Regularity of 3-Path Ideals of Trees and Unicyclic Graphs

Abstract

Let G be a simple graph and $I_3(G)$ be its 3-path ideal in the corresponding polynomial ring R. In this article, we prove that for an arbitrary graph G, ${\text {reg}}(R/I_3(G))$ is bounded below by $2\nu _3(G)$, where $\nu _3(G)$ denotes the 3-path induced matching number of G. We give a class of graphs, namely trees for which the lower bound is attained. Also, for a unicyclic graph G, we show that ${\text {reg}}(R/I_3(G))\le 2\nu _3(G)+2$ and provide an example that shows that the given upper bound is sharp.

1 Introduction

Let G be a finite simple graph with the vertex set V(G) and the edge set E(G). Let $R=\mathbb {K}[x:x\in V(G)]$ be the polynomial ring, where $\mathbb {K}$ is an arbitrary field. In [6], the notion of path ideals has been introduced. For a graph G, a square-free monomial ideal,

$$\begin{aligned} I_t(G):=\langle x_1\cdots x_t: \text { where } P:x_1,\dots ,x_t \text { is a } t\text {-path in } G\rangle \subseteq R \end{aligned}$$

is called t-path ideal of G. For the last few decades, researchers have been trying to establish the connection between the combinatorial invariants of graphs and the algebraic invariants of t-path ideal, see [1, 2, 4, 5, 7, 12, 13]. The Castelnuovo–Mumford regularity of an ideal is an important algebraic invariant that measures the complexity of the module. The Castelnuovo–Mumford regularity (or simply regularity) of a finitely generated graded R-module M, written ${\text {reg}}(M)$ is defined as ${\text {reg}}(M):=\max \{j-i: {\text {Tor}}_i(M,\mathbb {K})_j\ne 0\}.$ There are a few classes of graphs for which the explicit formula of the regularity of t-path ideal is known. In [5], Bouchat et al. studied the t-path ideal of rooted trees, and they provided a recursive formula for computing the graded Betti numbers of t-path ideals. Also, they gave a general bound for the regularity of t-path ideal of a rooted tree. In particular, for a rooted tree G, they proved that ${\text {reg}}(R/I_t(G))\le (t-1)[l_t(G)+p_t(G)],$ where $l_t(G)$ denotes the number of leaves in G, where level is at least $t-1$ and $p_t(G)$ denotes the maximal number of pairwise disjoint paths of length t in G. In [1] and [2], Alilooee and Faridi computed the regularity of t-path ideal of lines and cycles in terms of the number of vertices, respectively. Banerjee [3] studied the regularity of t-path ideal of gap free graphs and proved that the t-path ideals of gap free, claw free and whiskered-$K_4$ free graphs have linear minimal free resolutions for all $t\ge 3$. In this article, we restrict ourselves to $t=3$ and study the regularity of $I_3(G)$. It is important to note that for a 3-path ideal of gap free graph G, Banerjee proved that ${\text {reg}}(R/I_3(G))\le \max \{{\text {reg}}(R/I_2(G)),2\},$ where $I_2(G)$ is the 2-path ideal or monomial edge ideal of G, see [3, Theorem 3.3]. In [11], Katzman proved that for any graph G, ${\text {reg}}(R/I_2(G))\ge \nu (G)$, where $\nu (G)$ denotes the induced matching number of G. Motivated from the definition of $\nu (G)$, in this article, we define in an obvious way the 3-path induced matching number, denoted by $\nu _3(G)$ (see Sect 2 for the definition). Next, we prove that a similar lower bound can be obtained for the regularity of $I_3(G)$. More precisely, we prove ${\text {reg}}(R/I_3(G))\ge 2\nu _3(G)$. We also observe that this is a sharp lower bound for ${\text {reg}}(R/I_3(G))$. In fact, we show that if G is a tree, then ${\text {reg}}(R/I_3(G))=2\nu _3(G)$, see Theorem 4.4. It is desirable to answer the following problem:

Problem 1.1

Classify the classes of graphs G that satisfy the property

$$\begin{aligned} {\text {reg}}(R/I_3(G))=2\nu _3(G). \end{aligned}$$

Theorem 4.4 gives a class of graphs that satisfies the property ${\text {reg}}(R/I_3(G))=2\nu _3(G)$. On the other hand, Example 4.7 shows that some of the unicyclic graphs satisfy the desired property but not the whole class. More concretely, in Theorem 4.6, we prove that if G is a unicyclic graph, then ${\text {reg}}(R/I_3(G))\le 2\nu _3(G)+2$. Moreover, in Example 4.7, we give examples of unicyclic graphs showing that the regularity of 3-path ideal of a unicyclic graph can attain any of the values between the lower and upper bound of the regularity.

