^{a_{n}} plus the first l monomials in degree d in descending lexicographic order, where l is selected so that H(R/I, d) = H(R/L, d). Then H(R/I, d + 1) ≤ H(R/L, d + 1).

This conjecture is known when I contains powers of all the variables; it follows from the theorem of Clements and Lindström [6]. (For expositions of both the Kruskal-Katona and Clements-Lindström Theorems, see [25].) There has also been some progress when the a_{i} are all two; the result is known when the additional generators outside the regular sequence are generic quadrics [26, 18].

As Richert points out in [39], the reason for stating the EGH Conjecture in the way we have is to avoid trivial problems. For example, if I = (x_{1}, …, x_{n}), then I contains a regular sequence of length n in degree two, but with our definition of (a_{1}, …, a_{n})−LPP ideals, there is no (2, …, 2)−LPP ideal L with H(R/L, 1) = 0. Thus the most interesting case of the EGH Conjecture is when I contains a regular sequence of length n in degrees a_{1}, …, a_{n} and no smaller degrees; then we can always form the ideal L described above. To simplify the EGH Conjecture, we first borrow the following definition from Richert [39] that allows us a way to discuss a regular sequence of smallest degrees in an ideal.

Definition 4.7 Let I be an Artinian ideal in R. We say that I minimally contains an (a_{1}, …, a_{n})−regular sequence if I contains a regular sequence in degrees a_{1}, …, a_{n}, and I does not contain a regular sequence in degrees b_{1}, …, b_{n} for any sequence of b_{i}’s such that b_{i} ≤ a_{i} for each i with some b_{j} < a_{j}.

We now give the simplified version of the EGH Conjecture.

Conjecture 4.8 (Simplified EGH Conjecture) Fix a nonnegative integer d Let I be a homogeneous ideal minimally containing an (a_{1}, …, a_{n})−regular sequence, where a_{1} ≤ ⋯ ≤ a_{n}. Let L be an (a_{1}, …, a_{n})−LPP ideal such that all minimal generators of L in degrees higher than d are powers of variables, and H(R/I, d) = H(R/L, d). Then H(R/I, d + 1) ≤ H(R/L, d + 1).

By Corollary 4.4, any (a_{1}, …, a_{n})−LPP ideal satisfying the conditions in the Simplified EGH Conjecture will contain the same monomials in degrees d and d + 1, so any choice of such an L will provide the same upper bound. A canonical choice for L is to take as minimal generators x_{1}^{a_{1}}, …, x_{n}^{a_{n}} and then as many monomial generators in descending lexicographic order in degree d needed to make H(R/I, d) = H(R/L, d).

The Simplified EGH Conjecture is equivalent to the EGH Conjecture.

Theorem 4.9 The EGH Conjecture is true if and only if the Simplified EGH Conjecture is true.

Proof: Clearly, the EGH Conjecture implies the simplified version. Assume the Simplified EGH Conjecture is true. Suppose we have an ideal I containing a regular sequence in degrees a_{1}, …, a_{n} and minimally containing a regular sequence in degrees b_{1}, …, b_{n} with b_{i} ≤ a_{i} for each i. Fix some degree d. Let L be the (b_{1}, …, b_{n})−LPP ideal with all its generators other than the pure powers in degree d such that H(R/L, d) = H(R/I, d). By the Simplified EGH Conjecture, H(R/I, d + 1) ≤ H(R/L, d + 1).

If there does not exist a (a_{1}, …, a_{n})−LPP ideal L′ such that H(R/L′, d) = H(R/L, d), then the EGH Conjecture is vacuous. Suppose instead that there is such an L′, and we may assume that L′ has all its minimal generators besides the pure powers in degree d. Since L and L′ are both monomial ideals, the Clements-Lindström Theorem applies, and H(R/L, d+1) ≤ H(R/L′, d+1). Thus H(R/I, d+ 1) ≤ H(R/L′, d + 1), and the EGH Conjecture holds. □

In [39], Richert gives a different equivalent formulation of the EGH Conjecture. He approaches the question from a slightly different perspective, asking for which Hilbert functions one can have an ideal with regular sequence in prescribed degrees. We define two terms from [39] before stating Richert’s version of the EGH Conjecture.

Definition 4.10 Let H be a Hilbert functions, and let 1 ≤ a_{1} ≤ ⋯ ≤ a_{n} be integers. We call H an (a_{1}, …, a_{n})−valid Hilbert function if there exists an ideal I ⊂ R that minimally contains an (a_{1}, …, a_{n})−regular sequence, and H(R/I) = H. We call H an (a_{1}, …, a_{n})−LPP valid Hilbert function if there exists an (a_{1}, …, a_{n})−LPP ideal with H(R/L) = H.

Richert states the following conjecture in [39].

Conjecture 4.11 Let 1 ≤ a_{1} ≤ ⋯ ≤ a_{n} be integers. The set of (a_{1}, …, a_{n})−valid Hilbert functions is equal to the set of (a_{1}, …, a_{n})−LPP valid Hilbert functions.

One of Richert’s main results in [39] is the following theorem.

Theorem 4.12 The EGH Conjecture is equivalent to Conjecture 4.11.

We return to the conjectures about Hilbert functions in the final section of this chapter, relating them to the homological conjecture from the next section.

4.3 The Lex-Plus-Powers Conjecture

Macaulay’s Theorem shows that lexicographic ideals provide an upper bound for the growth of Hilbert functions of modules R/I. Theorem 1.14 gives the corresponding homological result, proving that lexicographic ideals have the largest graded Betti numbers among all ideals with the same Hilbert function. In the previous section, we discussed a number of conjectures on Hilbert functions. They assert that LPP ideals should give an upper bound for the growth of Hilbert functions of ideals with a maximal-length regular sequence in the same degrees. Therefore it is natural to wonder if there is a corresponding result about the graded Betti numbers of LPP ideals.

This relationship led to the formulation of the LPP Conjecture. Its history is murky but is perhaps best described as due to Evans or Charalambous and Evans and inspired by the work of Eisenbud, Green, and Harris. The conjecture first appeared in a paper of Evans and Richert [14]:

Conjecture 4.13 (LPP Conjecture) Let L ⊂ R be an (a_{1}, …, a_{n})−LPP ideal. Suppose I ⊂ R is a homogeneous ideal with the same Hilbert function that contains a regular sequence in degrees a_{1}, …, a_{n}. Then β^{R/I} ≤ β^{R/L}.

We give an example to illustrate the LPP Conjecture.

Example 4.14 Let S = k[a, b, c], and let I be an ideal generated by generic polynomials of degrees two, three, three, and four. Let L be the (2, 3, 3)−LPP ideal (a^{2}, b^{3}, c^{3}, ab^{2}c). Then H(S/I) = H(S/L) = (1, 3, 5, 5, 2). The LPP Conjecture asserts that β^{S/I} ≤ β^{S/L}. The Betti diagrams of the two modules are the following:

S/I: 0: 1 …  S/J: 0: 1 …
1: . 1 . .  1: . 1 . .
2: . 2 . .  2: . 2 . .
3: . 1 4 .  3: . 1 4 1
4: . . 1 2  4: . . 2 2

The diagrams are the same except on the degree six diagonal, where the entries in the diagram of S/J are larger, so β^{S/I} ≤ β^{S/L}.

There is substantial computational evidence for the LPP Conjecture, but proving it in its full generality is difficult. It is tempting to borrow from the proofs of Bigatti and Hulett in the lexicographic case; for example, instead of comparing an arbitrary ideal I to a lexicographic ideal, they consider the generic initial ideal of I. In characteristic zero, this gives a strongly stable ideal with graded Betti numbers the same or larger than those of I, and one has convenient formulas for the graded Betti numbers of strongly stable ideals. To use this approach with the LPP Conjecture, however, one needs some way to fix not only the Hilbert function but also the degrees of the regular sequence. These degrees can change when passing to the generic initial ideal. Another possibility is to use work of Charalambous and Evans that describes how to compute the minimal resolution for LPP ideals [5]. Unfortunately, this method can be hard to use to compare resolutions of LPP ideals to other ideals, partially because of some unpredictable ideal quotients that arise.

A number of people have worked on the LPP Conjecture recently. Richert [39] has proven the EGH and LPP Conjectures in the case n = 2 and for monomial ideals in three variables. He also has some reductions that may make it possible to understand connections between the EGH and LPP Conjectures more clearly in the future. We discuss one of these below. In addition, Gasharov, Hibi, and Peeva have introduced the idea of a-stable ideals, and they study their homological properties in [19]. Others, including Richert and Sabourin, have explored the Hilbert functions and socle dimensions of LPP ideals and are attempting to apply these ideas to the LPP Conjecture in three variables.

In another direction, in [39], Richert reduces the LPP Conjecture to a statement about the entries in a single row of Betti diagrams. First, he makes the following conjecture.

Conjecture 4.15 (LPP Conjecture for Last Rows) Let L be an (a_{1}, …, a_{n})- LPP ideal, and let r be the regularity of R/L. Then for all I containing a regular sequence in degrees a_{1}, …, a_{n} with H(R/I) = H(R/L), β_{r,r+j}^{R/I} ≤ β_{r,r+j}^{R/L} for each j.

Richert proves the following surprising result in [39].

Theorem 4.16 The LPP Conjecture for Last Rows is true if and only if the LPP Conjecture is true.

In other words, suppose the LPP Conjecture holds for the bottom row (the row of highest degree) in each Betti diagram. Then the LPP Conjecture is true.

As a corollary, Richert obtains a reduction of the EGH Conjecture [39].

Corollary 4.17 Let L be an (a_{1}, …, a_{n})−LPP ideal, and let I be an arbitrary ideal with the same Hilbert function that contains a regular sequence in degrees a_{1}, …, a_{n}. Suppose r is the regularity of R/L. If L has at least as many minimal generators in degree r + 1 as all such I, then L has at least as many minimal generators in each degree as all such I.

Consequently, it is enough to concentrate on a single row of the Betti diagram to prove the EGH and LPP Conjectures.

4.4 Relating the EGH Conjecture to the LPP Conjecture

In the final section of this chapter, we discuss the relationship between the EGH and LPP Conjectures. The intuition is that since the Betti diagram contains more information than the Hilbert function, it should be possible to prove the EGH Conjecture given the LPP Conjecture. It is straightforward to show that Theorem 1.14, the result of Bigatti, Hulett, and Pardue, implies Macaulay’s Theorem. We modify this argument to prove that the LPP Conjecture implies the Simplified EGH Conjecture. We note that this result has at least been observed by Richert; we include our proof of the implication simply to have one on record. No proof has appeared in the literature as of this writing.

We start with some notation from [39].

Definition 4.18 Let I be a homogeneous ideal in R. Define I_{(≤d)} to be the ideal generated by the elements of I of degree ≤ d.

We know that for each homogeneous ideal I ⊂ R, there exists a lexicographic ideal with the same Hilbert function. We now give the corresponding result for LPP ideals, given the LPP Conjecture.

Proposition 4.19 Suppose I ⊂ R minimally contains an (a_{1}, …, a_{n})−regular sequence. If the LPP Conjecture is true, then there exists an (a_{1}, …, a_{n})−LPP ideal L with the same Hilbert function.

Proof: Given I, we construct L degree-by-degree. If I_{1} = 0, then let L_{1} = 0. Otherwise, if dim_{k} I_{1} = l > 0, then there exists a regular sequence of l linear forms in I, so (a_{1}, …, a_{n}) = (1, …, 1, a_{l+1}, …, a_{n}), and x_{1}, …, x_{l} ∈ L. Thus H(R/I, 1) = H(R/L, 1).

Suppose now we have constructed L to include x_{1}^{a_{1}}, …, x_{n}^{a_{n}} and to match the Hilbert function of I in degrees ≤ d. That is, L has no nonpure power generators in degrees greater than d. Let 𝔪 = (x_{1}, …, x_{n}). Define Î = I + 𝔪^{d+1} and L̂ = L + 𝔪^{d+1}. Then Î and L̂ have the same Hilbert function, and both contain an (a_{1}, …, a_{n})−regular sequence, or both contain an (a_{1}, …, a_{r}, d + 1, …, d + 1)−regular sequence. Also, L̂ is an LPP ideal. Let A = #{a_{i} : a_{i} = d + 1}. Then

dim_{k}(R/I)_{d+1} − A = dim_{k}(R/I_{(≤d)})_{d+1} = β_{1,d+1}^{R/Î},

and by the LPP Conjecture,

β_{1,d+1}^{R/Î} ≤ β_{1,d+1}^{R/L̂} = dim_{k}(R/L_{(≤d)})_{d+1} = dim_{k}(R/L)_{d+1} − A

for the L we have created so far. Consequently,

dim_{k}(R/I)_{d+1} ≤ dim_{k}(R/L)_{d+1},

and therefore we can add monomial generators into L in degree d + 1 to make H(R/I, d + 1) = H(R/L, d + 1). □

As a corollary to Proposition 4.19, we have the main result of this section. Theorem 4.20 The LPP Conjecture implies the Simplified EGH Conjecture. Specifically, let I ⊂ R be a homogeneous ideal that minimally contains a regular sequence in degrees (a_{1}, …, a_{n}). Let L be any (a_{1}, …, a_{n})−LPP ideal in R such that H(R/I, d) = H(R/L, d), and all minimal generators of L that are not pure powers have degree at most d. If the LPP Conjecture is true, then H(R/I, d + 1) ≤ H(R/L, d + 1). Remark 4.21 Recall that we can always construct such an L in the statement of Theorem 4.20: Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) + the ideal generated by the greatest monomials in lexicographic order in degree d such that H(R/I, d) = H(R/L, d).

Proof of Theorem 4.20: We may replace the ideal L with a new ideal L′, an (a_{1}, …, a_{n})−LPP ideal such that H(R/I, s) = H(R/L′, s) for all s ≤ d and whose minimal generators that are not pure powers have degree at most d. Such an L′ exists by Proposition 4.19, and because L_{d} = L′_{d} and L_{d+1} = L′_{d+1} by Corollary 4.4, the Hilbert functions of L and L′ are the same. By the calculations in Propositions 4.19, dim_{k}(R/I)_{d+1} ≤ dim_{k}(R/L′)_{d+1}, which proves the theorem. □

Almost complete intersections and the

LPP Conjecture*

We continue our investigation of the EGH and LPP Conjectures. In this chapter, we prove some special cases of the conjectures in which the LPP ideal is an almost complete intersection. Additionally, we produce an explicit comparison map between two Koszul complexes of the same length. This map makes it easy to detect nonminimality in a mapping cone resolution.

5.1 Introduction

To attack the EGH and LPP Conjectures, one can investigate a number of different special cases. One possibility is to try to prove the conjectures for ideals in R = k[x_{1}, …, x_{n}] for small n, an approach Richert and Sabourin are pursuing. Another idea is to look at special kinds of ideals, such as the a-stable ideals of Gasharov, Hibi, and Peeva [19]. In this chapter, we take another direction to make progress on the LPP Conjecture, restricting the number of minimal generators the LPP ideal has.

