Breaking up simplicial homology and subadditivity of syzygies

Abstract

We consider the following question: If a simplicial complex $\Gamma$ has d-homology, then does the corresponding d-cycle always induce cycles of smaller dimension that are not boundaries? We provide an answer to this question in a fixed dimension. We use the breaking of homology to show the subadditivity property for the maximal degrees of syzygies of monomial ideals in a fixed homological degree.

1 Introduction

The motivation for this paper is the subadditivity property for the maximal degrees of syzygies of monomial ideals in polynomial rings. Let I be a homogeneous ideal in the polynomials ring $S=k[x_1,\ldots ,x_n]$ over a field k. Let $t_a$ denoted the maximum value of j such that the graded Betti number $\beta _{a,j}(S/I) \ne 0$. The ideal I satisfies the subadditivity property on the maximal degrees of its syzygies if

$$\begin{aligned} t_{a+b} \leqslant t_a + t_b \end{aligned}$$

(1)

where $a+b$ is not more than the projective dimension of the ideal.

The inequality in (1) arises most naturally in the context of (Castelnuovo–Mumford) regularity, which, for the ideal I, can be described as the maximum value of $t_a-a$, for all positive integers a. It has been shown to fail in general by Avramov, Conca and Iyengar [2], even if one restricts to Cohen–Macaulay or even Gorenstein settings (see [22] for examples and for a general survey on the topic). However, many special cases are known: certain algebras with codimension $\leqslant 1$ (Eisenbud, Huneke and Ulrich [11]), certain classes of Koszul rings (Avramov, Conca and Iyengar [2]), certain homological degrees for Gorenstein algebras (El Khoury and Srinivasan [12]), among others.

Avramov, Conca and Iyengar [2] conjectured that the subadditivity property holds for Kozul rings and for all monomial ideals (it is also open for toric ideals [22]). In the case of monomial ideals, there are special cases for which (1) has been verified: when $a=1$ (Herzog and Srinivasan [19]), when $a=1,2,3$ and I is generated in degree 2 (Fernández-Ramos and Gimenez [15], Abedelfatah and Nevo [3]), Cohen—Macaulay ideals generated by monomials of degree 2 when the base field has characteristic 0 [2], facet ideals of simplicial forests (Faridi [13]), ideals whose Betti diagram has a special “shape” (Bigdeli and Herzog [5]), several classes of edge ideals of graphs and path ideals of rooted trees (Jayanthan and Kumar [20]), and for a where the Stanley–Reisner complex of I has dimension bounded by $t_a-a$ (Abedelfatah [1]).

In the case of monomial ideals, the syzygies can be characterized as dimensions of homology modules of topological objects. This is one of the central themes of Stanley–Reisner Theory, connecting Commutative Algebra to Discrete Geometry and Topology. We refer the reader to the books [6, 24] for more details on these rich connections.

By viewing the subadditivity property as a geometric one, the inequality in (1) can be shown to follow from the following general type of question:

Does a topological object with d-homology break into sub-objects that have a-homology and b-homology, where a and b are related to d?

This approach was taken by the first author in [13], where the topological objects were atomic lattices (${{\mathrm{lcm}}}$ lattices of monomial ideals); see Question 2.1 and Question 2.2 below. In this paper, using Hochster’s formula (Equation (2)), we examine this problem from the point of view of the Stanley–Reisner complex, and we can provide a positive answer to the general question above for a fixed value of d. As a result, we show that subadditivity holds in a fixed homological degree for all monomial ideals. The last section interprets the square-free results of the paper for general monomial ideals.

2 Setup

2.1 The subadditivity property

Throughout the paper, let $S=k[x_1,\ldots ,x_n]$ be a polynomial ring over a field k. If I is a graded ideal of S with minimal free resolution

$$\begin{aligned} 0 \rightarrow \oplus _{j \in {\mathbb {N}}} S(-j)^{\beta _{p,j}}\rightarrow \oplus _{j \in {\mathbb {N}}} S(-j)^{\beta _{p-1,j}} \rightarrow \cdots \rightarrow \oplus _{j \in {\mathbb {N}}} S(-j)^{\beta _{1,j}} \rightarrow S, \end{aligned}$$

then for each i and j, the rank $\beta _{i,j}(S/I)$ of the free S-modules appearing above are called the graded Betti numbers of the S-module S/I.

If we let

$$\begin{aligned} t_a=\max \{j \ : \ \beta _{a,j}(S/I) \ne 0\}, \end{aligned}$$

a question is whether the $t_a$ satisfy the subadditivity property: $t_{a+b} \leqslant t_a + t_b$?

The answer is known to be negative for a general homogeneous ideal [2], and unknown in the case of monomial ideals. For the case of monomial ideals, there are special cases that are known [1, 3, 5, 13, 15, 19].

In the case of monomial ideals, Betti numbers can be interpreted as the homology of objects in discrete topology: simplicial complexes, order complexes of lattices, etc.; see, for example, [24] for a survey of this approach. As a result, the subadditivity question can be viewed as a question of breaking up homology in these objects. This idea was explored in [13] by the first author, where the subadditivity problem was solved for facet ideals of simplicial forests using homology of lattices.

By a method called polarization [16] (see Section. 5 for the definition), one can reduce questions regarding Betti numbers of monomial ideals to the class of square-free monomial ideals.

If $u \subset [n]=\{ 1,\ldots ,n\}$, then we define

$$\begin{aligned} {\mathbf {m}}_u= \Pi _{i \in u} x_i \end{aligned}$$

to be the square-free monomial with support u.

For our purposes it is useful to consider a finer grading of the Betti numbers by indexing the Betti numbers with monomials of the polynomial ring S. A multigraded Betti number of S/I is of the form $\beta _{i,{\mathbf {m}}}(S/I)$ where ${\mathbf {m}}$ is a monomial in S and

$$\begin{aligned} \beta _{i,j}(S/I)= \sum _{u \subseteq [n] \text{ and } |u|=j} \beta _{i,{\mathbf {m}}_u}(S/I). \end{aligned}$$

2.2 Simplicial complexes

A simplicial complex $\Gamma$ on a set W is a set of subsets of W with the property that if $F\in \Gamma$ then for every subset $G \subseteq F$ we have $G \in \Gamma$. Every element of $\Gamma$ is called a face, the maximal faces under inclusion are called facets, and a simplicial complex contained in $\Gamma$ is called a subcomplex of $\Gamma$. The set of all $v \in W$ such that $\{v\} \in \Gamma$ is called the vertex set of $\Gamma$, and is denoted by $V(\Gamma )$. The set of facets of $\Gamma$ is denoted by $\text{ Facets }(\Gamma )$. If $\text{ Facets }(\Gamma )=\{F_1,\ldots ,F_q\}$, then we denote $\Gamma$ by

$$\begin{aligned} \Gamma =\langle F_1,\ldots ,F_q \rangle . \end{aligned}$$

If $A \subset V(\Gamma )$, then the induced subcomplex $\Gamma _A$ is defined as

$$\begin{aligned} \Gamma _A=\{F \in \Gamma \ : \ F \subseteq A\}. \end{aligned}$$