2 Preliminaries

In this section, we recall all necessary definitions which will be used throughout the article.

Definition 2.1

Let G be a graph with the vertex set $V(G)=\{x_1,\dots ,x_n\}$ and the edge set E(G).

• (i) A subgraph H of a graph G is called an induced subgraph if for all $x_i,x_j\in V(H)$ such that $\{x_i,x_j\}\in E(G)$ implies that $\{x_i,x_j\}\in E(H)$. For a vertex $x\in V(G)$, let $G\setminus \{x\}$ denote the induced subgraph on the vertex set $V(G)\setminus \{x\}.$

• (ii) A path on n vertices is a graph whose vertices can be listed in the order $x_1,\dots ,x_n$ such that the edge set is $\{\{x_i,x_{i+1}\}:1\le i\le n\}$ and it is denoted by $P_n$. For $t\ge 2$, a path of length t in G is called a t-path.

• (iii) A cycle on n vertices $\{x_1,\dots ,x_n\}$, denoted by $C_n$, is the graph with the edge set $E(P_n)\cup \{\{x_1,x_n\}\}.$ A graph G is said to be tree if it does not contain any cycle. A disconnected tree is called a forest. A graph G is called unicyclic if G contains only one cycle.

• (iv) For a vertex $x\in V(G)$, the set $\{y\in V(G): \{x,y\}\in E(G)\}$ is called the neighborhood of x in G and it is denoted by $N_G(x)$. The set $N_G[x]$ denotes $N_G(x)\cup \{x\}$. For an edge $e=\{x,y\}\in E(G)$, the neighborhood of e is defined as

  $$\begin{aligned} N_G(e):=(N_G(x)\setminus \{y\})\cup (N_G(y)\setminus \{x\}). \end{aligned}$$

  $N_G[e]$ denotes the set $N_G(e)\cup \{x,y\}$, i.e., $N_G[e]=N_G[x]\cup N_G[y].$

• v) For a vertex $x\in V(G)$, the set $\{ \{y,z\}\in E(G): \{x,y,z\} \text { is a } 3\text {-path in }G \}$ is called the neighborhood edge set of x in G and it is denoted by $N_G^{edge}(x)$.

• (vi) A 3-path matching in a graph G is a subgraph consisting pairwise disjoint 3-paths. If the subgraph is induced, then 3-path matching is said to be a 3-path induced matching of G. The largest size of a 3-path induced matching is called the 3-path induced matching number, and it is denoted by $\nu _3(G)$.

The following remark shows the relationship between the 3-path induced matching number of a graph and its induced subgraph (Fig. 1).

Remark 2.2

One can observe that for an induced subgraph H of G, we have $\nu _3(H)\le \nu _3(G)$.

Example 2.3

Note that $\{x_1,x_2,x_3\},\{x_4,x_5,x_6\}$ is a 3-path matching but not a 3-path induced matching in G. Here, $\{x_1,x_2,x_7\},\{x_4,x_5,x_6\}$ is a 3-path induced matching in G and $\nu _3(G)=2$.

Fig. 1

G

3 A Lower Bound for the Regularity

For an arbitrary graph G, we first give a general lower bound for the regularity of 3-path ideal in terms of the regularity of 3-path ideal of its induced subgraph, and as a consequence, we prove that the regularity of $R/I_3(G)$ is bounded below by $2\nu _3(G)$.

