Geometric regularity of powers of two-dimensional squarefree monomial ideals

Abstract

Let I be a two-dimensional squarefree monomial ideal of a polynomial ring S. We evaluate the geometric regularity, $a_i$-invariants of $S/I^n$ for $i\ge 2$. It turns out that they are all linear functions in n from $n=2$. Also, it is shown that $\text{ g-reg }(S/I^n)={\text {reg}}(S/I^{(n)})$ for all $n\ge 1$.

1 Introduction

Let $S:=K[x_1,\ldots ,x_r]$ be the polynomial ring in variables $x_1,\ldots ,x_r$ over a field K and ${\mathfrak {m}}$ the maximal homogeneous ideal of S. Let M be a finitely generated graded S-module. For each $0\le i\le \dim M$, the $a_i$-invariant of M is defined by

$$\begin{aligned} a_i(M):=\max \{t: H_{{\mathfrak {m}}}^i(M)_t\ne 0\}, \end{aligned}$$

where $H_{{\mathfrak {m}}}^i(M)$ is the ith local cohomology module of M with support in ${\mathfrak {m}}$, and we understand $\max \emptyset =-\infty$. The regularity of M is defined by

$$\begin{aligned} {\text {reg}}(M):=\max \{a_i(M)+i: 0\le i\le \dim M\}. \end{aligned}$$

Let I be a homogeneous ideal of the polynomial ring S. It was proved that ${\text {reg}}(S/I^n)$ is a linear function in n for $n\gg 0$; see [6, 14, 22]. In other words, there exist integers d, e and $n_0$ such that ${\text {reg}}(S/I^n)=dn+e$ for all $n\ge n_0$. Based on this result, many authors have studied properties of the regularity of powers of homogeneous ideals. Roughly speaking, their researches fall into two classes. The one is devoted to understanding the nature of integers d, e and $n_0$ for some special or general ideals I; see e.g., [4, 7, 9]. The other is to computing explicitly or to bounding the regularity function ${\text {reg}}(S/I^n)$ for some special classes of ideals I; see e.g., [1, 2, 13].

Let $\Delta$ be a simplicial complex on $[r]:=\{1,2,\ldots ,r\}$. The Stanley–Reisner ideal of $\Delta$ is defined to be the ideal of S

$$\begin{aligned} I_{\Delta }:=({\mathbf {x}}_F: F \text{ is } \text{ a } \text{ minimal } \text{ non-face } \text{ of } \Delta ), \end{aligned}$$

where ${\mathbf {x}}_F$ is the squarefree monomial $\prod _{i\in F}x_i$. Every two-dimensional squarefree monomial ideal containing no variables is the Stanley–Reisner ideal of a simplicial complex of dimension one. Note that a simplicial complex of dimension one can be regarded as a simple graph that may contains isolated vertices. Two-dimensional squarefree monomial ideals attract many authors’ interests. For example, the Buchsbaum property of symbolic powers and ordinary powers of these ideals was studied in [15] and [16], respectively, and the Cohen–Macaulayness of symbolic powers and ordinary powers of such ideals was characterized in terms of the properties of their associated graphs in [17]. Recently, the regularity of symbolic powers of such ideals was computed explicitly in [12].

Inspired by the regularity for sheaves on projective spaces, M.E. Rossi et al. introduced the following weaker but natural notion of regularity in [20].

Definition 0.1

Let M be a finitely generated graded S-module. The geometric regularity of M is defined by

$$\begin{aligned} \text{ g-reg }(M):=\max \{a_i(M)+i: i>0\}. \end{aligned}$$

Let I be an arbitrary two-dimensional squarefree monomial ideal of S. In this paper, we evaluate the geometric regularity of $S/I^n$ (see Theorem 4.1) and then obtain the equality $\text{ g-reg }(S/I^n)={\text {reg}}(S/I^{(n)})$ for all $n\ge 1$.

The paper is structured as follows. In Sect. 1, we recall some concepts and results which we need in this paper. In Sect. 2, we consider the question: If I is a monomial ideal in S and $J:=(I, yx_1, \ldots , yx_r)$ is the ideal of $R:=S[y]$, could we compare the regularity or $a_i$-invariants of $S/I^n$ with the ones of $R/J^n$? This question is inspired by the following observation: If G is a simple graph on [r] and $G'$ is the graph obtaining from G by adding an isolated vertex $r+1$, then $I_{G'}=(I_G, x_{r+1}x_1,x_{r+1}x_2,\ldots ,x_{r+1}x_r )$. With S, I, R and J defined as above, we prove among other things that $a_1(R/J^n)=\max \{2n-2, a_1(S/I^{n-t})+t: 0\le t\le n-1\}$ if $\sqrt{I}\ne (x_1,\ldots ,x_n)$, and $a_i(R/J^n)=\max \{a_i(S/I^{n-t})+t: 0\le t\le n-1\}$ for $i\ge 2$; see Theorem 2.8.

In Sect. 3, we compute $a_i(S/I_G^n)$ for $i=1,2$ when G is a simple graph without any isolated vertices. We find that $a_2(S/I_G^n)=a_2(S/I_G^{(n)})$ holds for all such graphs G, but $a_1(S/I_G^n)$ and $a_1(S/I_G^{(n)})$ may be very different. Since $a_1(S/I_G^n)$ has been computed, we can clarify all graphs G and all $n>0$ for which $S/I_G^n$ is Cohen–Macaulay. This recovers two results of [17]. In the final section, by applying the aforementioned result obtained in Sect. 2, we can get the values of $\text{ g-reg } (S/I_G^n)$ for any graph G which may contain isolated vertices and for all $n\ge 1$. We conclude this paper by showing

$$\begin{aligned} \text{ g-reg }(S/I^n)={\text {reg}}(S/I^{(n)}) \end{aligned}$$

for all two-dimensional squarefree monomial ideals I.

2 Preliminaries

In this section, we fix notation and recall some concepts and results which will be used in this paper. Throughout this paper, we let $[r]:=\{1,2,\ldots ,r\}$ and $S:=K[x_1,\ldots ,x_r]$, the polynomial ring in variables $x_1,\ldots ,x_r$ over a field K.

2.1 Geometric regularity

We refer to [3] for the knowledge of local cohomology. It is known that $H_{{\mathfrak {m}}}^0(M/H_{{\mathfrak {m}}}^0(M))=0$ and $H_{{\mathfrak {m}}}^i(M)=H_{{\mathfrak {m}}}^i(M/H_{{\mathfrak {m}}}^0(M))$ for all $i>0$; see e.g., [3, Chapter 2]. In particular, we have

$$\begin{aligned} \text{ g-reg }(M)={\mathrm {reg}}(M/H_{{\mathfrak {m}}}^0(M)). \end{aligned}$$

Let I, J be graded ideals of S. We set as usual

$$\begin{aligned} I:J:=\{x\in S: xJ\subseteq I\} \end{aligned}$$

and

$$\begin{aligned} I:J^{\infty }:=\bigcup _{i\ge 1}I:J^i. \end{aligned}$$

The ideal $I:{\mathfrak {m}}^{\infty }$ is called the saturation of I. Note that $H_{{\mathfrak {m}}}^0(S/I)=(I:{\mathfrak {m}}^{\infty })/I$, we obtain

$$\begin{aligned} \text{ g-reg }(S/I)={\mathrm {reg}}(S/I:{\mathfrak {m}}^{\infty }). \end{aligned}$$

2.2 Simplicial complex

Recall from [10] that a simplicial complex $\Delta$ on [r] is a collection of subsets of [r] such that $\{i\}\in \Delta$ for each $i\in [r]$ and that if $\sigma \in \Delta$ and $\tau \subseteq \sigma$ then $\tau \in \Delta$. The elements $F\in \Delta$ are called faces of $\Delta$, and the dimension of each face $F\in \Delta$ is defined by $\dim F=|F|-1$, where |F| is the cardinality of A. Also, the dimension of $\Delta$, $\dim \Delta$, is given by $\max \{\dim F: F\in \Delta \}$. Hence, a simplicial complex of dimension one is a simple graph that may contain some isolated vertices. A facet is a maximal face of $\Delta$ (with respect to inclusion). Let ${\mathcal {F}}(\Delta )$ denote the set of $\Delta$. It is clear that ${\mathcal {F}}(\Delta )$ governs $\Delta$. When ${\mathcal {F}}(\Delta )=\{F_1,\ldots ,F_k\}$ , we write $\Delta =\langle F_1,\ldots ,F_k\rangle$.

A non-face of $\Delta$ is a subset F of [r] with $F\notin \Delta$, and let ${\mathcal {N}}(\Delta )$ denote the set of minimal non-face of $\Delta$. For any subset F of [r], we set

$$\begin{aligned} {\mathbf {x}}_{F}:=\prod _{i\in F}x_i. \end{aligned}$$

The Stanley–Reisner ideal of $\Delta$ is the ideal $I_{\Delta }$ which is generated by ${\mathbf {x}}_{F}$ with $F\notin \Delta$. In other words,

$$\begin{aligned} I_{\Delta }:=(x_F: F\in {\mathcal {N}}(\Delta )). \end{aligned}$$

By [10, Lemma 1.5.4], $I_{\Delta }$ has the following primary decomposition:

$$\begin{aligned} I_{\Delta }=\bigcap _{F\in {\mathcal {F}}(\Delta )} P_{{\overline{F}}}, \end{aligned}$$

where $P_{{\overline{F}}}:=(x_i: i\in [r]\setminus F).$ From this decomposition, we see that

$$\begin{aligned} {\mathrm {Krull-dim}} (S/I_{\Delta })=\dim \Delta +1. \end{aligned}$$

Thus, if G is a simple graph, then the Stanley–Reisner ideal $I_G$ is a two-dimensional squarefree monomial ideal of S. Conversely, any two-dimensional squarefree monomial ideal of S containing no variables arises in this fashion.

Recall from [10] that the augmented oriented chain complex $\widetilde{{\mathcal {C}}}(\Delta ; K)$ of $\Delta$ with respect to K is defined as follows. A $K-$basis of $\widetilde{{\mathcal {C}}}_j(\Delta ; K)$ is given by ${\mathbf {e}}_F$ with $\dim F=j$ and $F\in \Delta$. For $F=\{i_0<i_1<\cdots <i_j\}$, one denotes the element ${\mathbf {e}}_F$ by $[i_0,i_1,\ldots ,i_j]$. With this notation, the chain map $\partial : \widetilde{{\mathcal {C}}}_j(\Delta ; K)\rightarrow \widetilde{{\mathcal {C}}}_{j-1} (\Delta ; K)$ is given by

$$\begin{aligned} \partial ([i_0,i_1,\ldots ,i_j])=\sum _{k=0}^j(-1)^k[i_0,\ldots , i_{k-1},i_{k+1},\ldots ,i_j]. \end{aligned}$$

The simplicial homology ${\widetilde{H}}_j(\Delta ;K)$ is then defined as the jth homology group of the complex ${\widetilde{C}}(\Delta ;K)$. That is,

$$\begin{aligned} {\widetilde{H}}_j(\Delta ;K):=H_j(\widetilde{{\mathcal {C}}}(\Delta ; K)). \end{aligned}$$

We collect some easy facts on simplicial homology we need in the following lemmas. Recall that a simplicial complex is a simplex if it has a unique facet.

Lemma 1.1

• (1) ${\widetilde{H}}_{-1}(\Delta ;K)\ne 0$ if and only if $\Delta =\{\emptyset \}$.

• (2) If $\Delta$ is a simplex, then ${\widetilde{H}}_i(\Delta ;K)={\widetilde{H}}_i(\emptyset ;K)=0$ for all $i\in {\mathbb {Z}}$.

• (3) ${\widetilde{H}}_0(\Delta ;K)\ne 0$ if and only if $\Delta$ is disconnected.

• (4) Let $\Delta$ be a simplicial complex on [r] such that $\Delta \ne \emptyset$ and $\Delta \ne \{\emptyset \}$. Then, ${\widetilde{H}}_i(\Delta _1;K)={\widetilde{H}}_i(\Delta ;K)$ for $i\ne 0$ and $\dim _K{\widetilde{H}}_0(\Delta _1;K)=\dim _K{\widetilde{H}}_0(\Delta ;K)+1$, where $\Delta _1:=\Delta \cup \{r+1\}$ is the simplicial complex on $[r+1]$.

Proof

The statements (1), (3) and (4) are immediate from the definition. For (2), one may see e.g., [10, Example 5.1.9]. $\square$

The following is a copy of [12, Lemma 1.6].