Using this approach, the simplest case is when the LPP ideal has only the n generators x_{1}^{a_{1}}, …, x_{n}^{a_{n}} and thus is a complete intersection. In this case, the LPP Conjecture is trivial. The first nontrivial case is when the LPP ideal has one additional generator; that is, when L is an almost complete intersection. We prove

*Portions of this chapter are reprinted from “Almost complete intersections and the Lex-Plus-Powers Conjecture,” by Christopher A. Francisco, which appeared in the Journal of Algebra 276 (2004), no. 2, 737-760 [16]. this case in Theorem 5.20.

Theorem 5.20 Let L be an (a_{1}, …, a_{n})−LPP almost complete intersection. Let I be any ideal with the same Hilbert function as L that contains a regular sequence in degrees a_{1}, …, a_{n}. Then β^{R/I} ≤ β^{R/L}.

To resolve almost complete intersection ideals, we adopt an approach different from the technique Charalambous and Evans use to resolve LPP ideals. Instead of aiming for minimal free resolutions at the start, we form nonminimal resolutions and then find a systematic way to detect the nonminimality.

Our main interests in this chapter are the LPP Conjecture and related questions about Hilbert functions. We note that Migliore and Miró-Roig have done substantial work in a different direction in [35] to find the minimal graded free resolution of almost complete intersections whose generators are generic. Our results give sharp upper bounds for the graded Betti numbers of any ideal with the same Hilbert function as an LPP almost complete intersection, not only generic (n + 1)-generated ideals. However, the disadvantage is that many almost complete intersections do not have the same Hilbert function as an LPP almost complete intersection, and Migliore and Miró-Roig can, in a lot of cases, give the precise resolution of generic almost complete intersections.

This chapter is organized in the following way. In Section 2, we outline our strategy using almost complete intersection monomial ideals and prove some preliminary lemmas. We then modify the attack used on monomial ideals so that it works in the nonmonomial case in Section 3. In this section, we also give an explicit comparison map between two Koszul complexes that is vital in detecting nonminimality in the mapping cone resolutions we use. In Section 4, we identify complete intersection ideals that can have the Hilbert function of an LPP almost complete intersection and examine their resolutions. We handle final possibilities in Section 5, where we also obtain some bounds on the Hilbert functions of almost complete intersections. The proof of the main result on graded Betti numbers is in Section 6.

I thank the referee who read the version of this chapter that is to appear in the Journal of Algebra for his or her careful reading and helpful suggestions for improvement.

5.2 Resolutions of monomial almost complete intersections

To prove Theorem 5.20, we begin by comparing the resolutions of two almost complete intersection monomial ideals with the same Hilbert function. In this section, we assume that both ideals have minimal generators that form a regular sequence of maximal length in the same degrees. The hypothesis that both ideals are almost complete intersections gives us some extra structure to exploit.

Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{d_{1}} ⋯ x_{n}^{d_{n}}). We form the canonical short exact sequence

0 → R/(x_{1}^{a_{1}−d_{1}}, …, x_{n}^{a_{n}−d_{n}})(− ∑ d_{i}) → R/(x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) → R/L → 0.

We can find a resolution of R/L by taking the mapping cone induced by this short exact sequence. The situation here is particularly good because two of the three nonzero terms in the exact sequence are complete intersections, and therefore we can resolve these modules minimally by Koszul complexes.

The only difficulty is that this resolution of R/L might be nonminimal. Of course, the only places nonminimality may arise come from constants appearing in a comparison map between the resolutions of R/H and R/F, where H = (x_{1}^{a_{1}−d_{1}}, …, x_{n}^{a_{n}−d_{n}}), and F = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}). To see how to detect this nonminimality, we examine a comparison map closely in an example.

Example 5.1 Let L = (x_{1}^{2}, x_{2}^{3}, x_{3}^{3}, x_{1}x_{2}^{2}); this is a (2, 3, 3)−LPP ideal in S = k[x_{1}, x_{2}, x_{3}]. Let F = (x_{1}^{2}, x_{2}^{3}, x_{3}^{3}), and let

H = (x_{1}^{2}, x_{2}^{3}, x_{3}^{3}) : (x_{1}x_{2}^{2}) = (x_{1}, x_{2}, x_{3}^{3}).

We resolve S/L using the following diagram:

0 ⟶ S ⟶^{∂_{3}^{F}} S^{3} ⟶^{∂_{2}^{F}} S^{3} ⟶^{∂_{1}^{F}} S ⟶ S/F ⟶ 0
1↑ ^{C_{2}}↑ ^{C_{1}}↑ ↑^{x_{1}x_{2}^{2}} ↑^{x_{1}x_{2}^{2}}
0 ⟶ S ⟶_{∂_{3}^{H}} S^{3} ⟶_{∂_{2}^{H}} S^{3} ⟶_{∂_{1}^{H}} S ⟶ S/H ⟶ 0

Here, the ∂_{i}^{F} and ∂_{i}^{H} are the Koszul maps. We need to determine C_{2} and C_{1}. It is not hard to see that C_{2} should give the relationship between the generators of H and F, and from there, it is easy to compute the two maps. They are:

C_{2} = (\begin{matrix}1 & 0 & 0 \\ 0 & x_{2}^{2} & 0 \\ 0 & 0 & x_{1} \end{matrix}) and C_{1} = (\begin{matrix}x_{2}^{2} & 0 & 0 \\ 0 & x_{1} & 0 \\ 0 & 0 & x_{1}x_{2}^{2} \end{matrix}) .

Note that C_{2} is just a diagonal matrix with diagonal entries the powers of the x_{i} that appear in the additional generator x_{1}x_{2}^{2} of L. Moreover, C_{1} is just the matrix of 2 × 2 minors of C_{2}, and x_{1}x_{2}^{2} is the determinant of C_{2}.

This suggests a general strategy: Compute the penultimate vertical map C in the diagram, and fill in the other vertical maps with the appropriate exterior powers of C. We shall show that these maps form a comparison map between the two Koszul complexes.

Lemma 5.2 Let R = k[x_{1}, …, x_{n}] and L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{d_{1}} ⋯ x_{n}^{d_{n}}). Let F = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) and H = (x_{1}^{a_{1}−d_{1}}, …, x_{n}^{a_{n}−d_{n}}), and let ∂_{i}^{F} and ∂_{i}^{H} be the Koszul maps. Let C : ⋀^{n−1} R^{n} → ⋀^{n−1} R^{n} be given by e_{A} ↦ x_{i}^{d_{i}}e_{A}, where A = {1, 2, …, î, …, n}. Then the following diagram is commutative for all j = 1, …, n:

⋀^{j} R^{n} ⟶^{∂_{j}^{F}} ⋀^{j−1} R^{n}
⋀^{n−j} C↑ ↑^{⋀^{n−j+1} C}
⋀^{j} R^{n} ⟶_{∂_{j}^{H}} ⋀^{j−1} R^{n}

Proof: Let A = {i_{1}, …, i_{j}} ⊂ [n] = {1, …, n}, and let e_{A} = e_{i_{1}} ∧ ⋯ ∧ e_{i_{j}} be a basis element of ⋀^{j} R^{n} in the resolution of R/H. Note that C is just the diagonal matrix with diagonal entries x_{i}^{d_{i}} such that ∂_{n}^{F} ∘ 1 = C ∘ ∂_{n}^{H}. Also,

(⋀^{n−j}C) (e_{A}) = (∏_{l∉A} x_{l}^{d_{l}}) e_{A}.

We have

∂_{j}^{H}(e_{A}) = ∑_{i_{t}∈A}(−1)^{t}x_{i_{t}}^{a_{i_{t}}−d_{i_{t}}}e_{A\i_{t}}.

Applying ⋀^{n−j+1} C, we obtain

(⋀^{n−j+1}C ∘ ∂_{j}^{H}) (e_{A}) = ∑_{i_{t}∈A}(−1)^{t}x_{i_{t}}^{a_{i_{t}}−d_{i_{t}}} ( ∏_{l∉A\i_{t}} x_{l}^{d_{l}} ) e_{A\i_{t}}.

= ∑_{i_{t}∈A}(−1)^{t}x_{i_{t}}^{a_{i_{t}}} (∏_{l∉A} x_{l}^{d_{l}}) e_{A\i_{t}}.

Now, going the other direction,

(∂_{j}^{F} ∘ ⋀^{n−j}C) (e_{A}) = ∂_{j}^{F} ((∏_{l∉A} x_{l}^{d_{l}}) e_{A}) = ∑_{i_{t}∈A}(−1)^{t}x_{i_{t}}^{a_{i_{t}}} (∏_{l∉A} x_{l}^{d_{l}}) e_{A\i_{t}}.

The coefficients of each e_{A\i_{t}} are equal, and thus we are done. □

As a result, we can easily determine the nonminimality that occurs in the mapping cone resolution of an almost complete intersection monomial ideal. The matrix C in Lemma 5.2 has entry 1 only in the places corresponding to some d_{i} being 0, forcing cancellation in the tail of the resolution in degree a_{1} + ⋯ + â_{i} + ⋯ + a_{n}. Since the other vertical maps are matrices of minors of C, we can detect constants in the other vertical maps just by knowing C.

We use Lemma 5.2 to compare the resolutions of two monomial almost complete intersections. We show that if there is any nonminimality in the mapping cone resolution of an LPP almost complete intersection, there is corresponding nonminimality in the mapping cone resolution of the other ideal we are considering. Proposition 5.3 Let I be a monomial ideal containing minimal generators x_{1}^{a_{1}}, …, x_{n}^{a_{n}}. Suppose I has the same Hilbert function as an (a_{1}, …, a_{n})−LPP almost complete intersection L. Then β^{R/I} ≤ β^{R/L}. Proof: We may immediately reduce to the case in which we can apply Lemma 5.2: The theorem of Clements and Lindström in [6] shows that L has at least as many generators in each degree as I, and therefore I must also be an almost complete intersection. (Alternatively, see Lemma 5.17.) Hence we may assume that I = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{b_{1}} ⋯ x_{n}^{b_{n}}).

We are assuming that I and L have the same Hilbert function and have as minimal generators the same powers of the x_{i}. Because we can compute the Hilbert functions of I and L from the nonminimal mapping cone resolutions, the ideal quotients (x_{1}^{a_{1}−d_{1}}, …, x_{n}^{a_{n}−d_{n}}) and (x_{1}^{a_{1}−b_{1}}, …, x_{n}^{a_{n}−b_{n}}) have the same Hilbert function. We claim that the sets T_{d} = {a_{1} − d_{1}, …, a_{n} − d_{n}} and T_{b} = {a_{1} − b_{1}, …, a_{n} − b_{n}} are the same. To see this, reindex the a_{i} − d_{i} and a_{i} − b_{i} so that both sequences are both weakly increasing. If the sets are unequal, say j is the first index such that a_{j} − d_{j} ≠ a_{j} − b_{j}. Then the Hilbert functions of the ideal quotients differ in degree min(a_{j} − d_{j}, a_{j} − b_{j}), a contradiction.

We show that if some d_{i} is zero, then there exists an i′ (a different one for each i with d_{i} = 0) such that a_{i′} = a_{i} and b_{i′} = 0. Since any constant in the comparison map (aside from the far left vertical map, which always induces nonminimality) comes from some b_{i} or d_{i} being zero, this proves that any nonminimality in the mapping cone resolution of R/L occurs in the same degree in the resolution of R/I.

It is clear from the definition of an LPP ideal that d_{1} = a_{1} − 1, …, d_{j−1} = a_{j−1}−1, and d_{j+1} = ⋯ = d_{n} = 0 for some 1 < j ≤ n. Hence T_{d} = {1, …, 1, a_{j} −d_{j}, a_{j+1}, …, a_{n}} for some 1 < j ≤ n, and since the a_{i} are weakly increasing, 1 ≤ a_{j} − d_{j} ≤ a_{j+1} ≤ ⋯ ≤ a_{n}.

If no d_{i} = 0, then the mapping cone resolution of R/L is minimal (except for the obvious nonminimality from the far left vertical map), and we are done. Otherwise, d_{n} = 0. Since T_{b} = T_{d}, a_{n} ∈ T_{b}, and there is some r such that a_{n} = a_{r} −b_{r}. Because a_{n} ≥ a_{i} for all i, we must have a_{r} = a_{n} and b_{r} = 0.

Inductively, suppose that for s > l, the following holds: If d_{s} = 0, there is s′ such that a_{s′} = a_{s} and b_{s′} = 0. If d_{l} > 0, there is nothing to prove. Otherwise, d_{l} = 0, and then a_{l} ∈ T_{d} = T_{b}. By the induction hypothesis, the only way for a_{l} to be in T_{b} is for there to be an a_{l′} = a_{l} such that b_{l′} = 0 (for any a_{i} larger than a_{l} has b_{i} = 0 by induction), with l′ different from the other indices we have already used. This completes the induction.

Consequently, any nonminimality in the tail of the mapping cone resolution of L also occurs in the same degree in the mapping cone resolution of I. By Lemma 5.2, the comparison maps in the mapping cones are all exterior powers of the diagonal matrix with nonzero entries x_{i}^{a_{i}−d_{i}} or x_{i}^{a_{i}−b_{i}}. Hence the mapping cone resolution of I contains the same nonminimalities as the mapping cone resolution of L and possibly more. Thus β^{R/I} ≤ β^{R/L}. □

This establishes the inequality we want for the monomial case and suggests a method of attack for the nonmonomial case, which we explore in the next section.

5.3 The nonmonomial almost complete intersection case

It was particularly convenient to work with monomial ideals in the last section because the comparison map consists of diagonal matrices of a nice form. If one wishes to generalize Proposition 5.3 to nonmonomial ideals, the vertical maps will be much more complicated. We adapt the proof of the monomial case in this section.

Throughout, fix an LPP ideal L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{d_{1}} ⋯ x_{n}^{d_{n}}). Let I = (f_{1}, …, f_{n}, g) be an ideal with the same Hilbert function as L such that deg f_{i} = a_{i}, the f_{i} form a regular sequence, and (f_{1}, …, f_{n}) : (g) = (h_{1}, …, h_{n}) is a complete intersection. The reason for the last hypothesis is that we want to follow the monomial case; we get a resolution of R/I by taking the mapping cone of two Koszul complexes.

The goal in this section is Theorem 5.9, the LPP Conjecture in the setting outlined above. To compare the graded Betti numbers of R/I and R/L, we first find a convenient comparison map between the Koszul complexes on the f_{i} and h_{j}. Let F = (f_{1}, …, f_{n}) and H = (h_{1}, …, h_{n}). The strategy is to find a matrix C that writes the f_{i} in terms of the h_{j} and then to fill in the other vertical maps with the exterior powers of C. This gives us a comparison map in which we can detect the presence of nonzero constants in any of the vertical maps solely by examining C.

Lemma 5.4 Let ∂_{i}^{F} and ∂_{i}^{H} be the Koszul maps in the resolutions of R/F and R/H. Let C be any lift induced by g from the Comparison Theorem mapping ⋀^{n−1} R^{n} in the resolution of R/H to ⋀^{n−1} R^{n} in the resolution of R/F. Then the following diagram is commutative for all j = 1, …, n:

⋀^{j} R^{n} ⟶^{∂_{j}^{F}} ⋀^{j−1} R^{n}
⋀^{n−j} C↑ ↑^{⋀^{n−j+1} C}
⋀^{j} R^{n} ⟶_{∂_{j}^{H}} ⋀^{j−1} R^{n}

Proof: Let A = {i_{1}, …, i_{j}} ⊂ [n], and let e_{A} = e_{i_{1}} ∧ ⋯ ∧ e_{i_{j}} be a basis element of ⋀^{j} R^{n} in the resolution of R/H. Throughout, for X, Y ⊂ [n], let m_{(X,Y)} represent the minor of C obtained by omitting rows X and columns Y from C.