The Alexander dual $\Gamma ^{\vee }$ of $\Gamma$, if we set $F^c=V(\Gamma ) \setminus F$, is defined as

$$\begin{aligned} \Gamma ^{\vee }=\{F \subseteq V (\Gamma ) \ : \ F^c \notin \Gamma \} =\{V (\Gamma ) \setminus F \ : \ F \notin \Gamma \}. \end{aligned}$$

The link of a face F of $\Gamma$ is

$$\begin{aligned} {\mathrm{lk}}\ _\Gamma (F)=\{G \in \Gamma \ : \ F \cap G = \emptyset \text{ and } F \cup G \in \Gamma \}. \end{aligned}$$

If I is a square-free monomial ideal in S, it corresponds uniquely to a simplicial complex

$$\begin{aligned} {\mathcal {N}}(I)=\{u \subset [n] \ : \ m_u \notin I\} \end{aligned}$$

called the Stanley–Reisner complex of I. Conversely, if $\Gamma$ is a simplicial complex whose vertices are labelled with $x_{1},\ldots ,x_{n}$, then one can associate to it its unique Stanley–Reisner ideal

$$\begin{aligned} {\mathcal {N}}(\Gamma )=\{m_u \ : \ u \subset [n] \text{ and } u \notin \Gamma \}. \end{aligned}$$

The uniqueness of the Stanley–Reisner correspondence implies that

$$\begin{aligned} {\mathcal {N}}(\Gamma )=I \iff {\mathcal {N}}(I)=\Gamma . \end{aligned}$$

2.3 The LCM lattice

A lattice is a partially ordered set where every two elements have a greatest lower bound called their meet and a lowest upper bound called their join. A bounded lattice has an upper and a lower bound denoted by ${\hat{1}}$ and ${\hat{0}}$, respectively.

If L is a lattice with r elements, then the order complex of L is the simplicial complex on r vertices, where the elements of each chain in L form a face.

If I is a monomial ideal, then the lcm lattice of I, denoted by $\text{ LCM }(I)$, is a bounded lattice ordered by divisibility, whose elements are the generators of I and their least common multiples, and the meet of two elements is their least common multiple.

Two elements of a lattice are called complements if their join is ${\hat{1}}$ and their meet is ${\hat{0}}$. If the lattice is $\text{ LCM }(I)$, then it was shown in [13] that two monomials in $\text{ LCM }(I)$ are complements if their gcd is not in I and their lcm is the lcm of all the generators of I.

Gasharov, Peeva and Welker [17, 24] showed that multigraded Betti numbers of S/I can be calculated from the homology of (the order complex of) the lattice $\text{ LCM }(I)$: If ${\mathbf {m}}$ is a monomial in $L=\text{ LCM }(I)$, then

$$\begin{aligned} \begin{aligned} \beta _{i,{\mathbf {m}}}(S/I)=\dim _k {\tilde{H}}_{i-2} \left( (1, {\mathbf {m}})_{L}; k \right) \end{aligned} \end{aligned}$$

where $(1, {\mathbf {m}})_{L}$ refers to the subcomplex of the order complex consisting of all nontrivial monomials in L strictly dividing ${\mathbf {m}}$.

On the other hand, in a 1977 paper, Baclawski [4] showed that if L is a finite lattice whose proper part has nonzero homology, then every element of L has a complement.

The following question was raised in [13] as a potential way to answer the subadditivity question.

Question 2.1

If I is a square-free monomial ideal in variables $x_{1},\ldots ,x_{n}$, and $\beta _{i,n} (S/I)\ne 0$, $a,b >0$ and $i=a+b$, are there complements ${\mathbf {m}}$ and ${\mathbf {m}}'$ in $\text{ LCM }(I)$ with $\beta _{a,{\mathbf {m}}}(S/I) \ne 0$ and $\beta _{b,{\mathbf {m}}'}(S/I) \ne 0$?

Considering that it is enough to study the “top degree” Betti numbers (those of degree n, in this case) [9, 13], a positive answer to Question 2.1 will establish the subadditivity property for all monomial ideals, since

$$\begin{aligned} t_a+t_b \geqslant \deg ({\mathbf {m}})+ \deg ({\mathbf {m}}') \geqslant n=t_i. \end{aligned}$$

Question 2.1 can be written more generally as a question about the homology of the ${{\mathrm{lcm}}}$ lattice, or in fact, any finite lattice.

Question 2.2

If $L=\text{ LCM }(I)$ and ${\tilde{H}}_{i-2} \left( (1, x_1 \cdots x_n)_{L}; k \right) \ne 0$, $a,b>0$ and $i=a+b$, are there complements ${\mathbf {m}}$ and ${\mathbf {m}}'$ in $\text{ LCM }(I)$ with ${\tilde{H}}_{a-2} \left( (1,{\mathbf {m}})_{L}; k \right) \ne 0$ and ${\tilde{H}}_{b-2} \left( (1,{\mathbf {m}}')_{L}; k \right) \ne 0$?

With the same idea, one could translate Question 2.1 into breaking up simplicial homology using Hochster’s formula.

2.3.1 Hochster’s formula

Let $I=({\mathbf {m}}_1,\ldots , {\mathbf {m}}_q)$ be a square-free monomial ideal in the polynomial ring $S=k[x_{1},\ldots ,x_{n}]$. Hochster’s formula (see for example [18, Cor. 8.1.4 and Prop. 5.1.8]) states that if $I={\mathcal {N}}(\Gamma )$ and ${\mathbf {m}}_u$ a monomial, then

$$\begin{aligned} \beta _{i,{\mathbf {m}}_u}(S/I)=\dim _k \tilde{H}_{i-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}(u^c),k)=\dim _k\tilde{H}_{|u|-i-1}(\Gamma _u,k) \end{aligned}$$

(2)

where $u^c=[n]\setminus u$ is the set complement of u. We would now like to reinterpret Question 2.1 in the language of Hochster’s formula. To begin with, since we are dealing with square-free monomials, we can consider a monomial ${\mathbf {m}}_u$ equivalent to the set u and use intersections for $\gcd$, unions for ${{\mathrm{lcm}}}$, and ${\mathbf {m}}_u^c$ for $u^c$.

Suppose

$$\begin{aligned} \beta _{i,x_1\ldots x_n}(S/I)=\dim _k \tilde{H}_{i-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}(\emptyset ),k)=\dim _k \tilde{H}_{i-2}(\Gamma ^{\vee },k) \ne 0 \end{aligned}$$

and $i=a+b$ where $a,b >0$. We would like to know if there are complements ${\mathbf {m}}, {\mathbf {m}}' \in \text{ LCM }(I)$ such that

$$\begin{aligned} \beta _{a, {\mathbf {m}}}(S/I)\ne 0 \text{ and } \beta _{b, {\mathbf {m}}'}(S/I)\ne 0. \end{aligned}$$

First observe that, $\Gamma ^{\vee }=\langle {\mathbf {m}}_1^c,\ldots ,{\mathbf {m}}_q^c \rangle$ (e.g. [18] or [14, Prop. 2.4]).