Lemma 1.2

Let G be a simple graph considered as a simplicial complex of dimension one. Then, ${\widetilde{H}}_1(G;K)=0$ if and only if G contains no cycles.

2.3 Takayama’s formula

Let I be a monomial ideal of the polynomial ring S and ${\mathfrak {m}}$ the maximal homogeneous ideal of S. For any ${\mathbf {a}}=(a_1,\ldots ,a_r)\in {\mathbb {Z}}^r$, we put $G_{{\mathbf {a}}}=\{i\in [r]:a_i<0\}$. Let $\Delta (I)$ denote the simplicial complex of all $F\subseteq [r]$ such that $x_F\notin \sqrt{I}$. The famous Takayama’s formula [21] can be stated as follows.

Lemma 1.3

$\dim _K H^{i}_{{\mathfrak {m}}}(S/I)_{{\mathbf {a}}}$=$\left\{ \begin{array}{ll} \dim _K {\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\Delta _{{\mathbf {a}}}(I);K) , &{} G_{{\mathbf {a}}}\in \Delta (I); \\ 0, &{} \hbox {otherwise.} \end{array} \right.$

The simplicial complex $\Delta _{{\mathbf {a}}}(I)$ has several equivalent interpretations. Suppose that I is generated by monomials $(u_1,\ldots ,u_g)$. Then,

$$\begin{aligned} \begin{aligned} \Delta _{{\mathbf {a}}}(I)&=\{F\subseteq [r]\setminus G_{{\mathbf {a}}}: {\mathbf {x}}^{{\mathbf {a}}}\notin IS_{F\cup G_{{\mathbf {a}}}}\}\\&=\{F\subseteq [r]\setminus G_{{\mathbf {a}}}: \forall 1\le j\le g, \exists i\in [r]\setminus (F\cup G_{{\mathbf {a}}}) \text{ with } a_i<\deg _i (u_j)\}\\&=\{F\subseteq [r]\setminus G_{{\mathbf {a}}}: {\mathbf {x}}^{{\mathbf {a}}}{\mathbf {x}}_{F\cup G_{{\mathbf {a}}}}^t\notin I, \forall t\ge 1\}. \end{aligned} \end{aligned}$$

Here, $S_{F\cup G_{{\mathbf {a}}}}:=S[x_i^{-1}:i\in F\cup G_{{\mathbf {a}}}]$, and $\deg _i(u)$ is defined to be $a_i$ if $u=x_1^{a_1}\ldots x_r^{a_r}$.

The concept of monomial localization is introduced in [11] as a simplification of the localization. Fix a subset $F\subseteq [r]$. Let $\pi _F: S\rightarrow K[x_i: i\in [r]\setminus F]$ be the K-algebra homomorphism extended by the map sending $x_i$ to $x_i$ for $i\in [r]\setminus F$ and $x_i$ to 1 for $i\in F$. The image of a monomial ideal I of S under the map $\pi _F$ is called the monomial localization of I with respect to F, denoted by I[F]. It is clear that if I, J are monomial ideals of S, then $(IJ)[F]=I[F]J[F]$ and $(I\cap J)[F]=I[F]\cap J[F]$. Let ${\mathbf {a}}_{+}$ denote the nonnegative part of a vector ${\mathbf {a}}$. Under these notations, we have the following description of $\Delta _{{\mathbf {a}}}(I)$.

Lemma 1.4

In Lemma 1.3, $\Delta _{{\mathbf {a}}}(I)=\{F\subseteq [r]\setminus G_{{\mathbf {a}}}: {\mathbf {x}}^{{\mathbf {a}}_{+}}\notin I[F\cup G_{{\mathbf {a}}}]S\}.$

Proof

This is because $I[F\cup G_{{\mathbf {a}}}]S=IS_{F\cup G_{{\mathbf {a}}}}\cap S$.$\square$

We will use the above interpretation of $\Delta _{{\mathbf {a}}}(I)$ in this paper. Let $G\subseteq [r]$. Recall that ${\mathrm {Link}}_{\Delta }(G)$ is defined to be the subcomplex $\{F\setminus G:G\subseteq F\in \Delta \}$. The following result allows us to consider only the case when ${\mathbf {a}}\in {\mathbb {N}}^r$.

Lemma 1.5

Let ${\mathbf {a}}=(a_1,\ldots , a_r)$ be a vector in ${\mathbb {Z}}^r$ with $G_{{\mathbf {a}}}\ne \emptyset$. Then,

$$\begin{aligned} \Delta _{{\mathbf {a}}}(I)={\mathrm {Link}}_{\Delta _{\mathbf {a_+}}(I)}(G_{{\mathbf {a}}}). \end{aligned}$$

Proof

If F belongs to either $\Delta _{{\mathbf {a}}}(I)$ or ${\mathrm {Link}}_{\Delta _{\mathbf {a_+}}(I)}(G_{{\mathbf {a}}})$, then $F\subseteq [r]\setminus G_{{\mathbf {a}}}$. Now, let F be a subset of $[r]\setminus G_{{\mathbf {a}}}$. Then, $F\in \Delta _{{\mathbf {a}}}(I)$ if and only if ${\mathbf {x}}^{\mathbf {a+}}\notin I[F\cup G_{{\mathbf {a}}}]S$ if and only if $F\cup G_{{\mathbf {a}}}\in \Delta _{{\mathbf {a}}_+}(I)$ if and only if $F\in \mathrm {link}_{\Delta _{{\mathbf {a}}_+}}(G_{{\mathbf {a}}})$, as desired. $\square$

We close this section by giving a simplification of Takayama’s formula in the case that I is a power of a squarefree monomial ideal.

Lemma 1.6

Let I be a squarefree monomial ideal of S. Then, for all ${\mathbf {a}}\in {\mathbb {Z}}^r$ and $n\ge 1$, we have $\dim _K H^{i}_{{\mathfrak {m}}}(S/I^n)_{{\mathbf {a}}}=\dim _K {\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\Delta _{{\mathbf {a}}}(I^n);K).$

Proof

In view of Lemma 1.3 as well as Lemma 1.1.(2), it is enough to show that if $G_{{\mathbf {a}}}\notin \Delta (I^n)$, then $\Delta _{{\mathbf {a}}}(I^n)=\emptyset$.

Assume now that $G_{{\mathbf {a}}}\notin \Delta (I^n)$. Then, ${\mathbf {x}}_{G_{{\mathbf {a}}}}\in \sqrt{I^n}=I$ and so $1\in I[G_{{\mathbf {a}}}]$. From this, it follows that $1\in I^n[G_{{\mathbf {a}}}]$, and thus $\Delta _{{\mathbf {a}}}(I^n)=\emptyset$ by Lemma 1.4, as required. $\square$

3 Some comparisons

In this section, we always assume that I is a monomial ideal of S with $I\ne S$ and let

$$\begin{aligned} J:=(I,x_1y,\ldots ,x_ry) \end{aligned}$$

be the ideal of $R:=S[y]=K[x_1,\ldots ,x_r,y]$. We fix $n\ge 1$ and compare the regularity, g-regularity and $a_i$-invariants of $S/I^n$ with the ones of $R/J^k$.

We begin with basic lemmas on the regularity of graded modules.

Lemma 2.1

Let $0\rightarrow M\rightarrow N\rightarrow P\rightarrow 0$ be a short exact sequence of finitely generated graded S-modules. Then,

• (1) ${\text {reg}}(N)\le \max \{{\text {reg}}(M),{\text {reg}}(P)\}$;

• (2) ${\text {reg}}(M)\le \max \{{\text {reg}}(N),{\text {reg}}(P)+1\}$;

• (3) ${\text {reg}}(P)\le \max \{{\text {reg}}(N),{\text {reg}}(M)-1\}$;

• (4) ${\text {reg}}(N)={\text {reg}}(P)$ if ${\text {reg}}(M)\le {\text {reg}}(P).$

Proof

The first three statements are well known; see e.g., [5]. For the convenience of the readers, we give a proof of (4). Since ${\text {reg}}(M)\le {\text {reg}}(P)$, we obtain ${\text {reg}}(P)\le {\text {reg}}(N)$ by (3) and ${\text {reg}}(N)\le {\text {reg}}(P)$ by (1). Thus, ${\text {reg}}(N)={\text {reg}}(P)$, as desired. $\square$

Denote by ${\mathfrak {m}}$ the maximal homogenous ideal $(x_1,\ldots ,x_r)$ of S. The following lemma is a special case of [5, Theorem 2.2], but we include a short proof for the completeness.

Lemma 2.2

Let M be a finitely generated graded S-module. Then, ${\text {reg}}({\mathfrak {m}}M)\le {\text {reg}}(M)+1$.

Proof

Since $H_{{\mathfrak {m}}}^0(M/{\mathfrak {m}}M)=M/{\mathfrak {m}}M$, we have ${\text {reg}}(M/{\mathfrak {m}}M)=a_0(M/{\mathfrak {m}}M)$ and it is the largest degree of minimal generators of M, which, by e.g., [3, Theorem 15.3.1], is less than or equal to ${\text {reg}}(M)$. Thus, the desired inequality follows from the short exact sequence $0\rightarrow {\mathfrak {m}}M\rightarrow M\rightarrow M/{\mathfrak {m}}M\rightarrow 0$ together with Lemma 2.1.(2). $\square$

Proposition 2.3

Assume $I\subseteq {\mathfrak {m}}^2$. Then, the following statements hold.

• (1) If ${\text {reg}}(S/I^n)=dn+e$ for all $n\gg 0$ with $d\ge 3$, then ${\text {reg}}(R/J^n)={\text {reg}}(S/I^n)$ for all $n\gg 0$;

• (2) Given any positive integer $\lambda$, if $I^n$ has a linear resolution for all $n\le \lambda$, then ${\text {reg}}(R/J^n)={\text {reg}}(S/I^n)$ for all $n\le \lambda$.

Proof

We will use the following short exact sequence:

For this, we first look at the regularities of $(R/J^n:y)[-1]$ and $R/(J^n,y)$.

Note that $R=\bigoplus _{i\ge 0} S[y^i]$ and $J=(I,{\mathfrak {m}}y)$, it is not difficult to see

$$\begin{aligned} J^n=I^n\oplus I^{n-1}{\mathfrak {m}}y\oplus \ldots \oplus I{{\mathfrak {m}}}^{n-1}y^{n-1}\oplus \oplus _{i\ge 0}{{\mathfrak {m}}}^ny^{n+i}. \end{aligned}$$

This implies

$$\begin{aligned} J^n:y=I^{n-1}{\mathfrak {m}}\oplus I^{n-2}{\mathfrak {m}}^2y\oplus \ldots \oplus I{\mathfrak {m}}^{n-1}y^{n-2}\oplus \oplus _{i\ge 0}{\mathfrak {m}}^ny^{n-1+i} \end{aligned}$$

and so

$$\begin{aligned} R/J^n:y=S/I^{n-1}{\mathfrak {m}}\oplus (S/I^{n-2}{\mathfrak {m}}^2)y\oplus \ldots \oplus (S/I{\mathfrak {m}}^{n-1})y^{n-2} \oplus (S[y]/{\mathfrak {m}}^n)(y^{n-1}), \end{aligned}$$

where the last equality follows from the equality

$$\begin{aligned} (S/{\mathfrak {m}}^n)y^{n-1}\oplus (S/{\mathfrak {m}}^n)y^n\oplus \ldots =(S[y]/{\mathfrak {m}}^n)y^{n-1}. \end{aligned}$$

From this, it follows that

On the other hand, it is clear that $(J^n,y)=(I^n,y)$. Thus, ${\text {reg}}R/(J^n,y)={\text {reg}}(S/I^n)$.

• (1) We may assume ${\text {reg}}(S/I^n)=dn+e$ for all $n\ge n_0$. Then, by Lemma 2.2, one has

  $$\begin{aligned} {\text {reg}}(S/I^{n-k}{\mathfrak {m}}^k)+k\le \left\{ \begin{array}{ll} d(n-k)+e+2k, &{} \hbox {if }n_0\le n-k; \\ t+2k, &{} \hbox {if }1\le n-k\le n_0 \end{array} \right. , \end{aligned}$$

  where $t:=\max \{{\text {reg}}(S/I^n):n\le n_0\}$. Let $n_0':=n_0+t$. Since $d\ge 3$, it follows that ${\text {reg}}(S/I^{n-k}{\mathfrak {m}}^k)+k\le d(n-1)+e+2$ for all $1\le k\le n-1$ and for $n\ge n'_0$, and so ${\text {reg}}(R/J^n:y)[-1]\le d(n-1)+e+2$ for all $n\ge n_0'$ by ($\ddag$). This implies

  $$\begin{aligned} {\text {reg}}(R/J^n:y)[-1]\le {\text {reg}}(R/(J^n,y)) \end{aligned}$$

  for $n\gg 0.$ Now, the aforementioned short exact sequence ($\dag$) yields the desired equality in view of Lemma 2.1.(4).

