Edge ideals with almost maximal finite index and their powers

Abstract

A graded ideal I in $\mathbb {K}[x_1,\ldots ,x_n]$, where $\mathbb {K}$ is a field, is said to have almost maximal finite index if its minimal free resolution is linear up to the homological degree $\mathrm {pd}(I)-2$, while it is not linear at the homological degree $\mathrm {pd}(I)-1$, where $\mathrm {pd}(I)$ denotes the projective dimension of I. In this paper, we classify the graphs whose edge ideals have this property. This in particular shows that for edge ideals the property of having almost maximal finite index does not depend on the characteristic of $\mathbb {K}$. We also compute the nonlinear Betti numbers of these ideals. Finally, we show that for the edge ideal I of a graph G with almost maximal finite index, the ideal $I^s$ has a linear resolution for $s\ge 2$ if and only if the complementary graph $\bar{G}$ does not contain induced cycles of length 4.

1 Introduction

In this paper, we consider the edge ideals whose minimal free resolution has relatively large number of linear steps. Let I be a graded ideal in the polynomial ring $S={\mathbb {K}}[x_1,\ldots ,x_n]$, where ${\mathbb {K}}$ is a field, generated by homogeneous polynomials of degree d. The ideal is called r-steps linear, if I has a linear resolution up to the homological degree r, that is the graded Betti numbers $\beta _{i,i+j}(I)$ vanish for all $i\le r$ and all $j>d$. The number

$$\begin{aligned} \mathrm {index}(I)=\inf \{r: I\text { is not }r\text {-steps linear}\} \end{aligned}$$

is called the Green–Lazarsfeld index (or briefly index) of I. A related invariant, called the $N_{d,r}$-property, was first considered by Green and Lazarsfeld [16, 17]. In the paper [3], the index was introduced for the quotient ring S/I, where I is generated by quadratics, to be the largest integer r such that the $N_{2,r}$-property holds. It is in generally very hard to determine the value of the index. One reason is that this value, in general, depends on the characteristic of ${\mathbb {K}}$. The index of quadratic monomial ideals is more studied in the literature taking advantage of some combinatorial methods. Indeed, since the index is preserved passing through polarization, one may reduce to the case of squarefree quadratic monomial ideals which can be viewed as the edge ideals of simple graphs, and the index of these ideals is proved to be characteristic independent, see [10, Theorem 2.1].

The main question regarding the study of the index of edge ideals is to classify the graphs with respect to the index of their edge ideals, in particular, it is more interesting to see when the index attains its largest or smallest value. In 1990, Fröberg [14] classified the graphs whose edge ideals have a linear resolution. A graded ideal I is said to have a linear resolution if $\mathrm {index}(I)=\infty$. In fact, Fröberg showed that given a graph G, its edge ideal I(G) has a linear resolution over all fields if and only if the complement ${\bar{G}}$ of G is chordal, which means that all cycles in ${\bar{G}}$ of length $>3$ have a chord. In 2005, Eisenbud et al. [10] gave a purely combinatorial description of the index of edge ideals in terms of the size of the smallest cycle(s) of length $>3$ in the complementary graph, c.f. Theorem 1.1. This result shows that the index gets its smallest value 1 if and only if G admits a gap, i.e., ${\bar{G}}$ contains an induced cycle of length 4. If the index of I attains the largest finite value, we have $\mathrm {index}(I)=\mathrm {pd}(I)$, where $\mathrm {pd}(I)$ denotes the projective dimension of I. In this case, the ideal I is said to have maximal finite index, see [2]. In [2, Theorem 4.1], it was shown that the edge ideal I(G) has maximal finite index if and only if ${\bar{G}}$ is a cycle of length $>3$. In this paper, we proceed one more step and consider the edge ideals I(G) with $\mathrm {index}(I(G))=\mathrm {pd}(I(G))-1$. We call them edge ideals with almost maximal finite index. In Sect. 2 of this paper, we precisely determine the simple graphs whose edge ideals have this property, see Theorem 2.6. These graphs are presented in Figs. 1, 2, 3 and 4. In particular, it is deduced that the property of having almost maximal finite index is characteristic independent for edge ideals, though this is not the case for ideals generated in higher degrees, as discussed in the beginning of Sect. 2. It is also seen that the graded Betti numbers of these edge ideals do not depend on the characteristic of the base field. We will compute the Betti numbers in the nonlinear strands in Proposition 2.10. The main tool used throughout this section is Hochster’s formula, Formula (1).

In the second half of the paper, we study the index of powers of edge ideals with almost maximal finite index. Although, for arbitrary ideals, many properties such as depth, projective dimension or regularity stabilize for large powers (see e.g., [1, 5,6,7,8, 18, 20,21,22]), their initial behavior is often quite mysterious. However, edge ideals behave more controllable from the beginning. In the study of the index of powers of edge ideals, one of the main results is due to Herzog et al. [20, Theorem 3.2]. They showed that for a graph G, all powers of the edge ideal I(G) have a linear resolution if and only if so does I(G). On the other hand, it was shown in [2, Theorem 3.1] that all powers of I(G) have index 1 if and only if I(G) has also index 1. In the same paper, it was proved that if I(G) has maximal finite index $>1$, then $I(G)^s$ has a linear resolution for all $s\ge 2$. This shows that chordality of the complement of G is not a necessary condition on G so that all high powers of its edge ideal have a linear resolution. Francisco, Hà and Van Tuyl proved, in a personal communication, that being gap-free is a necessary condition for a graph G in order that a power of its edge ideal has a linear resolution (see also [24, Proposition 1.8]). However, Nevo and Peeva showed, by an example, that being gap-free alone is not a sufficient condition so that all high powers of the edge ideal have a linear resolution [24, Counterexample 1.10]. Later, Banerjee [1] and Erey [11, 12], respectively, proved that if a gap-free graph G is also cricket-free or diamond-free or $C_4$-free, then the ideal $I(G)^s$ has a linear resolution for all $s\ge 2$. The definition of these concepts are recalled in Sect. 3.

Section 3 is devoted to answer the question whether the high powers of edge ideals with almost maximal finite index have a linear resolution. Not all graphs whose edge ideals have this property are cricket-free or diamond-free. However, using some formulas for an upper bound of the regularity of either powers of edge ideals or in general monomial ideals offered in [1, Theorem 5.2] and [9, Lemma 2.10], respectively, we give a positive answer to this question in case the graphs are gap-free, see Theorem 3.1. We will prove this theorem in several parts, mainly in Theorems 3.7 and 3.11.

Theorem 3.1 together with [2, Theorem 4.1] yield the following consequence which is a partial generalization of the result of Herzog et al. [20, Theorem 3.2].

Theorem 0.1

Let G be a simple gap-free graph and let $I\subset S$ be its edge ideal. Suppose $\mathrm {pd}(I)\!-\!\mathrm {index}(I)\le \!1$. Then $I^s$ has a linear resolution over all fields for any $s\ge 2$.

One may ask which is the largest integer c such that Theorem 0.1 remains valid if one replaces $\mathrm {pd}(I) - \mathrm {index}(I) \le 1$ by $\mathrm {pd}(I) - \mathrm {index}(I) \le c$. Computation by Macaulay 2, [15], shows that in the example of Nevo and Peeva [24, Counterexample 1.10], $\mathrm {index}(I)=2$, and $\mathrm {pd}(I)=8$. Hence, c must be an integer with $1\le c\le 5$.

2 Preliminaries

In this section, we recall some concepts, definitions and results from Commutative Algebra and Combinatorics which will be used throughout the paper. Let $S={\mathbb {K}}[x_1, \ldots , x_n]$ be the polynomial ring over a field ${\mathbb {K}}$ with n variables, and let M be a finitely generated graded S-module. Let the sequence

$$\begin{aligned} 0\rightarrow F_p\rightarrow \cdots \rightarrow F_2 \rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0 \end{aligned}$$

be the minimal graded free resolution of M, where for all $i \ge 0$ the modules $F_i = \oplus _j S(-j)^{\beta _{i,j}^{{\mathbb {K}}}(M)}$ are free S-modules of rank $\beta _{i}^{{\mathbb {K}}}(M):=\sum _{j}\beta _{i,j}^{{\mathbb {K}}}(M)$. The numbers $\beta _{i,j}^{{\mathbb {K}}}(M) = \dim _{{\mathbb {K}}} \text{ Tor}^S_i(M, {\mathbb {K}})_j$ are called the graded Betti numbers of M and $\beta _i^{{\mathbb {K}}}(M)$ is called the ith Betti number of M. We write $\beta _{i,j}(M)$ for $\beta _{i,j}^{{\mathbb {K}}}(M)$ when the field is fixed. The projective dimension of M, denoted by $\mathrm {pd}(M)$, is the largest i for which $\beta _{i}(M)\ne 0$. The Castelnuovo–Mumford regularity of M, $\mathrm {reg}(M)$, is defined to be

$$\begin{aligned} \mathrm {reg}(M)=\sup \{j-i:\ \beta _{i,j}(M)\ne 0\}. \end{aligned}$$

Let I be a graded ideal of S generated in a single degree d. The Green–Lazarsfeld index (briefly index) of I, denoted by $\mathrm {index}(I)$, is defined to be

$$\begin{aligned} \mathrm {index}(I)=\inf \{i:\ \beta _{i,j}(I)\ne 0,\ \text {for some } j>i+d\}. \end{aligned}$$

Since $\beta _{0,j}(I)=0$ for all $j>d$, one always has $\mathrm {index}(I)\ge 1$. The ideal I is said to have a d-linear resolution if $\mathrm {index}(I)=\infty$. This means that for all i, $\beta _{i}(I)=\beta _{i,i+d}(I)$, and this is the case if and only if $\mathrm {reg}(I)=d$. Otherwise $\mathrm {index}(I)\le \mathrm {pd}(I)$. In case I has the largest possible finite index, that is $\mathrm {index}(I)=\mathrm {pd}(I)$, I is said to have maximal finite index.

In Sect. 2 of this paper, we deal with squarefree monomial ideals generated in degree 2. These ideals are the edge ideals of simple graphs. Recall that a simple graph is a graph with no loops and no multiple edges, and given a graph G on the vertex set $[n]:=\{1,\ldots ,n\}$, its edge ideal $I(G)\subset S$ is an ideal generated by all quadratics $x_ix_j$, where $\{i,j\}$ is an edge in G. We denote by E(G) the set of all edges of G, and by V(G) the vertex set of G. For a vertex $v\in V(G)$, the neighborhood $N_G(v)$ of v in G is defined to be

$$\begin{aligned} N_G(v)=\{u\in V(G):\ \{u,v\}\in E(G) \}. \end{aligned}$$

The complement ${\bar{G}}$ of G is a graph on V(G) whose edges are those pairs of V(G) which do not belong to E(G). The simplicial complex

$$\begin{aligned} \Delta (G)=\{F\subseteq V(G):\ \text {for all } \{i,j\}\subseteq F \text { one has } \{i,j\} \in E(G)\} \end{aligned}$$

is called the flag complex of G. The independence complex of G is the flag complex of ${\bar{G}}$. One can check that $I(G)=I_{\Delta ({\bar{G}})}$, where $I_{\Delta ({\bar{G}})}$ is the Stanley–Reisner ideal of $\Delta ({\bar{G}})$. We assume that the reader is familiar with the definition and elementary properties of simplicial complexes. For more details consult with [19].

The main tool used widely in Sect. 2 for the computation of the graded Betti numbers is Hochster’s formula [19, Theorem 8.1.1]. Let $\Delta$ be a simplicial complex on [n], and let ${\tilde{C}}(\Delta , {\mathbb {K}})$ be the augmented oriented chain complex of $\Delta$ over a field ${\mathbb {K}}$ with the differentials

$$\begin{aligned} \partial _i:&\bigoplus _{\begin{array}{c} F\in \Delta \\ \dim F=i \end{array}}{\mathbb {K}}F\rightarrow \bigoplus _{\begin{array}{c} G\in \Delta \\ \dim G=i-1 \end{array}}{\mathbb {K}}G,\\ \partial _i([v_0,\ldots ,v_{i}])=&\sum _{0\le j\le i}(-1)^{j}[v_0,v_1,\ldots , v_{j-1},v_{j+1}, \ldots , v_{i}], \end{aligned}$$

where by $[v_0,v_1,\ldots ,v_{i}]$ we mean the face $\{v_0,v_1,\ldots ,v_{i}\}\subseteq [n]$ of $\Delta$ with $v_0<v_1<\cdots <v_i$. Hochster’s formula states that for the Stanley–Reisner ideal $I:=I_{\Delta }\subset S$ one has

$$\begin{aligned} \beta _{i,j}(I)=\sum _{W\subseteq [n],\ |W|=j} \dim _{\mathbb {K}} {\widetilde{H}}_{j-i-2}(\Delta _W;{\mathbb {K}}), \end{aligned}$$

(1)

where $\Delta _W$ is the induced subcomplex of $\Delta$ on W and ${\widetilde{H}}_i(\Delta _W;{\mathbb {K}})$ is the ith reduced homology of the complex ${\widetilde{C}}(\Delta _W, {\mathbb {K}})$. We denote by $\partial _i^W$ the differentials of the chain complex ${\widetilde{C}}(\Delta _W, {\mathbb {K}})$.

Theorem 1.1 which is due to Eisenbud et al. [10] provides a combinatorial method for determining the index of the edge ideal of a graph. To this end, one needs to consider the length of the minimal cycles of the complementary graph. A minimal cycle is an induced cycle of length$>3$, and by an induced cycle we mean a cycle with no chord. The length of an induced cycle C is denoted by |C|.

Theorem 1.1

([10, Theorem 2.1]) Let I(G) be the edge ideal of a simple graph G. Then

$$\begin{aligned} \mathrm {index}(I(G))=\inf \{|C|: \ C \text { is a minimal cycle} \text { in } {\bar{G}}\}-3. \end{aligned}$$

3 Edge ideals with almost maximal final index

A graded ideal $I\subset S$ is said to have almost maximal finite index over ${\mathbb {K}}$ if $\mathrm {index}(I)=\mathrm {pd}(I)-1$. Since, in general, $\mathrm {pd}(I)$ and $\mathrm {index}(I)$ depend on the characteristic of the base field, the property of having almost maximal finite index may also be characteristic dependent. For example, setting $\Delta$ to be a triangulation of a real projective plane, the Stanley–Reisner ideal of $\Delta$ is generated in degree 3 and it has almost maximal finite index over all fields of characteristic 2, while it has a linear resolution over other fields (cf. [4, §5.3]). However, as we will see in Corollary 2.7, in the case of quadratic monomial ideals, having almost maximal finite index is characteristic independent. Note that, although by Theorem 1.1, the index of an arbitrary edge ideal does not depend on the base field, its projective dimension may depend. M. Katzman presents a graph in [23, Section 4] whose edge ideal has different projective dimensions over different fields.

In this section, we give a classification of the graphs whose edge ideals have almost maximal finite index. We will present this classification in Theorem 2.6, but before, we need some intermediate steps which give more insight about the complement of such graphs.

Unless otherwise stated, throughout this section, G is a simple graph on the vertex set [n] and ${\bar{G}}$ is its complement, $\Delta$ denotes the independence complex $\Delta ({\bar{G}})$, and $\partial$, $\partial ^W$ denote, respectively, the differentials of the augmented oriented chain complexes of $\Delta ({\bar{G}})$, $\Delta ({\bar{G}})_W$ over a fixed field ${\mathbb {K}}$.

First, in order to avoid repetition of some arguments, we gather some facts which will be used frequently in the sequel in the following Observation. Meanwhile, we also fix some notation.

Notation and Observation 2.1

Let G be a simple graph on the vertex set [n] and let $I:=I(G)\subset S$ be its edge ideal.

(O-1) The graph G is connected if and only if its flag complex $\Delta (G)$ is connected. On the other hand for an arbitrary simplicial complex $\varGamma$ and any field ${\mathbb {K}}$,

$$\begin{aligned} \dim _{\mathbb {K}}{\widetilde{H}}_0(\varGamma ;{\mathbb {K}})=(\text {Number of connected components of }\varGamma )-1, \end{aligned}$$

see [19, Problem 8.2]. Moreover, for any subset $W\subseteq [n]$ one has $\Delta (G_W)=\Delta (G)_W$, where $G_W$ is the induced subgraph of G on the vertex set W. It follows that $G_W$ is connected if and only if ${\widetilde{H}}_0(\Delta (G)_W;{\mathbb {K}})=0$. Now if $\beta _{i,i+2}(I)=0$ for some i, then by Hochster’s formula ${\widetilde{H}}_{0}(\Delta _W;{\mathbb {K}})=0$ and hence ${\bar{G}}_W$ is connected for all $W\subseteq [n]$ with $|W|=i+2$.

(O-2) Throughout, by $P=u_1-u_2-\cdots -u_r$ in G we mean a path in G on r distinct vertices with the set of edges $\bigcup _{1\le i\le r-1}\{\{u_i,u_{i+1}\}\}$. If, in addition $\{u_1,u_r\}\in E(G)$, then $C=u_1-u_2-\cdots -u_{r}-u_1$ is a cycle in G. Then

$$\begin{aligned} T(C):=\left( \sum _{1\le i\le r-1}[u_i,u_{i+1}]\right) -[u_1,u_{r}]\in \ker \partial ^{\Delta (G)}_1, \end{aligned}$$

(2)

where $\partial ^{\Delta (G)}$ denotes the differentials of the chain complex of $\Delta (G)$. It is shown in [7, Theorem 3.2] that ${\widetilde{H}}_1(\Delta (G); {\mathbb {K}})\ne 0$ if and only if there exists a minimal cycle C in G such that $T(C)\notin \mathrm {Im}\ \partial _2^{\Delta (G)}$. Indeed, it is proved that ${\widetilde{H}}_1(\Delta (G); {\mathbb {K}})$ is minimally generated by the nonzero homology classes $T(C)+\mathrm {Im}\ \partial _2^{\Delta (G)}$, where C is a minimal cycle in G.

If C is the base of a cone whose apex is the vertex $u_{r+1}$, then

$$\begin{aligned} T(C)=\partial _{2}^{\Delta (G)}\left( \left( \sum _{1\le i\le r-1}[u_{r+1},u_i,u_{i+1}]\right) -[u_{r+1},u_1,u_{r}]\right) \end{aligned}$$

which implies that $T(C)+\mathrm {Im}\ \partial _2^{\Delta (G)}=0$. Recall that an r-gonal cone with the apex a is a graph $G'$ with the vertex set $V(G')=V(C)\cup \{a\}$, where $a\notin V(C)$ and C is an r-cycle in $G'$ which is called the base of $G'$, and $E(G')=E(C)\cup \{\{a,u_i\}: u_i\in V(C)\}$.