We have

$$\begin{aligned}&{\mathbf {m}}\in \text{ LCM }(I) \\&\quad \iff {\mathbf {m}}={\mathbf {m}}_{i_1}\cup {\mathbf {m}}_{i_2}\cup \cdots \cup {\mathbf {m}}_{i_s} \text{ for } \text{ some } 1 \leqslant i_1< i_2< \cdots< i_s \leqslant q \\&\quad \iff {\mathbf {m}}^c = {\mathbf {m}}_{i_1}^c\cap {\mathbf {m}}_{i_2}^c\cap \cdots \cap {\mathbf {m}}_{i_s}^c \text{ for } \text{ some } 1 \leqslant i_1< i_2< \cdots < i_s \leqslant q \\&\quad \iff {\mathbf {m}}^c \text{ is } \text{ the } \text{ intersection } \text{ of } \text{ some } \text{ facets } \text{ of } \Gamma ^{\vee }. \end{aligned}$$

Moreover, if ${\mathbf {m}}, {\mathbf {m}}' \in \text{ LCM }(I)$, then

$$\begin{aligned} {\mathbf {m}}, {\mathbf {m}}' \text{ are } \text{ complements } \iff&{\mathbf {m}}\cup {\mathbf {m}}' =[n] \text{ and } {\mathbf {m}}\cap {\mathbf {m}}' \notin I\\ \iff&{\mathbf {m}}^c \cap {\mathbf {m}}'^c = \emptyset \text{ and } {\mathbf {m}}\cap {\mathbf {m}}' \in \Gamma \\ \iff&{\mathbf {m}}^c \cap {\mathbf {m}}'^c = \emptyset \text{ and } ({\mathbf {m}}\cap {\mathbf {m}}')^c \notin \Gamma ^{\vee }\\ \iff&{\mathbf {m}}^c \cap {\mathbf {m}}'^c = \emptyset \text{ and } {\mathbf {m}}^c \cup {\mathbf {m}}'^c \notin \Gamma ^{\vee }.\\ \end{aligned}$$

So we are looking for subsets $A,B \subseteq [q]$ such that

• 1. ${\mathbf {m}}^c=\bigcap _{j \in A} {\mathbf {m}}_j^c$ and ${\mathbf {m}}'^c=\bigcap _{j \in B} {\mathbf {m}}_j^c$

• 2. ${\mathbf {m}}^c \cap {\mathbf {m}}'^c = \emptyset$

• 3. ${\mathbf {m}}^c \cup {\mathbf {m}}'^c \notin \Gamma ^{\vee }$

• 4. $\tilde{H}_{a-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}({\mathbf {m}}^c),k) \ne 0$ and $\tilde{H}_{b-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}({\mathbf {m}}'^c),k) \ne 0$.

Now we can state Question 2.1 in the following form.

Question 2.3

If $\Gamma =\langle F_1,\ldots ,F_q \rangle$ is a simplicial complex with $\tilde{H}_{i-2}(\Gamma ,k) \ne 0$ and $i=a+b$ where $a,b >0$, can we find subsets $A,B \subseteq [q]$ such that

• 1. $F=\bigcap _{j \in A} F_j$ and $G=\bigcap _{j \in B} F_j$

• 2. $F \cap G = \emptyset$

• 3. $F \cup G \notin \Gamma$

• 4. $\tilde{H}_{a-2}({\mathrm{lk}}\ _{\Gamma }(F),k) \ne 0$ and $\tilde{H}_{b-2}({\mathrm{lk}}\ _{\Gamma }(G),k) \ne 0$?

Example 2.4

If ${\mathcal {N}}(I)^{\vee }=\Gamma =\langle xzu,xzv,xuv,yzu,yzv,yuv,xy \rangle$,

then $I=(xz,yz,xu,yu,xv,yv,zuv)$ has Betti table

$$\begin{aligned} \begin{array}{rlllll} &{}0 &{}1 &{}2 &{}3 &{}4\\ \text{ total }:&{}1 &{}7 &{}11 &{}6 &{}1\\ 0:&{}1 &{}. &{}. &{}. &{}.\\ 1:&{}. &{}6 &{}9 &{}5 &{}1 \\ 2:&{}. &{}1 &{}2 &{}1 &{}. \end{array} \end{aligned}$$

So $\beta _{i,xyzuv} \ne 0$ when $i=3,4$, which corresponds to nonvanishing of homology of links of faces of $\Gamma$ in dimensions 1, 2. We consider each case separately:

• 1. $i=3$, $a=1$, $b=2$. Then $\tilde{H}_{1}(\Gamma ,k)\ne 0$. Let $F=xy$ and $G=xuv \cap yuv =uv$, then $F\cap G = \emptyset$, $F \cup G=xyuv \notin \Gamma$, and

  $$\begin{aligned}&\tilde{H}_{a-2}({\mathrm {lk}}\ _{\Gamma }(F),k)=\tilde{H}_{-1}(\langle \emptyset \rangle ,k) \ne 0 \\&\text { and }\; \tilde{H}_{b-2}({\mathrm {lk}}\ _{\Gamma }(G),k)=\tilde{H}_{0}(\langle x,y \rangle ,k) \ne 0. \end{aligned}$$

• 2. $i=4$, $a=1$, $b=3$. Then $\tilde{H}_{2}(\Gamma ,k)\ne 0$. Let $F=yzu$ and $G=xzu \cap xuv \cap xzv \cap xy =x$, then $F\cap G = \emptyset$, $F \cup G=xyzu \notin \Gamma$, and

  $$\begin{aligned}&\tilde{H}_{a-2}({\mathrm {lk}}\ _{\Gamma }(F),k) =\tilde{H}_{-1}(\langle \emptyset \rangle ,k) \ne 0 \\&\text { and } \; \tilde{H}_{b-2}({\mathrm {lk}}\ _{\Gamma }(G),k)=\tilde{H}_{1}(\langle zu,uv,zv,y \rangle ,k) \ne 0. \end{aligned}$$

• 3. $i=4$, $a=2$, $b=2$. Then $\tilde{H}_{2}(\Gamma ,k)\ne 0$. Let $F=yzu \cap yuv=yu$ and $G=xzu \cap xzv =xz$, then $F\cap G = \emptyset$, $F \cup G=xyzu \notin \Gamma$, and

  $$\begin{aligned}&\tilde{H}_{a-2}({\mathrm {lk}}\ _{\Gamma }(F),k)=\tilde{H}_{0}(\langle z,v \rangle ,k) \ne 0 \\&\text { and } \;\tilde{H}_{b-2}({\mathrm {lk}}\ _{\Gamma }(G),k)=\tilde{H}_{0}(\langle u,v \rangle ,k) \ne 0. \end{aligned}$$

A dual version of Question 2.3 can be stated as follows (see Corollary 3.6 for the justification).

Question 2.5

If $\Gamma$ is a simplicial complex on the vertex set $\{x_{1},\ldots ,x_{n}\}$, and $\tilde{H}_{i-2}(\Gamma ,k) \ne 0$, and $n-i+1=a+b$, where a and b are positive integers, are there nonempty subsets $C, D \subseteq \{x_{1},\ldots ,x_{n}\}$ such that

• 1. $C\cup D =\{x_{1},\ldots ,x_{n}\}$

• 2. $C\cap D \in \Gamma$

• 3. $\tilde{H}_{|C|-a-1}(\Gamma _C,k) \ne 0$ and $\tilde{H}_{|D|-b-1}(\Gamma _D,k) \ne 0$?

Example 2.6

Let ${\mathcal {N}}(I)=\Gamma =\langle zwx,vwx,uvx,zux,zuy,uvy,vwy,zwy\rangle$.