• (2) We may assume that I is generated in a single degree $d\ge 2$. Then, ${\text {reg}}(S/I^n)=dn-1$ and this implies ${\text {reg}}(R/J^n:y)[-1]\le d(n-1)+1$ for $n\le \lambda$ by ($\ddag$). From this, it follows that ${\text {reg}}(R/J^n:y)[-1]\le {\text {reg}}(R/(J^n,y))$ for $n\le \lambda$, and then the desired equality follows by using the short exact sequence ($\dag$) and by Lemma 2.1.(4). $\square$

Remark 2.4

• (1) From this proof, the condition that I is a monomial ideal is not necessary in Proposition 2.3. That is, I could be any homogeneous ideal.

• (2) [18, Remark 5.7] shows that the condition that $d\ge 3$ in Proposition 2.3.(1) cannot be removed. But if assume further I is a squarefree monomial ideal, then the condition that $d\ge 3$ could be dropped, as shown by [18, Corollary 5.6].

Let ${\mathfrak {n}}$ denote the maximal homogeneous ideal $(x_1,\ldots ,x_r,y)$ of R.

Proposition 2.5

Suppose $I\subseteq {\mathfrak {m}}^2$. Then, $\text{ g-reg }(R/J^n)\ge n-1$ for all $n\ge 1$.

Proof

We claim that if ${\mathbf {x}}^{{\mathbf {a}}}y^t\in J^n: {\mathfrak {n}}^{\infty }$ for some $t\ge 0$, then $|{\mathbf {a}}|\ge n$. Let ${\mathbf {x}}^{{\mathbf {a}}}y^t\in J^n: {\mathfrak {n}}^{\infty }$. Then, there exists $k\ge 1$ such that ${\mathbf {x}}^{{\mathbf {a}}}y^{t+k}\in J^n$. It follows that

$$\begin{aligned} {\mathbf {x}}^{{\mathbf {a}}}y^{t+k}=u_{i_1}u_{i_2}\ldots u_{i_n}v, \end{aligned}$$

where $u_{i_j}$ are minimal generators of J for $j=1,\ldots ,n$ and v is a monomial. Since $\deg _{{\mathbf {x}}}(u_{i_j})\ge 1$ for all j, we have $|{\mathbf {a}}|=\deg _{{\mathbf {x}}}({\mathbf {x}}^{{\mathbf {a}}}y^{t+k})\ge \deg _{{\mathbf {x}}}(u_{i_1})+\cdots +\deg _{{\mathbf {x}}}(u_{i_n})\ge n$, as claimed. Here, $\deg _{{\mathbf {x}}} (u)$ is defined to be the integer $|{\mathbf {a}}|$ if $u={\mathbf {x}}^{{\mathbf {a}}}y^t$.

From this claim, it follows that

$$\begin{aligned} \text{ g-reg }(R/J^n)={\text {reg}}(R/J^n:{\mathfrak {n}}^{\infty })={\text {reg}}(J^n:{\mathfrak {n}}^{\infty })-1\ge n-1, \end{aligned}$$

as desired. Here, the last inequality follows from [3, Theorem 15.3.1]. $\square$

Example 2.6

Let I be the ideal ${\mathfrak {m}}^2$. Then,

• (1) ${\text {reg}}(R/J^n)={\text {reg}}(S/I^n)=2n-1$;

• (2) $\text{ g-reg }(R/J^n)=n-1$ and $\text{ g-reg }(S/I^n)=-\infty$.

Proof

• (1) It follows from Proposition 2.3.(2).

• (2) We first show that $J^n:{\mathfrak {n}}^{\infty }=(x_1,\ldots ,x_r)^n$. The inclusion has been proved in the proof of Proposition 2.5. Let $u={\mathbf {x}}^{{\mathbf {a}}}$ with $|{\mathbf {a}}|\ge n$. Then, for any ${\mathbf {x}}^{{\mathbf {b}}}y^t$ with $|{\mathbf {b}}|+t= n$, we may choose a vector ${\mathbf {c}}\in {\mathbb {N}}^r$ such that $|{\mathbf {c}}|=t$ and ${\mathbf {c}}\le \mathbf {a+b}$. Since ${\mathbf {x}}^{\mathbf {a+b-c}}\in (x_1,\ldots ,x_r)^{2n-2t}=I^{n-t}$ and $\mathbf {x^c}y^t\in (x_1y,\ldots ,x_ry)^t$, it follows that $u{\mathbf {x}}^{{\mathbf {b}}}y^t\in J^n$, and so $u\in J^n:{\mathfrak {n}}^{\infty }$, by noting that ${\mathfrak {n}}^n$ is generated by ${\mathbf {x}}^{{\mathbf {b}}}y^t$ with $|{\mathbf {b}}|+t=n$. Thus, we have shown $J^n:{\mathfrak {n}}^{\infty }=(x_1,\ldots ,x_r)^n$. Consequently,

  $$\begin{aligned} \text{ g-reg }(R/J^n)={\text {reg}}(R/J^n:{\mathfrak {n}}^{\infty })={\text {reg}}(J^n:{\mathfrak {n}}^{\infty })-1= n-1, \end{aligned}$$

  for all $n\ge 1$. Notice that $I^n:{\mathfrak {m}}^{\infty }=S$, the last statement follows. $\square$

This example suggests that the relationship between $\text{ g-reg }(R/J^n)$ and $\text{ g-reg }(S/I^n)$ is more subtle. In the rest part of this section, we will express the $a_i$-invariants of $R/J^n$ in terms of the ones of $S/I^{n-t}$ with $0\le t\le n-1$ and for $i\ge 1$ by using Takayama’s lemma. For this, we need some preparations describing the relations between simplicial complexes $\Delta _{({\mathbf {a}},t)}(J^n)$ and $\Delta _{{\mathbf {a}}}(I^n)$, where ${\mathbf {a}}$ is a vector in ${\mathbb {Z}}^r$ and $t\in {\mathbb {Z}}$. Note that $({\mathbf {a}},t)\in {\mathbb {Z}}^{r+1}$.

Proposition 2.7

• (1) Suppose $t<0$. Then, $\Delta _{({\mathbf {a}},t)}(J^n)$ is either $\emptyset$ or $\{\emptyset \}$. Moreover, $\Delta _{({\mathbf {a}},t)}(J^n)=\{\emptyset \}$ if and only if $|{\mathbf {a}}|\le n-1$ and $G_{{\mathbf {a}}}=\emptyset$.

• (2) Suppose that $0 \le t\le n-1$ and $F\subseteq [r]$. If either $G_{{\mathbf {a}}}\ne \emptyset$ or $F\ne \emptyset$, then $F\in \Delta _{({\mathbf {a}},t)}(J^n)\Longleftrightarrow F\in \Delta _{{\mathbf {a}}}(I^{n-t}).$

• (3) Suppose that $0 \le t\le n-1$ and $F\subseteq [r+1]$. If $r+1\in F$, then $F\in \Delta _{({\mathbf {a}},t)}(J^n)$ if and only if $F=\{r+1\}$, $G_{{\mathbf {a}}}=\emptyset$ and $|{\mathbf {a}}|\le n-1$.

• (4) Suppose $t\ge n$. Then, $\Delta _{({\mathbf {a}},t)}(J^n)$ is either $\emptyset$ or $\langle \{r+1\}\rangle$. Moreover, $\Delta _{({\mathbf {a}},t)}(J^n)=\langle \{r+1\}\rangle$ if and only if $G_{{\mathbf {a}}}=\emptyset$ and $|{\mathbf {a}}|\le n-1$.

Proof

• (1) Note that $J[r+1]=(x_1,\ldots ,x_r)$ and $G_{({\mathbf {a}},t)}=G_{{\mathbf {a}}}\cup \{r+1\}$. If $G_{{\mathbf {a}}}\ne \emptyset$, then $J^n[G_{({\mathbf {a}},t})]R=R$ and so $\Delta _{({\mathbf {a}},t)}(J^n)=\emptyset$. Now, assume $G_{{\mathbf {a}}}= \emptyset$. Then, for any $\emptyset \ne F\subseteq [r]$, we have $1\in J[F,G_{({\mathbf {a}},t)}]$, and so $F\notin \Delta _{({\mathbf {a}},t)}(J^n)$. This implies $\Delta _{({\mathbf {a}},t)}(J^n)$ is either $\emptyset$ or $\{\emptyset \}$. Moreover, $\emptyset \in \Delta _{({\mathbf {a}},t)}(J^n)$ if and only if ${\mathbf {x}}^{{\mathbf {a}}} \notin (x_1,\ldots ,x_r)^n$ if and only if $|{\mathbf {a}}|\le n-1$. This proves (1).

• (2) If either $G_{{\mathbf {a}}}\ne \emptyset$ or $F\ne \emptyset$, then $J[F\cup G_{{\mathbf {a}}}]=(I[F\cup G_{{\mathbf {a}}}], y)$. Thus, ${\mathbf {x}}^{{\mathbf {a}}}y^t\notin (J[F\cup G_{{\mathbf {a}}}])^n$ if and only if ${\mathbf {x}}^{{\mathbf {a}}}\notin (I[F\cup G_{{\mathbf {a}}}])^{n-t}$. This proves (2).

• (3) If either $F\supsetneq \{r+1\}$ or $G_{{\mathbf {a}}}\ne \emptyset$, then $J^n[F\cup G_{{\mathbf {a}}}]R=R$ and $F\notin \Delta _{({\mathbf {a}},t)}(J^n)$. Hence, if $F\in \Delta _{({\mathbf {a}},t)}(J^n)$, we must have $F=\{r+1\}$, $G_{{\mathbf {a}}}=\emptyset$. Now, assume that $F=\{r+1\}$ and $G_{{\mathbf {a}}}=\emptyset$. Then, $J[G_{({\mathbf {a}},t)}\cup F]=(x_1,\ldots ,x_r)$. Since ${\mathbf {x}}^{{\mathbf {a}}}\notin (x_1,\ldots ,x_r)^n$ if and only if $|{\mathbf {a}}|\le n-1$, the statement (3) has been proved.

• (4) Suppose either $F\cap [r]\ne \emptyset$ or $G_{{\mathbf {a}}}\ne \emptyset$. Since $y\in J[F\cup G_{{\mathbf {a}}}]$ and $t\ge n$, we have $y^t\in (J[F\cup G_{{\mathbf {a}}}])^n$ and $\mathbf {x^{a_+}}y^t\in (J[F\cup G_{{\mathbf {a}}}])^n$. Thus, if $G_{{\mathbf {a}}}\ne \emptyset$, then $\Delta _{({\mathbf {a}},t)}(J^n)=\emptyset$, and if $G_{{\mathbf {a}}}=\emptyset$ and $F\in \Delta _{({\mathbf {a}},t)}(J^n)$, then $F\cap [r]=\emptyset$.

Assume that $G_{{\mathbf {a}}}=\emptyset$ and $\Delta _{({\mathbf {a}},t)}(J^n)\ne \emptyset$. Then, $\emptyset \in \Delta _{({\mathbf {a}},t)}(J^n)$ and it follows that $\mathbf {x^{a}}y^t\notin J^n$. Since $t\ge n$, we have $|{\mathbf {a}}|\le n-1$, for otherwise, we have $\mathbf {x^{a}}y^t\subseteq (x_1y,\ldots ,x_ry)^n\subseteq J^n$, a contradiction. From this, it follows that $\mathbf {x^{a}}y^t\notin (J[r+1])^n=(x_1,\ldots ,x_r)^n$ and so $\{r+1\}\in \Delta _{({\mathbf {a}},t)}(J^n)$. Hence, we have $\Delta _{({\mathbf {a}},t)}(J^n)=\langle \{r+1\}\rangle$ and the first statement has been proved. The second one follows immediately by the fact ${\mathbf {x}}^{{\mathbf {a}}}y^t\notin (J[r+1])^n$ if and only if $|{\mathbf {a}}|\le n-1$. $\square$

We now present the main result of this section.