Computing first (⋀^{n−j+1} C ∘ ∂_{j}^{H})(e_{A}), we have

∂_{j}^{H}(e_{A}) = ∑_{l=1}^{j}(−1)^{l}h_{i_{l}}e_{A\i_{l}},

and applying ⋀^{n−j+1} C, we obtain

∑_{l=1}^{j}(−1)^{l}h_{i_{l}} ∑_{|B|=j−1} m_{(B,A\i_{l})}e_{B} = ∑_{|B|=j−1} ∑_{l=1}^{j}(−1)^{l}h_{i_{l}}m_{(B,A\i_{l})}e_{B},

where the summation is over all B ⊂ [n] of cardinality j − 1.

In the other direction,

(⋀^{n−j}C) (e_{A}) = ∑_{|D|=j} m_{(D,A)}e_{D},

with the summation over all D ⊂ [n] of cardinality j. Applying ∂_{j}^{F} yields

∑_{|D|=j} m_{(D,A)} ∑_{l=1}^{j}(−1)^{l}f_{D_{l}}e_{D\D_{l}},

where D_{l} represents the l-th element of D in increasing order.

To compare these two calculations, we convert the second to a summation over sets of cardinality j − 1. The messy part here is keeping track of the signs. Let p(y, Y) denote the position of y in the set Y ⊂ [n], where Y is ordered in the usual way. Then

(∂_{j}^{F} ∘ ⋀^{n−j}C) (e_{A}) = ∑_{|B|=j−1} ∑_{α∉B}(−1)^{p(α,B∪α)}f_{α}m_{(B∪α,A)}e_{B}.

We need to write the f_{i} in terms of the h_{j}. We index the entries of the matrix C unconventionally to keep the notation simpler. We have

(\begin{matrix}c_{nn} & ⋯ & c_{n1} \\ ⋮ & ... & ⋮ \\ c_{1n} & ⋯ & c_{11} \end{matrix}) ((−1)^{n}h_{n} ; ⋮ ; −h_{1}) = ((−1)^{n}f_{n} ; ⋮ ; −f_{1}) .

Thus (−1)^{i}f_{i} = (−1)^{n}c_{in}h_{n} + ⋯ + (−1)^{1}c_{i1}h_{1}. Hence

(∂_{j}^{F} ∘ ⋀^{n−j}C) (e_{A}) = ∑_{|B|=j−1} ∑_{α∉B}(−1)^{p(α,B∪α)}m_{(B∪α,A)}(−1)^{α} ∑_{r=1}^{n}(−1)^{r}c_{αr}h_{r}e_{B}.

Suppose first that r ∈ A, so r = i_{l} for some l. We show that the coefficients of h_{i_{l}}e_{B} in the two computations are identical by proving that

(−1)^{l}m_{(B,A\i_{l})} = ∑_{α∉B}(−1)^{p(α,B∪α)}m_{(B∪α,A)}(−1)^{α}(−1)^{i_{l}}c_{αi_{l}}.

Because

m_{(B,A\i_{l})} = ∑_{α∉B}(−1)^{p(α,B′)+p(i_{l},A′∪i_{l})}c_{αi_{l}}m_{(B∪α,A)},

where X′ is the complement of X in [n], it is enough to show that

l + p(α, B′) + p(i_{l}, A′ ∪ i_{l}) ≡ p(α, B ∪ α) + α + i_{l} (mod 2),

which would prove that the signs are consistent. Note that l = p(i_{l}, A), and for y ∈ Y, p(y, Y) + p(y, Y′ ∪ y) = y + 1, so the left-hand side is just

p(α, B′) + i_{l} + 1.

The right-hand side is

α + 1 − p(α, B′) + α + i_{l},

and thus the parities are equal.

Finally, if instead r ∉ A, we need the coefficient of h_{r}e_{B} in the expression for (∂_{j}^{F} ∘ ⋀^{n−j}C)(e_{A}) to be zero for all B. This follows because this formula includes the determinant of a submatrix of C with column r repeated. □

To prove that the ⋀^{j} C form a comparison map, we still need to show that the diagram

R ⟶^{π_{F}} R/F
det C↑ ↑^{g}
R ⟶_{π_{H}} R/H

is commutative, where π_{F} and π_{H} are the canonical projections. This follows from the next lemma.

Lemma 5.5 Let C be as above. Then det C = g in R/F.

Proof: We shall use a diagram chase to show that det C − g ∈ F, which gives the result.

Assume first that n ≥ 3. For j = 2, …, n−1, let A_{j} be vertical maps induced by g from ⋀^{n−j} R^{n} in the resolution of R/H to ⋀^{n−j} R^{n} in the resolution of R/F. Also, throughout, let M_{i} be matrices of appropriate size with their entries homogeneous polynomials. The following diagram is commutative:

⋀^{n−1} R^{n} ⟶^{∂_{n−1}^{F}} ⋀^{n−2} R^{n} ⟶^{∂_{n−2}^{F}} ⋯ ⟶^{∂_{2}^{F}} ⋀^{1} R^{n} ⟶^{∂_{1}^{F}} ⋀^{0} R^{n}
C−C↑ ^{⋀^{2} C−A_{2}}↑ ↑^{⋀^{n−1} C−A_{n−1}} ↑^{det C−g}
⋀^{n−1} R^{n} ⟶_{∂_{n−1}^{H}} ⋀^{n−2} R^{n} ⟶_{∂_{n−2}^{H}} ⋯ ⟶_{∂_{2}^{H}} ⋀^{1} R^{n} ⟶_{∂_{1}^{H}} ⋀^{0} R^{n}

Because (⋀^{2}C − A_{2}) ∘ ∂_{n−1}^{H} is zero,

im (⋀^{2}C − A_{2})^{T} ⊆ ker(∂_{n−1}^{H})^{T} = im(∂_{n−2}^{H})^{T} .

Hence (⋀^{2}C − A_{2}) = M_{2} ∘ ∂_{n−2}^{H} for some M_{2}. Commutativity of the next square to the right yields

(⋀^{3}C − A_{3}) ∘ ∂_{n−2}^{H} = ∂_{n−2}^{F} ∘ (⋀^{2}C − A_{2}) = ∂_{n−2}^{F} ∘ M_{2} ∘ ∂_{n−2}^{H}. Therefore,

im ((⋀^{3}C − A_{3}) − ∂_{n−2}^{F} ∘ M_{2})^{T} ⊆ ker(∂_{n−2}^{H})^{T} = im(∂_{n−3}^{H})^{T} .

Thus for some M_{3}, we have

(⋀^{3}C − A_{3}) = ∂_{n−2}^{F} ∘ M_{2} + M_{3} ∘ ∂_{n−3}^{H}.

Inductively, suppose that there exist M_{j−2} and M_{j−1} such that

⋀^{j−1}C − A_{j−1} = ∂_{n−j+2}^{F} ∘ M_{j−2} + M_{j−1} ∘ ∂_{n−j+1}^{H}.

By commutativity,

(⋀^{j}C − A_{j}) ∘ ∂_{n−j+1}^{H} = ∂_{n−j+1}^{F} ∘ (⋀^{j−1}C − A_{j−1}) ,

so

(⋀^{j}C − A_{j}) ∘ ∂_{n−j+1}^{H} = ∂_{n−j+1}^{F} ∘ (∂_{n−j+2}^{F} ∘ M_{j−2} + M_{j−1} ∘ ∂_{n−j+1}^{H}).

Thus, as in the arguments above,

(⋀^{j}C − A_{j}) − ∂_{n−j+1}^{F} ∘ M_{j−1} = M_{j} ∘ ∂_{n−j}^{H}

for some M_{j}. Consequently, ⋀^{n−1}C − A_{n−1} = M_{n−1} ∘ ∂_{1}^{H} + ∂_{2}^{F} ∘ M_{n−2} for some M_{n−2} and M_{n−1}. Commutativity of the rightmost square in the diagram gives

(det C − g) ∘ ∂_{1}^{H} = ∂_{1}^{F} ∘ (⋀^{n−1}C − A_{n−1}) ,

and so

(det C − g) ∘ ∂_{1}^{H} = ∂_{1}^{F} ∘ M_{n−1} ∘ ∂_{1}^{H}.

As a result, since ker(∂_{1}^{H})^{T} = 0,

(det C − g) = (f_{1} ⋯ f_{n})(p_{1} ⋯ p_{n})^{T}

for some homogeneous polynomials p_{i}, and therefore det C − g ∈ F.

If instead n = 2, then we have the following commutative diagram:

⋀^{1} R^{n} ⟶^{∂_{1}^{F}} ⋀^{0} R^{n}
C−C↑ ↑^{det C−g}
⋀^{1} R^{n} ⟶_{∂_{1}^{H}} ⋀^{0} R^{n}

Therefore (det C − g) ∘ ∂_{1}^{H} = 0. Since ker(∂_{1}^{H})^{T} = 0, det C = g. □

The two lemmas combine to give the following theorem about maps between two Koszul complexes of the same length.

Theorem 5.6 Let R = k[x_{1}, …, x_{n}], and suppose we have two complete intersection ideals M and N of the same codimension c such that we have a map ϕ : R/M ⟶ R/N. Let K^{M} and K^{N} be Koszul complexes resolving R/M and R/N. Let C be any lift induced by ϕ from the Comparison Theorem mapping K_{c−1}^{M} to K_{c−1}^{N}. Then the maps

⋀^{c−j}C : K_{j}^{M} ⟶ K_{j}^{N}

form a comparison map between the resolutions.

We now have a convenient, explicit comparison map, and therefore we can examine the mapping cone resolution of R/I and detect the nonminimality. First, we prove a lemma that shows that we can get the matrix C in a particularly helpful form.

Lemma 5.7 Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{d_{1}} ⋯ x_{r}^{d_{r}}) be an LPP ideal with d_{r} ≠ 0. Let I = (f_{1}, …, f_{n}, g) be an ideal with the same Hilbert function as L such that deg f_{i} = a_{i}, the f_{i} form a regular sequence, and (f_{1}, …, f_{n}) : (g) = (h_{1}, …, h_{n}) is a complete intersection.

Set F = (f_{1}, …, f_{n}) and H = (h_{1}, …, h_{n}), and if necessary, reindex so that deg h_{i} = a_{i} − d_{i}. There exists a map C lifting g from ⋀^{n−1}R^{n} in the resolution of R/H to ⋀^{n−1}R^{n} in the resolution of R/F such that

(\begin{matrix}c_{nn} & ⋯ & c_{n1} \\ ⋮ & ... & ⋮ \\ c_{1n} & ⋯ & c_{11} \end{matrix}) ((−1)^{n}h_{n} ; ⋮ ; −h_{1}) = ((−1)^{n}f_{n} ; ⋮ ; −f_{1}) ,

and the first n−r columns of C modulo 𝔪 = (x_{1}, …, x_{n}) are the first n−r columns of the n × n identity matrix.

Proof: We have indexed the h_{i} and f_{i} so that deg h_{i} ≤ deg f_{i} = a_{i} for each i, and the degrees of the h_{i} and f_{i} are weakly increasing. The entries of C are indexed unconventionally as in Lemma 5.4, so

c_{in}(−1)^{n}h_{n} + ⋯ + c_{ii}(−1)^{i}h_{i} + ⋯ c_{i1}(−1)h_{1} = (−1)^{i}f_{i}.

Note that deg f_{i} = deg h_{i} if and only if d_{i} = 0. In that case, c_{ii} is a constant, and since deg h_{i} = deg f_{i} ≥ ⋯ ≥ deg f_{1}, all of c_{ii}, …, c_{1i} must be constants. Thus a typical column of C has entries with positive degree at the top, possibly constant entries in the middle, and zeros at the bottom because deg h_{i} > deg f_{j} for small j.

Suppose that c_{ii} and c_{ji} are both nonzero constants in column i of C. Then d_{i} = 0, and deg f_{i} = deg h_{i} = deg f_{j}. We may perform row operations on C to change one of these nonzero constants to zero, modifying the generating set of F to be the appropriate linear combination of f_{i} and f_{j}. In this way, we may assume that each column of C contains at most one nonzero constant and that this constant is one. Moreover, by reindexing generators of F with the same degree, we may suppose that if d_{i} = 0 and if there is a nonzero constant in column i, it occurs on the diagonal in position c_{ii}. (Note that there may also be nonzero constants below the main diagonal in a column j of C when d_{j} ≠ 0; see Example 5.10 below.)

Let C̄ be the matrix C with its entries taken modulo 𝔪. We now prove that C̄ has the same first n − r columns as the n × n identity matrix.

If r = n, then all d_{i} > 0, and there is nothing to prove. If not, then d_{n} = 0, and so deg h_{n} = deg f_{n}. Therefore c_{nn} is a constant. Since deg h_{n} = deg f_{n} ≥ deg f_{j} for all j, c_{jn} is a constant for all j. Suppose all c_{jn} = 0. Then we can write all the f_{i} in terms of only h_{1}, …, h_{n−1}. Hence (f_{1}, …, f_{n}) ⊂ (h_{1}, …, h_{n−1}), meaning that a codimension n ideal is contained in a codimension n − 1 ideal, which is impossible. Thus some c_{jn} is a nonzero constant, and we may assume it is c_{nn} = 1 by the discussion above.

Now we proceed by induction. The base case is above: If only d_{n} = 0, c_{nn} = 1, and the rest of column n is zero. (Recall that the columns of C are indexed n through 1 from left to right.) For the induction hypothesis, suppose that for each j > v such that d_{j} = 0, column j of C̄ is column n − j + 1 of the n × n identity matrix. If d_{v} ≠ 0, there is nothing to prove. Suppose d_{v} = 0. Then deg h_{v} = a_{v} − 0 = deg f_{v}. Since

c_{vn}(−1)^{n}h_{n} + ⋯ + c_{vv}(−1)^{v}h_{v} + ⋯ + c_{v1}(−1)h_{1} = (−1)^{v}f_{v},

c_{vv} must be a constant. As before, deg h_{v} = deg f_{v} ≥ deg f_{i} for all i ≤ v, and hence c_{iv} is a constant for all i ≤ v. We may assume that c_{iv} = 0 for all i < v, and only c_{vv} is possibly nonzero.

We show that c_{vv} ≠ 0. By induction and the assumptions above, the entries of C look like the following:

(\begin{matrix}1 & ⋯ & c_{n,v+1} & c_{nv} & c_{n,v−1} & ⋯ & c_{n1} \\ ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋯ & ⋮ \\ 0 & ⋯ & 1 & c_{v+1,v} & c_{v+1,v−1} & ⋯ & c_{v+1,1} \\ 0 & ⋯ & 0 & c_{vv} & c_{v,v−1} & ⋯ & c_{v1} \\ 0 & ⋯ & 0 & 0 & c_{v−1,v−1} & ⋯ & c_{v−1,1} \\ ⋮ & … & ⋮ & ⋮ & ⋮ & ⋯ & ⋮ \\ 0 & ⋯ & 0 & 0 & c_{1,v−1} & ⋯ & c_{11} \end{matrix})

The entries on the main diagonal of columns n through v + 1 are one. Additionally, the entries below the main diagonal of C in columns n through v are zero, and the entries above the main diagonal in those columns have positive degree or are zero.