Proposition 3.1

Let H be an induced subgraph of G. Then,

$$\begin{aligned} \beta _{i,j}(R_H/I_3(H))\le \beta _{i,j}(R/I_3(G)) \end{aligned}$$

where $R_H=\mathbb {K}[x:x\in V(H)]$. Moreover, ${\text {reg}}(R_H/I_3(H))\le {\text {reg}}(R/I_3(G)).$

Proof

We prove that $R_H/I_3(H)$ is an algebra retract of $R/I_3(G)$. We first show that $I_3(H)=I_3(G)\cap R_H$. It is clear that $I_3(H) \subseteq I_3(G)\cap R_H$. For the converse part, let $f\in I_3(G)\cap R_H$. Suppose $f=\sum gh$ for $g\in R$ and $h\in I_3(G)$. We consider the mapping $\varphi :R\longrightarrow R_H$ by defining $\varphi (x)=x$ if $x\in V(H)$, otherwise $\varphi (x)=0$. Therefore,

$$\begin{aligned} \varphi (f)&= \sum \varphi (g)\varphi (h) \\&=\sum \varphi (g)h, \text { where } h\in I_3(H). \end{aligned}$$

This implies that $f\in I_3(H)$, and so $I_3(H)=I_3(G)\cap R_H$. Thus, $R_H/I_3(H)$ is a $\mathbb {K}$-subalgebra of $R/I_3(G)$. Now consider , where $\bar{\varphi }$ is the map induced by $\varphi$. It can be observed that $\bar{\varphi }\circ i$ is the identity map on $R_H/I_3(H)$. Hence, $R_H/I_3(H)$ is an algebra retract of $R/I_3(G)$. Now, the assertion follows from [14, Corollary 2.5]. $\square$

As an immediate consequence, we have a lower bound of the regularity of 3-path ideal of any graph G in terms of the combinatorial invariants of G.

Corollary 3.2

Let G be a simple graph and $I_3(G)$ be its 3-path ideal. Then,

$$\begin{aligned} {\text {reg}}(R/I_3(G))\ge 2\nu _3(G). \end{aligned}$$

Proof

For simplicity of notation, let $s=\nu _3(G)$. Suppose that $\{P_1,P_2,\dots ,P_{s}\}$ is a 3-path induced matching in G. Let H be the induced subgraph of G on the vertices $\cup _{i=1}^{s} V(P_i)$. Then, $I_3(H)$ is a complete intersection. Thus, by the Koszul complex, ${\text {reg}}(R_H/I_3(H))=2\nu _3(G)$. Hence, the assertion follows from Proposition 3.1. $\square$

4 Path Ideals of Trees and Unicyclic Graphs

In this section, we consider the 3-path ideal of trees and unicyclic graphs. In fact, we compute the exact regularity of $I_3(G)$ when G is a tree, and for unicyclic graphs, we give a sharp upper bound of ${\text {reg}}(I_3(G))$. We first prove some technical lemmas which will be needed to prove the main results.

Lemma 4.1

Let G be a simple graph and $I_3(G)$ be its 3-path ideal. Let $e=\{x,y\}\in E(G)$. Then, we have the followings:

• (1) $I_3(G):xy=L+J$, where $L=\left\langle N_G(e)\right\rangle$ and $J=I_3(G{\setminus } N_G[e])$.

• (2) $(I_3(G)+\langle xy \rangle ):x= \langle y \rangle +I_2(H)+I_3(G{\setminus } N_G[x]),$ where H is the union of $N_G^{{edge}}(x)$ and the complete graph on the vertex set $N_G(x)\setminus \{y\}$.

Proof

(1): Clearly, $L+J\subset I_3(G):xy$. On the other side, let $u\in I_3(G):xy$. This implies that $uxy\in I_3(G)$, and hence, there exists a minimal monomial generator $v\in I_3(G)$ such that $v\mid uxy$. If $v\not \mid u$, then $\gcd (v, xy)\ne 1$. This forces that ${\text {supp}}(u)\cap N_G(e)\ne \emptyset$. This gives $u\in L$, and hence, $I_3(G):xy=L+J$.