Theorem 2.8

• (1) If $i\ge 2$, then $a_i(R/J^n)=\max \{a_i(S/I^{n-t})+t: 0\le t\le n-1\}.$

• (2) If $I[j]\ne S[j]$ for some $j\in [r]$, then

  $$\begin{aligned} a_1(R/J^n)=\max \{2n-2, a_1(S/I^{n-t})+t: 0\le t\le n-1\}. \end{aligned}$$

Proof

Given any integer $i\ge 1$, we first show that $a_i(R/J^n)\ge a_i(S/I^{n-t})+t$ for all $0\le t\le n-1$. Fix $0\le t\le n-1$. If $a_i(S/I^{n-t})=-\infty$, there is nothing to prove. Now, we assume $a_i(S/I^{n-t})\ne -\infty$ and let ${\mathbf {a}}\in {\mathbb {Z}}^r$ such that $H_{{\mathfrak {m}}}^i(S/I^{n-t})_{{\mathbf {a}}}\ne 0$ with $|{\mathbf {a}}|=a_i(S/I^{n-t})$. Since ${\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\Delta _{{\mathbf {a}}}(I^{n-t});K)\ne 0$ (see Lemma 1.3), one has $\Delta _{{\mathbf {a}}}(I^{n-t})\ne \emptyset$. There are two cases to consider.

We first consider the case that $\Delta _{{\mathbf {a}}}(I^{n-t})=\{ \emptyset \}$. Since ${\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\{\emptyset \};K)\ne 0$, we have $|G_{{\mathbf {a}}}|=i\ge 1$ by Lemma 1.1.(1). From this, it follows that $r+1\notin F$ for any $F\in \Delta _{({\mathbf {a}},t)}(J^n)$ by Proposition 2.7.(3) and so $\Delta _{({\mathbf {a}},t)}(J^n)=\{\emptyset \}$ by Proposition 2.7.(2). Note that $\sqrt{J}=(\sqrt{I}, x_1y,\ldots ,x_ry)$, we have $\Delta (I^k)=\Delta (I)=\Delta (J) \cap 2^{[r]}=\Delta (J^\ell )$ for all $k,\ell \ge 1$. This implies $G_{({\mathbf {a}},t)}=G_{{\mathbf {a}}}\in \Delta {(J^n)}$, and so $H_{{\mathfrak {n}}}^i(R/J^n)_{({\mathbf {a}},t)}={\widetilde{H}}_{-1}(\{\emptyset \})\ne 0$ by Lemma 1.3. Thus, $a_i(R/J^n)\ge |{\mathbf {a}}|+t= a_i(S/I^{n-t})+t$.

We next consider the case when $\Delta _{{\mathbf {a}}}(I^{n-t})\supsetneqq \{ \emptyset \}$. In this case, we have either $\Delta _{({\mathbf {a}},t)}(J^n)=\Delta _{{\mathbf {a}}}(I^{n-t})\cup \{r+1\}$ and $G_{{\mathbf {a}}}=\emptyset$ or $\Delta _{({\mathbf {a}},t)}(J^n)=\Delta _{{\mathbf {a}}}(I^{n-t})$ by (2) and (3) of Proposition 2.7. From this, it follows that

$$\begin{aligned} \dim _K{\widetilde{H}}_{i-|G_{({\mathbf {a}},t)}|-1}(\Delta _{({\mathbf {a}},t)}(J^n);K)\ge \dim _K{\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\Delta _{{\mathbf {a}}}(I^{n-t});K)\ne 0 \end{aligned}$$

in view of Lemma 1.1.(4), and so $H_{{\mathfrak {n}}}^i(R/J^n)_{({\mathbf {a}},t)}\ne 0$. Thus, we also have $a_i(R/J^n)\ge a_i(S/I^{n-t})+t$.

Conversely, we may harmlessly assume that $a_i(R/J^n)\ne -\infty$. Let $({\mathbf {a}},t)\in {\mathbb {Z}}^{r+1}$ such that $H_{{\mathfrak {n}}}^i(R/J^n)_{({\mathbf {a}},t)}\ne 0$ and $a_i(R/J^n)=|{\mathbf {a}}|+t.$ It is clear that $\Delta _{({\mathbf {a}},t)}(J^n)\ne \emptyset$, and it is also clear that $t\le n-1$ by Proposition 2.7.(4).

If $i\ge 2$, then it must be $t\ge 0$, for otherwise, we have $\Delta _{({\mathbf {a}},t)}(J^n)=\{\emptyset \}$ and $G_{{\mathbf {a}}}=\emptyset$ by Proposition 2.7.(1), which implies $i=1$ by Lemma 1.1.(1), a contradiction. We also have either $G_{{\mathbf {a}}}\ne \emptyset$ or $\Delta _{({\mathbf {a}},t)}(J^n)\ne \{\emptyset \}$, since $H_{{\mathfrak {n}}}^i(R/J^n)_{({\mathbf {a}},t)}\ne 0$. From these, it follows that either $\Delta _{({\mathbf {a}},t)}(J^n)= \Delta _{{\mathbf {a}}}(I^{n-t})$ or $\Delta _{({\mathbf {a}},t)}(J^n)=\Delta _{{\mathbf {a}}}(I^{n-t})\cup \{r+1\}$ and $G_{{\mathbf {a}}}=\emptyset$ by Proposition 2.7.(2,3). Therefore, ${\widetilde{H}}_{i-|G_{{\mathbf {a}}}|-1}(\Delta _{{\mathbf {a}}}(I^{n-t});K)\ne 0$ in view of Lemma 1.1.(4). Since $G_{{\mathbf {a}}}=G_{({\mathbf {a}},t)}\in \Delta (J)\cap 2^{[r]}=\Delta (I)$, we have $H_{{\mathfrak {m}}}^i(S/I^{n-t})_{{\mathbf {a}}}\ne 0$ by Lemma 1.3 , and thus, $a_i(R/J^n)\le a_i(S/I^{n-t})+t$, completing the proof of (1).

Now, consider the case when $i=1$. If $t<0$, then it must be $\Delta _{({\mathbf {a}},t)}(J^n)=\{\emptyset \}$ and $G_{{\mathbf {a}}}=\emptyset$ with $|{\mathbf {a}}|\le n-1$ by Proposition 2.7.(1). Thus, $a_1(R/J^n)=|{\mathbf {a}}|-|t|\le n-2$.

Suppose that $t\ge 0$. If $\Delta _{({\mathbf {a}},t)}(J^n)=\{\emptyset \}$, then it must be $|G_{{\mathbf {a}}}|=1$ and so $\Delta _{{\mathbf {a}}}(I^{n-t})=\{\emptyset \}$ by Proposition 2.7.(2). Since $G_{({\mathbf {a}},t)}\in \Delta (J)$, we have $G_{{\mathbf {a}}}\in \Delta (I)$. From these, it follows that $H_{{\mathfrak {m}}}^1(S/I^{n-t})_{{\mathbf {a}}}\ne 0$ by Lemma 1.3 and so $a_1(R/J^n)\le a_1(S/I^{n-t})+t$.

If $\Delta _{({\mathbf {a}},t)}(J^n)\ne \{\emptyset \}$, then either $\Delta _{({\mathbf {a}},t)}(J^n)= \Delta _{{\mathbf {a}}}(I^{n-t})$ or $\Delta _{({\mathbf {a}},t)}(J^n)= \Delta _{{\mathbf {a}}}(I^{n-t})\cup \{r+1\}$ and $G_{{\mathbf {a}}}=\emptyset$ by (2) and (3) of Proposition 2.7. In the first case, we have $H_{{\mathfrak {m}}}^1(S/I^{n-t})_{{\mathbf {a}}}\ne 0$ and $a_1(R/J^n)\le a_1(S/I^{n-t})+t$; in the second case, it follows that $|{\mathbf {a}}|\le n-1$ by Proposition 2.7.(3) and so $a_1(R/J^n)=|{\mathbf {a}}|+t\le 2n-2$.

Finally, we show that $a_1(R/J^n)\ge 2n-2$. We may harmlessly assume that $I[1]\ne S[1]$. Let ${\mathbf {a}}$ denote the vector $(n-1,0,\ldots ,0)$ of ${\mathbb {Z}}^{r}$. Since

$$\begin{aligned} x_1^{n-1}y^{n-1}\notin (J[r+1])^nS=(x_1,\ldots ,x_r)^nS \end{aligned}$$

and

$$\begin{aligned} x_1^{n-1}y^{n-1}\notin (J[1])^n=(I[1],y)^nS, \end{aligned}$$

both $\{r+1\}$ and $\{1\}$ are faces of $\Delta _{({\mathbf {a}},n-1)}(J^n)$. On the other hand, for any $j\in [r]$, since $J[j,r+1]R=R$, one has $\{j,r+1\}\notin \Delta _{{\mathbf {a}}}(J^n)$. From this, it follows that $\Delta _{({\mathbf {a}},n-1)}(J^n)$ is disconnected and so $a_1(R/J^n)\ge |{\mathbf {a}}|=2n-2$, as required. $\square$

The condition that $I[j]\ne S[j]$ for some $j\in [r]$ is equivalent to that $\sqrt{I}\ne (x_1,\ldots , x_r)$. It is a weak requirement but necessary in Theorem 2.8.(2) in view of Example 2.6.

4 The $a_1$ and $a_2$ of $S/I_G^n$

By a simple graph, we mean an undirected graph having no loops and no parallel edges. In this section, we always assume that G is a simple graph without isolated vertex on vertex set [r]. Let E(G) denote the edge set of G, and let $I_G$ denote the Stanley–Reisner ideal of G when G is considered as an one-dimensional simplicial complex. This means

$$\begin{aligned} I_G=({\mathbf {x}}_F: F\subseteq [r], 2\le |F|\le 3, F\notin E(G) )=\bigcap _{e\in E(G)} P_e \subseteq K[x_1,\ldots ,x_r]. \end{aligned}$$

Here, ${\mathbf {x}}_F$ denotes the monomial $\prod _{i\in F}x_i$ and $P_e$ the monomial prime $(x_i:i\in [r]\setminus e)$. We will compute the values of $a_1(S/I_G^n)$ and $a_2(S/I_G^n)$ for $n\ge 1$. For the convenience, we set

$$\begin{aligned} a_i^j(S/I_G^n):=\max \{|{\mathbf {a}}|: {\mathbf {a}}\in {\mathbb {Z}}^r, H_{{\mathfrak {m}}}^i(S/I_G^n)_{{\mathbf {a}}}\ne 0, |G_{{\mathbf {a}}}|=j\} \end{aligned}$$

for $i,j\ge 0$. Here, $|{\mathbf {a}}|=a_1+\cdots +a_r$ and $|G_{{\mathbf {a}}}|$ is the cardinality of $G_{{\mathbf {a}}}$. Thus,

$$\begin{aligned} a_i(S/I_G^n)=\max \left\{ a_i^j(S/I_G^n):j\ge 0\right\} \end{aligned}$$

for all $i\ge 0$.

We recall some basic notions in graph theory. Let $p\in [r]$. We use $N_p$ to denote the neighborhood of p, that is, $N_p:=\{i\in [r]: \{i,p\} \text{ is } \text{ an } \text{ edge } \text{ of } G \}$. The degree of p, denoted by $\deg p$, is the cardinality of $N_p$. The maximal degree of vertices of G is denoted by $\deg (G)$. It is clear that if $\deg (G)=1$, then G is the disjoint union of some edges. Let $q\in [r]$. The path between p and q is a sequence of distinct vertices $p=p_0,p_1,\ldots ,p_{\ell }=q$ such that $\{p_i,p_{i+1}\}$ is an edge of G for $i=0,\ldots ,\ell -1$. We write this path as $p=p_0-p_1-\cdots -p_{\ell }=q$ and call $\ell$ to be its length. The distance between p and q, denoted by $d_G(p,q)$ or just d(p, q), is the minimal length of paths from p to q, with the convention that $d_G(p,q)=\infty$ if there is no paths connecting p and q. The maximum of $d_G(p,q)$ with $p,q\in [r]$ is called the diameter of G, denoted by $\text{ diam }(G)$. Thus, $\text{ diam }(G)=\infty$ if and only if G is disconnected.