Suppose c_{vv} = 0. Then (f_{1}, …, f_{v}) ⊂ (h_{1}, …, h_{v−1}) since the entries in the block from columns n to v and rows v to 1 are all zero. But then an ideal of codimension v is contained in an ideal of codimension v − 1, a contradiction. Hence c_{vv} is a nonzero constant, and we may take c_{vv} = 1 by rescaling. Therefore, by induction, C̄ has its first n − r columns equal to the first n − r columns of the n × n identity matrix. □

We give an example to illustrate how to put the matrix C in the form of Lemma 5.7.

Example 5.8 We consider the case of five quadrics in four variables. Let S = ℂ[a, b, c, d], and let f_{1} = ad + d^{2}, f_{2} = c^{2} − bd, f_{3} = ab − b^{2}, f_{4} = a^{2} − bc, and g = a^{2} + cd. The f_{i} form a regular sequence, and

(f_{1}, f_{2}, f_{3}, f_{4}) : g = (c, a − b + d, bd, b^{2} + d^{2})

is a complete intersection. Let I = (f_{1}, f_{2}, f_{3}, f_{4}, g) and L = (a^{2}, b^{2}, c^{2}, d^{2}, ab). Then L is a (2, 2, 2, 2)−LPP ideal, and H(S/I) = H(S/L).

We can write the f_{i} in terms of the h_{j} as

(\begin{matrix}1 & −2 & a + b − d & −b \\ 0 & −1 & b & 0 \\ 0 & −1 & 0 & c \\ 0 & 1 & d & 0 \end{matrix}) (\begin{matrix}h_{4} \\ h_{3} \\ h_{2} \\ h_{1} \end{matrix}) = (\begin{matrix}f_{4} \\ f_{3} \\ f_{2} \\ f_{1} \end{matrix}) .

We can convert the above 4 × 4 matrix into a matrix C that has at most one nonzero constant in each column. In the extra generator ab of L, the powers of c and d are both zero, so C should have at least two columns with nonzero constants. We accomplish this by changing the generating set of F to (f_{4}+2f_{1}, f_{1}, f_{1}+f_{2}, f_{1}+f_{3}), and we get

(\begin{matrix}1 & 0 & a + b + d & −b \\ 0 & 1 & d & 0 \\ 0 & 0 & d & c \\ 0 & 0 & b + d & 0 \end{matrix}) (\begin{matrix}h_{4} \\ h_{3} \\ h_{2} \\ h_{1} \end{matrix}) = (\begin{matrix}f_{4} + 2f_{1} \\ f_{1} \\ f_{1} + f_{2} \\ f_{1} + f_{3} \end{matrix}) .

The first two columns of C from the left agree with the leftmost columns of the identity matrix, and this agrees with what we expect from Lemma 5.7.

Theorem 5.9 Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{d_{1}} ⋯ x_{r}^{d_{r}}) be an LPP ideal (with d_{r} ≠ 0). Let I = (f_{1}, …, f_{n}, g) be an ideal with the same Hilbert function as L such that deg f_{i} = a_{i}, the f_{i} form a regular sequence, and (f_{1}, …, f_{n}) : (g) = (h_{1}, …, h_{n}) is a complete intersection. Then β^{R/I} ≤ β^{R/L}.

Proof: Let C^{I} be the matrix corresponding to I from Lemma 5.7. Then the first n − r columns of C^{I} with entries taken modulo 𝔪 = (x_{1}, …, x_{n}) are the first n − r columns of the n × n identity matrix. The matrix C^{L} that writes the x_{i}^{a_{i}} in terms of the x_{i}^{a_{i}−d_{i}} in the canonical way is the following:

(\begin{matrix}1 & 0 & … & 0 & 0 & … & 0 \\ 0 & 1 & … & 0 & 0 & … & 0 \\ ⋮ & ⋮ & ... & 0 & 0 & … & 0 \\ 0 & 0 & … & 1 & 0 & … & 0 \\ 0 & 0 & … & 0 & x_{r}^{d^{r}} & … & 0 \\ 0 & 0 & … & 0 & 0 & ... & 0 \\ 0 & 0 & … & 0 & 0 & 0 & x_{1}^{d_{1}} \end{matrix} )

The leftmost n − r columns of C^{I} and C^{L} are the same modulo 𝔪, and there are no nonzero constants in any of the rightmost r columns of C^{L}. Therefore any nonminimality C^{L} induces in the mapping cone resolution of R/L is also induced by C^{I} in the mapping cone resolution of R/I. Moreover, by Lemma 5.4, all the vertical maps in the comparison map from the resolution of R/(h_{1}, …, h_{n}) to that of R/(f_{1}, …, f_{n}) are the exterior powers of C^{I}. Similarly, all the vertical maps in the comparison map from the resolution of R/(x_{1}^{a_{1}−d_{1}}, …, x_{n}^{a_{n}−d_{n}}) to that of R/(x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) are the exterior powers of C^{L}. Because the leftmost n−r columns of C^{I} and C^{L} are the same modulo 𝔪, any nonzero constants in the matrices ⋀^{j}C^{L} will occur in the same places in the maps ⋀^{j}C^{I}. Hence any nonminimality that occurs in the mapping cone resolution of R/L occurs in the same degree in that of R/I, and thus β^{R/I} ≤ β^{R/L}. □

In Example 5.8, the 4 × 4 matrix C^{L} corresponding to the LPP ideal L = (a^{2}, b^{2}, c^{2}, d^{2}, ab) is

(\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & b \end{matrix}) .

This matrix and the matrix C for I have the same nonzero constants in the same positions, and thus L and I have the same graded Betti numbers. In fact, both ideals have the following Betti diagram:

S/L: total: 1 5 9 7 2
0: 1 ….
1: . 5 2 . .
2: . . 7 4 .
3: … 3 2

We give one additional example to illustrate how the graded Betti numbers of the LPP ideal can be strictly larger than those of the other ideal. Example 5.10 uses monomial ideals for simplicity.

Example 5.10 Let S = ℂ[a, b, c, d], L = (a^{2}, b^{3}, c^{4}, d^{5}, ab^{2}c^{2}), and I = (a^{2}, b^{3}, c^{4}, d^{5}, b^{2}c^{3}). Then L is a (2, 3, 4, 5)−LPP ideal, and L and I have the same Hilbert function. Note that

(a^{2}, b^{3}, c^{4}, d^{5}) : (ab^{2}c^{2}) = (a, b, c^{2}, d^{5}) and (a^{2}, b^{3}, c^{4}, d^{5}) : (b^{2}c^{3}) = (a^{2}, b, c, d^{5}).

We write a^{2}, b^{3}, c^{4}, d^{5} in terms of the generators of the ideal quotients, ordering everything as in the proof of Theorem 5.9, to give an example of the procedure followed there. The expression for L is on the left, and the one for I is on the right. We have

(\begin{matrix}1 & 0 & 0 & 0 \\ 0 & c^{2} & 0 & 0 \\ 0 & 0 & b^{2} & 0 \\ 0 & 0 & 0 & a \end{matrix}) (\begin{matrix}d^{5} \\ c^{2} \\ b \\ a \end{matrix}) = (\begin{matrix}d^{5} \\ c^{4} \\ b^{3} \\ a^{2} \end{matrix}) and (\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & c^{3} \\ 0 & 0 & b^{2} & 0 \\ 0 & 1 & 0 & 0 \end{matrix}) (\begin{matrix}d^{5} \\ a^{2} \\ b \\ c \end{matrix}) = (\begin{matrix}d^{5} \\ c^{4} \\ b^{3} \\ a^{2} \end{matrix}) .

There is a one in the upper left-hand corner of both matrices, which corresponds to the power of d being zero in ab^{2}c^{2}, the extra generator of L. Thus the same nonminimality from this nonzero constant will occur in both mapping cone resolutions. There is also a one in the matrix for I in the bottom row, and therefore there will be additional nonminimality in the mapping cone resolution of S/I. Hence we expect the ranks of the last three free modules in the minimal resolution of S/L to be respectively one, two, and one greater than the ranks in the minimal resolution of S/I.

The minimal resolutions of S/L and S/I are:

S/L: total: 1 5 10 9 3  S/I: total: 1 5 9 7 2
0: 1 ….  0: 1 ….
1: . 1 …  1: . 1 …
2: . 1 …  2: . 1 …
3: . 1 1 . .  3: . 1 1 . .
4: . 2 3 1 .  4: . 2 3 . .
5: . . 3 2 .  5: . . 2 2 .
6: . . 1 . .  6: . . 1 . .
7: . . 1 1 .  7: . . 1 1 .
8: . . 1 3 1  8: . . 1 3 .
9: … 2 2  9: … 1 2

There are two copies of S(−12) and S(−7) in the minimal resolution of S/L that do not appear in that of S/I, and otherwise the graded Betti numbers are the same.

In the next sections, we remove some of the restrictions we have placed on I.

5.4 Complete intersections

So far we have compared the graded Betti numbers of an LPP almost complete intersection L to another almost complete intersection I. In this section, we explore what happens when I is instead a complete intersection. As a consequence, we also obtain a better idea of what ideals have the same Hilbert function as an (a_{1}, …, a_{n})−LPP almost complete intersection.

Our objective is the following result:

Proposition 5.11 Let L be an (a_{1}, …, a_{n})−LPP almost complete intersection. Let I be any complete intersection ideal with the same Hilbert function as L that contains a regular sequence in degrees a_{1}, …, a_{n}. Then β^{R/I} ≤ β^{R/L}.

To prove this, we first determine what form L and I must have to satisfy the hypotheses of Proposition 5.11. We may assume that I is a monomial ideal since its resolution and Hilbert function depend only on the degrees of the minimal generators. Of course, for I to contain a regular sequence in degrees a_{1}, …, a_{n}, the minimal generators of I must be in degrees at most a_{1}, …, a_{n}.

Our first step is to compute the socle degree of R modulo an LPP almost complete intersection. Recall that the socle degree of an Artinian module R/L is the highest degree in which R/L is nonzero.

Lemma 5.12 Let L ⊂ R be an (a_{1}, …, a_{n})−LPP almost complete intersection. Then the socle degree of R/L is a_{1} + ⋯ + a_{n} − n − 1.

Proof: We have

L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{a_{1}−1} ⋯ x_{j−1}^{a_{j−1}−1}x_{j}^{b_{j}}).

The module R/(x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) has a single socle generator x_{1}^{a_{1}−1} ⋯ x_{n}^{a_{n}−1}, which has degree a_{1} + ⋯ + a_{n} − n. That element is zero in R/L, so the socle degree of R/L is at most a_{1} + ⋯ + a_{n} − n − 1.

To prove that the socle degree is exactly this quantity, we need only produce a monomial in degree a_{1} + ⋯ + a_{n} − n − 1 that is in the socle of R/L. The monomial x_{1}^{a_{1}−1}x_{2}^{a_{2}−2}x_{3}^{a_{3}−1} ⋯ x_{n}^{a_{n}−1} satisfies these conditions: Multiplying by any x_{i} for i ≠ 2 gives an element of (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}). Multiplying by x_{2} yields an element of L divisible by x_{1}^{a_{1}−1} ⋯ x_{j−1}^{a_{j−1}−1}x_{j}^{b_{j}} since j > 1. □

As a corollary, we can identify the complete intersections with the same Hilbert function as an LPP almost complete intersection.

Corollary 5.13 Let I be a monomial complete intersection that has the same Hilbert function as an (a_{1}, …, a_{n})−LPP almost complete intersection and that contains a regular sequence in degrees a_{1}, …, a_{n}. Then up to reindexing, I = (x_{1}^{a_{1}}, …, x_{i}^{a_{i}−1}, …, x_{n}^{a_{n}}) for some i.

Proof: Suppose I = (x_{1}^{a_{1}−c_{1}}, …, x_{n}^{a_{n}−c_{n}}) with all c_{i} ≥ 0. The socle degree of R/I must be a_{1} + ⋯ + a_{n} − n − 1. Therefore

−(c_{1} + ⋯ + c_{n}) − n = −n − 1,

and hence some c_{i} = 1, and the rest are zero. □

The following lemma describes which LPP almost complete intersections can have the same Hilbert function as a complete intersection.

Lemma 5.14 Let L be an (a_{1}, …, a_{n})−LPP almost complete intersection with the same Hilbert function as I = (x_{1}^{a_{1}}, …, x_{i}^{a_{i}−1}, …, x_{n}^{a_{n}}). Then

L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{a_{1}−1}x_{2}^{b_{2}})

for some 1 ≤ b_{2} ≤ a_{2} − 2.

Proof: We may suppose that a_{1} > 1 since otherwise, we may quotient by x_{1} and work in a ring with one fewer variable. Let

L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{a_{1}−1} ⋯ x_{j−1}^{a_{j−1}−1}x_{j}^{b_{j}}),

and let M = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}). We have the following two short exact sequences:

0 ⟶ R/(x_{1}^{a_{1}}, …, x_{i}, …, x_{n}^{a_{n}})(−a_{i}) ⟶ R/M ⟶ R/I ⟶ 0

0 ⟶ R/(x_{1}, …, x_{j−1}, x_{j}^{a_{j}−b_{j}}, x_{j+1}^{a_{j+1}}, …, x_{n}^{a_{n}})(−(a_{1} + ⋯ + a_{j−1} − (j − 1) + b_{j}))

⟶ R/M ⟶ R/L ⟶ 0

The middle term in both sequences is R/M, and R/I and R/L have the same Hilbert function by hypothesis. Therefore the two leftmost nonzero terms in the sequences have the same Hilbert function. Hence the shifts a_{i} and (a_{1}+⋯+a_{j−1}− (j − 1) + b_{j}) are equal because otherwise, the two leftmost nonzero terms in each sequence would have their first nonzero elements in different degrees. Since a_{1} > 1, we must have j = 2 because x_{i} is the only linear generator of (x_{1}^{a_{1}}, …, x_{i}, x_{n}^{a_{n}}). Also, b_{2} ≤ a_{2} − 2 because if b_{2} = a_{2} − 1, then a_{2} − b_{2} = 1, giving too many linear generators in M : (x_{1}^{a_{1}−1}x_{2}^{b_{2}}). □

We need one more lemma to describe the complete intersections I that have the same Hilbert function as an (a_{1}, …, a_{n})−LPP ideal.

Lemma 5.15 Let I be a monomial complete intersection ideal that contains a regular sequence in degrees a_{1}, …, a_{n}. Suppose that I has the same Hilbert function as an (a_{1}, …, a_{n})−LPP almost complete intersection. Then, up to reindexing, I = (x_{1}^{a_{1}}, x_{2}^{a_{2}−1}, x_{3}^{a_{3}}, …, x_{n}^{a_{n}}).