Then $I=(xy,zv,uw)$ has Betti table

$$\begin{aligned} \begin{array}{rllll} &{}0 &{}1 &{}2 &{}3 \\ \text{ total }:&{}1 &{}3 &{}3 &{}1\\ 0:&{}1 &{}. &{}. &{}. \\ 1:&{}. &{}3 &{}. &{}. \\ 2:&{}. &{}. &{}3 &{}.\\ 3:&{}. &{}. &{}. &{}1 \end{array} \end{aligned}$$

So $\beta _{3,xyzuvw}(S/I)\ne 0$ which corresponds to nonvanishing homology of $\Gamma$ in dimension 2 (i.e., $\tilde{H}_{2}(\Gamma ,k)\ne 0$). Let $a=1$ and $b=2$. Choose $C=\{x,y\}$ and $D=\{z,u,v,w\}$. Then $C\cup D=\{x,y,z,u,v,w\}$, $C\cap D=\emptyset \in \Gamma$ and

$\tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\langle x,y \rangle ,k) \ne 0$ and $\tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{1}(\langle zu,uv,vw,zw \rangle ,k)\ne 0$.

A positive answer to either Questions 2.3 or 2.5 would settle the subadditivity question for syzygies.

3 Main results

The following lemma is an easy exercise.

Lemma 3.1

$\Gamma$ simplicial complex and $A \in \Gamma$ and $B\in {\mathrm{lk}}\ _\Gamma (A)$, then

$$\begin{aligned} {\mathrm{lk}}\ _{{\mathrm{lk}}\ _\Gamma (A)}(B)={\mathrm{lk}}\ _\Gamma (A\cup B). \end{aligned}$$

In a simplicial complex $\Gamma$ we say a d-cycle $\Sigma$ is supported on faces $F_1,\ldots ,F_q$ if $\Sigma =a_1F_1+ \cdots + a_qF_q$ for nonzero scalars $a_1,\ldots ,a_q \in k$. We say that $\Sigma$ is a face-minimal cycle or minimally supported on $F_1,\ldots ,F_q$ if additionally no proper subset of $F_1,\ldots ,F_q$ is the support of a d-cycle. If $\Sigma$ is supported on $F_1\ldots ,F_q$, we call the simplicial complex $\langle F_1,\ldots ,F_q \rangle$ the support complex of $\Sigma$.

Example 3.3 can guide the reader through the statement of the theorem below, a variation of which appears as Theorem 4.2 of Theorem [25].

Theorem 3.2

Let k be a field, $\Gamma$ a d-dimensional simplicial complex, and

$$\begin{aligned} \Sigma =a_1F_1 + \cdots + a_qF_q, \; a_1, \ldots , a_q \in k \end{aligned}$$

a d-cycle in $\Gamma$ supported on $F_1,\ldots ,F_q$, which is not a boundary so that $\tilde{H}_{d}(\Gamma ,k) \ne 0$. Suppose A is a face of the support complex of $\Sigma$ such that for some $s \leqslant q$ we have

$$\begin{aligned} A \subseteq F_1\cap \ldots \cap F_s, \text{ and } A \not \subseteq F_j \text{ if } j>s \end{aligned}$$

and $0 \leqslant |A| \leqslant d+1$. Then

• 1. there are $\epsilon _i \in \{\pm 1\}$ for $i=1,\ldots ,s$ such that

  $$\begin{aligned} \Sigma _A=\epsilon _1a_1(F_1\setminus A)+\cdots + \epsilon _sa_s(F_s\setminus A) \end{aligned}$$

  is a $(d-|A|)$-cycle in ${\mathrm{lk}}\ _\Gamma (A)$ that is not a boundary in ${\mathrm{lk}}\ _\Gamma (A)$;

• 2. $\tilde{H}_{d-|A|}({\mathrm{lk}}\ _\Gamma (A),k)\ne 0$;

• 3. $A = F_1\cap \ldots \cap F_s$.

Proof

The case $|A|=d+1$ will result in ${\mathrm{lk}}\ _\Gamma (A)=\{\emptyset \}$ which has $(-1)$-homology. So we can assume that $0 \leqslant |A| \leqslant d$. To prove Statement 1, we will proceed using induction on $a=|A|$. If $a=0$, then ${\mathrm{lk}}\ _\Gamma (A)=\Gamma$, $\Sigma _A=\Sigma$ and there is nothing to prove.

Suppose $a>0$, $A=\{v_1,\ldots , v_a\}$, $A'=\{v_1,\ldots , v_{a-1}\}$ (or $A'=\emptyset$ when $a=1$) and $\Gamma '={\mathrm{lk}}\ _\Gamma (A')$, and suppose without loss of generality

$$\begin{aligned} A' \subseteq F_1\cap \ldots \cap F_t \text{ and } A' \not \subseteq F_j \text{ for } j>t\geqslant s. \end{aligned}$$

By the induction hypothesis, for some $\epsilon _i'\in \{\pm 1\}$ there is a $(d-(a-1))$-cycle

$$\begin{aligned} \Sigma _{A'}=a_1\epsilon _1'(F_1\setminus A')+\cdots + a_t\epsilon _t'(F_t\setminus A') \end{aligned}$$

in $\Gamma '$ that is not a boundary in $\Gamma '$ and $\tilde{H}_{d-(a-1)}(\Gamma ',k)\ne 0$. In particular, we must have $t\ne s$ as otherwise the support complex of $\Sigma _{A'}$ would be a cone with every facet containing $v_a$, a contradiction.

We know that $v_a \in (F_i\setminus A')$ if and only if $i\leqslant s$. Depending on the orientation of the faces of the complex $\Gamma '$, for some $\epsilon _{i}''\in \{\pm 1\}$, we can write

$$\begin{aligned} \begin{array}{rl} 0=&{}{}\partial (\Sigma _{A'})\\ &{}{}\\ =&{}{}\epsilon _1'a_1\partial (F_1\setminus A')+ \cdots + \epsilon _t'a_t\partial (F_t \setminus A')\\ &{}{}\\ =&{}{}\epsilon _1''\epsilon _1'a_1(F_1 \setminus A)+ \cdots + \epsilon _s''\epsilon _s'a_s(F_s \setminus A) + {\mathcal {U}} \\ &{}{}\\ &{}+ \partial (\epsilon _{s+1}'a_{s+1}F_{s+1} \setminus A'+ \cdots + \epsilon _t'a_tF_t \setminus A') \end{array} \end{aligned}$$

where ${\mathcal {U}}$ consists of all the summands above which contain the vertex $v_a$, and hence

$$\begin{aligned} {\mathcal {U}}=\sum _{j=1}^s \epsilon _j'a_j\left( \partial (F_j \setminus A') - \epsilon _j''F_j \setminus A \right) =0. \end{aligned}$$

If we set $\epsilon _i=\epsilon _i''\epsilon _i'$ and $\Sigma _A=\epsilon _1a_1(F_1 \setminus A) + \cdots + \epsilon _sa_s(F_s \setminus A)$ it follows that

$$\begin{aligned} \Sigma _A= -\partial (\epsilon _{s+1}'a_{s+1}(F_{s+1} \setminus A')+ \cdots + \epsilon _t'a_t(F_t \setminus A')) \end{aligned}$$

and

$$\begin{aligned} \partial (\Sigma _A)=- \partial ^2(\epsilon _{s+1}'a_{s+1}(F_{s+1} \setminus A')+ \cdots + \epsilon _t'a_t(F_t \setminus A'))=0. \end{aligned}$$

So $\Sigma _A$ is a $(d-a)$-cycle in ${\mathrm{lk}}\ _{\Gamma '}(v_a)={\mathrm{lk}}\ _\Gamma (A)$ by Lemma 3.1 (and since $v_a \in \Gamma '$). Since $\dim ({\mathrm{lk}}\ _\Gamma (A))=d-|A|$, the $(d-|A|)$-cycle $\Sigma _A$ is not a boundary in ${\mathrm{lk}}\ _\Gamma (A)$. Therefore, $\tilde{H}_{d-|A|}(\Gamma ,k)\ne 0$, proving Statement 2.