Let $\ell \ge 3$. By a cycle of length $\ell$, we mean a sequence of vertices $p_1,p_2,\ldots ,p_{\ell }$ such that $\{p_{i},p_{i+1}\}$ is an edge of G for $i=1,\ldots ,\ell -1$ with $p_1,\ldots ,p_{\ell -1}$ pairwise distinct and $p_1=p_{\ell }$. The girth of a simple graph G, denoted by $\text{ girth }(G)$, is the smallest length of cycles of G, with the convention that $\text{ girth }(G)=\infty$ if G contains no cycles. It is clear that $3\le \text{ girth }(G)\le \infty$ for any simple graph G, and $\text{ girth }(G)= \infty$ if and only if G is a forest. For unexplained terminology in graph theory, we refer to [23].

4.1 The computation of $a_1(S/I_G^n)$

In this subsection, we will compute $a_1(S/I_G^n)$.

It is clear that if I is generated by monomials $u_1, u_2, \ldots , u_k$, then the monomial localization $I[F]=\pi _F(I)$ is generated by $u_1[F],\ldots ,u_k[F]$, where $u_i[F]$ is the image of $u_i$ under the map $\pi _F$ for $i=1,\ldots ,k$. If $F=\{p_1,\ldots ,p_s\}$ we write $I[p_1,\ldots ,p_s]$ instead of $I[\{p_1,\ldots ,p_s\}]$. We begin with some descriptions of monomial localizations of $I_G$ and the simplicial complexes $\Delta _{{\mathbf {a}}}(I_G^n)$ for ${\mathbf {a}}\in {\mathbb {N}}^r$.

Lemma 3.1

Let G be a simple graph on [r], and let $\{p,q\}$ be an edge of G. Then,

$$\begin{aligned} I_G[p,q]=(x_i:i\in [r]\setminus \{p,q\}). \end{aligned}$$

Proof

It is immediate from the definition of $I_G[p,q]$. $\square$

Denote by E(G) the edge set of G. It is clear that if $\{p,q\}\notin E(G)$, then $1\in I[p,q]$, i.e., $I[p,q]=S[p,q]$, and so $\{p,q\}\notin \Delta _{{\mathbf {a}}}(I_G^n)$ for any ${\mathbf {a}}\in {\mathbb {Z}}^r$ and $n\ge 1$. We now describe the faces of dimension 1 of $\Delta _{{\mathbf {a}}}(I_G^n)$ for ${\mathbf {a}}\in {\mathbb {N}}^r$ and for $n\ge 1$.

Proposition 3.2

Let p, q be distinct vertices of G and ${\mathbf {a}}=(a_1,\ldots ,a_r)$ a vector of ${\mathbb {N}}^r$. Fix $n\ge 1$. Then, the following statements are equivalent:

• (1) $\{p,q\}\in \Delta _{{\mathbf {a}}}(I_G^n)$;

• (2) $\{p,q\}\in E(G)$ and $\sum _{i\in [r]\setminus \{p,q\}}a_i\le n-1$.

Proof

It is immediate from Lemma 3.1 together with Lemma 1.4. $\square$

For a simplicial complex $\Delta$ and an integer $i\ge 0$, recall from [10, Page 144] that the pure $i\hbox {th}$ skeleton of $\Delta$ is defined to be the pure simplicial complex $\Delta (i)$ whose facets are the faces F of $\Delta$ with $|F|=i+1$. By [17, Lemmas 1.3 and 2.1], we see that $\Delta _{{\mathbf {a}}}(I_G^n)(1)$ coincides with $\Delta _{{\mathbf {a}}}(I_G^{(n)})$.

Unlike the case of symbolic powers, $\Delta _{{\mathbf {a}}}(I_G^n)$ may contain a facet of dimension zero. Let $p\in [r]$. We say that p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$ if $\{p\}$ is a facet of $\Delta _{{\mathbf {a}}}(I_G^n)$. We want to know when a vertex p of G is an isolated one of $\Delta _{{\mathbf {a}}}(I_G^n)$. For this, we denote $M_p:=[r]\setminus (N_p\cup \{p\})$.

Lemma 3.3

Let G be a simple graph on [r] and p a vertex of degree $\ge 2$. Then,

$$\begin{aligned} I_G[p]=(x_ix_j:i\ne j \text{ and } i,j\in N_p)+(x_i: i\in M_p). \end{aligned}$$

Proof

It is clear from the definition of $I_G[p]$. $\square$

Proposition 3.4

Let p be a vertex of degree $\ge 2$, and ${\mathbf {a}}=(a_1,\ldots ,a_r)$ a vector in ${\mathbb {N}}^r$. Fix $n\ge 1$. Then, the following statements are equivalent:

• (1) The vertex p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$;

• (2) There exists $t\in \{0,\ldots ,n-1\}$ such that

• $\sum _{i\in M_p}a_i=t$; ➊

• $\sum _{i\in N_p\setminus \{j\}}a_i \ge n-t$ for all $j\in N_p$; ➋

• $\sum _{i\in N_p} a_i\le 2(n-t)-1$.➌

Proof

Set $J:=(x_ix_j:i\ne j \text{ and } i,j\in N_p)$ and $K:=(x_i: i\in M_p)$. Thus, $I_G[p]=J+K$ by Lemma 3.3. Since J and K have disjoint supports, we have ${\mathbf {x}}^{{\mathbf {a}}}\in (I_G[p])^n$ if and only if there exists $t\in \{0,\ldots , n\}$ such that ${\mathbf {x}}^{{\mathbf {a}}}\in J^{n-t}$ and ${\mathbf {x}}^{{\mathbf {a}}}\in K^t$. Denote by t the number $\sum _{i\in M_p}a_i$.

$\hbox {(1)}\Rightarrow$ (2) Assume that p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$. If $t\ge n$, then ${\mathbf {x}}^{{\mathbf {a}}}\in K^n\subseteq (I_G[p])^n$, and so $p\notin \Delta _{{\mathbf {a}}}(I_G^n)$, a contradiction. Thus, $t\in \{0,\ldots ,n-1\}$. For each $j\in N_p$, since $\{p,j\}\notin \Delta _{{\mathbf {a}}}(I_G^n)$, it follows that $\sum _{i\in [r]\setminus \{p,j\}}a_i\ge n$ by Proposition 3.2. This is equivalent to requiring $\sum _{i\in N_p\setminus \{j\}}a_i\ge n-t$. It remains to be shown the inequality ➌ holds.

First, we show that $a_i\le n-t-1$ for all $i\in N_p$. In fact, if $a_j\ge n-t$ for some $j\in N_p$, then, since $\sum _{i\in N_p\setminus \{j\}}a_i\ge n-t$, we have ${\mathbf {x}}^{\mathbf {a}}\in J^{n-t}$, and so ${\mathbf {x}}^{\mathbf {a}}\in I^n_G[p]$. This implies $p\notin \Delta _{{\mathbf {a}}}(I_G^n)$, contradicting to our assumption. Thus, we have $a_i\le n-t-1$ for all $i\in N_p$. Due to this fact, in order to obtain the inequality ➌ it is enough to prove the following statement:

Let $t\in \{0,\ldots ,n-1\}$. If $\sum _{i\in N_p}a_i\ge 2(n-t)$ and $a_i\le n-t-1$ for each $i\in N_p$, then $\mathbf {x^{a}}\in J^{n-t}.$

Set $k:=n-t$. We will proceed by induction on k. If $k=1$, there is nothing to prove; if $k=2$, then $\mathbf {x^{a}}$ can be written as $x_ix_jx_kx_{\ell }u$, where $i,j,k,\ell \in N_p$ are pairwise distinct and u is a monomial in S. It follows that $\mathbf {x^{a}}\in J^2$, as required. Suppose now that $k\ge 2$. Since $\sum _{i\in N_p}a_i\ge 2k$, we may write ${\mathbf {a}}={\mathbf {b}}+{\mathbf {c}}$ such that ${\mathbf {b}}=(b_1,\ldots ,b_r)\in {\mathbb {N}}^r,{\mathbf {c}}\in {\mathbb {N}}^r$ and $\sum _{i\in N_p}b_i=2k$. We may harmlessly assume further $N_p=\{1,2,\ldots ,s\}$ and $b_1\ge b_2\ge \cdots \ge b_s$. Since $b_i\le a_i\le k-1$ for $i=1,\ldots ,s$, we have both $b_1$ and $b_2$ as positive integers, and $b_i\le k-2$ for $i=3,\ldots ,s$. Let ${\mathbf {b}}'$ denote the vector $(b_1-1,b_2-1,b_3,\ldots ,b_s,b_{s+1},\ldots , b_r)\in {\mathbb {N}}^r$. Then, ${\mathbf {x}}^{{\mathbf {b}}'}\in J^{k-1}$ by the induction hypothesis. From this, it follows that ${\mathbf {x}}^{{\mathbf {a}}}={\mathbf {x}}^{{\mathbf {b}}'}x_1x_2{\mathbf {x}}^{{\mathbf {c}}}\in J^k$. Thus, the desired statement has been proved, and the proof of $\hbox {(1)}\Rightarrow$ (2) is now complete.

$\hbox {(2)}\Rightarrow$ (1) From the inequalities ➊ and ➋, it follows that $\sum _{i\in [r]\setminus \{p,j\}}a_i\ge n$ for any $j\in N_p$. Thus, $\{p,j\}$ is not an edge of $\Delta _{{\mathbf {a}}}(I_G^n)$ for any $j\in N_p$. Since $J^{n-t}$ is generated in degree $2(n-t)$, ${\mathbf {x}}^{{\mathbf {a}}}\notin J^{n-t}$ by the inequality ➌. This then implies ${\mathbf {x}}^{{\mathbf {a}}}\notin (I_G[p])^n$ and so p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$. $\square$

Corollary 3.5

• (1) Let $p\in [r]$. If $\deg p\le 2$, then p is never an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$ for any ${\mathbf {a}}\in {\mathbb {N}}^r$ and $n\ge 1$.

• (2) $\Delta _{{\mathbf {a}}}(I_G)$ contains no isolated vertices for any ${\mathbf {a}}\in {\mathbb {N}}^r.$

Proof

• (1) If $\deg p=1$, then $I_G[p]=I_G[p,q]=(x_i: i\in [r]\setminus \{p,q\})$, where q is the unique vertex belonging to $N_p$. If p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$, then it implies that ${\mathbf {x}}^{{\mathbf {a}}}\notin (I_G[p])^n$ and ${\mathbf {x}}^{{\mathbf {a}}}\in (I_G[p,q])^n$ at the same time, which is impossible. If $\deg p=2$, then the three conditions in Proposition 3.4 for p to be an isolated vertex cannot be fulfilled at the same time. In fact, if we write $N_p=\{1,2\}$, then $a_i\ge n-t$ for $i=1,2$ by ➋, and it follows that $\sum _{i\in N_p}a_i =a_1+a_2\ge 2(n-t)$. This is contradicted with ➌.

• (2) Assume that p is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G)$. Then, $\sum _{i\in N_p} a_i\le 1$, and $\sum _{i\in N_p\setminus \{j\}} a_i\ge 1$ for all $j\in N_p$ by Proposition 3.4. But this is impossible since $|N_p|\ge 3$ by (1). $\square$

Hereafter, we evaluate $a_1^i(S/I_G^n)$ for $i\ge 0$. If $i\ge 2$, then $a_1^i(S/I_G^n)=-\infty$ by Lemma 1.6 and by the obvious fact that ${\widetilde{H}}_i(\Delta ;K)=0$ for all $i\le -2$ and for all $\Delta$. In the following result, we compute $a_1^1(S/I_G^n).$

Proposition 3.6

(1) If either $\deg (G)\le 2$ or $n=1$, then $a_1^1(S/I_G^n)=-\infty$.

(2) If $\deg (G)\ge 3$ and $n\ge 2$, then $a_1^1(S/I_G^n)=2n-2$.

Proof

Let ${\mathbf {a}}\in {\mathbb {Z}}^r$ with $|G_{{\mathbf {a}}}|=1$. Then, $H_{{\mathfrak {m}}}^1(S/I_G^n)_{{\mathbf {a}}}\ne 0$ if and only if $\Delta _{{\mathbf {a}}}(I_G^n)=\{\emptyset \}$ by Lemma 1.6. The latter is equivalent to requiring p is an isolated vertex $\Delta _{{\mathbf {a}}_+}(I_G^n)$ by Lemma 1.5, where p is the unique element of $G_{{\mathbf {a}}}$. Thus, if either $\deg (G)\le 2$ or $n=1$, then $a_1^1(S/I_G^n)=-\infty$ by Corollary 3.5.