Proof: By Corollary 5.13 and Lemma 5.14, L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{a_{1}−1}x_{2}^{b_{2}}), with 1 ≤ b_{2} ≤ a_{2} − 2, and I = (x_{1}^{a_{1}}, …, x_{i}^{a_{i}−1}, …, x_{n}^{a_{n}}) for some i. Clearly, i ≠ 1 since L is first nonzero in degree a_{i}. Suppose i ≠ 2. The short exact sequences in Lemma 5.14 show that (x_{1}, x_{2}^{a_{2}−b_{2}}, x_{3}^{a_{3}}, …, x_{n}^{a_{n}}) and (x_{1}^{a_{1}}, …, x_{i}, …, x_{n}^{a_{n}}) have the same Hilbert function. Moreover, both I and L have x_{3}^{a_{3}}, …, x_{n}^{a_{n}} among their minimal generators (excluding any power of x_{i} in that list). Therefore (x_{1}, x_{2}^{a_{2}−b_{2}}, x_{i}^{a_{i}}) and (x_{1}^{a_{1}}, x_{2}^{a_{2}}, x_{i}) have the same Hilbert function. Since b_{2} > 0, we have a_{2}−b_{2} = a_{1}, and a_{i} = a_{2}. □

We can now give the proof of Proposition 5.11.

Proof of Proposition 5.11: Without loss of generality, we may assume that I = (x_{1}^{a_{1}}, x_{2}^{a_{2}−1}, x_{3}^{a_{3}}, …, x_{n}^{a_{n}}) and L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, x_{1}^{a_{1}−1}x_{2}^{a_{2}−a_{1}}). Both I and L contain x_{3}^{a_{3}}, …, x_{n}^{a_{n}} in their minimal generating set, and R/(x_{3}^{a_{3}}, …, x_{n}^{a_{n}}) is a complete intersection. Therefore we can obtain the resolutions of R/I and R/L over R by tensoring the resolutions of R/(x_{1}^{a_{1}}, x_{2}^{a_{2}−1}) = R/M_{I} and R/(x_{1}^{a_{1}}, x_{2}^{a_{2}}, x_{1}^{a_{1}−1}x_{2}^{a_{2}−a_{1}}) = R/M_{L} with the resolution of R/(x_{3}^{a_{3}}, …, x_{n}^{a_{n}}). Hence it suffices to check that the graded Betti numbers of R/M_{I} are at most those of R/M_{L}. By the Hilbert-Burch Theorem in the graded case, the resolutions are the following:

0 ⟶ R(−(a_{1} + a_{2} − 1)) ⟶ R(−a_{1}) ⊕ R(−(a_{2} − 1)) ⟶ R ⟶ R/M_{I} ⟶ 0

0 ⟶ R(−(a_{1} + a_{2} − 1)) ⊕ R(−a_{2}) ⟶ R(−a_{1}) ⊕ R(−(a_{2} − 1)) ⊕ R(−a_{2})

⟶ R ⟶ R/M_{L} ⟶ 0

Since β^{R/M_{I}} ≤ β^{R/M_{L}}, we are done. □

We use the information in the results above about the degrees of socle generators in LPP almost complete intersections in the following lemma.

Lemma 5.16 Let L be an (a_{1}, …, a_{n})−LPP almost complete intersection. Let I be an ideal containing a regular sequence f_{1}, …, f_{n} in degrees a_{1} − b_{1}, …, a_{n} − b_{n}, with at least one b_{i} > 0, among its minimal generators. Suppose that R/I is not a complete intersection. Then I and L do not have the same Hilbert function.

Proof: By Lemma 5.12, the socle degree of L is a_{1} + ⋯ + a_{n} − n − 1. The socle degree of a complete intersection in degrees a_{1} − b_{1}, …, a_{n} − b_{n}, with exactly one b_{i} > 0, is the same. If more than one b_{i} > 0, the socle degree is too low. Additionally, if we add another minimal generator to a complete intersection in degrees a_{1}, …, a_{i} −1, …, a_{n}, the socle degree will be less than a_{1} +⋯+a_{n} −n−1. Therefore, under the hypotheses of the lemma, if R/I is not a complete intersection, then I and L do not have the same Hilbert function. □

5.5 Upper bounds on some Hilbert functions

There are two cases left to consider to prove the LPP Conjecture when L is an (a_{1}, …, a_{n})-LPP almost complete intersection. We need to show that if I has the same Hilbert function as L and contains a regular sequence of minimal generators in the same degrees, then

(1) I cannot have more than n + 1 minimal generators, and

(2) the ideal quotient (maximal regular sequence of minimal generators of I) : (other minimal generator of I) must be a complete intersection.

The next lemma proves that if either of these conditions fails, I cannot have the same Hilbert function as an LPP almost complete intersection.

Lemma 5.17 Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, m) be an (a_{1}, …, a_{n})−LPP almost complete intersection. Let I = (f_{1}, …, f_{n}, g), where deg f_{i} = a_{i}, the f_{i} form a regular sequence, and deg g = deg m.

(1) If (f_{1}, …, f_{n}) : (g) is a complete intersection, then either H(R/I) = H(R/L), or there exists u > 0 such that H(R/I, u) < H(R/L, u).

(2) If (f_{1}, …, f_{n}) : (g) is not a complete intersection, then there exists u > 0 such that H(R/I, u) < H(R/L, u).

Proof: Write J = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}) and F = (f_{1}, …, f_{n}). Let C_{L} = J : (m), and let C_{I} = F : (g). Write d = deg g = deg m. The canonical short exact sequences show that

H(R/L, l) = H(R/J, l) − H(R/C_{L}, l − d),

and

H(R/I, l) = H(R/F, l) − H(R/C_{I}, l − d).

Of course, H(R/J) = H(R/F). Therefore it suffices to show that if H(R/I) ≠ H(R/L), then there is an u such that H(R/C_{I}, u) > H(R/C_{L}, u).

Since R/J and R/F both have a single socle generator in the same top degree, adding m and g respectively into those ideals kills the socle element, and thus H(R/C_{I}) and H(R/C_{L}) are both last nonzero in the same degree.

(1) Assume first that R/C_{I} is a complete intersection and that H(R/I) ≠ H(R/L). Then the degrees of the minimal generators of C_{I} and C_{L} have the same sum. The degrees of the minimal generators of C_{L} are 1, …, 1, a_{j+1} − b_{j+1}, a_{j+2}, …, a_{n}, where 0 ≤ b_{j+1} < a_{j+1}. Write the degrees of the minimal generators of C_{I} in weakly increasing order, and call them c_{i}. If the first j c_{i} are not all 1, we are done, for then H(R/C_{I}, 1) > H(R/C_{L}, 1). Otherwise, let r be the first index where the list of degrees differs.

If a_{r} < c_{r} (or if r = j + 1, then if a_{j+1} − b_{j+1} < c_{j+1}), we are done, since then H(R/C_{L}, a_{r}) < H(R/C_{I}, a_{r}). Otherwise, a_{r} > c_{r} (or a_{j+1} − b_{j+1} > c_{j+1}). Hence there exists s such that c_{r+s} > a_{r+s} because the lists of degrees have the same sum. By construction, F ⊂ C_{I} =: (h_{1}, …, h_{n}), where deg h_{i} = c_{i}. Comparing degrees of minimal generators of F and C_{I}, we find

(f_{1}, …, f_{r+s}) ⊂ (h_{1}, …, h_{r+s−1})

since deg h_{r+s} > deg f_{l} for l ≤ r + s (recall that the a_{i}, the degrees of the f_{i}, are weakly increasing). But this means that a depth r +s ideal is contained in an ideal of smaller depth, a contradiction.

Consequently, if R/C_{I} is a complete intersection, we cannot have H(R/C_{I}, l) ≤ H(R/C_{L}, l) for each l without equality occurring for every l. This gives the desired inequality for H(R/I) and H(R/L).

(2) Suppose instead that R/C_{I} is not a complete intersection but that H(R/C_{I}, l) ≤ H(R/C_{L}, l) for all l. C_{I} must have at least as many linear generators as C_{L}, or we are done. Moreover, since F ⊂ C_{I}, there is an (a_{1}, …, a_{n}) complete intersection inside C_{I}.

As before, the regularities of C_{I} and C_{L} are the same, and so the sum of the degrees of any maximal length regular sequence inside C_{I} must exceed, in the notation above, j + (a_{j+1} − b_{j+1}) + a_{j+2} + ⋯ + a_{n}. We form a minimal generating set of C_{I} as follows: Pick a linear form inside C_{I}, and call it h_{1}. Now pick another element of C_{I} of minimal degree, calling it h_{2}, such that h_{1} and h_{2} form a regular sequence. Continue until one has a regular sequence h_{1}, …, h_{n} of length n with deg h_{i} ≤ deg h_{i+1}. Write C_{I} = (h_{1}, …, h_{n}, h_{n+1}, …, h_{p}), where p ≥ n + 2 (since a theorem of Kunz shows that almost complete intersections are never Gorenstein [32]).

Let us consider the degrees of these generators. We know deg h_{i} = 1 for i ≤ j. Let c_{s} = deg h_{s} for j + 1 ≤ s ≤ n. Then

c_{j+1} + ⋯ + c_{n} > (a_{j+1} − b_{j+1}) + a_{j+2} + ⋯ + a_{n}.

If a_{j+1} − b_{j+1} < c_{j+1}, then we are done, for then H(R/C_{L}, a_{j+1} − b_{j+1}) < H(R/C_{I}, a_{j+1} − b_{j+1}). Otherwise, a_{j+1} − b_{j+1} ≥ c_{j+1}. Hence there exists r > j + 1 such that a_{r} < c_{r} because of the restriction on the sums of the degrees. Note that (f_{1}, …, f_{r}) must be contained in the portion of C_{I} in degrees at most a_{r}. Therefore, if {h_{q_{1}}, …, h_{q_{t}}} is the subset of {h_{n+1}, …, h_{p}} of polynomials of degree at most a_{r}, we have

(f_{1}, …, f_{r}) ⊂ (h_{1}, …, h_{r−1}, h_{q_{1}}, …, h_{q_{t}}).

But the left-hand side has depth r, while our construction guarantees that the right-hand side has depth only r − 1, a contradiction. Hence there must be some l such that H(R/C_{I}, l) > H(R/C_{L}, l), and we are done. □

Analyzing the preceding proof, we obtain a special case of the EGH Conjecture. Corollary 5.18 Let a_{1} ≤ ⋯ ≤ a_{n} be positive integers. Let I be an almost complete intersection generated by homogeneous polynomials f_{1}, …, f_{n}, and g, where the f_{i} form a regular sequence, deg f_{i} = a_{i}, and deg g = d. Assume d ≥ a_{1}. Let L = (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}, m), where m is the greatest monomial in lex order in degree d not in (x_{1}^{a_{1}}, …, x_{n}^{a_{n}}). Then H(R/I, d + 1) ≤ H(R/L, d + 1).

Given an ideal I as in Corollary 5.18, the idea is to take the appropriate LPP almost complete intersection L so that H(R/I, d) = H(R/L, d), where d is the degree of the extra generator g of I. Then we show that the Hilbert function of R/I cannot grow any faster than that of R/L in the next degree. Our assumption that d ≥ a_{1} is not a significant restriction; one can choose the generators of the maximal regular sequence to ensure that this is the case.

Proof: The proof is almost identical to that of Lemma 5.17. We outline the case in which (f_{1}, …, f_{n}) : (g) = (h_{1}, …, h_{n}) is a complete intersection. In the notation of Lemma 5.17, we need to show that H(R/C_{I}, 1) ≥ H(R/C_{L}, 1). If not, then there are more minimal linear generators in C_{I} than in C_{L}. Since the regularity of the ideals is the same, there is some r > j + 1 such that deg h_{r} > a_{r}, where we have ordered the h_{i} such that deg h_{i} ≤ deg h_{i+1} as above. Then (f_{1}, …, f_{r}) ⊂ (h_{1}, …, h_{r−1}), a contradiction. The proof of the case in which (f_{1}, …, f_{n}) : (g) is not a complete intersection is also essentially identical to the argument in Lemma 5.17. □

To conclude this section, we state two more results of this nature.

Proposition 5.19 Under the hypotheses of Corollary 5.18:

(1) If H(R/I) ≠ H(R/L), then let l_{0} be the first degree such that H(R/I, l) ≠ H(R/L, l). Then H(R/I, l_{0}) < H(R/L, l_{0}).

(2) If (f_{1}, …, f_{n}) : g is a complete intersection, then H(R/I, l) ≤ H(R/L, l) for all l.

Proof: (1) This is very similar to the proofs of the previous results. Suppose (f_{1}, …, f_{n}) : (g) is not a complete intersection (if it is, then (1) follows from (2)). As in Corollary 5.18, using the same notation, C_{I} has at most as many linear generators as C_{L}; if it has fewer, then H(R/I, d + 1) < H(R/L, d + 1), and we are done. Otherwise, choose a maximal length regular sequence h_{1}, …, h_{n} inside C_{I} as before, picking something of minimal degree each time. Suppose C_{L} and C_{I} have j linear generators. If deg h_{j+1} > a_{j+1} − b_{j+1}, we are done, for then H(R/I, d + a_{j+1} − b_{j+1}) < H(R/L, d + a_{j+1} − b_{j+1}). If not, then since the sum of the degrees of h_{1}, …, h_{n} is greater than the sum of the degrees of the minimal generators of C_{L}, for some n ≥ i > j + 1, deg h_{i} > a_{i}. But then, as before, we have an ideal (f_{1}, …, f_{i}) of depth i contained in an ideal of depth i − 1, a contradiction.

(2) Let C_{I} and C_{L} be as before. Call the minimal generators of C_{I} h_{1}, …, h_{n}, with their degrees weakly increasing. We need to show that H(R/C_{I}, l) ≥ H(R/C_{L}, l) for all l. To do this, we analyze how changing the degrees of the minimal generators of a complete intersection in a particular way affects the Hilbert series.

We need to determine the possible degrees v_{l} = (q_{1}, …, q_{n}) of the minimal generators of C_{I}, where q_{i} ≤ q_{i+1} for each i. By previous arguments, we know that q_{i} ≤ a_{i} for all i. We claim that we can start with the degree vector of the minimal generators of C_{L}, v_{0} = (1, …, 1, a_{j+1}−b_{j+1}, a_{j+2}, …, a_{n}), and move to any possible degree vector (q_{1}, …, q_{n}) of the minimal generators of C_{I} by a sequence of switches of degrees from some c_{j} and c_{i} to c_{j} −1 and c_{i} +1, where c_{j} > c_{i} +1. (Note that the regularities of C_{I} and C_{L} are the same, so the sums of the degrees of the minimal generators of each are the same.) To do this, we proceed in the following manner. Starting with v_{0}, take the greatest entry of v_{0} that is greater than its corresponding entry in v_{l}. Subtract one from it and add that to the lowest entry of v_{0} that needs to increase; then sort so that the new list is weakly increasing. Here, the ones in v_{0} cannot decrease, and the entries of a_{i} cannot increase, so we do not have to subtract from a lesser degree and add to a greater number.

Suppose we get to some v_{c} = (c_{1}, …, c_{n}) in the algorithm, where c_{i} ≤ c_{i+1} for each i. Say c_{j} is the greatest number that needs to decrease and c_{i} is the least number that needs to increase. We show that c_{j} > c_{i} + 1. If c_{j} = c_{i} + 1, then we are not changing the degrees by taking one from c_{j} and adding it to c_{i}. This is not a move the algorithm can require us to make because in this case, q_{i} > c_{i} = c_{j} − 1 ≥ q_{j}, a contradiction since j > i. If c_{j} < c_{i} + 1, then we previously needed an entry of c_{i} + 1 (or greater), and so we should not have decreased c_{j} (or something greater) earlier.