To see Statement 3, note that if $F_1\ldots ,F_s$ all contain a vertex outside A, then the support complex of $\Sigma _A$ would be a cone contradicting Statement 2. $\square$

Example 3.3

Let $\Gamma =\langle xy,zu,zv,uv \rangle$, which is the Alexander dual of the simplicial complex $\Gamma$ in Example 2.4.

As stated in Theorem 3.2, $\Gamma$ is a 1-dimensional simplicial complex and has $\Sigma =uz+zv+vu$ as a 1-cycle so that $\tilde{H}_{1}(\Gamma ,k) \ne 0$. Taking $A=\{z\}$, then $\Sigma _A=u-v$ is a 0-cycle in ${\mathrm{lk}}\ _{\Gamma }(A)=\langle u,v \rangle$ with $\tilde{H}_{0}({\mathrm{lk}}\ _{\Gamma }(A),k) \ne 0$.

Corollary 3.4

Let k be a field, $\Gamma$ a d-dimensional simplicial complex with $\tilde{H}_{d}(\Gamma ,k)\ne 0$, and let $\Sigma$ be a d-cycle in $\Gamma$ which is not a boundary. Let A be a face of the support complex of $\Sigma$, and suppose $F_1\ldots ,F_q$ are the facets of $\Gamma$ that contain A. Then

$$\begin{aligned} A=\displaystyle \bigcap _{j=1}^q F_j. \end{aligned}$$

Proof

Since ${\mathrm{lk}}\ _\Gamma (A)=\langle F_1 \setminus A, \ldots , F_q \setminus A \rangle$, if there is a vertex of $\displaystyle \bigcap _{j=1}^q F_j$ which is not in A, then ${\mathrm{lk}}\ _\Gamma (A)$ would be a cone and would therefore have no homology, contradicting Theorem 3.2. $\square$

Theorem 3.5 below is a formal statement on breaking homological cycles. We refer the reader to parts (2) and (3) of Example 2.4 where we demonstrated the theorem’s statement. Note also that the case in part (1) of Example 2.4 follows the same pattern, though a proof is not known yet.

Theorem 3.5

(Breaking up cycles on links). Let k be a field and $\Gamma =\langle F_1,\ldots ,F_r \rangle$ be a d-dimensional simplicial complex such that

$$\begin{aligned} \tilde{H}_{d}(\Gamma ,k) \ne 0 \text{ and } d+2=a+b \text{ for } \text{ some } a,b>0. \end{aligned}$$

Suppose $\Gamma$ contains a d-dimensional cycle

$$\begin{aligned} \Sigma =\displaystyle \sum _{j=1}^qa_jF_j \end{aligned}$$

supported on the facets $F_1,\ldots ,F_q$ of $\Gamma$, and $\Sigma$ is not a boundary in $\Gamma$. Then there are subsets $A,B \subseteq [q] \subseteq [r]$ with

$$\begin{aligned} F=\bigcap _{j \in A} F_j \text{ and } G=\bigcap _{j \in B} F_j \end{aligned}$$

such that

• 1. $F \cap G = \emptyset$;

• 2. $F \cup G \notin \Gamma$;

• 3. $\tilde{H}_{a-2}({\mathrm{lk}}\ _{\Gamma }(F),k) \ne 0$ and $\tilde{H}_{b-2}({\mathrm{lk}}\ _{\Gamma }(G),k) \ne 0$.

Moreover, if $a,b>1$, F and G and $\epsilon _j, \delta _j \in \{\pm 1\}$ could be chosen to additionally satisfy:

• 4. $|F|=b$ and $|G|=a$;

• 5. $\Sigma _F=\displaystyle \sum _{j \in A} \epsilon _ja_j \left( F_j \setminus F\right)$ is an $(a-2)$-cycle in ${\mathrm{lk}}\ _{\Gamma }(F)$ which is not a boundary ;

• 6. $\Sigma _G=\displaystyle \sum _{j \in B} \delta _j a_j \left( F_j \setminus G \right)$ is a $(b-2)$-cycle in ${\mathrm{lk}}\ _{\Gamma }(G)$ which is not a boundary.

Proof

Set $i=d+2$. We first consider the case $b=1$ and $a=i-1$. If $a=1$, then $d=0$ and $\Gamma$ is disconnected. Let F and G be two facets each belonging to a distinct connected component of $\Gamma$. Then we clearly have $F\cap G=\emptyset$ and $F \cup G \notin \Gamma$. Moreover, ${\mathrm{lk}}\ _{\Gamma }(F)={\mathrm{lk}}\ _{\Gamma }(G)=\{\emptyset \}$ and so

$$\begin{aligned} \tilde{H}_{a-2}({\mathrm{lk}}\ _{\Gamma }(F),k)=\tilde{H}_{b-2}({\mathrm{lk}}\ _{\Gamma }(G),k)= \tilde{H}_{-1}(\{ \emptyset \},k)\ne 0 \end{aligned}$$

as desired.

If $b=1$ and $a=i-1>1$, then $d=a+b-2>0$. By Theorem 3.2, if we take a vertex v in the support complex of $\Sigma$, then $\tilde{H}_{i-3}({\mathrm{lk}}\ _\Gamma (v),k) \ne 0$.

Since $\Sigma$ is a cycle, not all of $F_1, \ldots ,F_q$ contain v. Let G be one of the facets $F_1,\ldots , F_q$ that does not contain v. Then $F\cap G =\emptyset$ and $F\cup G \notin \Gamma$ (as G is a facet), and moreover

$$\begin{aligned} \tilde{H}_{a-2}({\mathrm {lk}}\ _\Gamma (F),k)=\tilde{H}_{i-3}({\mathrm {lk}}\ _\Gamma (v),k) \ne 0 \\ \text { and }\; \tilde{H}_{b-2}({\mathrm {lk}}\ _\Gamma (G),k)= \tilde{H}_{-1}(\{\emptyset \},k) \ne 0. \end{aligned}$$

Now suppose $a, b \geqslant 2$ and $a=i-b$. Suppose $F_1=\{w_1,v_1,\ldots ,v_{i-2}\}$. Then since $F_1$ is in the support of the $(i-2)$-cycle $\Sigma$, $\{w_1, v_2,\ldots ,v_{i-2}\}$ must appear in another one of the $F_j$ in the support of $\Sigma$, say $F_2$. Suppose $F_2=\{w_1,w_2,v_2,\ldots ,v_{i-2}\}$. Considering that $a=i-b \leqslant i-2$, let

$$\begin{aligned} G=\{v_1,\ldots ,v_a\} \text{ and } F=\{v_{a+1},\ldots ,v_{i-2},w_1,w_2\}. \end{aligned}$$

Then $|G|=a$ and $|F|=i-2+2-a=b$. Moreover $F \cap G =\emptyset$ by construction, and if $i-2=d$, then $F\cup G \notin \Gamma$ since $|F \cup G|=d+2$ which is larger than the size of any face of $\Gamma$.