If $n\ge 2$ and $\deg p\ge 3$ for some $p\in [r]$, we may harmlessly assume $\{1,2,3\}\subseteq N_p$. Then, p is an isolated vertex $\Delta _{{\mathbf {a}}_+}(I_G^n)$ by Proposition 3.4, where

$$\begin{aligned} {\mathbf {a}}=(n-1,n-1,1,0,\ldots ,-1,\ldots ,0) \end{aligned}$$

with “$-1$” appearing at the pth position. From this, it follows that $a_1^1(S/I_G^n)\ge 2n-2$. Finally, we obtain $a_1^1(S/I_G^n)\le 2n-2$ by the inequalities ➊ and ➌ in the statement of Proposition 3.4. $\square$

The value of $a_1(S/I_G^n)$ when $n=1$ is now clear.

Corollary 3.7

If G is disconnected, then $a_1(S/I_G)=0$; If G is connected, then $a_1(S/I_G)=-\infty$.

Proof

Let ${\mathbf {a}}\in {\mathbb {N}}^r$. If $H_{{\mathfrak {m}}}^1(S/I_G)_{{\mathbf {a}}}\ne 0$, then $\Delta _{{\mathbf {a}}}(I_G)$ is disconnected, and so $\Delta _{{\mathbf {a}}}(I_G)$ contain disjoint edges due to Corollary 3.5.(2). From this, it follows that ${\mathbf {a}}=(0,\ldots ,0)$ by Proposition 3.2 and $a_1^0(S/I_G)\in \{0,-\infty \}$.

Denote $(0,\ldots ,0)$ by ${\mathbf {0}}$. Since $\Delta _{{\mathbf {0}}}(I_G)=G$, $\Delta _{{\mathbf {0}}}(I_G)$ is disconnected if and only if G is disconnected. Thus, $a_1^0(S/I_G)=0$ if G is disconnected, and $a_1^0(S/I_G)=-\infty$ if G is connected. Now, the result follows from Proposition 3.6.(1). $\square$

We remark that Corollary 3.7 is also clear from [12, Lemma 2.1 and 2.2]. Next, we compute $a_1^0(S/I_G^n)$ for $n\ge 2$.

Definition 3.8

Let G be a simple graph on [r]. A vertex p of G is said to be compact if $\deg p\ge 3$ and it belongs to a triangle, where a triangle is a cycle of length 3.

One may look at Broom as shown in Fig. 1, where 3 is a compact vertex.

Proposition 3.9

Let G be a simple graph on [r] with $r\ge 3$. Suppose that G contains no compact vertices. Then, for all $n\ge 2$, we have $a_1^0(S/I_G^n)\le 2n-1$. Moreover, $a_1^0(S/I_G^n)= 2n-1$ if and only if there exist non-adjacent distinct vertices p and q of G such that $|N_p\cap N_q|\ge 3$.

Proof

We first prove $a_1^0(S/I_G^n)\le 2n-1$. If $a_1^0(S/I_G^n)=-\infty$, there is nothing to prove. Suppose now that $a_1^0(S/I_G^n)\ne -\infty$, and let ${\mathbf {a}}=(a_1,\ldots ,a_r)\in {\mathbb {N}}^r$ such that $H_{{\mathfrak {m}}}^1(S/I_G^{n})_{{\mathbf {a}}}\ne 0$ and $a_1^0(S/I_G^n)=|{\mathbf {a}}|$. Then, $\Delta _{{\mathbf {a}}}(I_G^n)$ is disconnected by Lemma 1.6 as well as Lemma 1.1.(3). We consider the following three cases.

Case 1: There exist disjoint edges of G, say $\{1,2\}$ and $\{3,4\}$, which belong to distinct connected components of $\Delta _{{\mathbf {a}}}(I_G^n)$. Then, by Proposition 3.2, we obtain

$$\begin{aligned} a_3+a_4+a_5+\cdots +a_r\le n-1 \end{aligned}$$

and

$$\begin{aligned} a_1+a_2+a_5+\cdots +a_r\le n-1. \end{aligned}$$

In particular, it follows that $|{\mathbf {a}}|\le 2(n-1)$.

Case 2: There exist an isolated vertex and an edge in $\Delta _{{\mathbf {a}}}(I_G^n)$. Let p be such an isolated vertex with $N_p$ and $M_p$ defined as before. Then, there exists $t\in \{0,\ldots ,n-1\}$ such that $\sum _{i\in M_p}a_i=t$, $\sum _{i\in N_p}a_i\le 2(n-t)-1$ and $a_i\le n-t-1$ for all $i\in N_p$ by Proposition 3.4.

Let $e=\{k,\ell \}$ be an edge of $\Delta _{{\mathbf {a}}}(I_G^n)$. Then, $\sum _{i\in [r]\setminus e}a_i\le n-1$ by Proposition 3.2. It is clear that $e\nsubseteq N_p$, for otherwise, we have p is a compact vertex by Corollary 3.5, a contradiction. If $e\subseteq M_p$, then $\sum _{i\in e}a_i \le t\le n-1$. From this, it follows that $|{\mathbf {a}}|\le 2n-2$. So we may assume $k\in N_p$ and $\ell \in M_p$. Then, $a_k\le n-t-1$ and $a_{\ell }\le t$, which also implies $|{\mathbf {a}}|\le 2n-2$.

Case 3: $\Delta _{{\mathbf {a}}}(I_G^n)$ contains distinct isolated vertices p and q. Then, by Proposition 3.4, there exist $s,t\in \{0,\ldots ,n-1\}$ such that

$$\begin{aligned} \begin{aligned}&\text{(1) }\qquad \sum _{i\in M_p}a_i=s, \qquad \sum _{i\in M_q}a_i=t;\\&\text{(2) }\qquad a_i\le n-s-1, \forall i\in N_p, \qquad a_i\le n-t-1, \forall i\in N_q;\\&\text{(3) }\qquad \sum _{i\in N_p}a_i\le 2(n-s)-1, \qquad \sum _{i\in N_q}a_i\le 2(n-t)-1. \end{aligned} \end{aligned}$$

Subcase 3.1: Suppose first that $\{p,q\}$ is an edge of G. Then, $p\in N_q$ and $q\in N_p$. We can write $N_p=A\sqcup \{q\}, N_q=B\sqcup \{p\}$. Here, $\sqcup$ means a disjoint union. Note that $A\cap B=\emptyset$ since G contains no compact vertices. It follows that $M_p=([r]\setminus \{p,q\})\setminus A$ and $M_q=([r]\setminus \{p,q\})\setminus B$, and so $M_p=B\sqcup C$ and $M_q=A\sqcup C$ for some $C\subseteq [r]$. Under these notions, we have $M_p=(N_q\sqcup C)\setminus \{p\}$ and so

$$\begin{aligned} \sum _{i\in N_q}a_i=s-\sum _{i\in C}a_i+a_p\le s-\sum _{i\in C}a_i+n-t-1. \end{aligned}$$

This implies

$$\begin{aligned} \begin{aligned} |{\mathbf {a}}|&= \sum _{i\in N_q}a_i+\sum _{i\in M_q}a_i+a_q\le s-\sum _{i\in C}a_i+n-t-1+t+n-s-1\\&=2n-2-\sum _{i\in C}a_i\le 2n-2. \end{aligned} \end{aligned}$$

Subcase 3.2: Suppose next that $\{p,q\}$ is not an edge of G. Since $p\in M_q$, it follows that $a_p\le t$ and so

Similarly,

Combining ① and ②, we conclude that $|{\mathbf {a}}|\le 2n-1$ and thus the first statement has been proved.

Now, assume that $a_1^0(S/I_G^n)= 2n-1$, and let ${\mathbf {a}}=(a_1,\ldots ,a_r)\in {\mathbb {N}}^r$ such that $H_{{\mathfrak {m}}}^1(S/I_G^n)_{{\mathbf {a}}}\ne 0$ and $|{\mathbf {a}}|=2n-1$. In view of the proof of the first statement, we have $\Delta _{{\mathbf {a}}}(I_G^n)$ containing distinct isolated vertices p and q, which are non-adjacent in G, and so there exist $s,t\in \{0,\ldots ,n-1\}$ satisfying the inequalities (1), (2), (3) stated before. Hence, since $p\in M_q$ and $q\in M_p$, we have

$$\begin{aligned} 2n-1=|{\mathbf {a}}|= \sum _{i\in N_p}a_i+\sum _{i\in M_p}a_i+a_p\le 2(n-t)-1+s+t \end{aligned}$$

and

$$\begin{aligned} 2n-1=|{\mathbf {a}}|=\sum _{i\in N_q}a_i+\sum _{i\in M_q}a_i+a_q\le 2(n-t)-1+s+t. \end{aligned}$$

From these, it follows that $s=t$ and so the inequalities above are actually equalities. Therefore, $a_p=a_q=t$ and $\sum _{i\in N_p}a_i=\sum _{i\in N_q}a_i=2(n-t)-1$.

We now show that $|N_p\cap N_q|\ge 3$. Set $A:=N_p\cap N_q$. Assume on the contrary that $|A|\le 2$. Then, $\sum _{i\in A}a_i\le 2(n-t)-2$. It follows that there exists $k\ge 0$ such that $\sum _{i\in A}a_i= 2(n-t)-2-k$ and $\sum _{i\in N_p\setminus A}a_i=\sum _{i\in N_q\setminus A}a_i=k+1.$ Since $(N_q\setminus A)\sqcup \{q\}\subseteq M_p$, we have $(N_q\setminus A)\sqcup N_p\sqcup \{p,q\} \subseteq [r]$ and this implies

$$\begin{aligned} |{\mathbf {a}}|\ge \sum _{i\in N_q\setminus A}a_i+\sum _{i\in N_p\setminus A}a_i+\sum _{i\in A}a_i+a_p+a_q= 2n+k, \end{aligned}$$

a contradiction. Thus, we have proved $|N_p\cap N_q|\ge 3$.

Conversely, suppose that there exist non-adjacent vertices p and q with $|N_p\cap N_q|\ge 3$. We may assume $\{1,2,3\}\subseteq N_p\cap N_q$ and let ${\mathbf {a}}$ denote the vector

$$\begin{aligned} (n-1,n-1,1,0,\ldots ,0)\in {\mathbb {N}}^r. \end{aligned}$$

It follows that p and q are isolated vertices of $\Delta _{{\mathbf {a}}}(I_G^n)$ by Proposition 3.4. This then implies $a_1^0(S/I_G^n)\ge |{\mathbf {a}}|=2n-1$, as required. $\square$

Proposition 3.10

If G contains a compact vertex, then $a_1^0(S/I_G^n)=3n-3$ for $n\ge 2$.

Proof

We may assume that G contains the Broom as shown in Fig. 1 as its subgraph. If we set ${\mathbf {a}}:=(n-1,n-1,n-2,1,0,\ldots ,0)$, then $\{1,2\}$ is an edge of $\Delta _{{\mathbf {a}}}(I_G^n)$ and 3 is an isolated vertex of $\Delta _{{\mathbf {a}}}(I_G^n)$ by Propositions 3.2 and 3.4. In particular, we have $\Delta _{{\mathbf {a}}}(I_G^n)$ is disconnected and $H_{{\mathfrak {m}}}^1(S/I_G^n)_{{\mathbf {a}}}\ne 0$ by Lemma 1.6. This implies $a_1^0(S/I_G^n)\ge |{\mathbf {a}}|=3n-3$.

It remains to be shown that $a_1^0(S/I_G^n)\le 3n-3$. This is equivalent to showing $|{\mathbf {a}}|\le 3n-3$ if $\Delta _{{\mathbf {a}}}(I_G^n)$ is disconnected (i.e., $H_{{\mathfrak {m}}}^0(S/I_G^n)_{{\mathbf {a}}}\ne 0$), where ${\mathbf {a}}\in {\mathbb {N}}^r$. Examining the proof of Proposition 3.9, we only need to consider the cases when $\Delta _{{\mathbf {a}}}(I_G^n)$ contains an isolated vertex p and an edge e with $e\subseteq N_p$ and when $\Delta _{{\mathbf {a}}}(I_G^n)$ contains two distinct isolated vertices p and q such that p is adjacent to q in G.

First, we consider the case when $\Delta _{{\mathbf {a}}}(I_G^n)$ contains an isolated vertex p and an edge e with $e\subseteq N_p$. Since $e\in \Delta _{{\mathbf {a}}}(I_G^n)$ and $e\subseteq N_p$, one has $|{\mathbf {a}}|-\sum _{i\in e}a_i \le n-1$ and $a_i\le n-1$ for $i\in e$. This immediately implies $|{\mathbf {a}}|\le 3n-3$.