For example, we might start with degrees (1, 1, 1, 5, 7, 9) and need to switch to (2, 2, 4, 6, 8). We would do this by changing (1, 1, 1, 5, 7, 9) to (1, 1, 2, 5, 7, 8) to (1, 2, 2, 5, 6, 8) to (2, 2, 2, 4, 6, 8).

Now, let us compute the effect of such switches in the degrees of minimal generators of complete intersection ideals on the Hilbert function. For ease in notation, without loss of generality, suppose we change a complete intersection with minimal generators in degrees c_{1}, …, c_{n} to one with minimal generators in degrees c_{1} + 1, c_{2} − 1, c_{3}, …, c_{n}. We may assume by the arguments above that c_{2} − c_{1} − 1 > 0. We compute the Hilbert series of R/(new ideal) minus the Hilbert series of R/(old ideal), which is:

\frac{(1 − t^{c_{1}+1})(1 − t^{c_{2}−1})∏_{i=3}^{n}(1 − t^{c_{i}})}{(1 − t)^{n}} − \frac{(1 − t^{c_{1}})(1 − t^{c_{2}})∏_{i=3}^{n}(1 − t^{c_{i}})}{(1 − t)^{n}} =

\frac{∏_{i=3}^{n}(1 − t^{c_{i}})}{(1 − t)^{n}} (t^{c_{1}} + t^{c_{2}} − t^{c_{2}−1} − t^{c_{1}+1}) = \frac{∏_{i=3}^{n}(1 − t^{c_{i}})}{(1 − t)^{n}}t^{c_{1}} (1 − t) (1 − t^{c_{2}−c_{1}−1})

= t_{1}^{c} (\frac{1 − t^{c_{2}−c_{1}−1}}{1 − t} ) (\frac{1 − t}{1 − t}) (\frac{1 − t^{c_{3}}}{1 − t} ) ⋯ (\frac{1 − t^{c_{n}}}{1 − t} ) .

Since c_{2} − c_{1} − 1 > 0, this expression is a polynomial with all nonnegative coefficients, so the Hilbert function of R/C_{L} is at most that of R/C_{I} in every degree. Hence H(R/I, l) ≤ H(R/L, l) for all r. □

We believe that the stronger statement of (2) in Proposition 5.19 should also hold when (f_{1}, …, f_{n}) : g is not a complete intersection. Finally, we note that if f_{1}, …, f_{n} and g are generic, then the inequality H(R/I, l) ≤ H(R/L, l) holds for all l by a result of Fröberg in [17]; one can also compute this directly without much difficulty.

5.6 Conclusion

We can now give our main result.

Theorem 5.20 Let L be an (a_{1}, …, a_{n})−LPP almost complete intersection. Let I be any ideal with the same Hilbert function as L that contains a regular sequence in degrees a_{1}, …, a_{n}. Then β^{R/I} ≤ β^{R/L}.

Proof: If I contains a complete intersection in degrees less than a_{1}, …, a_{n}, then by the results of Section 4, I is an (a_{1}, a_{2} − 1, a_{3}, …, a_{n}) complete intersection. The theorem then follows from Proposition 5.11. Otherwise, I has minimal generators f_{1}, …, f_{n} that form a regular sequence in degrees a_{1}, …, a_{n}. By Lemma 5.17, I is an almost complete intersection (f_{1}, …, f_{n}, g) with (f_{1}, …, f_{n}) : (g) a complete intersection. Then Theorem 5.9 gives the inequality for the Betti numbers. □

The extra structure of almost complete intersections is important in our arguments. All the mapping cones we have used come from two Koszul complexes, and one loses this structure if one has to consider ideals with n + 2 or more generators. In particular, it becomes much more difficult to detect nonminimality. We anticipate that other techniques will be necessary to handle LPP ideals with more than n + 1 generators.

Bounding the multiplicity of ideals

In our final chapter, we investigate a conjecture of Huneke and Srinivasan, generalized by Herzog and Srinivasan, on the multiplicity of ideals. The conjecture connects the multiplicity of an ideal with the shifts in its graded free resolution. We discuss a new approach to the problem, prove some special cases, and discuss why we think it has been so difficult to make general progress on this problem.

6.1 Introduction

One important invariant of a module that we have not discussed so far is the multiplicity. In this section, we review how to compute the multiplicity of an ideal and how the multiplicity is conjectured to be related to the graded free resolution.

We begin with a definition.

Definition 6.1 Let I be a homogeneous ideal in R = k[x_{1}, …, x_{n}]. Write the Hilbert series of R/I as \frac{g(t)}{(1−t)^{d}}, where g(1) ≠ 0. Then the multiplicity (or degree) of R/I, denoted e(R/I), is g(1).

There are several ways to compute the multiplicity. If R/I is Artinian, then the multiplicity is just dim_{k}(R/I), or the sum of the values of the Hilbert function of R/I. If R/I is not Artinian, let P = a_{d}t^{d} + ⋯ + a_{0}, where a_{d} ≠ 0, be its Hilbert polynomial. Then e(R/I) = a_{d}(d!).

Example 6.2 Let I = (a^{2}, ab^{3}, ab^{2}c, b^{4}, c^{5}) ⊂ S = k[a, b, c]. Then the Hilbert function of S/I is (1, 3, 5, 7, 6, 5, 3, 1), and e(S/I) = 31, the k-vector space dimension of S/I.

If we remove the generators b^{4} and c^{5}, we obtain an ideal J = (a^{2}, ab^{3}, ab^{2}c) that is not Artinian. The Hilbert series of S/J is

\frac{1 − t^{2} − 2t^{4} + 3t^{5} − t^{6}}{(1 − t)^{3}} = \frac{1 + t − 2t^{4} + t^{5}}{(1 − t)^{2}} ,

and thus e(S/J) = 1 + 1 − 2 + 1 = 1. Alternatively, the Hilbert polynomial of S/J is t + 3, and hence e(S/J) = 1(1!) = 1.

In [31], Huneke and Miller found a formula for the multiplicity of a Cohen-Macaulay ideal with a pure resolution. A pure resolution is one in which all minimal generators at a single step in the resolution have the same degree; that is, it has the form

0 ⟶ R(−d_{c})^{β_{c}} ⟶ ⋯ ⟶ R(−d_{2})^{β_{2}} ⟶ R(−d_{1})^{β_{1}} ⟶ R ⟶ R/I ⟶ 0. Huneke and Miller prove the following theorem.

Theorem 6.3 Let I be a homogeneous ideal of codimension c in R = k[x_{1}, …, x_{n}] such that R/I is Cohen-Macaulay and has a pure resolution. Then

e(R/I) = \frac{1}{c!} ∏_{i=1}^{c} d_{i}.

The proof is a clever use of the Residue Theorem from complex analysis and Vandermonde determinants. Theorem 6.3 is interesting in that it places a strong condition on the d_{i} because e(R/I) is always an integer.

In general, of course, there may be many different shifts at each step in the resolution. Suppose the i-th free module in the resolution of R/I is a direct sum of the form

F_{i} = R(−d_{i1}) ⊕ ⋯ ⊕ R(−d_{ib_{i}}).

For each i, let m_{i} = min{d_{i1}, …, d_{ib_{i}}}, and let M_{i} = max{d_{i1}, …, d_{ib_{i}}}. Huneke and Srinivasan propose the following generalization of Theorem 6.3.

Conjecture 6.4 Let R = k[x_{1}, …, x_{n}], and let I be a homogeneous ideal in R of codimension c such that R/I is Cohen-Macaulay. Then

\frac{1}{c!} ∏_{i=1}^{c} m_{i} ≤ e(R/I) ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}.

Example 6.5 Let I be the ideal (a^{2}, ab^{3}, ab^{2}c, b^{4}, c^{5}) from Example 6.2, where we computed that e(S/I) = 31. The graded free resolution of S/I is:

S/L: total: 1 5 7 3
0: 1 …
1: . 1 . .
2: ….
3: . 3 4 1
4: . 1 . .
5: . . 1 .
6: . . 1 1
7: . . 1 1

Therefore m_{1} = 2, m_{2} = 5, and m_{3} = 6; also, M_{1} = 5, M_{2} = 9, and M_{3} = 10. The conjectured bounds are

10 = \frac{1}{3!} (2)(5)(6) ≤ e(S/I) = 31 ≤ \frac{1}{3!} (5)(9)(10) = 75,

which hold in this case.

The lower bound of Conjecture 6.4 fails if we drop the hypothesis that R/I is Cohen-Macaulay. For example, let I = (a^{2}, ab) in S = k[a, b]. Then codim I = 1, e(S/I) = 1, and the minimal free resolution of S/I is

0 ⟶ S(−3) ⟶ S(−2)^{2} ⟶ S ⟶ S/I ⟶ 0.

Thus m_{1} = 2, and

\frac{1}{1!}(2) > 1 = e(S/I).

However, Herzog and Srinivasan conjecture in [28] that the upper bound should hold even without the Cohen-Macaulay hypothesis.

Conjecture 6.6 Let R = k[x_{1}, …, x_{n}], and let I be a homogeneous ideal in R of codimension c. Then

e(R/I) ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}.

When R/I is Cohen-Macaulay, its projective dimension is the same as the codimension of I. If R/I is not Cohen-Macaulay, then codim I < pd R/I, so Conjecture 6.6 is a strong statement in the sense that one only considers c shifts instead of the maximum shift from each step in the resolution.

We shall focus on the Cohen-Macaulay case in this chapter. There are a number of special cases of Conjecture 6.4 known. In [28], Herzog and Srinivasan generalize the pure resolution result in Theorem 6.3 to modules with a quasipure resolution; we shall discuss this case in more detail in Section 4. Herzog and Srinivasan also prove Conjecture 6.4 for modules R/I when I is stable, I is perfect of codimension two, I is squarefree strongly stable (for the definition, see [1]), and in some cases in which I is a codimension three Gorenstein ideal. They also prove the upper bound, Conjecture 6.6, for all Gorenstein codimension three ideals. Römer has recently generalized work of Herzog and Srinivasan in [28] and Gold [21] to prove Conjecture 6.6 for all codimension two Cohen-Macaulay ideals [40]. He also proves Conjecture 6.4 for componentwise linear ideals (see [40] for further details), which generalizes the stable case as well as a few others. Finally, Herzog and Srinivasan have a recent preprint [27] in which they prove weaker upper bounds for some monomial ideals using the Taylor resolution.

Despite the work on these many special cases, there has been little general progress on the conjectures bounding the multiplicity of an ideal. In Section 2, we introduce a new approach to the problem, relating it to the question of what resolutions occur for a given Hilbert function. We use this approach in Section 3 to discuss some computational results from the computer algebra system Macaulay 2 [23]. In the last section, we use the quasipure result of Herzog and Srinivasan to prove the upper bound of Conjecture 6.4 in some new cases, and we reduce the upper bound conjecture to a simpler question.

6.2 Relation to resolutions with a given Hilbert function

We investigate Conjecture 6.4 in this section by fixing a Hilbert function and determining which resolutions corresponding to modules with that Hilbert function satisfy the bounds. Recall that the lexicographic ideal has the largest graded Betti numbers for a given Hilbert function, and lexicographic ideals are stable. Hence by a theorem of Herzog and Srinivasan [28], lexicographic ideals satisfy the bounds of Conjecture 6.4. That is, the conjecture is true for the maximal element in the partial order on graded Betti numbers for a given Hilbert function. What about other elements in the partial order?

To consider this question, let I be a homogeneous ideal of codimension c in R = k[x_{1}, …, x_{n}], and assume that R/I is Cohen-Macaulay. Suppose R/I satisfies the bounds of Conjecture 6.4. Then, if m_{i}^{R/I} and M_{i}^{R/I} are the minimum and maximum shifts at each step of the resolution of R/I as in Section 1, we have

\frac{1}{c!} ∏_{i=1}^{c} m_{i}^{R/I} ≤ e(R/I) ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}^{R/I}.

Let J be an ideal of codimension c satisfying the following conditions: R/J is Cohen-Macaulay, J has the same Hilbert function as R/I, and β^{R/I} ≤ β^{R/J}. Then because β_{ij}^{R/I} ≤ β_{ij}^{R/J} for all i and j,

m_{i}^{R/J} ≤ m_{i}^{R/I} and M_{i}^{R/I} ≤ M_{i}^{R/J};

the resolution of R/J has all the terms of the resolution of R/I plus possibly more, so the minimum shifts can be lower, and the maximum shifts can be higher. In addition, because I and J have the same Hilbert function, and the Hilbert function determines the multiplicity, e(R/I) = e(R/J). Putting these facts together, we have

\frac{1}{c!} ∏_{i=1}^{c} m_{i}^{R/J} ≤ \frac{1}{c!} ∏_{i=1}^{c} m_{i}^{R/I} ≤ e(R/I) = e(R/J) ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}^{R/I} ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}^{R/J}.

Therefore R/J also satisfies the bounds of Conjecture 6.4. Consequently, we have proven the following proposition.

Proposition 6.7 Let I and J be homogeneous ideals of codimension c in R = k[x_{1}, …, x_{n}] such that R/I and R/J are Cohen-Macaulay. Suppose that I and J have the same Hilbert function, β^{R/I} ≤ β^{R/J}, and R/I satisfies the bounds in Conjecture 6.4. Then Conjecture 6.4 also holds for R/J. □

The condition that β^{R/I} ≤ β^{R/J} in Proposition 6.7 is actually a bit stronger than what we need. It is enough to have that β_{ij}^{R/I} ≠ 0 implies that β_{ij}^{R/J} ≠ 0.

Additionally, we can simplify our investigation: By Lemma 1.20, if R/I is a Cohen-Macaulay module, there exists an Artinian module R′/I′ with the same multiplicity and graded Betti numbers. Therefore, if there is a Cohen-Macaulay counterexample to Conjecture 6.4, there is an Artinian counterexample, and it suffices to work with Artinian ideals. We shall use this result frequently in the rest of the chapter.

The motivation for Proposition 6.7 is clear: Fix a Hilbert function. Suppose we can determine all the resolutions that are minimal in the partial order for that Hilbert function. If we can show that Conjecture 6.4 holds for all the minimal elements in the partial order, we can use Proposition 6.7 to “lift” the result to all the resolutions above the minimals, proving that the bounds of Conjecture 6.4 hold for every module with that Hilbert function.

Here is an example of this type of application.

Example 6.8 Let S = k[a, b, c], and let L ⊂ S be the lexicographic ideal such that S/L has Hilbert function H = (1, 3, 6, 9, 9, 6, 2). Let I be the ideal (a^{3}, b^{4}, c^{4}, b^{2}c^{2}). Then L and I have the same Hilbert functions, and the Betti diagrams of S/L and S/I are below:

S/L: total: 1 16 27 12  S/I: total: 1 4 5 2
0: 1 …  0: 1 …
1: ….  1: ….
2: . 1 . .  2: . 1 . .
3: . 3 5 2  3: . 3 . .
4: . 5 9 4  4: . . 2 .
5: . 5 9 4  5: . . 3 .
6: . 2 4 2  6: … 2

Note that by making all potentially possible cancellations in the Betti diagram of S/L, we obtain the Betti diagram of S/I. Therefore β^{S/I} ≤ β^{S/J} for all ideals J ⊂ S with the same Hilbert function as I. The bounds on S/I from Conjecture 6.4 are

27 = \frac{1}{3!}(3)(6)(9) ≤ e(S/I) = 36 ≤ \frac{1}{3!}(4)(7)(9) = 42,

and thus S/I satisfies the conjecture. By Proposition 6.7, the bounds of Conjecture 6.4 hold for all modules with the same Hilbert function as I.