(O-3) Now let D be an r-gonal dipyramid in G; that is a subgraph of G with the vertex set $V(D)=V(C)\cup \{a,b\}$ and $E(D)= \bigcup _{1\le i\le r}\left( \{a,u_i\}\cup \{b,u_i\}\right) \cup E(C)$ where C is an r-cycle as above which is called the waist of D. Then

$$\begin{aligned} T(D)&:=\left( \sum _{1\le i\le r-1}[a,u_i,u_{i+1}]-[b,u_i,u_{i+1}]\right) \nonumber \\&\,\quad -[a,u_1,u_{r}]+[b,u_1,u_{r}]\in \ker \partial _2^{\Delta (G)}. \end{aligned}$$

(3)

(O-4) Suppose $\mathrm {index}(I)=t$. By Theorem 1.1, ${\bar{G}}$ contains a minimal cycle $C=u_1-u_2-\cdots -u_{t+3}-u_1$ which has the smallest length among all minimal cycles of ${\bar{G}}$.

(i):

If $\beta _{t+1,t+4}(I)=0$, then ${\widetilde{H}}_{1}(\Delta _W;{\mathbb {K}})=0$ for all $W\subseteq [n]$ with $|W|=t+4$. Set $W=\{u_{t+4}\}\cup V(C)$ for an arbitrary vertex $u_{t+4}\in [n]{\setminus } V(C)$. Then C is a minimal cycle in ${\bar{G}}_W$ and $T(C)\in \ker \partial _1^{W}$ implies that $T(C)\in \mathrm {Im}\ \partial _2^{W}$. It follows that each edge e of C is contained in a 2-face $F_e$ of ${\Delta _W}$. Since C is minimal, we must have $F_e=e\cup \{u_{t+4}\}$ which means that $u_{t+4}$ is adjacent to all vertices of C in ${\bar{G}}$ and hence ${\bar{G}}_W$ is a cone.

(ii):

If $\beta _{t+2,t+5}(I)=0$, then ${\widetilde{H}}_{1}(\Delta _W;{\mathbb {K}})=0$ for all $W\subseteq [n]$ with $|W|=t+5$. Set $W=\{u_{t+4}, u_{t+5}\}\cup V(C)$ for arbitrary vertices $u_{t+4}, u_{t+5}\in [n]{\setminus } V(C)$. As in (i), $T(C)=\partial _2^{W}(L)$ for some $L\in \bigoplus _{\begin{array}{c} F\in \Delta _W\\ {\dim F=2} \end{array}} {\mathbb {K}}F$, and hence each edge of C is contained in a 2-face of $\Delta _W$. It follows that for each edge e of C either $\{u_{t+4}\}\cup e\in \Delta _W$ or $\{u_{t+5}\}\cup e\in \Delta _W$. If for all $e\in E(C)$ one has $\{u_{t+4}\}\cup e\in \Delta _W$, then $\Delta _W$ contains a cone. Same holds if we replace $u_{t+4}$ with $u_{t+5}$. Suppose $\{u_{t+4}\}\cup e, \{u_{t+5}\}\cup e' \notin \Delta _W$ for some $e, e'\in E(C)$, which implies that $u_{t+4}, u_{t+5}$ are not adjacent to all vertices of C in ${\bar{G}}$. Without loss of generality suppose $\{u_{t+5},u_1,u_2\}\notin \Delta _W$. It follows that $\{u_{t+4}\}\cup \{u_1,u_2\}\in \Delta _W$. If $\{u_1,u_2\}$ is the only edge e of C with $\{u_{t+4}\}\cup e\in \Delta _W$, then for all $e'\in E(C)$ with $e'\ne \{u_1,u_2\}$ one has $\{u_{t+5}\}\cup e'\in \Delta _W$. In particular, $\{u_1,u_{t+3}, u_{t+5}\},\{u_2,u_3,u_{t+5}\}\in \Delta _W$ which implies by the definition of $\Delta _W=\Delta ({\bar{G}}_W)$ that $\{u_{t+5},u_1,u_2\}\in \Delta _W$, a contradiction. Since $u_{t+4}$ is not adjacent to all vertices of C in ${\bar{G}}$, and since $\{u_{t+4},u_1\},\{u_{t+4},u_2\}\in E({\bar{G}})$, it follows that there exists $3\le j\le t+3$ such that $\{u_j, u_{t+4}\}\notin E({\bar{G}})$. Let a, b be, respectively, the biggest and the smallest integers with $2\le a<j<b\le t+3$ for which $\{u_a,u_{t+4}\},\{u_b,u_{t+4}\}\in E({\bar{G}})$. If such b does not exist we let $b=1$ which implies that $a\ne 2$ because otherwise $\{u_{t+5}\} \cup e' \in \Delta _W$ for all $e' \in E(C) {\setminus }\{\{u_1, u_2\}\}$, so $u_{t+5}$ is adjacent to all vertices of C in ${\bar{G}}$. Now if $b\ne 1$, then $C':=u_{t+4}-u_a-u_{a+1}-\cdots -u_{b}-u_{t+4}$ is a minimal cycle in ${\bar{G}}_W$ of length $b-a+2$, and if $b=1$ then $C':=u_{t+4}-u_a-u_{a+1}-\cdots -u_{t+3}-u_{1}-u_{t+4}$ is a minimal cycle of length $t+6-a$. Since $\mathrm {index}(I)=t$, we must have $|C'|\ge t+3$ in either case, and so $\{a,b\}=\{1,3\}$ if $b=1$, and $\{a,b\}=\{2,t+3\}$ if $b\ne 1$. In both cases, the vertex $u_{t+4}$ is adjacent to only three successive vertices of C in ${\bar{G}}$. Without loss of generality, we may assume that $u_1,u_2,u_3$ are these three vertices. Thus, $u_{t+4}$ is adjacent to only two edges $\{u_1,u_2\}, \{ u_2,u_3\}$ of C in ${\bar{G}}$ and hence $\{u_1,u_3,u_4\ldots , u_{t+3}\}\subseteq N_{{\bar{G}}}(u_{t+5})$. It follows that $\{u_{t+5}, u_2\}\notin E({\bar{G}})$ because $u_{t+5}$ is not adjacent to all vertices of C in ${\bar{G}}$. Thus, we get the minimal 4-cycle $C'':=u_{t+5}-u_1-u_2-u_3-u_{t+5}$. It follows that $t=1$ because $\mathrm {index}(I)=t$. Therefore, $|C|=4$ and the only 2-faces of $\Delta _W$ containing an edge of C are $\{u_1,u_2,u_5\}, \{u_2,u_3,u_5\}, \{u_3,u_4,u_6\}, \{u_1,u_4,u_6\}$. But no linear combination of theses faces will result in L with $\partial _{2}^{W}(L)=T(C)$. We need more 2-faces in $\Delta _W$. It follows that $\{u_i, u_5,u_6\}\in \Delta _W$ for some $1\le i\le 4$. In particular, $\{u_5,u_6\}\in E({\bar{G}})$. This forms a graph ${\bar{G}}_W$ which is drawn as the graph $G_{(d)_2}$ in Fig. 4.

(iii):

If $\beta _{t+2,t+6}(I)=0$, then ${\widetilde{H}}_{2}(\Delta ({\bar{G}})_W;{\mathbb {K}})=0$ for all $W\subseteq [n]$ with $|W|=t+6$. Suppose ${\bar{G}}$ contains a dipyramid D with the vertex set $\{u_{t+4}, u_{t+5}\}\cup V(C)$, where the waist C is a minimal cycle of length $t+3$. Set $W=\{u_{t+4}, u_{t+5},u_{t+6}\}\cup V(C)$ for arbitrary vertex $u_{t+6}\in [n]{\setminus }(V(C)\cup \{u_{t+4}, u_{t+5}\})$. Then by (O-3) one has $T(D)\in \ker \partial _2^{W}$ and hence $T(D)\in \mathrm {Im}\ \partial _3^W$. This implies that each 2-face of D is contained in a 3-face of ${\Delta _W}$. Since C is minimal, it follows that either $\{u_{t+4},u_{t+5}\}\in E({\bar{G}})$ or $\{u_{t+4}, u_{t+5}\}\cup V(C)\subseteq N_{{\bar{G}}}(u_{t+6})$.

Example 2.2

Here we give 7 types of the graphs G whose edge ideal $I:=I(G)$ has almost maximal finite index over all fields. Indeed, we present the complementary graphs ${\bar{G}}$ for which $\mathrm {pd}(I)=\mathrm {index}(I)+1$. Take $t = 1$ for the cases (c) and (d) below. Since the smallest minimal cycles in the following graphs ${\bar{G}}$ are of length $t+3\ge 4$, by Theorem 1.1 we have $\mathrm {index}(I)=t$. We show that $\mathrm {pd}(I)=t+1$. Note that as it is also clear from Hochster’s formula, $\beta _{i,j}(I)=0$ for all $j<i+2$ and hence, in order to show that $\mathrm {pd}(I)=t+1$ it is enough to prove $\beta _{t+1,j}(I)\ne 0$ for some $j\ge t+3$ and $\beta _{t+2,j}(I)= 0$ for all $t+4\le j\le n$. The argument below is independent of the choice of the base field.

(a) Let ${\bar{G}}$ be either of the graphs $G_{(a)_1}, G_{(a)_2}, G_{(a)_3}$ shown in Fig. 1 with $t\ge 1$. The two graphs $G_{(a)_1}, G_{(a)_2}$ have one minimal cycle $C=1-2-\cdots -(t+3)-1$, and the graph $G_{(a)_3}$, has two minimal cycles C and $C'=1-(t+4)-3-4-\cdots -(t+3)-1$. Setting $W=[t+4]$, we have $T(C)\in \ker \partial _1^W$ by (O-2). Since $t>0$, there are edges of C in all three graphs which are not contained in a 2-face of $\Delta _W$. In particular, $T(C)\notin \mathrm {Im}\ \partial _2^W$. Hence, ${\widetilde{H}}_1(\Delta _W;{\mathbb {K}})\ne 0$ which implies that $\beta _{t+1,t+4}(I)\ne 0$. Thus, $\mathrm {pd}(I)\ge t+1$. If $\beta _{t+2,j}(I)\ne 0$ for some j, then there exists $W\subseteq [t+4]$ with $|W|=j$ such that ${\widetilde{H}}_{|W|-t-4}(\Delta _W;{\mathbb {K}})\ne 0$. It then follows that $W=[t+4]$ and ${\widetilde{H}}_0(\Delta _W;{\mathbb {K}})\ne 0$. But ${{\bar{G}}}_W={\bar{G}}$ is connected meaning that $\Delta _W$ is connected, by (O-1). Hence, ${\widetilde{H}}_0(\Delta _W;{\mathbb {K}})=0$, a contradiction. Therefore, $\mathrm {pd}(I)=t+1$.

Fig. 1

The graphs $G_{(a)_i}$

(b) Let ${\bar{G}}$ be the graph $G_{(b)}$ in Fig. 2, where $t\ge 1$ and $\{i,t+4\}\in E({{\bar{G}}})$ for all $i\in [t+3]$. Then ${\bar{G}}$ has one minimal cycle $C=1-2-\cdots -(t+3)-1$ as in (a). Setting $W=[t+3]\cup \{t+5\}$, we have $T(C)\in \ker \partial _1^W{\setminus }\mathrm {Im}\ \partial _2^W$. It follows that $\beta _{t+1,t+4}(I)\ne 0$. Therefore, $\mathrm {pd}(I)\ge t+1$. For any $W\subseteq [t+5]$ with $|W|=t+4$, ${\bar{G}}_W$ is connected. So $\beta _{t+2,t+4}(I)=0$. Suppose $W=[t+5]$. Although $T(C)\in \ker \partial _1^W$ one also has $T(C)\in \mathrm {Im}\ \partial _2^W$, because C is the base of a cone with apex ${t+4}$. Hence, according to (O-2), $\beta _{t+2,t+5}(I)=0$. It follows that $\mathrm {pd}(I)=t+1$.

Fig. 2

The graph $G_{(b)}$

(c) Let ${\bar{G}}$ be the graph $G_{(c)}$ shown in Fig. 3. This graph consists of 3 minimal cycles of length 4. So $\mathrm {index}(I)=1$. Moreover, $\beta _{2,5}(I)\ne 0$ because $T(C)\in \ker \partial _1{\setminus }\mathrm {Im}\ \partial _2$, for all minimal cycles C in ${\bar{G}}$, and $\beta _{3,5}(I)=0$ because ${\bar{G}}$ is connected. Hence, $\mathrm {pd}(I)=\!2$.

Fig. 3

The graph $G_{(c)}$

(d) Let ${\bar{G}}$ be either of the graphs $G_{(d)_1}, G_{(d)_2}$ in Fig. 4. Both graphs have three minimal cycles of length 4. Since $G_{(d)_1}$ is a dipyramid, by (O-3) one has ${\widetilde{H}}_2(\Delta (\overline{G_{(d)_1}});{\mathbb {K}})\ne 0$ which implies that $\beta _{2,6}(I(\overline{G_{(d)_1}}))\ne 0$. Although, $G_{(d)_2}$ is not a dipyramid, it contains the minimal cycle $C=1-2-3-4-1$ which gives a nonzero homology class of ${\widetilde{H}}_1(\Delta (\overline{G_{(d)_2}})_W;{\mathbb {K}})\ne 0$, where $W=V(C)\cup \{5\}$. Hence, $\beta _{2,5}(I(\overline{G_{(d)_2}}))\ne 0$. Therefore, $\mathrm {pd}(I)\ge 2$ in both cases. To prove that $\mathrm {pd}(I)=2$ it is enough to show that $\beta _{3,5}(I)=\beta _{3,6}(I)=0$.

Considering any subset W of [6] with $|W|=5$, ${\bar{G}}_W$ and so $\Delta _W$ is connected in both cases. It follows that ${\widetilde{H}}_0(\Delta _W;{\mathbb {K}})=0$, and hence $\beta _{3,5}(I)=0$. Now ${\widetilde{H}}_1(\Delta ;{\mathbb {K}})=0$ because except for the cycle $C=1-2-3-4-1$ in $G_{(d)_2}$, all other minimal cycles in $G_{(d)_1}, G_{(d)_2}$ are bases of some cones and for the cycle C, we have

$$\begin{aligned} T(C)=\partial _2^W([1,2,5]+[2,3,5]+[3,4,6]-[1,4,6]-[3,5,6]+ [1,5,6]). \end{aligned}$$

Consequently, $\beta _{3,6}(I)=0$ and hence $\mathrm {pd}(I)=2$.

Fig. 4

The graphs $G_{(d)_i}$

Next lemma gives more intuition about the length of minimal cycles in ${\bar{G}}$, when I(G) has almost maximal finite index. For an integer k, we show by ${\overline{k}}$ the remainder of k modulo $t+3$, i.e., ${\overline{k}}\equiv k\pmod {t+3}$ with $0 \le {\overline{k}} < t + 3$, where $t\ge 1$ is an integer.

Lemma 2.3

Let G be a simple graph on [n]. Assume $I:=I(G)$ has almost maximal finite index. Then any minimal cycle in ${\bar{G}}$ is of length $\mathrm {index}(I)+3$.

Proof

Let $\mathrm {index}(I)=t$. Then $\mathrm {pd}(I)=t+1$ which means that $\beta _{i,j}(I)=0$ for all $i>t+1$ and all j. Using Theorem 1.1, there exists a minimal cycle C in ${\bar{G}}$ of length $t+3$ which has the smallest length among all the minimal cycles in ${\bar{G}}$. Let $C=u_1-u_2-\cdots -u_{t+3}-u_1$. Suppose $C'\ne C$ is a minimal cycle in ${\bar{G}}$ with $C'=v_1-v_2-\cdots -v_l-v_1$. Setting $W=V(C')$ and $T(C')$ as defined in (2) one has $T(C')\in \ker \partial _1^W$, while $\mathrm {Im}\ \partial _2^W=0$. Hence, ${\widetilde{H}}_{1}(\Delta _W;{\mathbb {K}})\ne 0$. Hochster’s formula implies that $\beta _{l-3,l}(I)\ne 0$. Since $\beta _{i,j}(I)=0$ for all $i>t+1$, we have $l\le t+4$. We claim that $l<t+4$. Since $t+3$ is the smallest length of a minimal cycle in ${\bar{G}}$, it follows that $l=t+3$, as desired.

Proof of the claim

Suppose $l=t+4$ and let $u\in [n]{\setminus }V(C')$. Note that such u exists since otherwise $V(C)\subset [n]=V(C')$ which implies that $C'$ is not minimal. Let $W=V(C')\cup \{u\}$. Since $\beta _{t+2,t+5}(I)= 0$, it follows that ${\widetilde{H}}_{1}(\Delta _W;{\mathbb {K}})= 0$. Therefore, $T(C')\in \ker \partial _1^W$ implies that $T(C')\in \mathrm {Im}\ \partial _2^W$. Hence, $\{u, v_i\}\in E({\bar{G}})$ for all $1\le i\le t+4$.

On the other hand, since $n>t+4$ there exist $v, v'\in [n]{\setminus } V(C)$ with $v\ne v'$. Setting $W=V(C)\cup \{v,v'\}$, by (O-4)(ii), either of the following cases happens:

• (i) either $\{v,u_i\}\in E({\bar{G}})$ for all $u_i\in V(C)$ or $\{v',u_i\}\in E({\bar{G}})$ for all $u_i\in V(C)$;

• (ii) else, $t=1$ and $\Delta _W$ is isomorphic to the graph $G_{(d)_2}$ in Fig. 4. In particular, $\{v,v'\}\in E({\bar{G}})$.

We show that $V(C)\cap V(C')= \emptyset$. Suppose on contrary that $V(C)\cap V(C')\ne \emptyset$, say $u_1\in V(C')$. Then $\{u_j, u_1\}\in E({\bar{G}})$ for all $u_j\in V(C){\setminus }V(C')$. Since C is minimal we conclude that $V(C){\setminus }V(C')\subseteq \{u_2,u_{t+3}\}$. Therefore, $\{u_1,u_3,\ldots ,u_{t+2}\}\subset V(C')$. Note that $V(C){\setminus }V(C')\ne \emptyset$ because otherwise $V(C)\subset V(C')$ which does not hold.

If $|V(C){\setminus }V(C')|=1$, without loss of generality we may suppose $V(C){\setminus }V(C')=\{u_{t+3}\}$. Then since $|V(C')|-|V(C)|=1$, we have $|V(C'){\setminus }V(C)|=2$. Let $v_{j_1},v_{j_2}\in V(C'){\setminus }V(C)$. Then $V(C')=\{v_{j_1},v_{j_2}\}\cup \{u_1,\ldots ,u_{t+2}\}$ with $\{v_{j_1},v_{j_2}\}\cap \{u_1,\ldots ,u_{t+2}\}=\emptyset$. Suppose (i) happens for $v_{j_1},v_{j_2}$. We may assume that $\{v_{j_1}, u_i\}\in E({\bar{G}})$ for all $1\le i\le t+3$. Since $t\ge 1$, $|\{u_1,\ldots ,u_{t+2}\}|\ge 3$ which implies that $v_{j_1}$ is adjacent to at least 3 vertices of $C'$ in ${\bar{G}}$ which contradicts the minimality of $C'$. So (i) cannot happen when $|V(C){\setminus }V(C')|=1$. Therefore, by (ii), $t=1$ and the induced subgraph of ${\bar{G}}$ on $V(C)\cup \{v_{j_1},v_{j_2}\}$ is isomorphic to $G_{(d)_2}$. Since $V(C')\subset V(C)\cup \{v_{j_1},v_{j_2}\}$, the cycle $C'$ which is of length 5 is an induced subgraph of $G_{(d)_2}$. This is a contradiction because all cycles in $G_{(d)_2}$ are of length 4. Therefore, $|V(C){\setminus }V(C')|=2$.