By Theorem 3.2, and noting that $i-2-|G|=b-2$ and $i-2-|F|=a-2$, we have

$$\begin{aligned} \tilde{H}_{a-2}({\mathrm{lk}}\ _\Gamma (F),k)\ne 0 \text{ and } \tilde{H}_{b-2}({\mathrm{lk}}\ _\Gamma (G),k)\ne 0, \end{aligned}$$

conditions 5 and 6 are satisfied, and if

$$\begin{aligned} A=\{ j\in [q] \ : \ F \subset F_j\} \text{ and } B=\{j \in [q] \ : \ G \subset F_j\} \end{aligned}$$

then

$$\begin{aligned} F=\bigcap _{j \in A} F_j \text{ and } G=\bigcap _{j \in B} F_j. \end{aligned}$$

$\square$

Another version of Theorem 3.5 below is one which gives lower-dimensional cycles in induced subcomplexes.

Corollary 3.6

(Breaking up cycles). Let $\Gamma$ be a simplicial complex on the vertex set $\{x_{1},\ldots ,x_{n}\}$, and suppose $\tilde{H}_{d-2}(\Gamma ,k) \ne 0$, where d is the smallest possible size of a nonface of $\Gamma$. Suppose $n-d+1=a+b$, where a and b are positive integers. Then there are nonempty subsets $C, D \subseteq \{x_{1},\ldots ,x_{n}\}$ such that

• 1. $C\cup D =\{x_{1},\ldots ,x_{n}\}$;

• 2. $C\cap D \in \Gamma$;

• 3. $\tilde{H}_{|C|-a-1}(\Gamma _C,k) \ne 0$ and $\tilde{H}_{|D|-b-1}(\Gamma _D,k) \ne 0$.

Proof

By Alexander duality—see Prop. 5.1.10 and the discussion preceding Prop. 5.1.8 in [18]—we have that $\tilde{H}_{n-d-1}(\Gamma ^{\vee },k) \ne 0$. Now d is the smallest possible size of a nonface of $\Gamma$, so by the definition of Alexander duals, $\dim (\Gamma ^{\vee })=n-d-1$.

Suppose $\Gamma ^{\vee }=\langle F_1,\ldots ,F_r \rangle$. If $n-d+1=a+b$, then, by Theorem 3.5, there are subsets A and B of [r] such that

$$\begin{aligned} F=\bigcap _{j \in A} F_j \text{ and } G=\bigcap _{j \in B} F_j \end{aligned}$$

and

• (i) $F \cap G = \emptyset$;

• (ii) $F \cup G \notin \Gamma ^{\vee }$;

• (iii) $\tilde{H}_{a-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}(F),k) \ne 0$ and $\tilde{H}_{b-2}({\mathrm{lk}}\ _{\Gamma ^{\vee }}(G),k) \ne 0$.

Now let

$$\begin{aligned} C=F^c=\bigcup _{j \in A} F_j^c \text{ and } D=G^c=\bigcup _{j \in B} F_j^c. \end{aligned}$$

Then by (i), $C\cup D=(F \cap G) ^c=\{x_{1},\ldots ,x_{n}\}$. By (ii), $(C\cap D)^c= F\cup G \notin \Gamma ^{\vee }$ so $C\cap D \in \Gamma$. Finally by (iii) and Equation (2), $\tilde{H}_{|C|-a-1}(\Gamma _C,k) \ne 0$ and $\tilde{H}_{|D|-b-1}(\Gamma _D,k) \ne 0$. $\square$

Theorem 3.7

(Subadditivity of syzygies of square-free monomial ideals). If I is a square-free monomial ideal in the polynomial ring $S=k[x_{1},\ldots ,x_{n}]$ where k is a field, and d is the smallest possible degree of a generator of I. Suppose $i=n-d+1$, $\beta _{i,n}(S/I) \ne 0$ and $i=a+b$, for some positive integers a and b. Then $t_i \leqslant t_a + t_b$.

Proof

By Hochster’s formula (Equation (2)), if $\Gamma ={\mathcal {N}}(I)$, then

$$\begin{aligned} \beta _{n-d+1,n}(S/I)=\beta _{n-d+1,x_1\cdots x_n}(S/I)=\dim _k \tilde{H}_{d-2}(\Gamma ,k)\ne 0. \end{aligned}$$

If $n-d+1=a+b$, then by Corollary 3.6, there are nonempty subsets $C, D \subseteq \{x_{1},\ldots ,x_{n}\}$ such that

$$\begin{aligned} C\cup D =\{x_{1},\ldots ,x_{n}\} \text{ and } C\cap D \in \Gamma , \end{aligned}$$

and

$$\begin{aligned} \tilde{H}_{|C|-a-1}(\Gamma _C,k) \ne 0 \text{ and } \tilde{H}_{|D|-b-1}(\Gamma _D,k) \ne 0. \end{aligned}$$

By Equation (2), this means that

$$\begin{aligned} \beta _{a,|C|}(S/I) \ne 0 \text{ and } \beta _{b,|D|}(S/I) \ne 0, \end{aligned}$$

so that $t_a \geqslant |C|$ and $t_b \geqslant |D|$. Putting this all together, we get

$$\begin{aligned} t_a + t_b \geqslant |C| + |D| \geqslant n = t_i, \end{aligned}$$

which settles our claim. $\square$

Discussion 3.8

Given a square-free monomial ideal I if we are looking for top degree Betti numbers, by Hochster’s formula (Equation (2))

$$\begin{aligned} \beta _{n-i-1,n}(S/I)=\dim _k\tilde{H}_{i}(\Gamma ,k). \end{aligned}$$

Now if d is the smallest possible degree of a generator of I, then all monomials of degree $\leqslant d-1$ are not in I, which means all possible faces of dimension $\leqslant d-2$ are in $\Gamma ={\mathcal {N}}(I)$. This means that the smallest index i with $\tilde{H}_{i}(\Gamma ,k)\ne 0$ is $d-2$, that is

$$\begin{aligned} \tilde{H}_{i}(\Gamma ,k)=0 \text{ for } i <d-2 \end{aligned}$$

and hence

$$\begin{aligned} \beta _{j,n}(S/I)=0 \text{ for } j=n-i-1>n-d+1. \end{aligned}$$

So $n-d+1$ is the maximum homological degree where we could have a nonvanishing top degree Betti number. We do not have an example of our setting where $n-d+1$ is not the projective dimension. After comparing with bounds on the projective dimension of S/I given by Dao and Schweig [8, Theorem 3.2, Remark 3.4] in terms of dominance parameters of clutters, we concluded that $n-d+1$ is often either the projective dimension of S/I or very close to it, though we were not able to determine how close.