We can use the same ideas to get a slight generalization of Theorem 6.3, the result of Huneke and Miller on the pure resolution case, in codimension three.

Proposition 6.9 Let R = k[x_{1}, …, x_{n}], and let I be a homogeneous ideal in R of codimension three such that R/I is Cohen-Macaulay and has a pure resolution.

Let J be any homogeneous ideal in R of codimension three with the same Hilbert function such that R/J is Cohen-Macaulay. Then Conjecture 6.4 holds for R/J.

Proof: By Lemma 1.20, we may assume that I and J are Artinian. Theorem 6.3 implies that R/I satisfies the bounds of Conjecture 6.4. Since the resolution of R/I is pure, I has minimal generators in only one degree, the lowest degree d for which H(R/I, d) < \binom{n−1+d}{d}. Since J has the same Hilbert function, J has the same number of generators in degree d and possibly more in higher degree. Hence β_{1j}^{R/I} ≤ β_{1j}^{R/J} for all j.

Because I is Artinian, there is some degree s in which H(R/I) is last nonzero, and β_{3,s+3}^{R/J} is the same for all ideals J with the same Hilbert function as I. Also, since R/I has a pure resolution, β_{3t}^{R/I} = 0 for all t ≠ s + 3. Thus β_{3j}^{R/I} ≤ β_{3j}^{R/J} for all j.

Next we consider the second place in the resolution. Suppose β_{2u}^{R/I} ≠ 0; then all other β_{2v}^{R/I} = 0. Let L be the lexicographic ideal of R with the same Hilbert function as I. Then

β_{2u}^{R/I} = β_{2u}^{R/L} − (β_{1u}^{R/L} + β_{3u}^{R/L})

since the resolution of R/I is pure. Because the resolution of R/J is also obtained from making cancellations in the resolution of R/L, we have

β_{2u}^{R/I} = β_{2u}^{R/L} − (β_{1u}^{R/L} + β_{3u}^{R/L}) ≤ β_{2u}^{R/J}.

Therefore β^{R/I} ≤ β^{R/J}. By Proposition 6.7, Conjecture 6.4 holds for R/J. □

6.3 Some computational work on the upper bound

We discuss some additional ways to use Proposition 6.7 in this section. The results we obtain give some insight into why Conjecture 6.4 is so difficult to prove in general, though to verify the conjecture for a particular Hilbert function, our techniques will often work reasonably well.

We used the computer algebra system Macaulay 2 [23] to investigate Conjecture 6.4 using Proposition 6.7. Our goal was to use Proposition 6.7 for Artinian ideals to verify the upper bound in Conjecture 6.4 for many Hilbert functions with small socle degree (the highest degree in which the Hilbert function is nonzero). We did most of our computations for ideals in S = k[a, b, c].

To prove the upper bound of Conjecture 6.4 for a particular Hilbert function, we began by computing the lexicographic ideal in S corresponding to that Hilbert function. The idea was to keep the multiplicity fixed but to lower the M_{i}, proving that the multiplicity is bounded above by a smaller bound. Starting with the Betti diagram of the lexicographic ideal, we made all potentially possible cancellations in the diagram to find minimal sets of graded Betti diagrams in the partial order. At this stage, we computed the upper bound from the resulting potential Betti diagram(s). If the upper bound of Conjecture 6.4 held for all the minimal diagrams, that proved the bound for all modules with that Hilbert function.

There are some complications to consider, however. First, as we have discussed in previous chapters, it is often difficult to determine whether a potential cancellation in a Betti diagram can occur. We decided to disregard this concern in our initial tests and return to it at the end; we shall discuss this in more detail later in the section. Second, there may be many incomparable minimal elements in the partial order for a given Hilbert function. To minimize the number of Betti diagrams we had to check, we made our cancellations in a particular way that we outline below, finding a particular minimal element in the partial order, and it sufficed to check that diagram to determine if the upper bound held.

In three variables, the process is easy. For each Hilbert function, we first resolved the lexicographic ideal corresponding to that Hilbert function. M_{3} is always fixed, for it is simply the maximal degree of a socle element plus three. (This is three plus the highest degree in which the Hilbert function is nonzero.) We lower M_{1} as much as possible by making all potentially possible cancellations of entries in the generators column of the Betti diagram with first syzygies. Finally, we lower M_{2}, if possible, by cancelling first syzygy entries in the Betti diagram with second syzygies. This yields a Betti diagram that may or may not represent the graded free resolution of an actual S-module. Regardless, we can still evaluate whether the bounds of Conjecture 6.4 hold, computing the multiplicity directly from the Hilbert function and reading the minimum and maximum shifts from the diagram with the cancellations.

To illustrate this, we give an example.

Example 6.10 Consider the Hilbert function (1, 3, 6, 7, 3, 1). The lexicographic ideal corresponding to this Hilbert function in S = k[x_{1}, x_{2}, x_{3}] has the following resolution:

S/L: total: 1 12 19 8
0: 1 …
1: ….
2: . 3 3 1
3: . 6 10 4
4: . 2 4 2
5: . 1 2 1

We first make all the cancellations potentially possible in the generators and first syzygies columns. This yields the Betti diagram below:

S/L: total: 1 6 13 8
0: 1 …
1: ….
2: . 3 . 1
3: . 3 8 4
4: . . 3 2
5: . . 2 1

We then cancel any remaining pairs in the first and second syzygies columns, and obtain the following diagram:

S/L: total: 1 6 7 2
0: 1 …
1: ….
2: . 3 . .
3: . 3 7 1
4: ….
5: … 1

The multiplicity corresponding to this Hilbert function is 1+3+6+7+3+1 = 21. The maximum shifts are 4, 5, and 8. Because

21 ≤ \frac{(4)(5)(8)}{3!} = \frac{80}{3},

the upper bound of Conjecture 6.4 holds for the diagram with all the cancellations. But we have lowered the maximum shifts as much as possible, and therefore the upper bound holds for all modules S/I with Hilbert function (1, 3, 6, 7, 3, 1).

Our computations were fruitful. In Macaulay 2 [23], we tested all Hilbert functions for modules S/I such that I contains no linear forms and S/I is zero in degree ten and higher; that is, we tested the Hilbert functions of the form (1, 3, h_{2}, …, h_{9}). We tested over 677,000 Hilbert functions with this technique, and only 197 returned potential Betti diagrams that did not satisfy the upper bound of Conjecture 6.4. For each of those 197 examples, we looked at the potential Betti diagram with the cancellations more carefully, and we found reasons the diagrams could not represent the resolution of a module. We proved that all but nine cannot exist using Theorem 1.18 and the fact that an Artinian ideal in three variables that is not a complete intersection must have at least four generators. For the other examples, we needed a few easy computations. We give three examples below to illustrate this process. The first uses Theorem 1.18.

Example 6.11 Consider the Hilbert function (1, 3, 6, 10, 15, 15, 11) for quotients of S = k[x_{1}, x_{2}, x_{3}]. The resolution of the lexicographic ideal corresponding to this Hilbert function is below:

S/L: total: 1 25 43 19
0: 1 …
1: ….
2: ….
3: ….
4: . 6 8 3
5: . 7 12 5
6: . 12 23 11

After making all the cancellations in the appropriate order, we obtain the following Betti diagram:

S/L: total: 1 6 19 14
0: 1 …
1: ….
2: ….
3: ….
4: . 6 1 3
5: ….
6: . . 18 11

The multiplicity corresponding to the Betti diagram is 1+3+6+10+15+15+11 = 61. The maximum shifts are 5, 8, and 9; note that the cancellations in degree eight are irrelevant to our computations. Since

61 > \frac{1}{3!}(5)(8)(9) = 60,

the upper bound of Conjecture 6.4 fails. We had to make the cancellations in degree six and seven to get maximum shifts for which the bound fails, so to show that the upper bound holds for all modules with this Hilbert function, it is enough to show that row four of this Betti diagram cannot exist. This is clear from Theorem 1.18 (or by simply observing that there cannot be three minimal second syzygies of degree seven when there is a single first syzygy of degree at most six).

We illustrate the technique used to eliminate the final nine potential counterexamples in the following two examples.

Example 6.12 Let H be the Hilbert function (1, 3, 6, 10, 15, 17, 17, 17, 15, 10), which yields a multiplicity of 111. The lexicographic ideal in S = k[x_{1}, x_{2}, x_{3}] corresponding to this Hilbert function has the following resolution:

S/L: total: 1 30 53 24
0: 1 …
1: ….
2: ….
3: ….
4: . 4 4 1
5: . 3 5 2
6: . 2 4 2
7: . 4 7 3
8: . 6 12 6
9: . 11 21 10

We first make all the potential cancellations between the generators and first syzygies column, and then we note that it will not be possible to lower M_{2} (or M_{3}, of course) since we have 12 first syzygies of degree 11 but only six second syzygies in that degree. The diagram below is the result of these cancellations.

S/L: total: 1 4 27 24
0: 1 …
1: ….
2: ….
3: ….
4: . 4 1 1
5: . . 3 2
6: … 2
7: . . 1 3
8: . . 1 6
9: . . 21 10

This gives maximum shifts of 5, 11, and 12, and the upper bound is thus 110, which is less than the multiplicity of 111. This Betti diagram cannot exist because of the single first syzygy of degree at most six and the second syzygy of degree seven. However, if we cancel the second syzygy of degree seven and a first syzygy of the same degree, Theorem 1.18 does not forbid the resulting Betti diagram. Therefore we need to be a bit more creative.

We shall show that an almost complete intersection I of four degree five polynomials cannot have Hilbert function H. Call the generators of I f_{1}, f_{2}, f_{3}, and f_{4}, and assume, reindexing if necessary, that the first three form a regular sequence. Let J = (f_{1}, f_{2}, f_{3}) : (f_{4}). We have the following short exact sequence:

0 ⟶ S/J(−5) ⟶ S/(f_{1}, f_{2}, f_{3}) ⟶ S/I ⟶ 0.

We are given the Hilbert function of S/I, and the Hilbert function of the complete intersection S/(f_{1}, f_{2}, f_{3}) is easy to compute. The Hilbert function of S/J(−5) is the difference between the two:

(1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1)
- (1, 3, 6, 10, 15, 17, 17, 17, 15, 10)
(0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 6, 3, 1)

Therefore the Hilbert function of S/J should be (1, 2, 1, 0, 0, 6, 3, 1), which is clearly not an allowable Hilbert function. Hence no almost complete intersection of four degree five polynomials exists with Hilbert function H, and thus the upper bound holds for all modules S/I with this Hilbert function.

Example 6.13 In this example, we follow the same procedure as in Example 6.12, but the situation is more complicated. Consider the Hilbert function (1, 3, 6, 10, 15, 21, 23, 24, 22, 15). The resolution of the corresponding lexicographic ideal L is:

S/L: total: 1 38 68 31
0: 1 …
1: ….
2: ….
3: ….
4: ….
5: . 5 6 2
6: . 3 5 2
7: . 5 9 4
8: . 9 17 8
9: . 16 31 15

Both M_{2} = 11 and M_{3} = 12 are fixed, and the multiplicity is 140. The only way to violate the upper bound of Conjecture 6.4 is to have M_{1} = 6. After making all potentially possible cancellations in columns one and two, we obtain:

S/L: total: 1 5 35 31
0: 1 …
1: ….
2: ….
3: ….
4: ….
5: . 5 3 2
6: … 2
7: … 4
8: . . 1 8
9: . . 31 15

It looks as though this is unlikely to be the Betti diagram of a module, but it does not violate Theorem 1.18. We proceed by analyzing what Hilbert functions are possible for an Artinian ideal I in S with five generators in degree six. Suppose I = (f_{1}, …, f_{5}), where f_{1}, f_{2}, and f_{3} form a regular sequence. We have the following two exact sequences:

0 ⟶ S/((f_{1}, f_{2}, f_{3}) : (f_{4}))(−6) ⟶ S/(f_{1}, f_{2}, f_{3}) ⟶ S/(f_{1}, f_{2}, f_{3}, f_{4}) ⟶ 0

0 ⟶ S/((f_{1}, f_{2}, f_{3}, f_{4}) : (f_{5}))(−6) ⟶ S/(f_{1}, f_{2}, f_{3}, f_{4}) ⟶ S/I ⟶ 0 From the exact sequences, it follows that

H(S/(f_{1}, f_{2}, f_{3}), d) − H(S/I, d) =
H(S/((f_{1}, f_{2}, f_{3}) : (f_{4})), d − 6) + H(S/((f_{1}, f_{2}, f_{3}, f_{4}) : (f_{5})), d − 6).

The Hilbert functions on the left side of the equation are easy to compute, and the difference is

(0, 0, 0, 0, 0, 0, 2, 3, 5, 10, 21, 15, 10, 6, 3, 1),

the sum of the Hilbert functions on the right side of the equation. We argue that the Hilbert functions of two modules S/J_{1} and S/J_{2} can never yield this sum. Either H(R/J_{1}, 1) = 3 and H(R/J_{2}, 1) = 0, or H(R/J_{1}, 1) = 2 and H(R/J_{2}, 1) = 1 (reindexing the J_{i} if necessary). In the first case, H(R/J_{1}, 4) ≤ 15, and H(R/J_{2}, 4) = 0, so their sum in degree four cannot be 21. In the second case, H(R/J_{1}, 4) ≤ 5, and H(R/J_{2}, 4) ≤ 1, so the values cannot add up to 21.

Thus R/I cannot have the Hilbert function (1, 3, 6, 10, 15, 21, 23, 24, 22, 15), and there are no counterexamples to the upper bound of Conjecture 6.4 with this Hilbert function. Applying this technique eliminates the eight potential counterexamples remaining after Example 6.12.

After ruling out the 197 possible counterexamples, we have the following theorem from the computations in Macaulay 2 [23].

Theorem 6.14 Let I be an Artinian ideal in S = k[x_{1}, x_{2}, x_{3}] such that S/I is zero in degree ten and higher. Then S/I satisfies the upper bound of Conjecture 6.4; that is,

e(S/I) ≤ \frac{1}{3!}M_{1}M_{2}M_{3}.

□

This process is more complicated in more variables, but it is still possible to check only a single Betti diagram. We begin with a proposition that gives lower bounds for the maximal shifts in the resolution of a Cohen-Macaulay module.

Proposition 6.15 Suppose I is a homogeneous ideal in R = k[x_{1}, …, x_{n}] of codimension c such that R/I is Cohen-Macaulay. Then in the resolution of R/I, M_{i} ≥ M_{i−1} + 1.

Proof: Suppose R/I has regularity d − c. Since R/I is Cohen-Macaulay, the dual of the resolution of R/I is the minimal graded free resolution of the module N = Ext_{R}^{c}(R/I, R)(−d). (The shift of −d comes from the fact that the highest degree generator of a free module in the minimal free resolution of R/I has degree d.) The Betti diagram of M is obtained by rotating the Betti diagram of R/I 180 degrees and shifting the degree.