(2): It can be easily seen that $\langle y \rangle +I_2(H)+I_3(G{\setminus } N_G[x])\subset (I_3(G)+\langle xy \rangle ):x$. For the converse part, let $u \in (I_3(G)+\langle xy \rangle ):x$. Then, $ux\in I_3(G)+\langle xy \rangle$. Assume that $y\not \mid u$. Then, $ux\in I_3(G)$, and hence, there exists a minimal monomial generator $v\in I_3(G)$ such that $v\mid ux$. If $v\not \mid u$, then $x\mid v$. Let $v=xv_1$. Since v is 3-path in G, $x\in N_G(v_1)$. Now, if $v_1$ is an edge in G, then $v_1\in N^\text {edge}_G(x)$, on the other hand ${\text {supp}}(v_1)\subset N_G(x)$ which further implies that $u\in I_2(H)$. This yields that $(I_3(G)+\langle xy \rangle ):x\subset \langle y \rangle +I_2(H)+I_3(G{\setminus } N_G[x])$, and hence,

$$\begin{aligned} (I_3(G)+\langle xy \rangle ):x= \langle y \rangle +I_2(H)+I_3(G\setminus N_G[x]) \end{aligned}$$

which completes the proof. $\square$

We recall a property satisfied by a tree from [10, Proposition 4.1].

Remark 4.2

If G is a tree containing a vertex of degree at least two, then there exists a vertex $v\in V(G)$ with $N_G(v)=\{v_1,\dots ,v_r\}$, where $r\ge 2$ and $\deg _G(v_i)=1$ for $i< r$.

Lemma 4.3

Let G be a tree and $v\in V(G)$ with the notation as in Remark 4.2. Then,

$$\begin{aligned} \nu _3(G\setminus N_G[e])\le \nu _3(G)-1,\text { where } e=\{v,v_r\} \end{aligned}$$

Proof

Let $\nu _3(G{\setminus } N_G[e])=s$ and $P_1,\dots ,P_s$ be a 3-path induced matching in $G{\setminus } N_G[e]$. Since $\deg _G(v_1)=1$, $N_G(v_1)\cap V(P_i)=\emptyset$ for $1\le i\le s$. Also, note that for $1\le i\le s$, $N_G(v)\cap V(P_i)=\emptyset =N_G(v_r)\cap V(P_i)$. Therefore, $\{v_1,v,v_r\},P_1,\dots ,P_s$ is a 3-path induced matching in G, and hence, $\nu _3(G)\ge s+1$. $\square$

Now, we are ready to prove the exact regularity formula for the 3-path ideals of trees.

Theorem 4.4

Let G be a tree and $I_3(G)$ be its 3-path ideal. Then, ${\text {reg}}(R/I_3(G))=2\nu _3(G)$.

Proof

By Corollary 3.2, it is enough to prove that ${\text {reg}}(R/I_3(G))\le 2\nu _3(G)$.

We proceed by induction on |V(G)|. By Remark 4.2, let v be such a vertex of G such that $N_G(v)=\{v_1,\dots ,v_r\}$ and $\deg _G(v_i)=1$ for $i<r$, where $r\ge 2$. Set $u=v_r$, and let $N_G(u)=\{u_1=v,u_2,\dots ,u_s\}$ for $s\ge 1$. Consider the following short exact sequence:

$$\begin{aligned} 0\longrightarrow \frac{R}{I_3(G):uv}(-2){\mathop {\longrightarrow }\limits ^{\cdot uv}} \frac{R}{I_3(G)}\longrightarrow \frac{R}{\langle uv,I_3(G)\rangle }\longrightarrow 0. \end{aligned}$$

(1)

Using Lemma 4.1(1), we have

$$\begin{aligned}{} & {} I_3(G):uv=\langle v_i:1\le i\le r-1\rangle +\langle u_j:2\le j\le s\rangle +I_3(G\setminus N_G[e]), \\{} & {} \text { where } e=\{u,v\}. \end{aligned}$$

By virtue of Lemma 4.3, we have $\nu _3(G\setminus N_G[e]) \le \nu _3(G)-1$. Therefore, by inductive hypothesis, ${\text {reg}}(R/(I_3(G):uv))\le 2\nu _3(G{\setminus } N_G[e])\le 2\nu _3(G)-2.$ Now, set $J=\langle uv,I_3(G)\rangle$ and consider the following short exact sequence:

$$\begin{aligned} 0\longrightarrow \frac{R}{J:u}(-1){\mathop {\longrightarrow }\limits ^{\cdot u}} \frac{R}{J}\longrightarrow \frac{R}{\langle u, J\rangle }\longrightarrow 0. \end{aligned}$$