It follows from $V(C){\setminus }V(C')=\{u_2,u_{t+3}\}$ that $V(C')=\{v_{j_1},v_{j_2}, v_{j_3}\}\cup \{u_1,u_3,\ldots ,u_{t+2}\}$ with $\{v_{j_1},v_{j_2}, v_{j_3}\}\cap \{u_1,u_3,\ldots ,u_{t+2}\}=\emptyset$. If at least two of the vertices $v_{j_1},v_{j_2},v_{j_3}$, say $v_{j_1},v_{j_2}$, are adjacent to all vertices of C in ${\bar{G}}$, then $v_{j_1}-u_1-v_{j_2}-u_3-v_{j_1}$ is a 4-cycle in $C'$ which contradicts the minimality of $C'$, because $|C'|\ge 5$. Hence, at most one vertex from $v_{j_1},v_{j_2},v_{j_3}$ is adjacent to all vertices of C in ${\bar{G}}$. If none of them is adjacent to all $u_i$ in ${\bar{G}}$, by (ii), we have $\{v_{j_1},v_{j_2}\},\{v_{j_2},v_{j_3}\},\{v_{j_1},v_{j_3}\}\in E({\bar{G}})$ and hence $C'$ contains a triangle which is a contradiction. Therefore, exactly one vertex among $v_{j_1},v_{j_2},v_{j_3}$, say $v_{j_1}$, is adjacent to all vertices $u_i$ in ${\bar{G}}$. Now $\{u_1,u_3,\ldots ,u_{t+2}\}\subset V(C')$ and minimality of $C'$ imply that $t=1$ and that $v_{j_1}$ is not adjacent to $v_{j_2},v_{j_3}$.

Setting $W=\{v_{j_2},v_{j_3}\}\cup V(C)$, since (i) does not happen for this W, one concludes that $\Delta _W$ is isomorphic to $G_{(d)_2}$. Therefore, $\{v_{j_2},v_{j_3}\}\in E({\bar{G}})$ and, in ${\bar{G}}$ the vertex $v_{j_2}$ is adjacent to three successive vertices $u_{\overline{i-1}}, u_{i}, u_{\overline{i+1}}$ of C, and the vertex $v_{j_3}$ is adjacent to $u_{\overline{i+1}}, u_{\overline{i+2}}, u_{\overline{i-1}}$, where $1\le i\le 4$. Since $\{v_{j_2},v_{j_3}\}\in E(C')$, and since $V(C')=\{v_{j_1}, v_{j_2},v_{j_3},u_1,u_3\}$, it follows that either $i=1$ or $i=3$, otherwise $C'$ is not minimal. Without loss of generality suppose $i=1$. Thus, $v_{j_2}$ is adjacent to $u_1,u_2,u_4$ but not to $u_3$, and $v_{j_3}$ is adjacent to $u_2,u_3,u_4$ but not to $u_1$. Setting $W=V(C)\cup \{v_{j_1},v_{j_2},v_{j_3}\}$ one has

$$\begin{aligned} T'&=\left( \sum _{1\le i\le 3}[v_{j_1},u_i,u_{i+1}]\right) -[v_{j_1},u_1,u_4]- [v_{j_2},u_1,u_{2}]\\&\quad +[v_{j_2},u_1,u_{4}]-[v_{j_3},u_2,u_{3}]\\&\quad -[v_{j_3},u_3,u_4]+[v_{j_2},v_{j_3},u_2]-[v_{j_2},v_{j_3},u_4]\in \ker \partial _2^W. \end{aligned}$$

Since $\beta _{3,7}(I)=0$, we have $T'\in \mathrm {Im}\ \partial _3^W$ which requires that $\Delta _W$ contains faces of dimension 3 which is not the case here, a contradiction. Consequently, $V(C)\cap V(C')= \emptyset$, as desired.

Setting $W=\{u_j\}\cup V(C')$ for some $1\le j\le t+3$, since $T(C')\in \ker \partial _1^W$ and $\beta _{t+2,t+5}(I)=0$, we conclude that $u_j$ is adjacent to all vertices of $C'$ in ${\bar{G}}$. In particular, $\{u_1,v_i\}, \{u_3,v_i\}\in E({\bar{G}})$ for all $1\le i\le t+4$. Let $W=V(C')\cup \{u_1,u_3\}$. Then $\Delta _W$ consists of an induced dipyramid D. Thus, $T(D)\in \ker \partial _2^W$, while $\mathrm {Im}\ \partial _3^W=0$, where T(D) is defined in (3). It follows that ${\widetilde{H}}_{2}(\Delta _W;{\mathbb {K}})\ne 0$ and so $\beta _{t+2,t+6}(I)\ne 0$, a contradiction. Therefore, $l<t+4$ and the claim follows. $\square$ $\square$

In the next corollary, we highlight some information obtained from Observation 2.1 about the vertices not belonging to a minimal cycle.

Corollary 2.4

Let G be a simple graph on [n]. Assume $I:=I(G)$ has almost maximal finite index. Let C be a minimal cycle in ${\bar{G}}$. Then

• (a) all vertices in $[n]{\setminus }V(C)$ are adjacent to some vertex in V(C) in the graph ${\bar{G}}$.

• (b) For any pair of vertices $v,v'\in [n]{\setminus }V(C)$ whenever $|N_{{\bar{G}}}(v)\cap V(C)|\le 2$, then $V(C)\subseteq N_{{\bar{G}}}(v')$.

• (c) If $\mathrm {index}(I)=1$, then there are at most two vertices in $[n]{\setminus }V(C)$ which are not adjacent to all vertices of C in ${\bar{G}}$.

• (d) If $\mathrm {index}(I)>1$, then there is at most one vertex in $[n]{\setminus }V(C)$ which is not adjacent to all vertices of C in ${\bar{G}}$.

Proof

Let $\mathrm {index}(I)=t$. By assumption $\mathrm {pd}(I)=t+1$. By Lemma 2.3, all minimal cycles of ${\bar{G}}$ are of length $t+3$. Let $C=u_1-u_2-\cdots -u_{t+3}-u_1$ be a minimal cycle of ${\bar{G}}$.

(a) Let $u_{t+4}\in [n]{\setminus }V(C)$, and set $W=V(C)\cup \{u_{t+4}\}$. Since $\beta _{t+2,t+4}(I)=0$ we conclude that ${\bar{G}}_W$ is connected using (O-1). It follows that $u_{t+4}$ is adjacent to some vertex of C in the graph ${\bar{G}}$.

(b) If $|[n]{\setminus }V(C)|\le 1$, then there is nothing to prove. Suppose $u_{t+4},u_{t+5}\in [n]{\setminus }V(C)$. Set $W= V(C)\cup \{u_{t+4},u_{t+5}\}$. Since $\beta _{t+2,t+5}(I)=0$, (O-4)(ii) implies that for each edge e of C we either have $e\cup \{u_{t+4}\}\in \Delta _W$ or $e\cup \{u_{t+5}\}\in \Delta _W$. This in particular shows that if $u_{t+4}$ is adjacent to at most 2 vertices of C in ${\bar{G}}$, then $u_{t+5}$ is adjacent to all of them in ${\bar{G}}$.

(c) Suppose $u_{5},u_{6}\in [n]{\setminus }V(C)$ are not adjacent to all vertices of C in ${\bar{G}}$. The argument in (O-4)(ii) shows that we may assume that $\{u_1,u_2,u_3\}\subseteq N_{{\bar{G}}}(u_5)$ but $u_4\notin N_{{\bar{G}}}(u_5)$ and $\{u_1,u_3,u_4\}\subseteq N_{{\bar{G}}}(u_6)$ but $u_2\notin N_{{\bar{G}}}(u_6)$. Now suppose $u_7\in [n]{\setminus }V(C)$ is not adjacent to all vertices of C in ${\bar{G}}$. By replacing $u_6$ with $u_7$ in (O-4)(ii) one sees that $u_7$ is not adjacent to $u_2$ in ${\bar{G}}$, and replacing $u_5$ with $u_7$ in the same argument shows that $u_7$ is adjacent to $u_2$ in ${\bar{G}}$, a contradiction.

(d) Suppose $u_{t+4},u_{t+5}$ are two vertices in $[n]{\setminus } V(C)$ which are not adjacent to all vertices of C in ${\bar{G}}$. The argument in (O-4)(ii) shows that $t=1$, a contradiction. $\square$

The crucial point in the classification of the edge ideals with almost maximal finite index is to determine the number of the vertices of the graph with respect to the index of the ideal. In the following, we compute this number.

Proposition 2.5

Let G be a simple graph on [n] with no isolated vertex such that $I=I(G)$ has almost maximal finite index t. Then G has either $n=t+4$ or $n=t+5$ vertices.

Proof

Since $\mathrm {index}(I)=t$ there is a minimal cycle $C=u_1-u_2-\cdots -u_{t+3}-u_1$ in ${\bar{G}}$. Moreover, ${\bar{G}}\ne C$, because otherwise $\mathrm {pd}(I)=\mathrm {index}(I)$ by [2, Theorem 4.1]. Since C is a minimal cycle, ${\bar{G}}\ne C$ means that there exists $v\in [n]{\setminus }V(C)$. Therefore, $n\ge t+4$.

Suppose on contrary that $n> t+5$. So $n-|V(C)|>2$.

Suppose first $t>1$. It follows from Corollary 2.4(d) that there exist $u_{t+4},u_{t+5}\in [n]{\setminus }V(C)$ such that $u_{t+4},u_{t+5}$ are adjacent to all vertices of C in ${\bar{G}}$. Therefore, $C':=u_{t+4}-u_1-u_{t+5}-u_3-u_{t+4}$ is a 4-cycle. Since $t>1$, $C'$ is not minimal and hence $\{u_{t+4},u_{t+5}\}\in E({\bar{G}})$.

Since $u_{t+4}, u_{t+5}$ are not isolated in G, there exist $v_1, v_2\in [n]{\setminus }(V(C)\cup \{ u_{t+4}, u_{t+5}\})$ such that $\{v_1, u_{t+4}\}, \{v_2,u_{t+5}\}\notin E({\bar{G}})$. By Corollary 2.4(a), $v_1, v_2$ are adjacent to some vertices of C in ${\bar{G}}$. If $v_1$ is adjacent to at least two vertices in ${\bar{G}}$, say $u_a, u_b\in V(C)$ such that $b\ne \overline{a+1}$ and $a\ne \overline{b+1}$, then we will have a minimal 4-cycle $v_1-u_a-u_{t+4}-u_b-v_1$ which contradicts $t>1$. Thus, $v_1$ is adjacent to either one vertex $u_a$ or two vertices $u_a, u_{\overline{a+1}}$ of C in ${\bar{G}}$. In particular, $v_1$ is not adjacent to all vertices of C in ${\bar{G}}$. Same holds for $v_2$. Corollary 2.4(d) implies that $v_1=v_2$. If $v_1$ is adjacent to only one vertex $u_a$ of C in ${\bar{G}}$, setting $W=\{u_{t+4},v_1\}\cup V(C){\setminus }\{u_a\}$, $\Delta _W$ is not connected and so $\beta _{t+2,t+4}(I)\ne 0$, a contradiction. Therefore, $v_1$ is adjacent to $u_a, u_{\overline{a+1}}$ in ${\bar{G}}$. Setting $W=\{u_{t+4},u_{t+5},v_1\}\cup V(C){\setminus } \{u_a, u_{\overline{a+1}}\}$, $\Delta _W$ is not connected and so $\beta _{t+2,t+4}(I)\ne 0$, a contradiction. Consequently, $n\le t+5$ when $t>1$.

Now suppose $t=1$. Since $n-|V(C)|>2$ we have $n\ge 7$. By Corollary 2.4(c), at least one vertex, say $v_1$ in $[n]{\setminus }V(C)$ is adjacent to all vertices of C in ${\bar{G}}$. Since $v_1$ is not isolated in G, there exists $v_2\in [n]{\setminus }(V(C)\cup \{v_1\})$ such that $\{v_1,v_2\}\notin E({\bar{G}})$. We claim that $v_2$ is not adjacent to some vertex of C in ${\bar{G}}$.

Proof of the claim

Suppose on contrary that $v_2$ is adjacent to all vertices of C in ${\bar{G}}$. Then we get an induced dipyramid D on the vertex set $V(C)\cup \{v_1,v_2\}$. Now set $W=V(C)\cup \{v_1,v_2,v_3\}$ for some $v_3\in [n]{\setminus }(V(C)\cup \{v_1,v_2\})$. Since $\beta _{3,7}(I)=0$ we have $T(D)\in \mathrm {Im}\ \partial _{3}^W$, with T(D) similar to the one in (3), which implies that each 2-face of D is contained in a 3-face of $\Delta _W$ and hence $v_3$ is adjacent to all vertices of $V(C)\cup \{v_1,v_2\}$ in ${\bar{G}}$. As $v_3$ is not isolated in G, there exists $v_4\in [n]{\setminus }(V(C)\cup \{v_1,v_2,v_3\})$ such that $\{v_3,v_4\}\notin E({\bar{G}})$. Replacing $v_3$ with $v_4$ in the above argument we conclude that $v_4$ is also adjacent to all vertices of $V(C)\cup \{v_1,v_2\}$ in ${\bar{G}}$. Now set $W=\{v_1,v_2,v_3,v_4\}\cup V(C)$. Then

$$\begin{aligned} T=&\sum _{1\le i\le 3}\left( [v_1,v_3,u_i,u_{i+1}]-[v_1,v_4,u_i,u_{i+1}]-[v_2,v_3,u_i,u_{i+1}]\right. \\&\qquad \left. +[v_2,v_4,u_i,u_{i+1}]\right) \\&-[v_1,v_3,u_1,u_4]+[v_1,v_4,u_1,u_4]+[v_2,v_3,u_1,u_4]-[v_2,v_4,u_1,u_4]\in \ker \partial _3^W \end{aligned}$$

while $T\notin \mathrm {Im}\ \partial _{4}^W$, because $\Delta _W$ contains no 4-face. This implies that $\beta _{3,8}(I)\ne 0$ which is a contradiction. So the claim follows.

Without loss of generality suppose $\{v_2,u_4\}\notin E({\bar{G}})$. Now consider $v'_3\in [n]{\setminus } (V(C)\cup \{v_1,v_2\})$. We show that $v'_3$ is adjacent to all vertices of C in ${\bar{G}}$. Otherwise, setting $W=\{v_2,v'_3\}\cup V(C)$ the same discussion as in (O-4)(ii) shows that ${\bar{G}}_W$ is isomorphic to the graph $G_{(d)_2}$ in Fig. 4, where $\{v'_3,u_2\}\notin {\bar{G}}_W$. Hence, setting $W=\{v_1,v_2,v'_3\}\cup V(C)$, we have

$$\begin{aligned} T'&=\left( \sum _{1\le i\le 3}[v_1,u_i,u_{i+1}]\right) -[v_1,u_1,u_4]- [v_2,u_1,u_{2}]-[v_2,u_2,u_{3}]\\&\quad -[v'_3,u_3,u_{4}]\\&\quad +[v'_3,u_1,u_4]-[v_2,v'_3,u_1]+[v_2,v'_3,u_3]\in \ker \partial _2^W, \end{aligned}$$

while $T'\notin \mathrm {Im}\ \partial ^W_3$ because $\{v_2,u_1,u_2\}$ is not contained in a 3-face of $\Delta _W$, and we get a contradiction. Thus, $v'_3$ is adjacent to all vertices of C in ${\bar{G}}$. It follows that setting $W=\{v_1,v_2,v'_3\}\cup V(C)$, a dipyramid D with the vertex set $V(C)\cup \{v_1,v'_3\}$ lies in $\Delta _W$ and so $T(D)\in \ker \partial _{2}^W$ which implies that $T(D)\in \mathrm {Im}\ \partial _{3}^W$. Thus, each 2-face of D is contained in a 3-face of $\Delta _W$. Since $v_2$ is not adjacent to $u_4$ in ${\bar{G}}$, we conclude that $\{v_1,v'_3\}\in E({\bar{G}})$. Note that by $\beta _{3,5}(I)= 0$, setting $W=V(C)\cup \{v_2\}$, the vertex $v_2$ is adjacent to some vertex $u_i$ of V(C) in ${\bar{G}}$. Now setting $W=\{v_1,v_2\}\cup V(C){\setminus }\{u_i\}$, the same reason implies that $v_2$ is adjacent to some vertex $u_j$ in $V(C){\setminus }\{u_i\}$ in the graph ${\bar{G}}$. Finally, setting $W=\{v_1,v_2, v'_3\}\cup V(C){\setminus }\{u_i,u_j\}$ indicates that in the graph ${\bar{G}}$ the vertex $v_2$ is adjacent to either three vertices $u_i,u_j,u_k$ of C or it is adjacent to the two vertices $u_i,u_j$ of C and to $v'_3$. We show that in the first case $v_2$ is also adjacent to $v'_3$ in ${\bar{G}}$. Suppose the first case happens. Since $\{v_2,u_4\}\notin E({\bar{G}})$, setting $W=\{v_1,v_2,v'_3,u_1,u_3,u_4\}$ we have a minimal cycle $C':=v_2-u_3-u_4-u_1-v_2$ in ${\bar{G}}_W$ with $T(C')\in \ker \partial _1^W$. Since $\beta _{3,6}(I)=0$, any edge of $C'$ must be contained in a 2-face of $\Delta _W$ and since $\{v_1,v_2\}\notin E({\bar{G}})$ it follows that $v_2$ is adjacent to $v'_3$ in ${\bar{G}}$.

Now since $v'_3$ is not isolated in G, there exists $v'_4\in [n]{\setminus }(V(C)\cup \{v_1,v_2,v'_3\})$ with $\{v'_3,v'_4\}\notin E({\bar{G}})$. Replacing $v'_3$ with $v'_4$ in the above discussion, we see that $v'_4$ is adjacent to all vertices of C in ${\bar{G}}$. Setting $W=V(C)\cup \{v_2,v'_3,v'_4\}$, we have an induced dipyramid D on the vertex set $V(C)\cup \{v'_3,v'_4\}$ with $T(D)\in \ker \partial ^W_{2}$ and since $\{v_2,u_4\}\notin {\bar{G}}_W$ we have $T(D)\notin \mathrm {Im}\ \partial ^W_{3}$ that is a contradiction with $\beta _{3,7}(I)=0$. Thus, $n\le t+5$ when $t=1$. $\square$

Now we are ready to state the main result of this section which determines the graphs whose edge ideals have almost maximal finite index.