Example 3.9

Let $I=(xyz,xzv,xuv,yzu,yuv)$ be an ideal of $S=k[x,y,z,u,v]$ in 5 variables. Here the smallest degree of a generator of I is $d=3$, so $n-d+1=3$, so we pick $a=1$ and $b=2$. According to Macaulay2 [23], the Betti table of S/I is

$$\begin{aligned} \begin{array}{rlllll} &{}0 &{}1 &{}2 &{}3\\ \text{ total }:&{}1 &{}5 &{}5 &{}1\\ 0:&{}1 &{}. &{}. &{}.\\ 1:&{}. &{}. &{}. &{}. \\ 2:&{}. &{}5 &{}5 &{}1 \end{array} \end{aligned}$$

which verifies that

$$\begin{aligned} t_3=5, t_2=4, t_1=3 \Longrightarrow t_3 < t_1+t_2 =7. \end{aligned}$$

Example 3.10

In Example 2.4, $I=(xz,yz,xu,yu,xv,yv,zuv)$ is a square-free monomial ideal in 5 variables where $d=2$ and $n-d+1=4$. According to the Betti table of I, $t_4=5$, $t_3=5$, $t_2=4$ and $t_1=3$. Here $t_4<t_1+t_3=8$ and $t_4<2t_2=8$. Note that we also have $\beta _{3,5}(S/I)\ne 0$ where $3 <4=n-d+1$ while still we have $t_3 < t_1+t_2=7$.

4 Special cases of breaking up simplicial homology

In this section, we consider breaking up special classes of cycles, where we can provide a combinatorial description for the lower-dimensional cycles.

4.1 The case of a disconnected simplicial complex

We begin with an example.

Example 4.1

Let ${\mathcal {N}}(I)=\Gamma =\langle uv,xy,yz,xz \rangle$ be a simplicial complex on $n=5$ vertices.

Here $\tilde{H}_{0}(\Gamma ,k)\ne 0$ and hence $\beta _{4,uvxyz}(S/I)\ne 0$. If $4=a+b$, then using Corollary 3.6 we have the following two cases to consider.

• 1. $a=1$ and $b=3$. Let $C=\{u,x\}$ and $D=\{u,v,y,z\}$. Then $C\cup D= \{u,v,x,y,z\}$, $C \cap D=\{u\}\in \Gamma$ and

  $$\begin{aligned} \tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\langle u,x \rangle ,k)\ne 0 \\ \text { and }\; \tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{0}(\langle uv,yz \rangle ,k)\ne 0. \end{aligned}$$

• 2. $a=b=2$. Let $C=\{u,x,v\}$ and $D=\{u,y,z\}$. Then $C\cup D= \{u,v,x,y,z\}$, $C \cap D=\{u\}\in \Gamma$ and

  $$\begin{aligned} \tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\langle uv,x \rangle ,k)\ne 0 \\ \text { and } \; \tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{0}(\langle u,yz \rangle ,k)\ne 0. \end{aligned}$$

In general, if $\Gamma$ is a disconnected complex on n vertices with Stanley–Reisner ideal I, then $\beta _{n-1,n}(S/I) \ne 0$, and if $n-1=a+b$ for some $a,b>0$, then we can always find disconnected induced subcomplexes $\Gamma _C$ and $\Gamma _D$ where $C=a+1$ and $D=b+1$, as in the example above. Below we demonstrate how this can be done.

If $\Gamma$ is disconnected, then it has the form

$$\begin{aligned} \Gamma =\Gamma _1 \cup \cdots \cup \Gamma _t \end{aligned}$$

where $\Gamma _1,\ldots ,\Gamma _t$ are connected components and $t > 1$. In this case, $|V(\Gamma _i)| \geqslant 1$ for all $1 \leqslant i \leqslant t$, $V(\Gamma )=V(\Gamma _1) \cup \cdots \cup V(\Gamma _t)$ and $V(\Gamma _k) \cap V(\Gamma _l) = \emptyset$ for all $1 \leqslant k < l \leqslant t$.

Without loss of generality and up to renaming the variables, we can assume the following:

• $|V(\Gamma _1)| \leqslant |V(\Gamma _2)| \leqslant \cdots \leqslant |V(\Gamma _t)|$,

• $x_k \in V(\Gamma _k)$ for $1 \leqslant k \leqslant t$,

• $V(\Gamma _1)=\{x_1,x_{t+1},\ldots ,x_{t+|V(\Gamma _1)|-1}\}$

• $V(\Gamma _k)=\{x_k,x_{{\tiny (t+|V(\Gamma _1)|+\cdots +|V(\Gamma _{k-1})|-k+2)}},\ldots ,x_{{\tiny (t+|V(\Gamma _1)|+\cdots +|V(\Gamma _{k})|-k)}} \}$ for each $1 < k \leqslant t$.

Example 4.2

The simplicial complex $\Gamma$ in Example 4.1 can be relabeled and written as $\Gamma =\Gamma _1 \cup \Gamma _2$ where $\Gamma _1=\langle x_1 x_3 \rangle$ and $\Gamma _2=\langle x_2x_4,x_4x_5,x_2x_5\rangle$.

For each $1 \leqslant a< n-1$, define

$$\begin{aligned} C=\{x_1,x_2, \ldots ,x_{a+1}\} \text{ and } D=\{x_1,x_{a+2},\ldots ,x_n\}. \end{aligned}$$

Clearly $C\cup D=\{x_1,\ldots ,x_n\}$, $|C|=a+1$, $|D|=n-a$ and $C\cap D=\{x_1\}\in \Gamma$. Moreover, it is easy to see that both $\Gamma _C$ and $\Gamma _D$ are disconnected induced subcomplexes of $\Gamma$ on the subsets $\{x_1,x_2,\ldots ,x_{a+1}\}$ and $\{x_1,x_{a+2},\ldots ,x_n\}$, respectively. Therefore, if $b=n-a-1$

$$\begin{aligned} \tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\Gamma _C,k)\ne 0 \text{ and } \tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{0}(\Gamma _D,k)\ne 0. \end{aligned}$$

4.2 The case of a graph cycle

Recall that a cycle in a graph G is an ordered list of distinct vertices $x_1,\ldots ,x_n$ where the edges are $x_{i-1}x_i$ for $2\leqslant i \leqslant n$ and $x_nx_1$. Graph cycles characterize nontrivial 1-homology in simplicial complexes; see, for example, Theorem 3.2 in [7].

Suppose $\Gamma$ is a simplicial complex on the set $\{x_1,\ldots ,x_n\}$ that is the support complex of a face-minimal graph cycle, so that $\tilde{H}_{1}(\Gamma ,k)\ne 0$. This means that $\beta _{n-2,n}(S/I) \ne 0$. Suppose $n-2=a+b$ for some $a,b>0$.

Without loss of generality, $\Gamma$ can be written in the form

$$\begin{aligned} \Gamma =\langle x_1x_2,x_2x_3, \ldots , x_{n-1}x_n,x_nx_1 \rangle . \end{aligned}$$

For $1 \leqslant a < n-2$, define

$$\begin{aligned} C=\{x_1,x_3,x_4,\ldots ,x_{a+2}\} \text{ and } D=\{x_2,x_{a+3},\ldots ,x_n\}. \end{aligned}$$

Clearly, $C\cup D=\{x_1,\ldots ,x_n\}$, $|C|=a+1$, $|D|=n-a-1$ and $C\cap D=\emptyset \in \Gamma$. Moreover, it is easy to see that both $\Gamma _C$ and $\Gamma _D$ are disconnected induced subcomplexes of $\Gamma$ on the subsets $\{x_1,x_3,x_4,\ldots ,x_{a+2}\}$ and $\{x_2,x_{a+3},\ldots ,x_n\}$, respectively. Therefore,

$$\begin{aligned} \tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\Gamma _C,k)\ne 0 \end{aligned}$$

and

$$\begin{aligned} \tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{0}(\Gamma _D,k)\ne 0 \end{aligned}$$

where $b=n-a-2$.