For the case when $\Delta _{{\mathbf {a}}}(I_G^n)$ contains two distinct isolated vertices p and q such that p is adjacent to q in G, we note that $N_p,N_q,M_p,M_q$ and s, t satisfy the inequalities as given in the proof of Case 3. Then,

$$\begin{aligned} |{\mathbf {a}}|=\frac{1}{2}\left( \sum _{i\in N_p}a_i+\sum _{i\in M_p}a_i+\sum _{i\in N_p}a_i+\sum _{i\in M_p}a_i+a_p+a_q\right) \le 3n-3. \end{aligned}$$

This completes the proof. $\square$

Proposition 3.11

Let G be a simple graph with ${\mathrm {diam}}(G)\ge 3$. Then, $a_1^0(S/I_G^n)\ge 2n-2$ for all $n\ge 2$.

Proof

Since ${\mathrm {diam}}(G)\ge 3$, there exist a pair of vertices, say 1, 2, such that $d_G(1,2)\ge 3$. Set ${\mathbf {a}}:=(n-1,n-1,0,\ldots ,0)$. We claim that $\Delta _{{\mathbf {a}}}(I_G^n)$ is disconnected. In fact, it is not difficult to see that

$$\begin{aligned} {\mathbf {x}}^{\mathbf {a}}=x_1^{n-1}x_2^{n-1}\notin (I^n_G[i])S \end{aligned}$$

for $i=1,2$ and so both $\{1\}$ and $\{2\}$ are faces of $\Delta _{{\mathbf {a}}}(I_G^n)$. If $\Delta _{{\mathbf {a}}}(I_G^n)$ is connected, then there exists a path $1=p_0-p_1-p_2-\cdots -p_{\ell }=2$ of $\Delta _{{\mathbf {a}}}(I_G^n)$, connecting 1 and 2. Note that this is also a path of G by Proposition 3.2. Since $d_G(1,2)\ge 3$, $\{p_1,p_2\}\cap \{1,2\}=\emptyset$. This implies that $\{p_1,p_2\}\notin \Delta _{{\mathbf {a}}}(I_G^n)$ by Proposition 3.2 again, a contradiction. From this, our claim follows. Hence, $H_{{\mathfrak {m}}}^1(S/I_G^n)_{{\mathbf {a}}}\ne 0$ and $a_1^0(S/I_G^{n})\ge |{\mathbf {a}}|=2n-2$. $\square$

We are now in the position to present the main result of this subsection. For the simplicity of the statement of this result, we use ${\mathfrak {C}}_i, i=1,2,\ldots , 5$ to indicate the following five conditions on a simple graph G, respectively.

${\mathfrak {C}}_1$: G contains a compact vertex;

${\mathfrak {C}}_2$: G contains no compact vertices, but it contains two non-adjacent vertices such that the intersection of their neighborhoods contains at least three vertices;

${\mathfrak {C}}_3$: G satisfies neither ${\mathfrak {C}}_1$ nor ${\mathfrak {C}}_2$, but $\deg (G)\ge 3$;

${\mathfrak {C}}_4$: ${\mathrm {diam}}(G)\ge 3$ and $\deg (G)\le 2$;

${\mathfrak {C}}_5$: ${\mathrm {diam}}(G)\le 2$ and $\deg (G)\le 2$.

Note that if $\deg (G)\le 2$, then G is the disjoint union of some paths and some cycles. Denote by $C_{\ell }$ and $P_{\ell }$ a cycle and a path of length $\ell$, respectively. Thus, if $G\in {\mathfrak {C}}_5$, then $G\in \{P_1,P_2, C_4,C_5\}$.

Proposition 3.12

Let G be a simple graph on [r] with $r\ge 3$. Then, for all $n\ge 2$, we have

$$\begin{aligned} a_1(S/I_G^n)=\left\{ \begin{array}{ll} 3n-3, &{} G\in {\mathfrak {C}}_1; \\ 2n-1, &{} G\in {\mathfrak {C}}_2; \\ 2n-2, &{} G\in {\mathfrak {C}}_3; \\ 2n-2, &{} G\in {\mathfrak {C}}_4; \\ 2n-2, &{}G=C_5\hbox { and }n\ge 3;\\ -\infty , &{}G=C_5\hbox { and }n= 2;\\ -\infty , &{} G\in \{P_2,C_4\}. \end{array} \right. \end{aligned}$$

Proof

If $G\in {\mathfrak {C}}_1$, then $a_1(S/I_G^n)=3n-3$ by Proposition 3.10. If $G\in {\mathfrak {C}}_2$, then $a_1(S/I_G^n)=2n-1$ by Proposition 3.9. If either $G\in {\mathfrak {C}}_3$ or $G\in {\mathfrak {C}}_4$, then $a(S/I_G^n)=2n-2$ by Lemmas 3.6 and 3.11 together with Proposition 3.9. It is left to consider the case when $G\in {\mathfrak {C}}_5$, namely $G\in \{P_2,C_4,C_5\}$. The case when $G\in \{P_2,P_4\}$ is proved in a similar way as in the case when $G=C_5$ and we omit its proof.

Suppose now that G is 5-cycle, that is, G is the Pentagon in Fig. 1. Then, $H_{{\mathfrak {m}}}^1(S/I_G^n)_{{\mathbf {a}}}\ne 0$ if and only if ${\mathbf {a}}\in {\mathbb {N}}^r$ and there exist two disjoint edges of G, which belongs to distinct connected components of $\Delta _{{\mathbf {a}}}(I_G^n)$. This implies $|{\mathbf {a}}|\le 2n-2$, by the same argument using in the proof of Case 1 of Proposition 3.9 and so $a_1(S/I_G^n)\le 2n-2$. If $n\ge 3$, we set ${\mathbf {a}}:=(1,n-2,0,n-2,1)$. Then, $\Delta _{{\mathbf {a}}}(I_G^n)=\langle \{1,2\},\{4,5\}\rangle$ is disconnected and so $a_1(S/I_G^n)= 2n-2$. It remains to be shown that if $n=2$, then $\Delta _{{\mathbf {a}}}(I_G^n)$ is connected for any ${\mathbf {a}}\in {\mathbb {N}}^r$. If not, we may harmlessly assume that $\{1,2\}$ and $\{3,4\}$ belong to different connected components of $\Delta _{{\mathbf {a}}}(I_G^n)$. Note that $\{2,3\}\notin \Delta _{{\mathbf {a}}}(I_G^n)$, we have $a_1=a_4=1$ and $a_2=a_3=a_5=0$ by Lemma 3.2. From this, it follows that both $\{1,5\}$ and $\{4,5\}$ belong to $\Delta _{{\mathbf {a}}}(I_G^n)$ and thus $\Delta _{{\mathbf {a}}}(I_G^n)$ is connected, a contradiction. This completes the proof. $\square$

Since $\text{ Krull- }\dim S/I_G^n=2$, we see that $S/I_G^n$ is Cohen–Macaulay if and only if $H^0_{{\mathfrak {m}}}(S/I_G^n)=H^1_{{\mathfrak {m}}}(S/I_G^n)=0$, namely $a_0(S/I_G^n)=a_1(S/I_G^n)=-\infty$. The following result recovers [17, Corollaries 3.4 and 3.5].

Corollary 3.13

Let G be a simple graph on [r] with $r\ge 3$ and n an integer $\ge 1$. Then, the following statements are equivalent:

• (1) $S/I_G^n$ is Cohen–Macaulay;

• (2) Either $G\in \{P_2,C_4\}$ or $G=C_5$ and $n=2$.

Proof

If (1) holds, then $a_1(S/I_G^n)=-\infty$, and thus we obtain (2) by Proposition 3.12. Conversely, we choose any G and n satisfying (2). Then, it is not difficult to check that $I_G^n=I_G^{(n)}$ and so $S/I_G^n$ has a positive depth. From this, it follows that $H^0_{{\mathfrak {m}}}(S/I_G^n)=0$. Since $H^1_{{\mathfrak {m}}}(S/I_G^n)=0$ by Proposition 3.12, the result follows. $\square$

To illustrate the difference between $a_1(S/I_G^n)$ and $a_1(S/I_G^{(n)})$, we recall a well-known concept in the combinatorial theory.

Definition 3.14

A simple complex $\Delta$ is called a matroid provided that whenever F and G are faces of $\Delta$ with $|F|<|G|$, there exists $x\in G\setminus F$ such that $F\cup \{x\}$ is also a face of $\Delta$.

According to [19, Corollary 2.6], a graph G, considered as a simple complex, is a matroid if and only if every pair of disjoint edges of G is contained in a 4-cycle. It was proved in [17] that if G is a matroid, then $H_{{\mathfrak {m}}}^1(S/I_G^{(n)})=0$, i.e., $a_1(S/I_G^{(n)})=-\infty$. In contrast, the values of $a_1(S/I_G^n)$ behave very differently when G is a matroid.

Example 3.15

Let G be a matroid on [r] with $r\ge 3$. Then, $a_1(S/I_G^n)$ could be any number of $\{3n-3, 2n-1, 2n-2, -\infty \}$, depending on the structure of G. In fact, according to Proposition 3.12, we see that if G is a complete graph on [4], then $a_1(S/I_G^n)=3n-3$; if G is the graph on [5] with $E(G)=\{\{4,i\}, \{5,i\}: i=1,2,3\}$ (such a graph is called a diamond in the lattice theory), then $a_1(S/I_G^n)=2n-1$; if G is the graph on [4] with $E(G)=\{\{4,i\}:i=1,2,3\}$, then $a_1(S/I_G^n)=2n-2$. Finally, if G is a 4-cycle, then $a_1(S/I_G^n)=-\infty$.

We end this subsection by a characterization when a graph is a matroid, which may be of some independent interest. It is clear that every matroid has its diameter $\le 2$. We now consider obstructions for a graph of $\text{ diam }(G)\le 2$ to be a matroid. Let $V\subseteq [r]$. Recall that the induced subgraph of G on V is the subgraph G[V] on vertex set V such that for any $i,j\in V$, i is adjacent to j in G[V] if and only if i is adjacent to j in G.

Proposition 3.16

Let G be a simple graph with ${\mathrm {diam}}(G)\le 2$. Then, G is not a matroid if and only if it has an induced subgraph which is isomorphic to either a Broom or a Pentagon; see Fig. 1.

Fig. 1

Obstructions to be a matroid

Proof

If G has an induced subgraph isomorphic to either Broom or Pentagon, as shown in Fig. 1, then $\{1,2\}$ and $\{3,4\}$ are disjoint edges that do not belong to any 4-cycles. Thus, G is not a matroid. For the proof of the converse, it is enough to show that there is no a graph G of ${\mathrm {diam}}(G)\le 2$ such that G is not a matroid and that G has no induced subgraph isomorphic to one of the graphs as shown in Fig. 1.

Assume on the contrary that such a graph exists. Let G be such a graph. Since G is not matroid, there are disjoint edges, say $\{1,2\}$ and $\{3,4\}$, which does not belong to any 4-cycle. We consider the following cases:

Suppose that $d_G(i,j)=2$ for any $i\in \{1,2\}$ and $j\in \{3,4\}$. (This implies none of $\{i,j\}$ with $i\in \{1,2\}$ and $j\in \{3,4\}$ is an edge of G.) Let $1-p_1-3$ and $1-p_2-4$ be two paths. If $p_1=p_2$, then, since $\{1,3\}$ and $\{1,4\}$ are not edges of G, the subgraph induced on $\{1,3,4, p_1=p_2\}$ is isomorphic to a Broom, a contradiction. Thus, $p_1\ne p_2$. Since G contains no induced subgraphs isomorphic to a pentagon, at least one of the pairs $\{3,p_2\},\{4,p_1\},\{p_1,p_2\}$ is an edge of G. If $\{3,p_2\}\in E(G)$, then the subgraph induced on $\{1,p_2,3,4\}$ is isomorphic to a Broom, a contradiction again. From this, it follows that $\{3,p_2\}\notin E(G)$. With the same reason $\{4,p_1\}\notin E(G)$. It must be $\{p_1,p_2\}$ is an edge of G. Thus, the subgraph of G induced on $\{1,p_1,p_2,3\}$ is isomorphic to a Broom. This yields a contradiction again.