Theorem 2.6

Let G be a simple graph on [n] with no isolated vertex and let $I=I(G)\subset \!S$. Then I has almost maximal finite index if and only if ${\bar{G}}$ is isomorphic to one of the graphs given in Example 2.2.

Proof

The “if” implication follows from Example 2.2. We prove the converse. Suppose $\mathrm {index}(I)=t$. Then there is a minimal cycle $C:=u_1-u_2-\cdots -u_{t+3}-u_1$ in ${\bar{G}}$. Moreover, by Proposition 2.5 there exists $u_{t+4}\in [n]{\setminus }V(C)$ which by Corollary 2.4(a) is adjacent to some vertex $u_i$ of V(C) in ${\bar{G}}$. Without loss of generality we may assume that $i=1$. By Proposition 2.5, we have $n-(t+3)\le 2$. We consider two cases:

Case (i) Suppose $[n]{\setminus }V(C)=\{u_{t+4}\}$ and let $1\le l\le t+3$ be the largest integer such that $\{u_{l},u_{t+4}\}\in E({\bar{G}})$.

$\bullet$:

If $l=1$, then ${\bar{G}}=G_{(a)_1}$ in Fig. 1.

$\bullet$:

If $l=2$, then ${\bar{G}}=G_{(a)_2}$ in Fig. 1.

$\bullet$:

If $3\le l<t+3$, then there is a minimal cycle $C'=u_1-u_{t+4}- u_l- u_{l+1}-\cdots -u_{t+3}-u_1$ of length $t+6-l$. By Lemma 2.3, $t+6-l=t+3$ which implies $l=3$. If $\{u_2,u_{t+4}\}\notin E({\bar{G}})$, then we will have a minimal 4-cycle on the vertex set $\{u_1,u_2,u_3,u_{t+4}\}$. It follows from Lemma 2.3 that $|C|=4$. Hence, ${\bar{G}}$ is isomorphic to $G_{(c)}$ in Fig. 3. If $\{u_2,u_{t+4}\}\in E({\bar{G}})$, then ${\bar{G}}=G_{(a)_3}$ in Fig. 1.

$\bullet$:

If $l=t+3$, then since G does not have isolated vertices, there exists $1<j<t+3$, such that $\{u_{t+4},u_j\}\notin E({\bar{G}})$. Let $k,k'$ with $1\le k< j<k'\le t+3$ be, respectively, the largest index and the smallest index such that $\{u_k,u_{t+4}\}, \{u_{k'},u_{t+4}\}\in E({\bar{G}})$. It follows that $C'=u_{t+4}-u_k- u_{k+1}-\cdots -u_{k'}-u_{t+4}$ is a minimal cycle and hence $|C'|=k'-k+2=t+3$. Therefore, we have either $(k,k')=(1,t+2)$ or $(k,k')=(2,t+3)$. In both cases, ${\bar{G}}$ is isomorphic to $G_{(a)_3}$.

Case (ii) Suppose $[n]{\setminus }V(C)=\{u_{t+4},u_{t+5}\}$. By Corollary 2.4(a) both $u_{t+4},u_{t+5}$ are adjacent to at least one vertex of C in ${\bar{G}}$.

$\bullet$:

Suppose in the graph ${\bar{G}}$ the vertex $u_{t+4}$ is adjacent to at most 2 vertices of C one of which is $u_1$. Then $u_{t+5}$ is adjacent to all vertices of C in ${\bar{G}}$ by Corollary 2.4(b). Since $\Delta _W$ is connected for $W=[n]{\setminus }\{u_1\}$, we conclude that $u_{t+4}$ is adjacent to some vertex in $[n]{\setminus }\{u_1\}$ in ${\bar{G}}$ and since $u_{t+5}$ is not isolated in G, $u_{t+5}$ is not adjacent to $u_{t+4}$ in ${\bar{G}}$ and hence $u_{t+4}$ is adjacent to some $u_j\in V(C)$ with $j\ne 1$ in ${\bar{G}}$. We show that either $j=2$ or else $j=t+3$. Otherwise there is a minimal cycle $C':=u_{t+4}-u_1-u_2-\cdots -u_j-u_{t+4}$ of length $j+1$ which is equal to $t+3$, by Lemma 2.3. Thus, $j=t+2$ which implies that $C'':=u_{t+4}-u_{t+2}-u_{t+3}-u_1-u_{t+4}$ is a minimal 4-cycle and hence $t=1$. But $T(C''')\in \ker \partial _1{\setminus }\mathrm {Im}\ \partial _{2}$, where $C''':=u_{5}-u_3-u_{6}-u_1-u_{5}$ is a minimal 4-cycle in ${\bar{G}}$. It follows that $\beta _{3,6}(I)\ne 0$, a contradiction. Thus, either $j=2$ or else $j=t+3$ and hence ${\bar{G}}$ is isomorphic to $G_{(b)}$ in Fig. 2. Same holds if we interchange $u_{t+4}$ and $u_{t+5}$ in the above argument.

$\bullet$:

Now suppose $u_{t+4}, u_{t+5}$ are adjacent to at least 3 vertices of C in ${\bar{G}}$. If $u_{t+4}$ as well as $u_{t+5}$ is not adjacent to some vertices of C in ${\bar{G}}$, then as seen in the argument of (O-4)(ii), the graph ${\bar{G}}$ is isomorphic to $G_{(d)_2}$ in Fig. 4.

Now consider the case that at least one of the vertices $u_{t+4}, u_{t+5}$, say $u_{t+5}$, is adjacent to all vertices of C in ${\bar{G}}$. The argument below also works if we interchange $u_{t+4}, u_{t+5}$.

Suppose $u_{t+4}$ is adjacent to (at least) three vertices $u_1, u_k, u_j$ of C with $1<k<j\le t+3$ in the graph ${\bar{G}}$. Since $u_{t+5}$ is not isolated in G, we have $\{u_{t+4},u_{t+5}\}\notin E({\bar{G}})$. If $(k,j)\ne (2,t+3)$, then we get the minimal 4-cycle $C'=u_{t+4}-u_{1}-u_{t+5}-u_l-u_{t+4}$, where $l=k$ if $k\ne 2$, and else $l=j$, and hence $t=1$. If $(k,j)= (2,t+3)$, then we get the minimal 4-cycle $C''=u_{t+4}-u_{2}-u_{t+5}-u_{t+3}-u_{t+4}$ and so $t=1$ also in this case. From $t=1$ we conclude that $u_1, u_k, u_j$ are successive vertices in C. Without loss of generality we may assume that $(k,j)=(2,3)$. Since $\{u_5,u_6\}\notin E({\bar{G}})$, in case $\{u_{5}, u_4\}\notin E({\bar{G}})$, the graph ${\bar{G}}$ is isomorphic to $G_{(d)_2}$, and in case $\{u_{5}, u_4\}\in E({\bar{G}})$, it is isomorphic to $G_{(d)_1}$ in Fig. 4. This completes the proof. $\square$

All the arguments so far in this section were characteristic independent; consequently

Corollary 2.7

The property of having almost maximal finite index for edge ideals is independent of the characteristic of the base field. In other words, given a simple graph G, its edge ideal I(G) has almost maximal finite index over some field if and only if it has this property over all fields.

Corollary 2.8

Let G be a simple graph on [n] with no isolated vertex such that $I=I(G)\subset \!S$ has almost maximal finite index. Then over all fields

$$\begin{aligned} \mathrm {pd} (I)={\left\{ \begin{array}{ll} n-3 \quad \text {if } {\bar{G}}=G_{(c)} \text { or } G_{(a)_i},\ i=1,2,3,\\ n-4 \quad \text {if } {\bar{G}}=G_{(b)}\text { or } G_{(d)_i}, \ i=1,2.\end{array}\right. } \end{aligned}$$

In particular, $3\le \mathrm {depth} (I)\le 4$.

Proof

Let $\mathrm {index}(I)=t$. By Theorem 2.6, ${\bar{G}}\in \{G_{(a)_1}, G_{(a)_2}, G_{(a)_3}, G_{(b)}, G_{(c)}, G_{(d)_1}, G_{(d)_2}\}$. It follows that

$$\begin{aligned} n={\left\{ \begin{array}{ll} t+4\quad \text {if } {\bar{G}}=G_{(c)}, G_{(a)_i},\ i=1,2,3,\\ t+5\quad \text {if } {\bar{G}}=G_{(b)}, G_{(d)_i},\ i=1,2.\end{array}\right. } \end{aligned}$$

Since $\mathrm {pd}(I)=t+1$, the assertion follows. The last assertion follows from the Auslander–Buchsbaum formula. $\square$

In the rest of this section, we study the last graded Betti numbers of edge ideals with almost maximal finite index. We first see in the following lemma that the graded Betti numbers of the edge ideals with this property are independent of the characteristic of the base field. The proof takes a great benefit of Katzman’s paper [23].

Lemma 2.9

Let $I\subset S$ be the edge ideal of a simple graph with almost maximal finite index. Then the Betti numbers of I are characteristic independent.

Proof

[23, Theorem 4.1] states that the Betti numbers of the edge ideals of the graphs with at most 10 vertices are independent of the characteristic of the base field. It follows that the graded Betti numbers of $I=I(G)$ with $\mathrm {index}(I)=t$ are characteristic independent when ${\bar{G}}\in \{G_{(c)},G_{(d)_1}, G_{(d)_2}\}$.

By [23, Corollary 1.6, Lemma 3.2(b)], if G has a vertex v of degree 1 or at least $|V(G)|-4$, then the Betti numbers of I(G) are characteristic independent if and only if the Betti numbers of $I(G-\{v\})$ are characteristic independent. Here $G-\{v\}$ is the induced subgraph of G on $V(G){\setminus }\{v\}$. Since the vertex $t+4$ is of degree one in $\overline{G_{(b)}}$, it follows that the Betti numbers of $I(\overline{G_{(b)}})$ are characteristic independent if and only if so are the Betti numbers of $I(\overline{G_{(a)_2}})$. For ${\bar{G}}\in \{G_{(a)_i}:\ 1\le i\le 3\}$, since $t+4$ is adjacent to at least t vertices in the graph G and since $|V(G)|=t+4$, it is enough to show that the Betti numbers of $I(G-\{t+4\})$ are characteristic independent. But $G-\{t+4\}$ is the complement of a minimal cycle of length $t+3$. Note that by Hochster’s formula, all the linear Betti numbers $\beta _{i,i+2}(I)$ are obtained from computing the dimension of ${\widetilde{H}}_0(\Delta ({\bar{G}})_W;{\mathbb {K}})$ with $W\subseteq V(G)$ and $|W|=i+2$, and this dimension equals the number of connected components of ${\bar{G}}_W$ minus one. Therefore these Betti numbers do not depend on the characteristic of the base field, see also [23, Corollary 1.2(b)]. Moreover, as seen in [2, Theorem 4.1, Proposition 4.3], the edge ideal of the complement of a minimal cycle has one nonzero non-linear Betti number $\beta _{t,t+3}(I)=1$ over all fields. Therefore, the Betti numbers of $I(G-\{t+4\})$ are characteristic independent, as desired. $\square$

For the edge ideals with linear resolution all nonlinear Betti numbers are zero. For the edge ideals with maximal finite index t, it is seen in [2, 10] that there is only one nonzero non-linear Betti number $\beta _{t,t+3}(I)=1$ over all fields. In the case of ideals with almost maximal finite index with $\mathrm {index}(I)=t$, the nonlinear Betti numbers appear in the last two homological degrees of the minimal free resolution. By the arguments that we had so far, it is easy to compute the $(t+1)$th graded Betti numbers and also tth non-linear Betti numbers, where I is the edge ideal with almost maximal finite index. Nevertheless, in the cases ${\bar{G}}=G_{(c)}$ and ${\bar{G}}=G_{(d)_i}$ for $i=1,2$ one can see the whole Betti table, using Macaulay 2, [15]. Note that since all the graphs in Example 2.2 have at most $t+5$ vertices, where $\mathrm {index}(I(G))=t$, and since the edge ideals are generated in degree 2, by Hochster’s formula it is enough to consider $\beta _{i,j}(I(G))$ for $i+2\le j\le t+5$.

Proposition 2.10

Let G be a graph such that $I:=I(G)$ has almost maximal finite index t. Then over all fields, $\beta _{t,t+4}(I)=\beta _{t,t+5}(I)=0$ and

$$\begin{aligned} \beta _{t,t+3}(I)&={\left\{ \begin{array}{ll} 1\quad \text {if } {\bar{G}}=G_{(a)_1} \text { or } G_{(a)_2}\text { or } G_{(b)},\\ 2\quad \text {if } {\bar{G}}=G_{(a)_3},\\ 3\quad \text {otherwise},\end{array}\right. } \\ \beta _{t+1,t+3}(I)&={\left\{ \begin{array}{ll} 1\quad \text {if } {\bar{G}}=G_{(a)_1} \text { or } G_{(b)},\\ 0\quad \text {otherwise,}\end{array}\right. }\\ \beta _{t+1,t+4}(I)&={\left\{ \begin{array}{ll} 2\quad \text {if } {\bar{G}}=G_{(c)} \text { or } G_{(d)_2},\\ 0\quad \text {if }{\bar{G}}=G_{(d)_1},\\ 1\quad \text {otherwise,}\end{array}\right. }\\ \beta _{t+1,t+5}(I)&={\left\{ \begin{array}{ll} 1\quad \text {if } {\bar{G}}=G_{(d)_1},\\ 0\quad \text {otherwise.}\end{array}\right. } \end{aligned}$$

In particular,

$$\begin{aligned} \mathrm {reg} (I)={\left\{ \begin{array}{ll} 4\quad \text {if } {\bar{G}}=G_{(d)_1},\\ 3\quad \text {otherwise.}\end{array}\right. } \end{aligned}$$

Proof

All the equalities are straightforward consequences of the use of Hochster’s formula and Observation 2.1. However, the Betti number $\beta _{t,t+3}(I)$ can be also deduced from [13, Theorem 4.6]. It is worth to emphasize that although ${\widetilde{H}}_1(\Delta ,{\mathbb {K}})$ is spanned by the set of $0\ne T(C)+\mathrm {Im}\ \partial _2$ for all minimal cycles C of ${\bar{G}}$, this set may not be a basis. In case ${\bar{G}}=G_{(c)}$, the graph ${\bar{G}}$ has three minimal cycles C of length 4 with $T(C)\notin \mathrm {Im}\ \partial _{2}$, but for the cycle $C=1-2-3-4-1$, T(C) is a linear combination of $T(C'), T(C'')$, where $C', C''$ are the two other cycles of $G_{(c)}$. Hence $\dim _{{\mathbb {K}}}{\widetilde{H}}_1(\Delta (G_{(c)}),{\mathbb {K}})=2$. In the case of $G_{(a)_3}$, we have $0\ne T(C)+ \mathrm {Im}\ \partial _{2}$, where C is the minimal cycle on $[t+3]$, but T(C) is a linear combination of $T(C'), T(C''),T(C''')$, where $C'$ is the minimal cycle on $[t+4]{\setminus } \{2\}$, and $C'', C'''$ are the two triangles in $G_{(a)_3}$ and hence $\dim _{\mathbb {K}}{\widetilde{H}}_1(\Delta (G_{(a)_3}),{\mathbb {K}})=1$. $\square$

4 Powers of edge ideals with large Index

Due to a result of Herzog et al. [20, Theorem 3.2], if the edge ideal $I:=I(G)$ has a linear resolution, that is $\mathrm {index}(I)=\infty$, then all of its powers have a linear resolution as well. In case I has maximal finite index $t>1$, then by [2, Corollary 4.4] the ideal $I^s$ has a linear resolution for all $s\ge 2$. Note that in general for any edge ideal I with $\mathrm {index}(I)=1$, one has $\mathrm {index}(I^s)=1$ for all $s\ge 2$, see Remark 3.3. In this section, we investigate when the higher powers of the edge ideal I with almost maximal finite index have a linear resolution. Indeed, the aim of this section is to prove the following:

Theorem 3.1

Let G be a simple graph with no isolated vertex whose edge ideal $I(G)\subset S$ has almost maximal finite index. Then $I(G)^s$ has a linear resolution for all $s\ge 2$ if and only if G is gap-free.

Theorem 3.1 follows from Remarks 3.3 and 3.5, and Theorems 3.7 and 3.11. Indeed, we will see in Remark 3.5 that this theorem holds for G with $G\in \{\overline{G_{(a)_1}}, \overline{G_{(a)_2}}\}$ with $t>1$. For $G=\overline{G_{(a)_3}}$ with $t>1$ and $G=\overline{G_{(b)}}$ we will prove the assertion in Theorems 3.7 and 3.11, respectively. As it is mentioned in Remark 3.3, all other graphs whose edge ideals have almost maximal finite index contain a gap. Recall that a gap in a graph G is an induced subgraph on 4 vertices and a pair of edges with no vertices in common which are not linked by a third edge; see the graph $G_1$ in Fig. 5. The graph G is called gap-free if it does not admit a gap; equivalently if ${\bar{G}}$ does not contain an induced 4-cycle. This property plays an important role in the study of the resolution of powers of edge ideals; for example

Proposition 3.2

(Francisco-Hà-Van Tuyl; unpublished, see [24, Proposition 1.8] and [2, Theorem 3.1]) Let G be a simple graph. If $I(G)^s$ has a linear resolution for some $s\ge 1$, then G is gap-free.

On the other hand,

Remark 3.3

A more precise statement about the gap-free graphs is given in [2, Theorem 3.1] which says that for a graph G the following are equivalent:

• (a) G admits a gap;

• (b) $\mathrm {index}(I(G)^s)=1$ for all $s\ge 1$;

• (c) there exists $s\ge 1$ with $\mathrm {index}(I(G)^s)=1$.

If G is the graph whose complement is one of $G_{(a)_1}, G_{(a)_2}, G_{(a)_3}, G_{(b)}$ with $t=1$, or one of $G_{(c)}, G_{(d)_1}, G_{(d)_2}$, then G has a gap. So by the above equivalence $\mathrm {index}(I(G)^s)=1$ for all $s\ge 1$ in this case.

In order to prove Theorem 3.1 for $G=\overline{G_{(a)_3}}$ with $t>1$, we need the following result of Banerjee [1].

Theorem 3.4

([1, Theorem 5.2]) Let G be a simple graph and let $I:=I(G)$ be its edge ideal. Let ${\mathcal {G}}(I^s)=\{m_1,\ldots , m_r\}$. Then for all $s\ge 1$

$$\begin{aligned} \mathrm {reg}(I^{s+1})\le \max \{\mathrm {reg}(I^s),\ \mathrm {reg}(I^{s+1}: m_k) + 2s \text { for } 1\le k\le r \}, \end{aligned}$$

where $(I^{s+1}: m_k)$ denotes the colon ideal, i.e., $(I^{s+1}: m_k)=\{f\in S:\ fm_k\in I^{s+1}\}$.