(2)

Observe that, $\langle u, J\rangle =\langle u \rangle +I_3(G{\setminus }\{u\})$. By Remark 2.2, $\nu _3(G\setminus \{u\})\le \nu _3(G).$ Thus, it follows from inductive hypothesis that ${\text {reg}}(R/\langle u, J\rangle )\le 2\nu _3(G{\setminus } \{u\})\le 2\nu _3(G).$ On the other hand, by Lemma 4.1(2), $J:u=\langle v\rangle +I_2(H)+I_3(G{\setminus } N_G[u]),$ where H is the union of $N_G^{\text{ e }dge}(u)$ and the complete graph on the vertex set $N_G(u)\setminus \{v \}$.

Case-I: If $s=1$, then $N_G^{\text{ e }dge}(u)=\{\{v,v_i\}:1\le i\le r-1 \}$. Also, note that the graph $G{\setminus } N_G[u]$ does not have an edge. This implies that $J:u=\langle v\rangle$, and hence ${\text {reg}}\left( S/(J:u)\right) =0\le 2\nu _3(G)-1$.

Case-II: Suppose $s\ge 2$ and $\deg (u_i)=1$ for $2\le i\le s$. Then, H is the union of $N_G^{\text{ e }dge}(u)=\{\{v,v_i\}:1\le i\le r-1 \}$ and the complete graph on the vertex set $\{u_2,\dots , u_s\}$. In this case, the graph $G{\setminus } N_G[u]$ has no edge. Thus, $J:u=\langle v\rangle +I_2(H)$. Since H is co-chordal, by [8, Theorem 1], $I_2(H)$ has a linear resolution. Hence, we get ${\text {reg}}(R/(J:u))=1\le 2\nu _3(G)-1$.

Therefore, it follows from [15, Corollary 18.7] applying to the short exact sequences (1) and (2) that

$$\begin{aligned} {\text {reg}}(R/I_3(G))\le \max \{ {\text {reg}}(R/(I_3(G):uv))+2,{\text {reg}}(R/(J:u))+1,{\text {reg}}(R/\langle u,J\rangle )\}. \end{aligned}$$

Hence, ${\text {reg}}(R/J)\le 2\nu _3(G)$.

Case-III: Suppose now $s\ge 2$ and $\deg _G(u_i)\ge 2$ for some $2\le i\le s$. Without loss of generality, we assume that $\deg _G(u_i)\ge 2$ for all $2\le i\le t$ and $\deg (u_i)=1$ for all $t+1\le i\le s$. In this case, H is a union of the edges $N_G^{\text {edge}}(u)$ and the complete graph on the vertex set $\{u_2,\dots , u_s\}$, i.e.,

$$\begin{aligned} J:u= & {} \langle v\rangle +\langle u_iu_j: 2\\\le & {} i< j\le s\rangle +\sum \limits _{i=2}^t\left\langle u_ix: x\in N_G(u_i)\setminus \{u\}\right\rangle +I_3\left( G\setminus N_G[u]\right) . \end{aligned}$$

Set $J_1=J:u$ and $J_{i}=J_{1}+\langle u_2,\dots ,u_{i}\rangle$ for $2\le i\le t$. For $2\le i\le t$, consider the following short exact sequence:

$$\begin{aligned} 0\longrightarrow \frac{R}{J_{i-1}:u_i}(-1){\mathop {\longrightarrow }\limits ^{\cdot u_i}} \frac{R}{J_{i-1}}\longrightarrow \frac{R}{J_{i}}\longrightarrow 0. \end{aligned}$$

(3)

By Lemma 4.1, $J:uu_i=\langle u_j:1\le j\le s \text { and } j\ne i\rangle + \langle w: w\in N_G(u_i){\setminus } \{u\}\rangle +I_3(G{\setminus } N_G[\{u,u_i\}]$. Therefore,

$$\begin{aligned} J_{i-1}:u_i&= (J_{1}+\langle u_2,\dots ,u_{i-1}\rangle ):u_i= (J:uu_i)+\langle u_2,\dots ,u_{i-1}\rangle \\&= \langle u_j:1\le j\le s, \ j\ne i\rangle +\langle w: w\in N_G(u_i)\setminus \{u\}\rangle +I_3(G\setminus N_G[\{u,u_i\}]. \end{aligned}$$