From this, it is easy to see that −m_{c−i}^{N} = M_{i}^{R/I}. Since N is a graded module, m_{c−i}^{N} + 1 ≤ m_{c−i+1}^{N}. Therefore

−M_{i}^{R/I} + 1 ≤ −M_{i−1}^{R/I}, and M_{i}^{R/I} ≥ M_{i−1}^{R/I} + 1.
□

Here is a brief example to illustrate Proposition 6.15.

Example 6.16 Consider the following resolution:

total: 1 4 4 1
0: 1 …
1: . 2 1 .
2: . 1 2 1
3: . 1 1 .

This cannot be the resolution of a Cohen-Macaulay module because there is a syzygy of degree five in column two but no syzygy of degree at least six in the third column. (This is the resolution of k[a, b, c]/(a^{2}, ab, ac^{2}, b^{4}), a quotient by a stable ideal, but the module is not Cohen-Macaulay.)

We show now that for each Hilbert function, there exists a Betti diagram D corresponding to that Hilbert function such that if D satisfies the upper bound of Conjecture 6.4, then all modules with the same Hilbert function do as well.

Theorem 6.17 Let H be a Hilbert function for an Artinian module of the form R/I. Let L be the lexicographic ideal such that H(R/L) = H, and let D_{L} be the Betti diagram of R/L. Then there exists a Betti diagram D corresponding to the Hilbert function H and below D_{L} in the partial order such that the following holds: Let J be any ideal such that H(R/J) = H. Then the product of the maximal shifts M_{1} ⋯ M_{n} of D is at most M_{1}^{R/J} ⋯ M_{n}^{R/J}.

In other words, for each Artinian Hilbert function, there is a Betti diagram below that of the lexicographic ideal with that Hilbert function that yields the minimum product of maximal shifts.

Proof: Our algorithm for producing D is the following: We lower M_{1} as much as possible by making all potentially possible cancellations between the generator and first syzygy column in the Betti diagram of the lexicographic ideal corresponding to H. Then we try to lower M_{2} by cancelling first syzygies with second syzygies, and so on, proceeding left to right across the Betti diagram. Note that M_{n} is fixed by the Hilbert function. Also, there may be many such D with the same Hilbert function that have minimal M_{i}.

The question is whether this gives us the lowest possible value of M_{1}M_{2} ⋯ M_{n}. Suppose we have choice of cancellation. Consider the diagram below of a lexicographic ideal in columns i−1 through i+1. Suppose that the entries of a, b, c ≠ 0 in degree d; the entries e, f are nonzero by Proposition 6.15. The asterisks represent possibly nonzero entries.

\begin{matrix}column #: & 0 & … & i − 1 & i & i + 1 & … \\ d − i − 1: & . & … & * & * & c & … \\ d − i: & . & … & * & b & * & … \\ d − i + 1: & . & … & a & e & f & … \end{matrix}

We have a choice: We could cancel the a and b in columns i−1 and i, attempting to lower M_{i−1} and/or M_{i}, or we could cancel the b and c in columns i and i + 1, attempting to lower M_{i} and/or M_{i+1}. We claim that we must eventually make the cancellations in columns i − 1 and i to have any hope of lowering M_{i} below d + 1.

If we do not cancel the entry a down to zero, we have M_{i−1} ≥ d, and then Proposition 6.15 implies that M_{i} ≥ d + 1 and M_{i+1} ≥ d + 2. Therefore cancelling the entries b and c will not change M_{i} and M_{i+1}.

Hence making all possible cancellations in the first two columns, then the second and third columns, etc., will lead to the minimum value of M_{1}M_{2} ⋯ M_{n}. □

When we used our algorithm in four variables, we found examples of potential Betti diagrams that do not satisfy the upper bound of Conjecture 6.4, and we were not able to show that all cannot occur. The Betti diagrams get much more complicated as one adds variables, and the dearth of theorems like Theorem 1.18 precludes further progress with this technique right now. The main problem is that one cannot use this approach to get results in general without being able to tell what resolutions actually occur at the bottom of the partial order on the resolutions of modules with a given Hilbert function. This helps us to understand one reason Conjecture 6.4 has been so difficult; it is closely related to the problem of finding exactly what resolutions are the bottom of the posets, something we are a long way from being able to do.

6.4 Truncation and a reduction in the Cohen-Macaulay case

In this section, we discuss another way to investigate the upper bound of Conjecture 6.4. Throughout, I will be a homogeneous ideal in R = k[x_{1}, …, x_{n}] such that R/I is Cohen-Macaulay.

The technique we will use is to “truncate” I in an appropriate degree to get rid of parts of the resolution that are irrelevant to the conjecture. Let I_{≥d} be the ideal in R consisting of all elements of I of degree d or higher. Instead of working with R/I, we will work with modules of the form R/I_{≥d}.

Example 6.18 Let I = (a^{3}, b^{3}, c^{3}, ab, bc) ⊂ S = k[a, b, c]. Truncating in degree three, we have I_{≥3} = (a^{3}, b^{3}, c^{3}, a^{2}b, abc, ab^{2}, b^{2}c, bc^{2}). Note that I contains two monomials of degree two that I_{≥3} does not, and therefore, since both ideals are Artinian, we have e(S/I) + 2 = e(S/I_{≥3}) = 13.

In Example 6.18, the multiplicity increased when we truncated I. The next lemma shows how the multiplicity changes in general after truncation.

Lemma 6.19 Let I be a homogeneous ideal in R = k[x_{1}, …, x_{n}]. Then e(R/I) ≤ e(R/I_{≥d}) for all d.

Proof: Fix d. Suppose first that R/I is not Artinian. Then R/I has a nonzero Hilbert polynomial. Since truncation only affects the dimension in finitely many degrees (that is, only finitely many fewer monomials are in in_{>}(I_{≥d}) than are in in_{>}(I)), R/I and R/I_{≥d} have the same Hilbert polynomial. Thus they have the same multiplicity.

If R/I is Artinian, then e(R/I) = dim_{k}(R/I), and e(R/I_{≥d}) = dim_{k}(R/I_{≥d}). Since dim_{k}(R/I)_{t} ≤ dim_{k}(R/I_{≥d})_{t} for all t, e(R/I) ≤ e(R/I_{≥d}). □

Thus the multiplicity will always increase or stay the same after truncating. We are also interested in how the graded Betti numbers of R/I are related to those of R/I_{≥d}. We find this relation in the following lemma.

Lemma 6.20 Let I be a homogeneous ideal in R = k[x_{1}, …, x_{n}], and let d be a positive integer. Then for each integer l ≥ 0,

β_{i,i+d+l}^{R/I} = β_{i,i+d+l}^{R/I_{≥d}}.

That is, rows d and higher of the Betti diagrams of R/I and R/I_{≥d} are the same.

Moreover, if I has its highest degree minimal generator in degree ≥ d, and R/I is Cohen-Macaulay, then M_{i}^{R/I} = M_{i}^{R/I_{≥d}}.

Proof: We have the short exact sequence

0 ⟶ I/I_{≥d} ⟶ R/I_{≥d} ⟶ R/I ⟶ 0.

This induces a long exact sequence in Tor: For all l ≥ 0,

⋯ ⟶ Tor_{i}(I/I_{≥d}, k)_{i+d+l} ⟶ Tor_{i}(R/I_{≥d}, k)_{i+d+l} ⟶ Tor_{i}(R/I, k)_{i+d+l} ⟶

Tor_{i−1}(I/I_{≥d}, k)_{i+d+l} ⟶ ⋯

is an exact sequence of k-vector spaces. Moreover, I/I_{≥d} has finite length; it is zero in degree d and higher and has highest degree socle generator in degree d − 1. Therefore I/I_{≥d} has regularity d − 1, meaning β_{i,i+d+r}^{I/I_{≥d}} = 0 for all i and all r ≥ 0.

For l ≥ 0, because d + l > d − 1,

dim_{k} Tor_{i}(I/I_{≥d}, k)_{i+d+l} = β_{i,i+d+l}^{I/I_{≥d}} = 0.

Similarly,

dim_{k} Tor_{i−1}(I/I_{≥d}, k)_{i+d+l} = β_{i−1,i+d+l}^{I/I_{≥d}} = 0.

Consequently, as k-vector spaces,

Tor_{i}(R/I_{≥d}, k)_{i+d+l} ≅ Tor_{i}(R/I, k)_{i+d+l}.

Hence their dimensions over k are equal, and thus for all l ≥ 0,

β_{i,i+d+l}^{R/I} = β_{i,i+d+l}^{R/I_{≥d}}.

The final statement of the lemma follows from Proposition 6.15. □ Example 6.21 We illustrate Lemma 6.20 in an example. Let S = k[a, b, c], and let I = (a^{2}, b^{4}, c^{5}, ac, bc, ab^{2}) ⊂ S. The Betti diagrams of S/I and S/I_{≥3} are below.

S/I: total: 1 6 8 3  S/I_{≥3}: total: 1 10 15 6
0: 1 …  0: 1 …
1: . 3 2 .  1: ….
2: . 1 2 1  2: . 8 11 4
3: . 1 2 1  3: . 1 2 1
4: . 1 2 1  4: . 1 2 1

We are truncating in degree three, and the Betti diagrams in rows three and below are identical, as we would expect from Lemma 6.20. Apart from the one in the (0,0) place, there are only zeros in the Betti diagram of S/I_{≥3} before row two since we have no generators until degree three. The Betti diagram of S/I_{≥3} has more generators (and syzygies) in row two than are in the Betti diagram of S/I because all the monomials of degree three in I are minimal generators of I_{≥3}.

We use Lemmas 6.19 and 6.20 to reduce the upper bound portion of Conjecture 6.4 to the case of ideals whose minimal generators are all in a single degree.

Proposition 6.22 Let I be an Artinian homogeneous ideal in R = k[x_{1}, …, x_{n}]. Let d be the highest degree in which I has a minimal generator. If R/I_{≥d} satisfies the upper bound of Conjecture 6.4, then so does R/I.

Proof: By Lemma 6.19, e(R/I) ≤ e(R/I_{≥d}), and we are assuming that R/I_{≥d} satisfies the upper bound of Conjecture 6.4, so

e(R/I) ≤ e(R/I_{≥d}) ≤ \frac{1}{n!} ∏_{i=1}^{n} M_{i}^{R/I_{≥d}}.

By Lemma 6.20, M_{i}^{R/I} = M_{i}^{R/I_{≥d}} for each i, and thus R/I satisfies the upper bound of Conjecture 6.4. □

As a consequence, we can reduce the upper bound question for Cohen-Macaulay ideals to a simpler case.

Corollary 6.23 If the upper bound of Conjecture 6.4 holds for all Artinian ideals in R = k[x_{1}, …, x_{n}] whose minimal generators are all in a single degree, then the upper bound of Conjecture 6.4 holds for all homogeneous ideals I ⊂ R such that R/I is Cohen-Macaulay.

Proof: We may immediately reduce to the Artinian case using Lemma 1.20. The reduction to ideals with minimal generators all in a single degree follows from Proposition 6.22: Given a homogeneous ideal I with its highest degree generator in degree d, one can work with R/I_{≥d} instead of R/I. □

We use Corollary 6.23 in conjunction with a result of Herzog and Srinivasan in [28]. Let m_{i} and M_{i} be the minimal and maximal shifts respectively at step i of the graded free resolution of a Cohen-Macaulay module R/I. We say that R/I has a quasipure resolution if M_{i−1} ≤ m_{i} for all i ≥ 1. This means that the maximal shift at step i − 1 is bounded above by the minimal shift at step i. For example, a Betti diagram corresponding to a quasipure resolution might look like this:

total: 1 11 20 14 4
0: 1 ….
1: . 4 6 4 .
2: . 7 14 6 .
3: …. 2
4: … 3 1

Herzog and Srinivasan prove the following theorem in [28].

Theorem 6.24 Let I be a homogeneous ideal of codimension c in R such that R/I is Cohen-Macaulay. If R/I has a quasipure resolution, then R/I satisfies both bounds of Conjecture 6.4.

Their proof is numerical: Any potential quasipure resolution below that of a lexicographic ideal satisfies the bounds; there is no need for there to exist a module with that resolution. We exploit this result by combining it with Proposition 6.15 and Lemmas 6.19 and 6.20.

Theorem 6.25 Let I be a homogeneous ideal of codimension c in R such that R/I is Cohen-Macaulay of regularity d. Suppose I contains a minimal generator of degree d or d + 1. Then R/I satisfies the upper bound of Conjecture 6.4; that is,

e(R/I) ≤ \frac{1}{c!} ∏_{i=1}^{c} M_{i}.

Proof: We may again assume that I is Artinian and that c = n. Note that if R/I has regularity d, then degree d + 1 is the highest degree in which I can have a minimal generator. By Proposition 6.15, the resolution of R/I_{≥d} is concentrated in rows d − 1 and d of the Betti diagram, the bottom two rows. Therefore R/I_{≥d} has a quasipure resolution, and it satisfies the bounds of Conjecture 6.4. We use this to show that R/I satisfies the upper bound of Conjecture 6.4. By Lemmas 6.19 and 6.20, we have

e(R/I) ≤ e(R/I_{≥d}) ≤ ∏_{i=1}^{c} M_{i}^{R/I_{≥d}} = ∏_{i=1}^{c} M_{i}^{R/I}.

□

This result gives, for example, an easy proof of the upper bound for stable Cohen-Macaulay ideals; Herzog and Srinivasan prove both the upper and lower bounds for stable ideals without the Cohen-Macaulay hypothesis in [28]. The examples in Section 3 of potential Betti diagrams with three nonzero rows (instead of just two) that do not satisfy the upper bound of Conjecture 6.4 show that no further reduction like the one in Theorem 6.25 is possible.

We illustrate Theorem 6.25 with an example. Example 6.26 Let I = (a^{3}, b^{4}, c^{4}, ab^{2}, a^{2}bc^{3}) ⊂ S = k[a, b, c]. Then S/I has regularity six, and I has a minimal generator in degree 6. We resolve S/I and S/I_{≥6} below.

S/I: total: 1 5 8 4  S/I_{≥6}: total: 1 27 46 20
0: 1 …  0: 1 …
1: ….  1: ….
2: . 2 . .  2: ….
3: . 2 2 .  3: ….
4: ….  4: ….
5: . 1 5 3  5: . 27 45 19
6: . . 1 1  6: . . 1 1

Note that S/I does not have a quasipure resolution, so Theorem 6.24 does not apply. However, the truncation S/I_{≥6} does have a quasipure resolution, and S/I and S/I_{≥6} have the same maximum shifts at each step in the resolution. Also, e(S/I) = 31 ≤ 57 = e(S/I_{≥6}), and thus S/I satisfies the upper bound of Conjecture 6.4 because S/I_{≥6} does.

It would be interesting to have structure theorems for Artinian ideals whose minimal generators are all in one degree, the situation in Corollary 6.23. For example, can we say anything about their Hilbert functions, perhaps giving upper bounds in the spirit of the Hilbert function results in Chapter 4? Even more ambitiously, what can we say about the graded free resolutions of such ideals? These questions are difficult, but perhaps it would be possible to make some progress in the codimension three case. The upper bound of Conjecture 6.4 is wide open even in the Cohen-Macaulay codimension three case, so such results would represent substantial progress.

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