As a consequence of this theorem, Banerjee showed in [1, Theorem 6.17] that for any gap-free and cricket-free graph G, the ideal $I(G)^s$ has a linear resolution for all $s\ge 2$. A cricket is a graph isomorphic to the graph $G_2$ in Fig. 5, and a graph G is called cricket-free if G contains no cricket as an induced subgraph.

Two other classes of graphs which produce edge ideals whose higher powers have linear resolution were given by Erey. She proved in [11, 12] that $I(G)^s$ has a linear resolution for all $s\ge 2$ if G is gap-free and also it is either diamond-free or $C_4$-free. A diamond is a graph isomorphic to the graph $G_3$ in Fig. 5, and a diamond-free graph is a graph with no diamond as its induced subgraph. A $C_4$-free graph is a graph which does not contain a 4-cycle as an induced subgraph, i.e., its complement is gap-free.

Fig. 5

$G_1$ a gap, $G_2$ a cricket, $G_3$ a diamond

Remark 3.5

Clearly, the graphs $\overline{G_{(a)_1}}, \overline{G_{(a)_2}}$ are cricket-free and hence the statement of Theorem 3.1 holds in these two cases using [1, Theorem 6.17]. Note that these graphs are gap-free for $t\ge 2$.

On the other hand, the graphs $\overline{G_{(a)_3}}$ and $\overline{G_{(b)}}$ contain crickets for large enough t. Indeed, if $t\ge 3$, then the induced subgraph of $\overline{G_{(a)_3}}$ on the vertex set $\{1,2,3, 5,t+4\}$, and if $t\ge 2$, then the induced subgraph of $\overline{G_{(b)}}$ on $\{3,4,5,t+4,t+5\}$ are isomorphic to a cricket. These graphs are not diamond-free in general as well, because for $t\ge 3$, the induced subgraphs on the vertex sets $\{2,4,6,t+4\}$ and $\{3,5,6,t+5\}$ form, respectively, diamonds in $\overline{G_{(a)_3}}$ and $\overline{G_{(b)}}$. They are not even $C_4$-free for $t\ge 3$ since ${G_{(a)_3}}$ and ${G_{(b)}}$ contain gaps. Therefore, when $G\in \{\overline{G_{(a)_3}},\overline{G_{(b)}}\}$ and t is large enough, one cannot take benefit of the results of Banerjee or Erey to deduce Theorem 3.1.

It is shown in [1, Section 6] that for the edge ideal I of a simple graph G and the minimal generator $m_k$ of $I^s$, $s\ge 1$, the ideal $(I^{s+1}: m_k)$ is a quadratic monomial ideal whose polarization coincides with the edge ideal of a simple graph with the construction explained in Lemma 3.6. For the details about the polarization technique, the reader may consult with [19]. Throughout this section, for an edge $e=\{i,j\}$ of a graph G its associated quadratic monomial $x_ix_j$ is denoted by $\mathbf{x}_e$.

Lemma 3.6

[1, Lemma 6.11] Let G be a simple graph with the edge ideal $I:=I(G)$, and let $m_k=\mathbf{x}_{e_1}\cdots \mathbf{x}_{e_s}$ be a minimal generator of $I^s$, where $e_1,\ldots , e_s$ are some edges of G. Then the polarization $(I^{s+1}: m_k)^\mathrm{pol}$ of the ideal $(I^{s+1}: m_k)$ is the edge ideal of a new graph $G_{e_1\ldots e_s}$ with the following structure:

• (1) $V(G)\subseteq V(G_{e_1\ldots e_s})$, $E(G)\subseteq E(G_{e_1\ldots e_s})$.

• (2) Any two vertices u, v, $u\ne v$, of G that are even-connected with respect to $e_1\cdots e_s$ are connected by an edge in $G_{e_1\ldots e_s}$.

• (3) For every vertex u which is even-connected to itself with respect to $e_1\cdots e_s$ there is a new vertex $u'\notin V(G)$ which is connected to u in $G_{e_1\ldots e_s}$ by an edge and not connected to any other vertex (so $\{u,u'\}$ is a whisker in $G_{e_1\ldots e_s}$ ).

In [1], two vertices u and v of a graph G (u may be same as v) are said to be even-connected with respect to an s-fold product $e_1\cdots e_s$ in G if there is a path $P=p_0-p_1-\cdots -p_{2k+1}$, $k\ge 1$, in G such that:

• (1) $p_0=u, p_{2k+1}=v.$

• (2) For all $0\le l \le k-1$, $\{p_{2l+1}, p_{2l+2}\}=e_i$ for some $1\le i\le s$.

• (3) For all i, $|\{l:\ 0\le l\le k-1, \ \{p_{2l+1}, p_{2l+2} \}=e_i \} | \le | \{j :\ 1\le j\le s, \ e_j=e_i \} |$.

• (4) For all $0 \le r \le 2k$, $\{p_r, p_{r+1}\}$ is an edge in G.

Now we are ready to prove Theorem 3.1 for $G=\overline{G_{(a)_3}}$ with $t>1$.

Theorem 3.7

Let G be a graph on $n\ge 6$ vertices such that $G_{(a)_3}$ is its complement. Let $I:=I(G)$ be the edge ideal of G. Then $I^s$ has a linear resolution for $s\ge 2$.

Proof

Note that $t+4=n\ge 6$ implies that $t>1$. We first show that for any $s\ge 1$ and any s-fold product $e_1\cdots e_s$ of the edges in G, the graph $\overline{G_{e_1\cdots e_s}}$ is chordal, where $G_{e_1\cdots e_s}$ is a simple graph explained in Lemma 3.6 with the edge ideal $I(G_{e_1\cdots e_s})=(I^{s+1}:\ \mathbf{x}_{e_1}\cdots \mathbf{x}_{e_s})^\mathrm{pol}$.

Since by [1, Lemmas 6.14, 6.15], any induced cycle of $\overline{G_{e_1\cdots e_s}}$ is an induced cycle of ${\bar{G}}$, we conclude that if $\overline{G_{e_1\cdots e_s}}$ contains an induced cycle C of length $>3$, then $C\in \{C_1, C_2\}$, where $C_1=1-2-\cdots -(t+3)-1$ and $C_2=1-(t+4)-3-4-\cdots -(t+3)-1$. Thus, in order to prove that $\overline{G_{e_1\cdots e_s}}$ is chordal, we need to show that $C_1,C_2$ are not induced cycles in $\overline{G_{e_1\cdots e_s}}$ for $s\ge 1$.

We claim that there exist $k,l\in V(G_{e_1\cdots e_s})$ such that $\{k,l\}\in E(G_{e_1\cdots e_s})\cap E(C_r)$, $r=1,2$. It follows that $C_r$ is not a subgraph of $\overline{G_{e_1\cdots e_s}}$, as desired.

Proof of the claim

Let $e_1=\{i,j\}$ with $i<j$ and let $s\ge 1$. We choose $\{k,l\}\in E(C_1)$ as follows:

(a):

If $e_1=\{4,t+4\}$, then let $k=1$ and $l=t+3$;

(b):

if $e_1=\{1,t+2\}$, then let $k=3$ and $l= 2$;

(c):

if $e_1=\{2,t+3\}$, then let $k=4$ and $l= 3$;

(d):

otherwise, let $k=\overline{i-2}$ and $l=\overline{i-1}$.

Since $C_2$ is obtained from $C_1$ by replacing 2 with $t+4$, in order to find $\{k,l\}\in E(C_2)$, we choose $\{k,l\}\in E(C_2)$ as suggested in $(a)-(d)$ with an extra condition that if $\{k,l\}$ is obtained from (b), (d) and it contains 2, then we replace 2 with $t+4$ in this pair.

By the above choices of k, l, although $\{k,l\}\notin E(G)$, we have $\{k,i\}, \{j, l\}\in E(G)$. It follows that $k-i-j-l$ is a path in G and hence, by definition, k and l are even-connected with respect to $e_1\cdots e_s$. Therefore, $\{k,l\}\in E(G_{e_1\cdots e_s})$. This completes the proof of the claim. $\square$

Now since $\overline{G_{e_1\cdots e_s}}$ is chordal for $s\ge 1$, by [14, Theorem 1], $I(G_{e_1\cdots e_s})$ has a 2-linear resolution for $s\ge 1$. It follows that for any choice of the edges $e_1,\ldots , e_s$ of G one has

$$\begin{aligned} \mathrm {reg}((I^{s+1}:\ \mathbf{x}_{e_1}\cdots \mathbf{x}_{e_s}))=\mathrm {reg}((I^{s+1}:\ \mathbf{x}_{e_1}\cdots \mathbf{x}_{e_s})^\mathrm{pol})=\mathrm {reg}(I(G_{e_1\cdots e_s}))=2. \end{aligned}$$

The first equality follows from [19, Corollary 1.6.3]. By Proposition 2.10, we have $\mathrm {reg}(I)=3$. Theorem 3.4 implies that $\mathrm {reg}(I^2)\le 4$. Since $I^2$ is generated in degree 4 we conclude that $\mathrm {reg}(I^2)=4$. Now induction on $s>1$ and using Theorem 3.4 yield the assertion. $\square$

Now it remains to prove Theorem 3.1 for $\overline{G_{(b)}}$. The crucial point in the proof of Theorem 3.1 for $\overline{G_{(a)_3}}$ was to show that $\mathrm {reg}((I(\overline{G_{(a)_3}})^{s+1}:\ m_k))=2$ for all minimal generators $m_k$ of $I(\overline{G_{(a)_3}})^{s}$. Having proved this statement, we deduced that the upper bound of $\mathrm {reg}(I(\overline{G_{(a)_3}})^{s+1})$ in Theorem 3.4 is $2s+2$ and hence the desired conclusion was followed. The same method will not work for $\overline{G_{(b)}}$. Indeed, computations by Macaulay 2, [15], show that $\mathrm {reg}((I(\overline{G_{(b)}})^{s+1}:\ m_k))=3$, where $s\ge 1$ and $m_k=x_{t+5}^sx_{t+4}^s$ is a minimal generator of $I(\overline{G_{(b)}})^{s}$. Hence, the upper bound of $\mathrm {reg}(I(\overline{G_{(b)}})^{s+1})$ in Theorem 3.4 is at least $2s+3$ which is greater than the degree of the generators of $I(\overline{G_{(b)}})^{s+1}$ and consequently one cannot deduce that this ideal has a linear resolution only by computing the upper bound in Theorem 3.4. Therefore, in order to prove Theorem 3.1 for $\overline{G_{(b)}}$ we need some other tool. This tool is provided in the following result of Dao et al.

Lemma 3.8

[9, Lemma 2.10] Let $I\subset S$ be a monomial ideal, and let x be a variable appearing in some generator of I. Then

$$\begin{aligned} \mathrm {reg}(I)\le \max \{\mathrm {reg}((I:x)) + 1,\mathrm {reg}(I+(x))\}. \end{aligned}$$

Moreover, if I is squarefree, then $\mathrm {reg}(I)$ is equal to one of these terms.

We will apply this result for $I:=I(\overline{G_{(b)}})^{s+1}$, $s\ge 1$, and $x:=x_{t+5}$. In Theorem 3.10, we will compute the regularity of the ideal $(I(\overline{G_{(b)}})^{s+1}:x_{t+5})$, $s\ge 1$, by showing that it has linear quotients. Recall that a graded ideal I is said to have linear quotients if there exists a homogeneous generating set of I, say $\{f_1,\ldots ,f_m\}$, such that the colon ideal $\left( (f_1,\ldots ,f_{i-1}):f_i\right)$ is generated by variables for all $i>1$. By [19, Theorem 8.2.1] equigenerated ideals with linear quotients have a linear resolution.

In the next result, Proposition 3.9, we provide a step of the proof of Theorem 3.10 which is a bit long yet easy to follow. In the proof of this proposition and also Theorem 3.10, we need to order the generators of the given ideals. To this end, we should first order the multisets of edges of the associated graphs. We will use the following order in both proofs:

Let G be a simple graph. For $e=\{i,j\}\in E(G)$ with $i<j$ and $e'=\{i',j'\}$ with $i'<j'$, we let $e<e'$ if either $i<i'$ or $i=i'$ with $j<j'$. Let $r\ge 1$. We denote by ${\mathbf {e}}:=(e_{i_1},\ldots ,e_{i_{r}})$ the multiset $\{e_{i_1},\ldots ,e_{i_{r}}\}$ of the edges of G, where $e_{i_1}\le \cdots \le e_{i_{r}}$. If $\mathbf {e'}:=(e_{i'_1},\ldots ,e_{i'_{r}})$ is another ordered multiset in E(G) of the same size r, we let ${\mathbf {e}}\le \mathbf {e'}$ if either ${\mathbf {e}}=\mathbf {e'}$ or else there exists $1\le j\le r$ such that $e_{i_l}=e_{i'_l}$ for all $l<j$ and $e_{i_j}<e_{i'_j}$.

For the ordered multiset ${\mathbf {e}}:=(e_{i_1},\ldots ,e_{i_{r}})$ of the edges of the graph G, we denote by $\mathbf {x}_{{\mathbf {e}}}$ the monomial $\mathbf {x}_{e_{1}}\cdots \mathbf {x}_{e_{r}}$. Moreover, we denote by $\mathrm {supp}(m)$ the set of all variables dividing the monomial $m\in S$ and also denote by $\deg _{m}x_{i}$ the largest integer d such that $x_i^d$ divides m. We use the notation $m|m'$ ( resp.) when a monomial m divides (does not divide resp.) a monomial $m'$.

Proposition 3.9

Let $C=1-2-\cdots -(t+3)-1$, $t\ge 1$, be a cycle graph and let $t+4$ be a vertex not belonging to C. Then for $s\ge 0$ the ideal $L=I({\bar{C}})^{s}(x_3,\ldots ,x_{t+4})$ has linear quotients.

Proof

For $s=0$ the assertion is obvious. Suppose $s\ge 1$. We order the edges of $G:={\bar{C}}$ as described above. Each element m in the minimal generating set ${\mathcal {G}}(L)$ of L can be written as $m=\mathbf {x}_{\mathbf {e}_{}}x_k$, where $\mathbf {e}_{}=(e_{1},\ldots , e_{{s}})$ is an ordered multiset of the edges of ${\bar{C}}$ and $x_{k}\in \{x_3,\ldots ,x_{t+4}\}$. Note that there may be different multisets associated to m and hence different presentations of m as above. In this case, we consider the presentation of m whose associated ordered multiset of the edges is the smallest. This means that if there is another presentation of m as $\mathbf {x}_{\mathbf {e}_{}'}x_{k'}$ with , then we consider the presentation $\mathbf {x}_{\mathbf {e}_{}}x_k$ for m if $\mathbf {e}_{}<\mathbf {e}_{}'$. By this setting, each minimal generator $m_l$ of L has a unique smallest presentation $m_l=\mathbf {x}_{\mathbf {e}_{l}}x_k$, where $\mathbf {e}_{l}$ denotes the smallest multiset of the edges associated to $m_l$.

Now we order the generators of L as follows: for $m_q,m_l\in {\mathcal {G}}(L)$ with $m_q=\mathbf {x}_{\mathbf {e}_{q}}x_{k'}$, $m_l=\mathbf {x}_{\mathbf {e}_{l}}x_{k}$, we let $m_q<m_l$ if either ${\mathbf {e}}_q<{\mathbf {e}}_{l}$ or ${\mathbf {e}}_q={\mathbf {e}}_{l}$ with $k'<k$.

Suppose ${\mathcal {G}}(L)=\{m_1,\ldots ,m_r\}$ with $m_1<\!\cdots \!<m_r$. We show that for any $m_l\in \!{\mathcal {G}}(L)$ with $l\!>1$, the ideal $\left( (m_1,\ldots , m_{l-1}):m_l\right)$ is generated by some variables. Set $J_l:=(m_1,\ldots , m_{l-1})$. By [19, Proposition 1.2.2], the ideal $(J_l:m_l)$ is generated by the elements of the set $\{m_q/\gcd (m_q,m_l):\ 1\le q\le l-1\}$. Let $m_{q,l}:=m_q/\gcd (m_q,m_l)$ for $m_q<m_l$. Suppose $m_l={\mathbf {x}}_{{\mathbf {e}}_l} x_k$, $m_q={\mathbf {x}}_{{\mathbf {e}}_q} x_{k'}$ with ${\mathbf {e}}_l:=(e_{1},\ldots ,e_{s})$, ${\mathbf {e}}_q:=(e'_{1},\ldots ,e'_{s})$ and $3\le k, k'\le t+4$. Let $e_{i}=\{a_i, b_i\}$, $e'_{i}=\{a'_i, b'_i\}$ with $1\le a_i<b_i-1\le t+2$ and $1\le a'_i<b'_i-1\le t+2$ for $1\le i\le s$.

In order to show that $(J_l:m_l)$ is generated by variables, we show that for each $q<l$, there exists $p<l$ such that $m_{p,l}$ is of degree one and it divides $m_{q,l}$. If $\deg m_{q,l}=1$, then we set $p:=q$ and so we are done. Assume that $\deg m_{q,l}>1$. First suppose $q=1$. Then $m_q=x_1^sx_3^{s+1}$. If $x_1| m_{q,l}$, then there exists $1\le i\le s$ with $1\notin e_i=\{a_i,b_i\}$. If $b_i\ne t+3$, then set $e:=\{1,b_i\}$ and if $b_i=t+3$ with $a_i\ne 2$, set $e:=\{1,a_i\}$. Now set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_i})x_k$. Since $e<e_i$, we have $p<l$. Moreover, $m_{p,l}=x_1$ and so we are done in this case. Suppose $e_i=\{2,t+3\}$ for all $e_i\in \mathbf {e}_{l}$ with $1\notin e_i$. It follows that $x_3|m_{q,l}$. If $k\ne 3$, then set $m_p:=\mathbf {x}_{\mathbf {e}_{l}}x_{3}$. Since $3<k$, we have $p<l$ and $m_{p,l}=x_3$. If $k=3$, then set $e:=\{1,3\}$ and $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_i})x_{t+3}$. Now suppose . Then $x_3|m_{q,l}$ and $1\in e_i$ for all $1\le i\le s$, and since $\deg m_{q,l}>1$ there exists $e_i\in \mathbf {e}_{l}$ with $3\notin e_i$. Set $e:=\{1,3\}$ and $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_i})x_k$. So we are done if $q=1$. Now suppose $q>1$ and for all $q'<q$, there is $p'<l$ with $\deg m_{p',l}=1$ and $m_{p',l}|m_{q',l}$. We prove the assertion by induction on q.