Thus, $d_G(i,j)=1$ for some $i\in \{1,2\}$ and $j\in \{3,4\}$. Say $d(1,3)=1$, or equivalently, $\{1,3\}$ is an edge of G. Then, $\{2,4\}$ is not an edge of G. Let $2-p-4$ be a path. If $p\in \{1,3\}$, say $p=1$, then since the graph of G induced on $\{1,2,3,4\}$ can not be a Broom, $\{2,3\}$ must be an edge of G. Thus, $1-2-3-4-1$ is a 4-cycle of G, a contradiction. Consequently, $p\notin \{1,3\}$

Since the subgraph of G induced on $\{1,2,3,4,p\}$ is not a pentagon, at least one of $\{1,4\},\{2,3\}, \{1,p\},\{3,p\}$ is an edge of G. If $\{1,4\}$ is an edge, since the subgraph of G induced on $\{2,1,3,4\}$ is not a Broom, we have $\{2,3\}\in E(G)$. This implies $1-2-3-4-1$ is a 4-cycle containing $\{1,2\},\{3,4\}$, a contradiction. Thus, $\{1,4\}$ is not an edge. Similarly, $\{2,3\}\notin E(G)$. If $\{1,p\}\in E(G)$, then subgraph induced on $\{3, 1,2,p\}$ is isomorphic to a Boom, thus $\{1,p\}\notin E(G)$. At last, $\{3,p\}\notin E(G)$. Thus, in all cases, our assumption leads to a contradiction. $\square$

Corollary 3.17

Let G be a simple graph on [r] with $r\ge 3$. Then, the following statements are equivalent

• (1) G is a matroid;

• (2) $\text{ diam }(G)\le 2$ and G contains neither Broom nor Pentagon as its induced graphs.

4.2 The computation of $a_2(S/I_G^n)$

In this subsection, we will prove that $a_2(S/I_G^n)=a_2(S/I_G^{(n)})$ and give the formula of $a_2(S/I_G^n)$.

Proposition 3.18

Let G be a simple graph on [r] with $r\ge 3$. Then,

$$\begin{aligned} a_2^j(S/I_G^n)=a_2^j(S/I_G^{(n)}) \end{aligned}$$

for all $j\ge 0$ and $n\ge 1$. In particular, $a_2(S/I_G^n))=a_2(S/I_G^{(n)})$ for $n\ge 1$.

Proof

We have known that $\Delta _{{\mathbf {a}}}(I_G^n)(1)=\Delta _{{\mathbf {a}}}(I_G^{(n)})$ by Lemma 3.2. Let $T\in \{I_G^n, I_G^{(n)}\}$. If ${\mathbf {a}}$ is a vector in ${\mathbb {Z}}^r$ with $|G_{{\mathbf {a}}}|=2$, then

$$\begin{aligned} H_{{\mathfrak {m}}}^2(S/T)_{{\mathbf {a}}}\ne 0\Longleftrightarrow \Delta _{{\mathbf {a}}}(T)=\{\emptyset \}\Longleftrightarrow G_{{\mathbf {a}}}\in \Delta _{{\mathbf {a}}_{+}}(T). \end{aligned}$$

Here, the first equivalence follows from Lemma 1.6 as well as Lemma 1.1.(1), and the second one is from Lemma 1.5. On the other hand, we have $G_{{\mathbf {a}}}\in \Delta _{{\mathbf {a}}_{+}}(I_G^n)$ if and only if $G_{{\mathbf {a}}}\in \Delta _{{\mathbf {a}}_{+}}(I_G^{(n)})$ by Lemma 3.2. This proves $a_2^2(S/I_G^n)=a_2^2(S/I_G^{(n)})$.

Let ${\mathbf {a}}$ be a vector in ${\mathbb {Z}}^r$ with $|G_{{\mathbf {a}}}|=1$. Say $G_{{\mathbf {a}}}=\{p\}$. Then, $H_{{\mathfrak {m}}}^2(S/T)_{{\mathbf {a}}}\ne 0$ if and only if $\Delta _{{\mathbf {a}}}(T)$ is disconnected. This is equivalent to requiring that there exist $i\ne j\in [r]\setminus \{p\}$ such that $\{p,i\}$ and $\{p,j\}$ belong to $\Delta _{{\mathbf {a}}_+}(T)$ by Lemma 1.5. Since $\{p,k\}\in \Delta _{{\mathbf {a}}_{+}}(I_G^n)$ if and only if $\{p,k\}\in \Delta _{{\mathbf {a}}_{+}}(I_G^{(n)})$ for $k=i,j$ by Lemma 3.2, the equality $a_2^1(S/I_G^n)=a_2^1(S/I_G^{(n)})$ follows.

Let ${\mathbf {a}}\in {\mathbb {N}}^r$. Then, $H_{{\mathfrak {m}}}^2(S/T)_{{\mathbf {a}}}\ne 0\Longleftrightarrow \Delta _{{\mathbf {a}}}(T) \text{ contains } \text{ a } \text{ cycle }$ by Lemma 1.2. It is clear that $\Delta _{{\mathbf {a}}}(I_G^n)$ contains a cycle if and only if $\Delta _{{\mathbf {a}}}(I_G^{(n)})$ contains a cycle. From this, it follows that $a_2^0(S/I_G^n)=a_2^0(S/I_G^{(n)})$. Note that $a_2^j(S/I_G^n)=a_2^j(I_G^{(n)})=-\infty$ for all $j\ge 3$, our proof completes. $\square$

We can obtain the following formula on $a_2(S/I_G^n)$ immediately by combining Proposition 3.18 with [12, Theorem 2.8].

Proposition 3.19

Let G be a simple graph on vertex set [r] with $r\ge 3$. Then, for all $n\ge 1$,

$$\begin{aligned} a_2(S/I_G^n)=\left\{ \begin{array}{ll} 3n-3, &{} \text{ girth }(G)=3; \\ 2n-2, &{} \text{ girth }(G)=4; \\ 2n-3, &{} 5\le \text{ girth }(G)< \infty \hbox { and }n\ge 2; \\ 0, &{} 5\le \text{ girth }(G)< \infty \hbox { and }n=1;\\ 2n-3, &{} \text{ girth }(G)=\infty \hbox { and }\deg (G)\ge 2; \\ n-3, &{} \text{ girth }(G)=\infty \hbox { and }\deg (G)=1. \end{array} \right. \end{aligned}$$

5 Conclusions

If I is a nonzero two-dimensional squarefree monomial ideal of $S=K[x_1,\ldots ,x_r]$ containing no variables, then $r\ge 3$ and $I=I_G$, where G is a simple graph on [r] with $E(G)\ne \emptyset$ that may contain isolated vertices. We now can state and prove the main result of this paper.

Theorem 4.1

Let $r\ge 3$, and let G be a simple graph on vertex set [r] with $E(G)\ne \emptyset$ that may contain some isolated vertices. Then, for all $n\ge 1$, we have

$$\begin{aligned} \qquad \text{ g-reg } (S/I_G^{n})= \left\{ \begin{array}{ll} 3n-1, &{} \text{ girth }(G)=3; \\ 2n, &{} \text{ girth }(G)=4; \\ 2n-1, &{} 5\le \text{ girth }(G)< \infty \hbox { and }n\ge 2; \\ 2, &{} 5\le \text{ girth }(G)< \infty \hbox { and }n=1; \\ 2n-1, &{} \text{ girth }(G)=\infty \hbox { and }\deg (G)\ge 2; \\ 2n-1, &{} \deg (G)=1. \end{array} \right. \end{aligned}$$

Proof

In the process of our proof, Theorem 2.8 and Propositions 3.12, 3.19 will be used many times without referring to them each time.

Note that $\text{ g-reg } (S/I_G^{n})=\max \{a_i(S/I_G^n)+i:i=1,2\}$. If $\text{ girth }(G)=3$, then $a_1(S/I_G^n)\le 3n-3$ for all $n\ge 1$ by Proposition 3.12, Corollary 3.7 and Theorem 2.8 and it follows that $\text{ g-reg }(S/I_G^n)=3n-1$ by Proposition 3.19 and Theorem 2.8. If $\text{ girth }(G)=4$, then G contains no compact vertices. From this, it follows that $a_1(R/I_G^n)\le 2n-1$ by Propositions 3.12 and so $\text{ g-reg }(S/I_G^n)=2n$ by Proposition 3.19.

Assume that $5\le \text{ girth }(G)<\infty$. If $n\ge 2$, then, since G does not satisfy ${\mathfrak {C}}_2$, we have $a_1(S/I_G^n)\le 2n-2$. This then implies $\text{ g-reg }(S/I_G^n)=2n-1$. If $n=1$, then $a_1(S/I_G)\le 0$ by Corollary 3.7 and so $\text{ g-reg }(S/I_G^n)=2$.

If $\text{ girth }(G)=\infty$ and $\deg (G)\ge 2$, we also have $a_1(S/I_G^n)\le 2n-2$ and so $\text{ g-reg }(S/I_G^n)=2n-1$.

Now, consider the case when $\deg (G)=1$. Since $r\ge 3$, either G contains at least two disjoint edges or G contains at least an edge and an isolated vertex. In the former case, we have $\text{ diam }(G)=\infty \ge 3$. From this, it follows that $a_1(S/I_G^n)=2n-2$ and so $\text{ g-reg }(S/I_G^n)=2n-1$. In the latter case, we have $a_1(S/I_G^n)=2n-2$ by Theorem 2.8. This also implies $\text{ g-reg }(S/I_G^n)=2n-1$. $\square$

One special case of Theorem 4.1 can be achieved directly: If G is a triangle, then $I_G=(x_1x_2x_3)$ and so $I_G^n$ has a linear resolution for all $n\ge 1$. Since $S/I_G^n$ has a positive depth, we have

$$\begin{aligned} \text{ g-reg }(S/I_G^n)={\text {reg}}(S/I_G^n)={\text {reg}}(I_G^n)-1=3n-1. \end{aligned}$$

The following result follows immediately by comparing Theorem 4.1 with [12, Theorem 2.9].

Corollary 4.2

Let G be a simple graph on vertex set [r] with $r\ge 3$ that may contain some isolated vertices. If G contains at least one edge, then, for all $n\ge 1$, we have

$$\begin{aligned} \text{ g-reg }(S/I_G^n)={\text {reg}}(S/I_G^{(n)}). \end{aligned}$$

In general, a two-dimensional squarefree monomial ideal may contain some variables. More precisely, if I is a two-dimensional squarefree monomial ideal which is not generated by variables, then $I=(I_G,y_1,\ldots ,y_k)$, where $k\ge 0$ and G is again a simple graph on vertex set [r] with $r\ge 3$ and $E(G)\ne \emptyset$ that may contain isolated vertices. We now consider this case. It is not difficult to see that

$$\begin{aligned} S[y]/(I_G,y)^n=S/I_G^n\oplus (S/I_G^{n-1})y\oplus \cdots \oplus (S/I_G)y^{n-1}. \end{aligned}$$

By [8, Theorem 3.4], we have

$$\begin{aligned} S[y]/(I_G,y)^{(n)}=S/I_G^{(n)}\oplus (S/I_G^{(n-1)})y\oplus \cdots \oplus (S/I_G)y^{n-1}. \end{aligned}$$

From these, it follows that

$$\begin{aligned} \text{ g-reg }(S[y]/(I_G,y)^n)=\max \{\text{ g-reg }(S/I_G^i)+n-i:1\le i\le n\} \end{aligned}$$

and

$$\begin{aligned} \text{ reg }(S[y]/(I_G,y)^{(n)})=\max \{\text{ reg }(S/I_G^{(i)})+n-i:1\le i\le n\}. \end{aligned}$$

Combining these equalities with Corollary 4.2, we get the last result of this paper.

Corollary 4.3

Let I be a two-dimensional squarefree monomial ideal. Then, for all $n\ge 1$, we have

$$\begin{aligned} \text{ g-reg }(S/I^n)={\text {reg}}(S/I^{(n)}). \end{aligned}$$

In view of this result, one may guess the equality $I^{(n)}=I^n:{\mathfrak {m}}^{\infty }$ holds for $n\ge 1$ if I is a two-dimensional squarefree monomial ideal. But this is not the case since $a_1(S/I_G^n)\ne a_1(S/I_G^{(n)})$ in general.

In order to obtain a formula for ${\text {reg}}(S/I_G^n)$, we have to obtain the value for $a_0(S/I_G^n)$, which seems much more difficult. As a first step to compute ${\text {reg}}(R/I_G^n)$, it would be of interest to compute the depth function $\text{ depth }( S/I_G^n)$.