Example 4.3

Let ${\mathcal {N}}(I)=\Gamma =\langle x_1x_2,x_2x_3,x_3x_4,x_4x_5,x_1x_5 \rangle$.

Then $\tilde{H}_{1}(\Gamma ,k)\ne 0$ and hence $\beta _{3,x_1\cdots x_5}(S/I)\ne 0$. Taking $a=1$ and $b=2$, set $C=\{x_1,x_3\}$ and $D=\{x_2,x_4,x_5\}$. Then

$\tilde{H}_{|C|-a-1}(\Gamma _C,k)=\tilde{H}_{0}(\langle x_1,x_3 \rangle ,k)\ne 0$ and $\tilde{H}_{|D|-b-1}(\Gamma _D,k)=\tilde{H}_{0}(\langle x_2,x_4x_5 \rangle ,k)\ne 0$.

5 The case of general monomial ideals

The polarization [16] of a monomial ideal I is a method to transform I to a square-free monomial ideal, by adding new variables to the polynomial ring. The procedure is described below.

Definition 5.1

(Polarization). Let I be minimally generated by monomials ${\mathbf {m}}_1,\ldots {\mathbf {m}}_q$ in the polynomial ring $R=k[x_{1},\ldots ,x_{n}]$. For $i\in \{1 ,\ldots ,n\}$, let

$$\begin{aligned} p_i = {\left\{ \begin{array}{ll} 1 &{} \text{ if } x_i \not \mid {\mathbf {m}}_u \text{ for } \text{ every } u \in [q]\\ \max \left\{ j \ : \ x_i^j \mid {\mathbf {m}}_u \text{ for } \text{ some } u \in [q] \right\} &{} \text{ otherwise. } \end{array}\right. } \end{aligned}$$

Let S be the polynomial ring in ${\mathbf {p}}=p_1 +\cdots +p_n$ variables

$$\begin{aligned} S=k[x_{i,j} \ : \ 1 \leqslant i \leqslant n, \ 1 \leqslant j \leqslant p_i] \end{aligned}$$

and let the polarization of I be the square-free monomial ideal

$$\begin{aligned} \mathcal {P}(I)= \left( \mathcal {P}({\mathbf {m}}_1),\ldots ,\mathcal {P}({\mathbf {m}}_q) \right) \end{aligned}$$

where, if ${\mathbf {m}}={x_{a_1}}^{b_1}\cdots {x_{a_c}}^{b_c}$ where the $a_i$ are distinct integers in $\{1,\ldots ,n \}$ and $1 \leqslant b_i \leqslant p_i$ for $1\leqslant i \leqslant c$, then

$$\begin{aligned} \mathcal {P}({\mathbf {m}})=x_{a_1,1}\cdots x_{a_1,b_1} x_{a_2,1}\cdots x_{a_2,b_2} \cdots x_{a_c,1}\cdots x_{a_c,b_c}. \end{aligned}$$

Example 5.2

If $I=(x^2,xy^3z^2) \subseteq k[x,y,z]$, then its polarization is the square-free monomial ideal $\mathcal {P}(I)=(x_1x_2,x_1y_1y_2y_3z_1z_2)$ in the polynomial ring $k[x_1,x_2,y_1,y_2,y_3,z_1,z_2]$.

Corollary 5.3

(Subadditivity of syzygies of monomial ideals). If I is a monomial ideal in the polynomial ring $R=k[x_{1},\ldots ,x_{n}]$ where k is a field, d is the smallest possible degree of a generator of I, and ${\mathbf {p}}$ is defined as in Definition 5.1. Suppose $i={\mathbf {p}}- d + 1$, $\beta _{i,{\mathbf {p}}}(R/I) \ne 0$ and $i=a+b$, for some positive integers a and b. Then $t_i \leqslant t_a + t_b$.

Proof

Let $I=({\mathbf {m}}_1,\ldots ,{\mathbf {m}}_q)$, whose polarization is the square-free monomial ideal $\mathcal {P}(I)$ in the polynomial ring S in ${\mathbf {p}}$ variables in Definition 5.1. Since $\beta _{i,{\mathbf {p}}}(R/I)\ne 0$, we must have $\beta _{i,{\mathbf {m}}}(R/I)\ne 0$ for some ${\mathbf {m}}\in \text{ LCM }(I)$. On the other hand, ${\mathbf {p}}$ is the largest possible degree for a monomial in $\text{ LCM }(I)$, and so ${\mathbf {m}}= {{\mathrm{lcm}}}({\mathbf {m}}_1,\ldots ,{\mathbf {m}}_q)$, the top monomial in the lcm lattice of I.

Now the two lcm lattices $\text{ LCM }(I)$ and $\text{ LCM }(\mathcal {P}(I))$ are isomorphic ([17]), and the degree ${\mathbf {p}}$ square-free monomial $\mathcal {P}({\mathbf {m}})$ sits on top of the lattice $\text{ LCM }(\mathcal {P}(I))$, and so $\beta _{i,{\mathbf {p}}}(S/\mathcal {P}(I)) \ne 0$. Now since $\deg ({\mathbf {m}}_i)=\deg (\mathcal {P}({\mathbf {m}}_i))$ for all $1 \leqslant i \leqslant q$, the conditions for Theorem 3.7 hold, and therefore $t_i \leqslant t_a + t_b$ holds for the ideal $\mathcal {P}(I)$. But as the graded Betti numbers of I and $\mathcal {P}(I)$ are equal, the inequality also holds for I, and we are done. $\square$

Example 5.4

Let $I=(xy^2,xyz,y^3,y^2z)$ be an ideal of $R=k[x,y,z]$. Here ${\mathbf {p}}=5$ and the smallest degree of a generator is $d=3$, so ${\mathbf {p}}-d+1=3$. We pick $a=1$ and $b=2$. According to Macaulay2 [23] the Betti table of R/I is

$$\begin{aligned} \begin{array}{rlllll} &{}0 &{}1 &{}2 &{}3\\ \text{ total }:&{}1 &{}4 &{}4 &{}1\\ 0:&{}1 &{}. &{}. &{}.\\ 1:&{}. &{}. &{}. &{}. \\ 2:&{}. &{}4 &{}4 &{}1 \end{array} \end{aligned}$$

which verifies that

$$\begin{aligned} t_3=5, t_2=4, t_1=3 \Longrightarrow t_3 < t_1+t_2 =7. \end{aligned}$$

6 Final remarks

Questions 2.1, 2.2, 2.3, and 2.5 are all equivalent, though their different settings allow the application of different (inductive) tools. All of them are open in their full generality as far as we know, though each can be answered positively for certain classes of ideals or combinatorial objects. A positive answer to either would settle the subadditivity question for monomial ideals in a polynomial ring.