Suppose there exist $e'_i\in {\mathbf {e}}_q$ and $e_j\in {\mathbf {e}}_l$ with $e'_i=e_j$. The monomials $m'_{l}:= m_l/{\mathbf {x}}_{e_j}, m'_q:=m_q/{\mathbf {x}}_{e'_i}$ belong to $I({\bar{C}})^{s-1}(x_3,\ldots ,x_{t+4})$ and $m'_q<m'_l$ and $m_{q,l}=m'_q/\gcd (m'_q,m'_l)$. If there exists $m'_p\in I({\bar{C}})^{s-1}(x_3,\ldots ,x_{t+4})$ with $m'_p<m'_l$, where $m'_p/\gcd (m'_p,m'_l)$ is of degree one dividing $m'_q/\gcd (m'_q,m'_l)$, then setting $m_p:=m'_p{\mathbf {x}}_{e_i}$ one has $m_p\in J_l$ and $\deg m_{p,l}=1$, where $m_{p,l}$ divides $m_{q,l}$, as desired. So it is enough to prove the assertion for $m'_q,m'_l$. Consequently, from now on we may suppose that ${\mathbf {e}}_q\cap {\mathbf {e}}_l=\emptyset$. In particular, ${\mathbf {e}}_q\ne {\mathbf {e}}_l$ and hence ${\mathbf {e}}_q<{\mathbf {e}}_l$. Since ${\mathbf {e}}_q, {\mathbf {e}}_l$ do not share an edge, it follows that $e'_1<e_1$ which means that either $a'_1<a_1$ or $a'_1=a_1$ with $b'_1<b_1$.

Case (i) $a'_1<a_1$. If $a'_1=k$, then $3\le k<a_1<b_1-1$ implies that $e:=\{k, b_1\}\in E({\bar{C}})$ with $e<e_1$ and hence by interchanging $x_{a_1}$ in $\mathbf {x}_{e_{1}}$ and $x_k$ we get a smaller presentation for $m_l$, a contradiction. Therefore, $a'_1\ne k$. Note that $a'_1<a_1\le a_i<b_i$ for all i. Thus, which implies that $x_{a'_1}|m_{q,l}$.

Since $a'_1<b_i-1$ for all i, we have $e:=\{a'_1,b_i\}\in E({\bar{C}})$, unless $\{a'_1,b_i\}=\{1,t+3\}$. If $\{a'_1,b_i\}=\{1,t+3\}$ for all i and if there exists i with $a_i\ne 2$, then set $e:=\{a'_1,a_i\}$. In both cases, we have $e<e_i$ and hence $m_p:= (\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_i})x_k<m_l$ with $m_{p,l}=x_{a'_1}$. Suppose $a'_1=1$ and $e_i=\{2,t+3\}$ for all i. We have $k\in \{3,t+3,t+4\}$, because otherwise by interchanging $x_{t+3}$ in $\mathbf {x}_{e_i}$ and $x_k$ we get a smaller presentation for $m_l$, a contradiction. If $k=3$, then set $e:=\{1,3\}$ and $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_i})x_{t+3}$. If $k\in \{t+3,t+4\}$, since $b'_1\notin \{2,t+3,t+4\}$, we have and hence $x_{b'_1}|m_{q,l}$. Set $m_p:=\mathbf {x}_{\mathbf {e}_{l}}x_{b'_1}$.

Case (ii) $a'_1=a_1$ and $b'_1<b_1$. If $x_{b'_1}|m_{q,l}$ then set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_1}/\mathbf {x}_{e_1})x_k$. Suppose and hence $x_{b'_1}|m_l$. If $k=b'_1$, then interchanging $x_{b_1}$ in $\mathbf {x}_{e_1}$ and $x_k=x_{b'_1}$ will result in a smaller presentation for $m_l$, a contradiction. Therefore, $k\ne b'_1$ and hence $b'_1\in e_h$ for some $e_h\in \mathbf {e}_{l}$. Thus, $e_h\ne e_1$. If b is another vertex of $e_h$, then $b\in \{b_1,b_1-1,\overline{b_1+1}\}$ since otherwise we get a smaller presentation of $m_l$ by interchanging $b_1$ in $e_1$ and $b'_1$ in $e_h$, a contradiction.

Since $\deg m_{q,l}>1$, we have $\mathrm {supp}(m_{q,l})\ne \{x_{t+4}\}$. Thus, $x_a|m_{q,l}$ for some $a\ne t+4$. If $b_1\notin \{a,\overline{a-1},\overline{a+1}\}$ ($b\notin \{a,\overline{a-1},\overline{a+1}\}$ resp.), then set $e:=\{a,b_1\}$ ($e:=\{a,b\}$ resp.) and $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_1}\mathbf {x}_{e}/(\mathbf {x}_{e_1}\mathbf {x}_{e_h}))x_k$. Suppose $b_1, b\in \{a,\overline{a-1},\overline{a+1}\}$ for all $a\ne t+4$ with $x_a|m_{q,l}$.

If $b_1= \overline{a+1}$, since $b_1>1$, we have $a\ne t+3$ and $b_1=a+1$. Since $a_1<b'_1-1<b_1-1=a$, one has $e:=\{a_1,a\}\in E({\bar{C}})$. Set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_1})x_k$. Now suppose $b_1\in \{a,\overline{a-1}\}$ for all a with $x_a|m_{q,l}$ and $a\ne t+4$. If $a=1$, then since $a'_1=a_1$ is the smallest vertex in $\mathbf {e}_{q}$ one has $a_1=1$ and $b_1\in \{1,t+3\}$ which is a contradiction. If $a\in \{2,3\}$, then $b_1\in \{1,2,3\}$ which is again a contradiction because $1\le a_1<b'_1-1<b_1-1$. Therefore, $a\ge 4$ for all a with $x_a|m_{q,l}$. In particular, $\overline{a-i}=a-i$ for $a\ne t+4$ and $i=1,2,3$. If $a<k$, then set $m_p:=\mathbf {x}_{\mathbf {e}_{l}}x_a$ and so we are done. Suppose $a\ge k$ for all a with $x_a|m_{q,l}$. Since $\deg m_{q,l}>1$, there exists a with $x_a\in \mathrm {supp}(m_{q,l})\cap \mathrm {supp}(\mathbf {x}_{\mathbf {e}_{q}})$. Suppose $e'_{i_1}=\{a,c\}\in \mathbf {e}_{q}$.

If $x_{c}|m_{q,l}$, then $c\ne t+4$ implies that $b_1\in \{a,a-1\}\cap \{c,{c-1}\}$. But $c\notin \{a,a-1,\overline{a+1}\}$ and hence $\{a,a-1\}\cap \{c,{c-1}\}=\emptyset$, a contradiction. Thus, and consequently $x_{c}|m_l$.

Suppose $c=k$. Then $c\le a$ and since $\{a,c\}\in E({\bar{C}})$ we have $c<a-1$. If $c\ne {a-2}$, then $b\in \{a,{a-1},\overline{a+1}\}$ implies that $e:=\{c, b\}\in E({\bar{C}})$. Set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}\mathbf {x}_{e'_1}/(\mathbf {x}_{e_1}\mathbf {x}_{e_h}))x_a$. We have $m_p\le m_l$. If $m_p=m_l$, then $b_1=a$ which implies that $(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}\mathbf {x}_{e'_1}/(\mathbf {x}_{e_1}\mathbf {x}_{e_h}))x_a$ is a smaller presentation for $m_l$, a contradiction. Thus, $m_p<m_l$ and $m_{p,l}=x_a$, as desired. Now suppose $k=c={a-2}$. Then $a\ge 5$, and $a_1\in \{a-1,a-2,{a-3}\}$ since otherwise $\{a_1,a-2\}\in E({\bar{C}})$ and by interchanging $x_{b_1}$ in $\mathbf {x}_{e_1}$ and $x_k=x_{a-2}$ in the presentation of $m_l$ we get a smaller presentation which is a contradiction. From $a_1\in \{a-1,a-2,{a-3}\}$, and $a_1+1<b'_1<b_1\in \{a,a-1\}$, we conclude that $a_1={a-3}$, $b'_1=a-1$ and $b_1=a$. Thus, $b\in \{a,a-1,\overline{a+1}\}$ implies that $b=\overline{a+1}$ and therefore interchanging $x_{b'_1}$ in $\mathbf {x}_{e_h}$, where $e_h=\{b'_1,b\}=\{a-1,\overline{a+1}\}$, and $x_k=x_{a-2}$ will give a smaller presentation, a contradiction. Note that since $a\ge 5$, we have $\{b, a-2\}=\{\overline{a+1}, a-2\}\in E({\bar{C}})$.

Assume now that $c\ne k$. Then there is $e_{i_2}\in \mathbf {e}_{l}$ with $c\in e_{i_2}$. If d is another vertex of $e_{i_2}$, then we may assume that $d<a$, because $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ and if $d>a$, then we can set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_{i_1}}/\mathbf {x}_{e_{i_2}})x_k$ which yields the result.

First assume that $b_1=a$. If $b'_1\le d$, since $a_1+1<b'_1\le d<a=b_1$, we have $e:=\{a_1,d\}\in E({\bar{C}})$ with $e<e_1$ and hence interchanging a in $e_1$ and d in $e_{i_2}$ will give a smaller presentation for $m_l$, a contradiction. Thus, $b_1=a$ implies that $b'_1>d$ and since $d<b'_1<b_1$ we have $e:=\{d, b_1\}\in E({\bar{C}})$, because otherwise $d=1$ and $b_1=t+3$ and since $a_1\le d$ we have $a_1=1$ and $e_1=\{1,t+3\}$, a contradiction. Moreover, $e_{i_2}\ne e_1,e_h$. Set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_1}\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}/(\mathbf {x}_{e_1}\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_h}))x_k$. We have $m_p\le m_l$. If $b=a$, then $(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_1}\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}/(\mathbf {x}_{e_1}\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_h}))x_k$ is a smaller presentation for $m_l$, a contradiction. Hence, $b\ne a$ and thus $m_p<m_l$ with $m_{p,l}=x_a$.

Now assume that $b_1= a-1$ which implies that $b\in \{a,a-1\}$, because $b\in \{b_1-1, b_1,\overline{b_1+1}\}\cap \{a-1,a,\overline{a+1}\}$. If $d<b'_1$, then $d<b'_1<b_1=a-1\le b$. If $\{d,b\}= \{1,t+3\}$, then $d=1$ implies that $a_1=1$ and since $c\ne a-1$ we have $e_1\ne e_{i_2}$ which implies that $\{1,a-1\}=e_1<e_{i_2}=\{1,c\}$ and hence $b_1=a-1<c$. Moreover, $b=t+3\in \{a-1,a\}$ implies that $a=t+3$ and hence $c=t+3$, a contradiction to $\{a,c\}\in E({\bar{C}})$. Thus, $\{d,b\}\ne \{1,t+3\}$ which implies that $e:=\{d,b\}\in E({\bar{C}})$. Set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}\mathbf {x}_{e'_1}\mathbf {x}_{e'_{i_1}}/(\mathbf {x}_{e_h}\mathbf {x}_{e_1}\mathbf {x}_{e_{i_2}}))x_k$. Thus, we may suppose that $b'_1\le d$. If $d<a-1$, then set $e:=\{a_1,d\}$ and since $e_1\ne e_{i_2}$ we set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}/(\mathbf {x}_{e_1}\mathbf {x}_{e_{i_2}}))x_k$. Suppose now that $d=a-1$.

Note that from $k\le a$ we conclude that $k\in \{a-1,a\}$ since otherwise if $k=a-2$, then by interchanging $x_{b_1}=x_{a-1}$ in $\mathbf {x}_{e_1}$ and $x_{k}=x_{a-2}$ we get a smaller presentation for $m_l$, and if $k<a-2$, setting $e:=\{k,b\}$, we again get $(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e'_1}\mathbf {x}_{e}/(\mathbf {x}_{e_1}\mathbf {x}_{e_h}))x_{a-1}$ as a smaller presentation for $m_l$.

If $x_{a-1}|m_{q,l}$, or if $\deg _{m_q}x_{a-1}< \deg _{m_l}x_{a-1}$, then set $m_{q'}:=(\mathbf {x}_{\mathbf {e}_{q}}\mathbf {x}_{e_{i_2}}/\mathbf {x}_{e'_{i_1}})x_{k'}$. Since $m_{q'}<m_q$ and since $\mathrm {supp}(m_{q',l})\subseteq \mathrm {supp}(m_{q,l})$, by induction hypothesis we are done. Suppose $\deg _{m_q}x_{a-1}= \deg _{m_l}x_{a-1}$. Since $x_{a-1}|m_l$, we have $x_{a-1}|m_q$ as well. Note that $k'\ne a-1$, otherwise interchanging $x_{k'}$ and $x_a$ in $\mathbf {x}_{e'_{i_1}}$ will result in a smaller presentation for $m_q$, a contradiction. It follows that there exists $e'_{i_3}=\{a-1,f\}\in \mathbf {e}_{q}$ for some f with $f\ne c, a_1$ because $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$.

If $x_f|m_{q,l}$, then we must have $a-1=b_1\in \{f,\overline{f-1}, \overline{f+1}\}$, a contradiction. Thus, . As $f\ne a,a-1$, we have $f\ne b$ and also $f\ne k$ which implies that f appears in an edge of $\mathbf {e}_{l}$. If $f=b'_1$, then again $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ implies that $b=a$. Therefore, we have $e'_{i_1}=\{a,c\}, e'_{i_3}=\{a-1, b'_1\}\in \mathbf {e}_{q}$ and $e_{i_2}=\{a-1,c\}, e_h=\{a,b'_1\}\in \mathbf {e}_{l}$ which contradict the fact that $\mathbf {e}_{l}$ and $\mathbf {e}_{q}$ are the smallest multisets associated to $m_l$ and $m_q$, respectively. Thus, $f\ne b'_1$ and hence $f\notin e_1\cup e_h\cup e_{i_2}\cup \{k\}$. It follows that there exists $e_{i_4}\ne e_1,e_{i_2}, e_h$ such that $e_{i_4}=\{f,g\}\in \mathbf {e}_{l}$ for some g with $g\ne a-1$. If $g>a$, then $a\le t+2$ and hence $\overline{a+1}=a+1$. We have $f\ne a+1$ because otherwise we will have $g>a+2$ and since $k\in \{a,a-1\}$ by interchanging $x_f$ in $\mathbf {x}_{e_{i_4}}$ and $x_k$ one gets a smaller presentation for $m_l$ which is a contradiction. Note that assuming $f=a+1$ one deduces from $g>a$ and $g\notin \{a,a+1,\overline{a+2}\}$ that $a+1<g\le t+3$ and hence $\overline{a+2}=a+2$. Now set $e:=\{f,a\}\in E({\bar{C}})$ and $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_{i_4}})x_k$. If $g=a$, then we have $e'_{i_1}=\{a,c\}, e'_{i_3}=\{a-1, f\}\in \mathbf {e}_{q}$ and $e_{i_2}=\{a-1,c\}, e_{i_4}=\{a,f\}\in \mathbf {e}_{l}$ which again contradict the fact that $\mathbf {e}_{l}$ and $\mathbf {e}_{q}$ are the smallest multisets associated to $m_l$ and $m_q$, respectively. Thus, $g\ne a$ which implies that $g<a-1$. If $g\notin \{a_1,a_1+1, \overline{a_1-1}\}$, then interchanging $a-1$ in $e_1$ and g in $e_{i_4}$ will give a smaller presentation for $m_l$, a contradiction. Thus, $g\in \{a_1,a_1+1, \overline{a_1-1}\}$. Since $g\ge a_1$, we have $g\in \{a_1, a_1+1\}$. If $g=a_1$, then $e_1\le e_{i_4}$ implies that $f\ge a-1$. Since $f\ne a, a-1$ we have $f>a$. This in particular implies that $a\ne t+3$ and hence it follows from $a_1+1<a-1$ that $e:=\{a_1, a\}\in E({\bar{C}})$. Now set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}/\mathbf {x}_{e_{i_4}})x_k$. Suppose $g=a_1+1$. Then $e:=\{a_1, f\}\in E({\bar{C}})$ because $a_1\le f$ and $f\notin \{a_1,a_1+1\}$ by $e_{i_4}=\{f, a_1+1\}\in \mathbf {e}_{l}$. Moreover, $e':=\{a_1+1, a-1\}\in E({\bar{C}})$ because $a_1+1<b'_1<a-1$. If $f<a-1$, then $(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e}\mathbf {x}_{e'}/(\mathbf {x}_{e_1}\mathbf {x}_{e_{i_4}})x_k$ is a smaller presentation for $m_l$ which is a contradiction. Thus, $f>a$ and hence set $m_p:=(\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{e''}/\mathbf {x}_{e_{i_4}})x_k$, where $e'':=\{a_1+1, a\}$. This completes the proof. $\square$

Now we extend the ideal L of Proposition 3.9 to another ideal which contains L and has linear quotients.

Theorem 3.10

Let $I\subset S$ be the edge ideal of the graph $G=\overline{G_{(b)}}$, with $t\ge 1$. Then the ideal $(I^{s+1}:x_{t+5})$ has linear quotients for all $s\ge 0$.

Proof

Set $J:= (I^{s+1}:x_{t+5})$. We first determine the minimal generating set ${\mathcal {G}}(J)$ of J. Note that $E(G)=E({\bar{C}})\cup \{x_{t+5}x_i:\ 3\le i\le t+4\}$, where $C=1-2-\cdots -(t+3)-1$ is the unique induced cycle of $G_{(b)}$ of length$>3$. Hence,

$$\begin{aligned} I^{s+1}=\sum _{k=0}^{s+1}I({\bar{C}})^{s+1-k}(x_{t+5})^k(x_3,\ldots ,x_{t+4})^k. \end{aligned}$$

By [19, Proposition 1.2.2], the ideal J is generated by monomials $m/\gcd (m, x_{t+5})$, where $m\in I^{s+1}$. It follows that

$$\begin{aligned} J=I({\bar{C}})^{s+1}+\sum _{k=0}^{s}I({\bar{C}})^{s-k}(x_{t+5})^k(x_3,\ldots ,x_{t+4})^{k+1}. \end{aligned}$$

Since each edge of ${\bar{C}}$ contains a vertex in $\{3,\ldots ,t+3\}$, we have $I({\bar{C}})^{s+1}\subset I({\bar{C}})^{s}(x_3,\ldots ,x_{t+4})$. Therefore,

$$\begin{aligned} J=\sum _{k=0}^{s}I({\bar{C}})^{s-k}(x_{t+5})^k(x_3,\ldots ,x_{t+4})^{k+1}. \end{aligned}$$

For $0\le k\le s$, let $L_{k}:=I({\bar{C}})^{s-k}(x_{t+5})^k(x_3,\ldots ,x_{t+4})^{k+1}$. Clearly, for $0\le k, k'\le s$ with $k\ne k'$ we have ${\mathcal {G}}(L_k)\cap {\mathcal {G}}(L_{k'})=\emptyset$, where ${\mathcal {G}}(L_{k})$ denotes the minimal generating set of $L_k$. Therefore, ${\mathcal {G}}(J)$ is the disjoint union of all ${\mathcal {G}}(L_k)$ for $0\le k\le s$. In particular, J is generated by monomials of degree $2s+1$. For $s=0$, J is generated by variables and hence we have the assertion. Suppose $s\ge 1$.

We order the multisets of the edges of ${\bar{C}}$ as described before Proposition 3.9. Each element $m_l$ of ${\mathcal {G}}(L_k)$ can be written as $m_l=\mathbf {x}_{\mathbf {e}_{l}}{x_{t+5}}^k\mathbf {x}_{l}$, where $\mathbf {e}_{l}=(e_{1},\ldots ,e_{{s-k}})$ is an ordered multiset of the edges of ${\bar{C}}$ of size $s-k$ with $e_i=\{a_i,b_i\}$, $a_i<b_i$, and $\mathbf {x}_{l}:=x_{j_1}\cdots x_{j_{k+1}}$ with $3\le j_1\le \cdots \le j_{k+1}\le t+4$. Similar to the proof of Proposition 3.9, we consider the smallest presentation for $m_l$, i.e., the one in which $\mathbf {e}_{l}$ is the smallest possible multiset associated to $m_l$. So this presentation is unique.

Now we give an order on the generators of J. To this end, we use the lexicographic order $<_{lex}$ on the monomials of the ring S induced by $x_1<x_2<\cdots <x_{t+5}$; see [19, Section 2.1.2] for the definition of the lexicographic order.

For $m_q,m_l\in {\mathcal {G}}(J)$, we let $m_q<m_l$ in the following cases:

• $m_q\in L_{k'}$ and $m_l\in L_{k}$ with $0\le k'<k\le s$;

• $m_q,m_l\in L_k$ for some $0\le k\le s-1$ and either ${\mathbf {e}}_q<{\mathbf {e}}_{l}$ or ${\mathbf {e}}_q={\mathbf {e}}_{l}$ with ${\mathbf {x}}_q<_{lex}{\mathbf {x}}_l$;

• $m_q,m_l\in L_s$, and either $m_q\ne x_{t+5}^sx_i^{s+1}, m_l\ne x_{t+5}^sx_j^{s+1}$ for all $3\le i, j\le t+4$ with $m_q<_{lex}m_l$, or $m_q= x_{t+5}^sx_i^{s+1}$ and $m_l= x_{t+5}^sx_j^{s+1}$ for some $3\le i<j\le t+4$, or $m_q\ne x_{t+5}^sx_i^{s+1}$ for all $3\le i\le t+4$ and $m_l= x_{t+5}^sx_j^{s+1}$ for some $3\le j\le t+4$.

Suppose ${\mathcal {G}}(J)=\{m_1,\ldots ,m_r\}$ with $m_1<\!\cdots \!<m_r$. We show that for any $m_l\in \!{\mathcal {G}}(J)$ with $l\!>1$, the ideal $\left( (m_1,\ldots , m_{l-1}):m_l\right)$ is generated by some variables. Set $J_l:=(m_1,\ldots , m_{l-1})$. By [19, Proposition 1.2.2], the ideal $(J_l:m_l)$ is generated by the elements of the set $\{m_q/\gcd (m_q,m_l):\ 1\le q\le l-1\}$. Let $m_{q,l}:=m_q/\gcd (m_q,m_l)$ for $m_q<m_l$.

Suppose $m_q=\mathbf {x}_{\mathbf {e}_{q}}x_{t+5}^{k'}\mathbf {x}_{q}\in {\mathcal {G}}(L_{k'})$ with $0\le k'\le s$ and ${\mathbf {e}}_{q}=(e'_1,\ldots ,e'_{s-k'})\subseteq E({\bar{C}})$ with $e'_i=\{a'_i,b'_i\}$, $a'_i<b'_i$, and ${\mathbf {x}}_{q}=x_{j'_1}\cdots x_{j'_{k'+1}}$ with $3\le j'_1\le \cdots \le j'_{k'+1}\le t+4$, and suppose $m_l\in {\mathcal {G}}(L_{k})$ with $k'\le k\le s$. Suppose $\deg m_{q,l}>1$. We show that there is $1\le p<l$ such that $\deg m_{p,l}=1$ and $m_{p,l}| m_{q,l}$. This will imply that J has linear quotients. We may assume $k\ge 1$ because by Proposition 3.9, $L_0=I({\bar{C}})^{s}(x_3,\ldots ,x_{t+4})$ has linear quotients. By the same argument as in the proof of Proposition 3.9, we may assume that $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$. First assume $q=1$. Then $m_q=x_1^sx_3^{s+1}$ and $x_1|m_{q,l}$, because $k\ge 1$. If $x_3|\mathbf {x}_{l}$, then set $e:=\{1,3\}$ and $m_p:=\mathbf {x}_{e}m_l/(x_3x_{t+5})\in L_{k-1}$. Otherwise, we have $x_3|m_{q,l}$ and we may set $m_p:=m_lx_3/x_{j_i}$ for some $j_i$. Suppose now that $q>1$ and suppose that for all $m_{q'}$ with $q'<q$ there exists $m_{p'}<m_l$ with $\deg m_{p',l}=1$ and $m_{p',l}|m_{q',l}$. We prove the assertion by induction on q.

Note that

• (a) $x_{t+5}\notin \mathrm {supp}(m_{q,l})$, because $\deg _{m_q}x_{t+5}\le \deg _{m_l}x_{t+5}$.

• (b) Assume $a\in \{3,4,\ldots , j_{k+1}-1\}$. Except for the case where $k=s$ with $m_l= x_{t+5}^sx_a^{s}x_{j_{s+1}}$, one has $x_a\in (J_l:m_l)$, because $m_p:=x_am_l/x_{j_{k+1}}\in J_l$ and $m_{p,l}=x_a$.

• (c) If $k=s$ and $m_l=x_{t+5}^s x_a^{s}x_{j_{s+1}}$, where $3\le a<j_{s+1}-1<t+3$, then $e\!:=\{a,j_{s+1}\}\in E({\bar{C}})$ and by setting $m_p:=\mathbf {x}_{e}m_l/(x_{j_{s+1}}x_{t+5})$ one has $m_p\in L_{s-1}\subseteq J_l$ and $x_a\in (J_l:m_l)$.

• (d) For any a with $j_{k+1} + 1< a < t + 4$, we have $x_a \in (J_l : m_l)$ because $m_p:={\mathbf {x}}_em_l/(x_{t+5}x_{j_{k+1}})\in L_{k-1}\subseteq J_l$.

If $\mathrm {supp}(m_{q,l}) \cap \{x_3, x_4, \ldots , x_{j_{k+1}-1}\}\ne \emptyset$, then we are done. Indeed, assume that there exists $x_a \in \mathrm {supp}(m_{q,l})\cap \{x_3, x_4, \ldots , x_{j_{k+1}-1}\}$. Then by (b) and (c), it is sufficient to check only the case where $k = s$, $m_l = x^s_{t+5}x^s_ax_{j_{s+1}}$, and $j_{s+1} \in \{a + 1, t + 4\}$. Since $\deg _{m_q}x_a\ge s+1$, if $m_q\in L_s$, then $m_q=x_{t+5}^sx_a^{s+1}>m_l$, a contradiction. Thus, $m_q\notin L_s$ and hence there exists $e'_i\in \mathbf {e}_{q}$ with $a\in e'_i$. Suppose d is another vertex of $e'_i$. Since $d\notin \{a,a+1, t+4,t+5\}$, we have $x_d|m_{q,l}$. Set $m_p:=\mathbf {x}_{e'_i}m_l/(x_ax_{t+5})$. Then, $m_p<m_l$ and $m_{p,l}=x_d$ and so we are done.

Moreover, if $x_a\in \mathrm {supp}(m_{q,l})\cap \{x_{j_{k+1}+2}, \ldots , x_{t+3}\}$, then we are again done by (d). Thus, using (a) and the above discussion, we may suppose that

$$\begin{aligned} \mathrm {supp}(m_{q,l})\subseteq \{x_1,x_2,x_{j_{k+1}},x_{j_{k+1}+1}, x_{t+4}\}{\setminus }\{x_{t+5}\}. \end{aligned}$$

(4)

Note that

• (i) if $x_1\in \mathrm {supp}(m_{q,l})$ and $j_1\ne {t+3}, t+4$, then set $e:=\{1,j_1\}$ and $m_p:=\mathbf {x}_{e}m_l/(x_{j_1}x_{t+5})$;

• (ii) if $x_2\in \mathrm {supp}(m_{q,l})$ and there exists $j_i\ne 3,{t+4}$, $1\le i\le k+1$, then set $e:=\{2,j_i\}$ and $m_p:=\mathbf {x}_{e}m_l/(x_{j_i}x_{t+5})$;

• (iii) if $x_{j_{k+1}}\in \mathrm {supp}(m_{q,l})$ and $j_1<j_{k+1}-1{ <t+3}$, then set $e:=\{j_1,j_{k+1}\}$ and $m_p:=\mathbf {x}_{e}m_l/(x_{j_1}x_{t+5})$;

• (iv) if $x_{j_{k+1}+1}\in \mathrm {supp}(m_{q,l})$ and $j_1<j_{k+1}{ <t+3}$, then set $e:=\{j_1,j_{k+1}+1\}$ and $m_p:=\mathbf {x}_{e}m_l/(x_{j_1}x_{t+5})$.

Thus, by (4) it remains to find $m_p$ in the following cases:

• (v) $x_1\in \mathrm {supp}(m_{q,l})$ and $j_1\in \{{t+3}, t+4\}$;

• (vi) $x_2\in \mathrm {supp}(m_{q,l})$; and $j_i\in \{3,{t+4}\}$ for all $1\le i\le k+1$;

• (vii) $x_{j_{k+1}}\in \mathrm {supp}(m_{q,l})$ and either $j_1\in \{ j_{k+1}-1,j_{k+1}\}$ or $j_{k+1}=t+4$;

• (viii) $x_{j_{k+1}+1}\in \mathrm {supp}(m_{q,l})$ and either $j_1=j_{k+1}$ or $j_{k+1}=t+3$;

• (ix) $x_{t+4}\in \mathrm {supp}(m_{q,l})$.

In (viii), we have $j_{k+1}\ne t+4$ because $x_{t+5}\notin \mathrm {supp}(m_{q,l})$. Since we will check the case $x_{t+4}\in \mathrm {supp}(m_{q,l})$ in (ix), we may suppose in Case (vii) that $j_{k+1}\ne t+4$ and in Case (viii) that $j_{k+1}\ne t+3$. Moreover, having $j_1\in \{j_{k+1}-1, j_{k+1}\}$ in (vii) we have either $x_{j_1}\in \mathrm {supp}(m_{q,l})$ with $\mathrm {supp}(\mathbf {x}_{l})= \{x_{j_1}\}$ or $x_{j_1+1}\in \mathrm {supp}(m_{q,l})$ with $\mathrm {supp}(\mathbf {x}_{l})= \{x_{j_1}, x_{j_1+1}\}$. In Case (viii), since $j_1=j_{k+1}$ we get $x_{j_1+1}\in \mathrm {supp}(m_{q,l})$ and $\mathrm {supp}(\mathbf {x}_{l})=\{x_{j_1}\}$. So combining the two cases (vii) and (viii), we will end up with the following ones:

• (vii’) $x_{j_1}\in \mathrm {supp}(m_{q,l})$ and $\mathrm {supp}(\mathbf {x}_{l})=\{x_{j_{1}}\}$;

• (viii’) $x_{j_1+1}\in \mathrm {supp}(m_{q,l})$ and either $\mathrm {supp}(\mathbf {x}_{l})=\{x_{j_{1}}\}$ or $\mathrm {supp}(\mathbf {x}_{l})=\{x_{j_{1}}, x_{j_1+1}\}$.

So we replace (vii), (viii) with (vii$'$), (viii$'$). Now we prove the assertion in the above five cases. Note that since $m_q\ne m_l$ there exists $1\le b\le t+5$ such that $\deg _{m_q}x_b<\deg _{m_l}x_b$. Suppose B is the set of all such b.

Case (v) Since $j_1\in \{t+3,t+4\}$, we have $\mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_{t+3},x_{t+4}\}$ and hence $j_{k+1}\in \{t+3,t+4\}$ implies that $\mathrm {supp}(m_{q,l})\subseteq \{x_1,x_2,x_{t+3}{, x_{t+4}}\}$ by (4). Since $x_1\in \mathrm {supp}(m_{q,l})$, there exists $e'_i\in \mathbf {e}_{q}$ with $e'_i=\{1,b'_i\}$ for some $3\le b'_i\le t+2$. Since $x_{b'_i}\notin \{x_1,x_2,x_{t+3},x_{t+4}\}$, we have $x_{b'_i}|m_l$, and it follows from $\mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_{t+3},x_{t+4}\}$ that . Thus, there exists $e_j\in \mathbf {e}_{l}$ with $b'_i\in e_j$. If d is another vertex of $e_j$, then $d> 1$ because $\mathbf {e}_{l}\cap \mathbf {e}_{q}=\emptyset$. Set $m_p:=\mathbf {x}_{e'_i}m_l/\mathbf {x}_{e_j}$ and so we are done in this case. This case together with (i) imply that if $x_1|m_{q,l}$, then we have the desired $m_p$. Suppose in the remaining cases that $x_1\notin \mathrm {supp}(m_{q,l})$.

Case (vi) Since $x_{j_{k+1}}\in \mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_3,x_{t+4}\}$, we have $\mathrm {supp}(m_{q,l})\subseteq \{x_2,x_{3},x_4{, x_{t+4}}\}$ by (4). Since $x_2\in \mathrm {supp}(m_{q,l})$, there exists $e'_i\in \mathbf {e}_{q}$ with $e'_{i}=\{2,b'_{i}\}$ for some $4\le b'_{i}\le t+3$.

First suppose $m_l\in L_s$. Then $m_l=x_{t+5}^s\mathbf {x}_{l}$ and because $\mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_3,x_{t+4}\}$. Thus, $x_{b'_{i}}|m_{q,l}$ and therefore $b'_{i}=4$. In case $x_{t+4}|\mathbf {x}_{l}$ we set $m_p:=x_{4}m_l/x_{t+4}$. Otherwise, we have $m_l=x_{t+5}^sx_3^{s+1}$, and hence we can set $m_p:=x_{t+5}^sx_3^{s}x_{4}$.

Suppose now that $m_l\notin L_s$. There exists $e_{j}=\{a_{j},b_{j}\}\in \mathbf {e}_{l}$ with $a_{j}\ne 2$. If $a_{j}\ne 1$ then set $e:=\{2,b_{j}\}$ and $m_p:=\mathbf {x}_{e}m_l/\mathbf {x}_{e_{j}}$. Suppose that $e_{j}=\{1,b_{j}\}$ for all $e_{j}\in \mathbf {e}_{l}$ with $2\notin e_{j}$.

If $x_{b'_{i}}|m_{q,l}$, then $b'_{i}=4$. If $x_{t+4}|\mathbf {x}_{l}$, then set $m_p:=x_{4}m_l/x_{t+4}$. Otherwise, we have $\mathrm {supp}(\mathbf {x}_{l})=\{x_3\}$. Then $b_{j}=3$ for all $e_j = \{1, b_j\} \in \mathbf {e}_{l}$, since otherwise we get a contradiction to the fact that we have considered the smallest presentation for $m_l$. If $x_2|m_l$, then there exists $e_{r}=\{2,b_{r}\}\in \mathbf {e}_{l}$, where $b_{r}>4$ because $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$. Then set $m_p:=\mathbf {x}_{e'_{i}}m_l/\mathbf {x}_{e_{r}}$. If , then $m_l=x_{t+5}^kx_1^{s-k}x_3^{s+1}$. Thus, $3\in B$ because $3\notin e'_i = \{2, 4\}\in \mathbf {e}_{q}$. If $1\in B$, then set $e:=\{1,4\}$ and $m_{q'}=\mathbf {x}_{e}m_q/\mathbf {x}_{e'_i}$, and if $1\notin B$, then there exists $e'_f=\{1,b'_f\}\in \mathbf {e}_{q}$ with $b'_f>3$ because $e_j=\{1, 3\} \in \mathbf {e}_{l}$ and $\mathbf {e}_{q} \cap \mathbf {e}_{l} = \emptyset$, so set $m_{q'}=\mathbf {x}_{e_j}m_q/\mathbf {x}_{e'_f}$ and use induction.

Now suppose . Since $\mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_3,x_{t+4}\}$, we have $e_r:=\{1,b'_i\}\in \mathbf {e}_{l}$ because all edges in $\mathbf {e}_{l}$ contain either 1 or 2 and $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$. Since $b'_i>3$ we have $\mathrm {supp}(\mathbf {x}_{l})=\{x_{t+4}\}$ because otherwise we get a smaller presentation for $m_l$. If $1\notin B$, then $e'_f:=\{1,b'_f\}\in \mathbf {e}_{q}$ with $b'_f\ne b'_i$ because $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$. If $x_{b'_f}|m_{q,l}$, then set $m_p:=x_{b'_f}m_l/x_{t+4}$. If , then we have $\{2,b'_f\}\in \mathbf {e}_{l}$ since $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ and each edge of $m_l$ contains either 1 or 2. Now we have $\{1,b'_i\}, \{2,b'_f\}\in \mathbf {e}_{l}$ and $\{1,b'_f\}, \{2,b'_i\}\in \mathbf {e}_{q}$ which contradict the fact that both $\mathbf {e}_{q}, \mathbf {e}_{l}$ are the smallest multisets associated to $m_q,m_l$ respectively. Suppose $1\in B$. Then we can use inductive hypothesis for $m_{q'}:=\mathbf {x}_{e_r}m_q/\mathbf {x}_{e'_i}$. So we are done in this case too. By settling this case and according to Case (ii), we have the desired $m_p$ if $x_2|m_{q,l}$. Suppose in the remaining cases that $\mathrm {supp}(m_{q,l})\subseteq \{x_{j_{k+1}}, x_{{j_{k+1}}+1}{ , x_{t+4}}\}$.

Case (vii’) Since $\mathrm {supp}(\mathbf {x}_{l})=\{x_{j_1}\}$, we have $\mathrm {supp}(m_{q,l})\subseteq \{x_{j_1}, x_{j_1+1}, { x_{t+4}\}}$. There exists $e'_{i_1}=\{j_1,c\}\in \mathbf {e}_{q}$ for some c because otherwise $\deg _{m_q}x_{j_1}\le k'+1\le k+1=\deg _{m_l}x_{j_1}$, a contradiction. It follows that $j_1\le t+3$. Since we have $x_c|m_l$ and since $c\ne j_1$, there exists $e_{i_2}=\{c,d\}\in \mathbf {e}_{l}$ for some d. By $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$, we have $d\ne j_1$. Note that $d<j_1$, because otherwise interchanging $x_d$ in $\mathbf {x}_{e_{j}}$ and $x_{j_1}$ in $\mathbf {x}_{l}$ will result in a smaller presentation of $m_l$.

Suppose $d\!<\!j_1\!-\!1$. If $\{d,\!j_1\}\!\ne \!\{1,\!t+3\}$, then set $e\!:=\!\{d,\!j_1\}$ and $m_p\!:=\!\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}\!m_l/(\mathbf {x}_{e_{i_2}}\!x_{j_1}\!x_{t+5})$. If $\{d, j_1\}= \{1,t+3\}$, then in case $1\in B$, set $m_{q'}:=\mathbf {x}_{e_{i_2}}m_q/\mathbf {x}_{e'_{i_1}}<m_q$ and use induction. In case $1\notin B$, there exists $e'_{i_3}=\{1,f\}\in \mathbf {e}_{q}$ for some $f\notin \{ c, 1,2,t+3, t+4\}$ which implies that . Hence, $x_f|m_l$ and thus $e_{i_4}=\{f,g\}\in \mathbf {e}_{l}$ for some $g\ne 1$. If $g=t+3$, then $\{c,t+3\}, \{1,f\}\in \mathbf {e}_{q}$ and $\{1,c\}, \{f,t+3\}\in \mathbf {e}_{l}$ which contradict the fact that $\mathbf {e}_{q}$ and $\mathbf {e}_{l}$ both have minimum presentations. Thus, $g\ne t+3$. If $g\ne t+2$, set $e:=\{g,t+3\}$ and $m_p:=\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}\mathbf {x}_{e'_{i_3}}m_l/(\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_{i_4}}x_{t+3}x_{t+5})$, and if $g=t+2$, set $e=\{1,t+2\}, e'=\{f, t+3\}$ and $m_p:=\mathbf {x}_{e}\mathbf {x}_{e'}\mathbf {x}_{e'_{i_1}}m_l/(\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_{i_4}}x_{t+3}x_{t+5})$.

Suppose $d=j_1-1$. If $j_1-1\in B$, then set $m_{q'}:=\mathbf {x}_{e_{i_2}}m_q/\mathbf {x}_{e'_{i_1}}$ and use induction hypothesis. If $j_1-1\notin B$, then $x_{j_1-1}|m_q$. If $x_{j_1-1}|\mathbf {x}_{q}$, then interchanging $x_{j_1-1}$ in $\mathbf {x}_{q}$ and $x_{j_1}$ in $\mathbf {x}_{e'_{i_1}}$ will give a smaller presentation of $m_q$, a contradiction. Therefore, there exists $e'_{i_3}=\{j_1-1, f\}\in \mathbf {e}_{q}$ for some $f\ne c$. Then $f\notin \{j_1-2,j_1-1, j_1\}$. If $x_f|m_{q,l}$, then $f={j_1+1}$. Set $m_p:=\mathbf {x}_{e'_{i_1}}\mathbf {x}_{e'_{i_3}}m_l/(\mathbf {x}_{e_{i_2}}x_{j_1}x_{t+5})$. Suppose .

It follows that there exists $e_{i_4}=\{f, g\}\in \mathbf {e}_{l}$ for some $g\ne j_1-1$. We have $f\ne {j_1+1}$ because otherwise, it follows that $g\notin \{j_1,j_1-1,{j_1+1}\}$ which implies that $m_l$ will have a smaller presentation by interchanging $x_f$ in $\mathbf {x}_{e_{i_4}}$ and $x_{j_1}$ in $\mathbf {x}_{l}$. Since $f\ne j_1+1$ we have either $g< j_1-1$ or $g=j_1$ because otherwise one can interchange $x_g$ in $\mathbf {x}_{e_{i_4}}$ and $x_{j_1}$ in $\mathbf {x}_{l}$ to get a smaller presentation. If $g=j_1$, then we have $\{j_1,c\}, \{j_1-1, f\}\in \mathbf {e}_{q}$ and $\{j_1-1, c\}, \{j_1,f\}\in \mathbf {e}_{l}$ which contradict the fact that both $\mathbf {e}_{q}, \mathbf {e}_{l}$ are the smallest multisets associated to $m_q, m_l$, respectively. Thus, $g<j_1-1$. In case $\{g, j_1\}\ne \{1,t+3\}$, set $e:=\{g, j_1\}$ and $m_p:=\mathbf {x}_{e'_{i_1}}\mathbf {x}_{e'_{i_3}}\mathbf {x}_{e}m_l/(\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_{i_4}}x_{j_1}x_{t+5})$. In case $\{g, j_1\}= \{1,t+3\}$, set $e:=\{1,t+2\}, e':=\{f,t+3\}$ and $m_p:=\mathbf {x}_{e'_{i_1}}\mathbf {x}_{e}\mathbf {x}_{e'}m_l/(\mathbf {x}_{e_{i_2}}\mathbf {x}_{e_{i_4}}x_{t+3}x_{t+5})$. Note that since $e'_{i_3}, e_{i_4}\in E({\bar{C}})$ we have $f\notin \{1,t+2,t+3\}$ and hence $e'\in E({\bar{C}})$. Thus, we are done in this case too.

In general, if $x_{j_1}\in \mathrm {supp}(m_{q,l})$, then by (4) we have ${j_1}\in \{j_{k+1}, j_{k+1}+1, t+4\}$. But ${j_1}\le j_{k+1}$ implies that $j_1=j_{k+1}$ and hence $\mathrm {supp}{(\mathbf {x}_{l})}=\{x_{j_1}\}$. Thus, by the discussion in (vii’) we are done if $x_{j_1}\in \mathrm {supp}(m_{q,l})$. Therefore, we may assume in the rest of the proof that $x_{j_1}\notin \mathrm {supp}(m_{q,l})$.

Case (viii’) Since $\mathrm {supp}(\mathbf {x}_{l})\subseteq \{x_{j_1},x_{j_1+1}\}$, we have $\mathrm {supp}(m_{q,l})\subseteq \{x_{j_1+1},x_{j_1+2},x_{t+4}\}$ by (4). Moreover, if $j_1+1= t+4$, then $\mathrm {supp}(m_{q,l})=\{x_{t+4}\}$ and since this case will be discussed in (ix) we may assume here that $j_1+1\ne t+4$. Assume first that $x_{j_1}|\mathbf {x}_{\mathbf {e}_{q}}$. Then $e'_{i_1}=\{j_1,c\}\in \mathbf {e}_{q}$ for some c. Since $c\notin \{j_1, j_1+1\}$, we have $e_{i_2}=\{c,d\}\in \mathbf {e}_{l}$ for some d. Since $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ and since we have the smallest presentation of $m_l$, we have $d<j_1$. If $\{d, j_1+1\}\ne \{1,t+3\}$, then set $e:=\{d,j_1+1\}$ and $m_p:=\mathbf {x}_{e'_{i_1}}\mathbf {x}_{e}m_l/(\mathbf {x}_{e_{i_2}}x_{j_1}x_{t+5})$. If $\{d, j_1+1\}=\{1,t+3\}$, then set $e:=\{c, t+3\}, e':=\{1,t+2\}$ and $m_p:=\mathbf {x}_{e}\mathbf {x}_{e'}m_l/(\mathbf {x}_{e_{i_2}}x_{t+2}x_{t+5})$. Note that $e:=\{c, t+3\}\in E({\bar{C}})$ because $c\notin \{1,t+2,t+3\}$.

Now assume . Suppose . Since $x_{j_1+1}|m_{q,l}$, we have $x_{j_1+1}|\mathbf {x}_{q}$. If $x_{j_1+2}|\mathbf {x}_{\mathbf {e}_{q}}$, then $e:=\{a,j_1+2\}\in \mathbf {e}_{q}$ for some $a\notin \{j_1+1, j_1+2, \overline{j_1+3}\}$. Since , we have $a\ne j_1$ too. Thus, $\{a, j_1+1\}\in E({\bar{C}})$ and hence one can interchange $x_{j_1+2}$ in $\mathbf {x}_{e}$ and $x_{j_1+1}$ in $\mathbf {x}_{q}$ to get a smaller presentation for $m_q$, a contradiction. Thus, . Therefore, $\mathrm {supp}(\mathbf {x}_{\mathbf {e}_{q}})\cap \mathrm {supp}(m_{q,l})=\emptyset$ which implies that $\mathbf {x}_{\mathbf {e}_{q}}|\mathbf {x}_{\mathbf {e}_{l}}$ and since $k'\le k$ we have $\mathbf {e}_{q}=\mathbf {e}_{l}$. But $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ implies that $\mathbf {e}_{q}=\emptyset =\mathbf {e}_{l}$. Thus, $m_q, m_l\in L_s$. If $m_l=x_{t+5}^sx_{j_1}^{s+1}$, then set $m_p:=x_{t+5}^sx_{j_1}^{s}x_{j_1+1}$. Suppose $m_l= x_{t+5}^sx_{j_1}^{r}x_{j_1+1}^{s+1-r}$, where $0<r<s+1$. By the order of the generators of $L_s$ we have $m_q\ne x_{t+5}^sx_{j_1+1}^{s+1}$. Since $\mathrm {supp}(m_q)\subseteq \mathrm {supp}(m_{q,l})\cup \mathrm {supp}(m_l)\subseteq \{x_{j_1}, x_{j_1+1},x_{j_1+2},x_{t+4}, x_{t+5}\}$ and $m_q<m_l$, we have $x_{j_1}\in \mathrm {supp}(m_q)$. If $x_{j_1+2}\in \mathrm {supp}(m_q)$ ($x_{t+4}\in \mathrm {supp}(m_q)$ resp.), then set $m_{q'}:=x_{j_1+1}m_q/x_{j_1+2}$ ($m_{q'}:=x_{j_1+1}m_q/x_{t+4}$ resp.) and use induction. Otherwise, we have $m_q=x_{t+5}^sx_{j_1}^{r'}x_{j_1+1}^{s+1-r'}$ with $0<r'<r$ because $m_q<m_l$. Set $m_{q'}:=x_{j_1}m_q/x_{j_1+1}$ and use induction. Suppose now that $x_{j_1+1}|\mathbf {x}_{\mathbf {e}_{q}}$. There exists $e'_{i_1}=\{j_1+1, c\}\in \mathbf {e}_{q}$ for some c. Since $x_c\notin \mathrm {supp}(\mathbf {x}_{l})\cup \mathrm {supp}(m_{q,l})$, there exists $e_{i_2}=\{c,d\}\in \mathbf {e}_{l}$ for some d with $d\ne j_1+1$. If $d>j_1+1$, then set $m_p:=\mathbf {x}_{e'_{i_1}}m_l/\mathbf {x}_{e_{i_2}}$. If $d<j_1-1$, then set $e:=\{d,j_1\}$ which is an edge of ${\bar{C}}$ because $j_1\ne t+3$. Now set $m_p:=\mathbf {x}_{e}\mathbf {x}_{e'_{i_1}}m_l/(\mathbf {x}_{e_{i_2}}x_{j_1}x_{t+5})$. If $d=j_1-1$, then $c\ne j_1-1$ and hence one can set $e:=\{j_1-1,j_1+1\}, e':=\{c,j_1\}$ which are edges of ${\bar{C}}$. Now set $m_p:=\mathbf {x}_{e}\mathbf {x}_{e'}m_l/(\mathbf {x}_{e_{i_2}}x_{j_1}x_{t+5})$. Suppose $d=j_1$ which implies that $c\ne j_1-1$. It follows that because otherwise, interchanging $x_{j_1}$ in $\mathbf {x}_{q}$ and $x_{j_1+1}$ in $\mathbf {x}_{e'_{i_1}}$ will give a smaller presentation for $m_q$. Since , we have ${j_1}\in B$. Set $m_{q'}=\mathbf {x}_{e_{i_2}}m_q/\mathbf {x}_{e'_{i_1}}$ and use induction. So we are done also in this case.

Now by (iii), (iv), (vii$'$) and (viii$'$) we may assume in the remaining case that $\mathrm {supp}(m_{q,l})=\{x_{t+4}\}$.

Case (ix) If there exists $b\in B{\setminus }\{1,2, t+5\}$, then setting $m_{q'}:=x_bm_q/x_{t+4}$, we are done by induction hypothesis. Suppose $B\subseteq \{1,2,t+5\}$.

If $1\in B$, then $m_q, m_l\notin L_s$ and there exists $e'_{i}=\{a'_{i},b'_{i}\}\in \mathbf {e}_{q}$ with $1\notin e'_{i}$. If $a'_{i}\ne 2$ set $e:=\{1,a'_{i}\}$, else if $b'_{i}\ne t+3$ set $e:=\{1,b'_{i}\}$. Then $m_{q'}:=\mathbf {x}_{e}m_q/\mathbf {x}_{e'_{i}}<m_q$ and $m_{q',l}$ divides $m_{q,l}$. By induction hypothesis, we are done. Suppose for all $e'_{i}\in \mathbf {e}_{q}$ with $1\notin e'_{i}$ one has $e'_{i}=\{2,t+3\}$. Since $\deg _{m_q}{x_1}<\deg _{m_l}{x_1}$ there exists $e_j=\{1,b_{j}\}\in \mathbf {e}_{l}$ with $b_{j}\ne 1,2,t+3$ and since $\mathbf {e}_{q}\cap \mathbf {e}_{l}=\emptyset$ we have $\deg _{\mathbf {x}_{e_q}}{x_{b_{j}}}=0$ for all such $b_{j}$. Since $b_{j}\notin B$, we must have $x_{b_{j}}|\mathbf {x}_{q}$. If $b_{j}\ne 3$ by interchanging $x_{b_{j}}$ in $\mathbf {x}_{q}$ and $x_{t+3}$ in $\mathbf {x}_{e'_{i}}$ we get a smaller presentation of $m_q$ which is a contradiction. Thus $b_{j}=3$. Set $m_{q'}:=\mathbf {x}_{e_j}x_{t+3}m_q/(\mathbf {x}_{e'_{i}}x_3)$. Then $m_{q'}<m_q$ and $m_{q',l}|m_{q,l}$ and so we are done by induction hypothesis.

Now assume $1\notin B$ and $2\in B$. Again $m_q, m_l\notin L_s$ and there exists $e'_{i}=\{a'_{i},b'_{i}\}\in \mathbf {e}_{q}$ with $2\ne a'_{i}<b'_{i}$. If $e'_{i}=\{1,b'_{i}\}$ for all $e'_i\in \mathbf {e}_{q}$ with $2\notin e'_{i}$, then $x_1|m_{q,l}$ because otherwise $s-k'=\deg _{m_q}x_1+\deg _{m_q}x_2<\deg _{m_l}x_1+\deg _{m_l}x_2\le s-k$ and hence $k'>k$, a contradiction. But $x_1\in \mathrm {supp}(m_{q,l})=\{x_{t+4}\}$ is also a contradiction. Therefore, there exists $e'_{i}=\{a'_{i},b'_{i}\}\in \mathbf {e}_{q}$ with $2< a'_{i}<b'_{i}$. Set $e:=\{2,b'_{i}\}$ and $m_{q'}:=\mathbf {x}_{e}m_q/\mathbf {x}_{e'_{i}}$. Then $m_{q'}<m_q$ and $m_{q',l}|m_{q,l}$ and so we are again done by induction hypothesis.

Suppose now that $B=\{t+5\}$. Since $\mathrm {supp}(m_{q,l})=\{x_{t+4}\}$ we have

$$\begin{aligned} \mathbf {x}_{\mathbf {e}_{q}}(\mathbf {x}_{q}/m_{q,l})=\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{l}. \end{aligned}$$

(5)

Since $\deg _{m_q}x_{t+5}<\deg _{m_l}x_{t+5}$, we have $k'<k$. Moreover, $\deg _{m_l}x_1+\deg _{m_l}x_2\le s-k$ which implies by (5) that at most $s-k$ edges of $\mathbf {e}_{q}$ contain either 1 or 2. Now we choose $s-k+1$ edges $e''_1,\ldots , e''_{s-k+1} \in \mathbf {e}_{q}$ with the property that no edge in $\mathbf {e}_{q}{\setminus } \{e''_1,\ldots , e''_{s-k+1}\}$ contains 1 or 2. Set $\mathbf {e}_{p}:=\{{e''_1}, \ldots , e''_{s-k+1}\}$ and $\mathbf {x}_{p}:=x_{t+4}\mathbf {x}_{\mathbf {e}_{q}}\mathbf {x}_{q}/(\mathbf {x}_{\mathbf {e}_{p}}m_{q,l})$. It follows from the choice of $\mathbf {e}_{p}$ that neither $x_1$ nor $x_2$ divides $\mathbf {x}_{p}$. Hence, $m_p:=\mathbf {x}_{\mathbf {e}_{p}}x_{t+5}^{k-1}\mathbf {x}_{p}\in L_{k-1}$. Since by (5), we have $\mathbf {x}_{\mathbf {e}_{p}}\mathbf {x}_{p}/x_{t+4}=\mathbf {x}_{\mathbf {e}_{l}}\mathbf {x}_{l}$, we conclude that $m_{p,l}=x_{t+4}$. This completes the proof. $\square$

Now we use Theorem 3.10 to show that $I(\overline{G_{(b)}})^k$ has a linear resolution for $s\ge 2$ when $G_{(b)}$ does not have an induced 4-cycle, that is the number of its vertices is more than or equal to 7.

Theorem 3.11

Let G be a graph on $n\ge 7$ vertices such that $G_{(b)}$ is its complement. Let $I:=I(G)$ be the edge ideal of G. Then $I^s$ has a linear resolution for $s\ge 2$.

Proof

By construction, $n=t+5$, where $t\ge 2$. We apply Lemma 3.8 for $I^s$ and $x:=x_{t+5}$ to prove the assertion. To this end, we first compute $\mathrm {reg}(I^s+( x_{t+5}))$. Setting $C=1-2-\cdots -(t+3)-1$, for all $s\ge 1$ we have

$$\begin{aligned} I^s+( x_{t+5})&=(I({\bar{C}})+(x_{t+5})(x_3,\ldots , x_{t+4}))^s+(x_{t+5}) =I({\bar{C}})^s+(x_{t+5}). \end{aligned}$$

Since $x_{t+5}$ does not appear in the support of the generators of $I({\bar{C}})^s$, we have

$$\begin{aligned} \mathrm {reg}(I^s+(x_{t+5}))=\mathrm {reg}(I({\bar{C}})^s+(x_{t+5}))=\mathrm {reg}(I({\bar{C}})^s). \end{aligned}$$

It is proved in [2, Corollary 4.4] that $I({\bar{C}})^s$ has a linear resolution for $s\ge 2$ when $|C|>4$, which is the case here because $t+3>4$. Thus, $\mathrm {reg}(I^s+(x_{t+5}))= 2s$ for $s\ge 2$. On the other hand $(I^s:x_{t+5})$ has linear quotients by Theorem 3.10, and it is seen in its proof that $(I^s:x_{t+5})$ is generated in degree $2s-1$ for $s\ge 1$. Therefore, $(I^s:x_{t+5})$ has a $(2s-1)$-linear resolution for $s\ge 1$, see [19, Theorem 8.2.1], and hence $\mathrm {reg}((I^s:x_{t+5}))=2s-1$. Now using Lemma 3.8 we have $\mathrm {reg}(I^s)\le 2s$ for $s\ge 2$. Since $I^s$ is generated in degree 2s we conclude that $I^s$ has a linear resolution for $s\ge 2$. $\square$