Further, we can write

$$\begin{aligned} J_{i-1}:u_i= & {} \langle u_j:1\le j\le s, \ j\ne i\rangle +\langle w: w\in N_G(u_i)\setminus \{u\}\rangle \\{} & {} \quad +I_3(G\setminus \{N_G[v]\cup N_G[u]\cup N_G[u_i]\}) \end{aligned}$$

for $2\le i\le t$. Also, $J_{t}=\langle v,u_2,\dots ,u_t\rangle +I_3(G{\setminus } \{N_G[v]\cup N_G[u]\})$. Now, it follows from [15, Corollary 18.7] applying to the short exact sequence (3) that

$$\begin{aligned} {\text {reg}}(R/J_1)\le \max \{{\text {reg}}(R/(J_{i-1}:u_i))+1, {\text {reg}}(R/J_{t}):2\le i\le t \}. \end{aligned}$$

Since $G\setminus \{N_G[v]\cup N_G[u]\cup N_G[u_i]\}$ is an induced subgraph of $G\setminus N_G[e]$, by Remark 2.2 and Lemma 4.3, we have $\nu _3(G{\setminus } \{N_G[v]\cup N_G[u]\cup N_G[u_i]\})\le \nu _3(G{\setminus } \{N_G[v]\cup N_G[u]\})\le \nu _3(G)-1$.

By inductive hypothesis, ${\text {reg}}(R/(J_{i-1}:u_i))\le 2\nu _3(G{\setminus } \{N_G[v]\cup N_G[u]\cup N_G[u_i]\})\le 2(\nu _3(G)-1)$ and ${\text {reg}}(R/J_{t})\le 2\nu _3(G{\setminus } \{N_G[v]\cup N_G[u]\})\le 2\nu _3(G)-2$. By applying [15, Corollary 18.7] to the above short exact sequences, we have

$$\begin{aligned}&{\text {reg}}(R/I_3(G)) \le \max \{{\text {reg}}(R/(I_3(G):uv))+2,{\text {reg}}(R/J) \} \\&\quad \le \max \{{\text {reg}}(R/(I_3(G):uv))+2,{\text {reg}}(R/\langle u,J\rangle ),{\text {reg}}(R/J_1)+1 \} \\&\quad \le \max \{{\text {reg}}(R/(I_3(G):uv))+2,{\text {reg}}(R/\langle u,J\rangle ),\\&\quad {\text {reg}}(R/(J_1:u_2))+2,{\text {reg}}(R/J_2)+1 \} \\&\quad \le \max _{2\le i\le t} \{{\text {reg}}(R/(I_3(G):uv))+2,{\text {reg}}(R/\langle u,J\rangle ),\\&\qquad {\text {reg}}(R/(J_{i-1}:u_i))+2,{\text {reg}}(R/J_{t})+1\}. \end{aligned}$$

Hence, the assertion follows. $\square$

Now, we proceed to study the regularity of 3-path ideal of unicyclic graphs. If G is a cycle, then the regularity of $R/I_3(G)$ has been computed in [2]. So, we assume that G is not a cycle. We give a sharp upper bound for the regularity of $R/I_3(G)$. The idea of the proof is kind of similar to the proof of Theorem 4.4. We fix the following notation for unicyclic graphs.

Notation 4.5

Let G be a unicyclic graph with the induced cycle C. Then, trees are attached to at least one vertex of C, say $u\in V(C)$. Let $v\in N_G(u){\setminus } V(C)$ and $e=\{u,v\}$. Clearly, $N_G(u)\setminus \{v\}$ contains at least 2 vertices and set $N_G(u)=\{u_1=v,u_2,\dots ,u_t\}$ for $t\ge 3$.

Theorem 4.6

Let G be a unicyclic graph and $I_3(G)$ be its 3-path ideal. Then,

$$\begin{aligned} 2\nu _3(G)\le {\text {reg}}(R/I_3(G))\le 2\nu _3(G)+2. \end{aligned}$$

Proof

Let G be a unicyclic graph with the notation as in Notation 4.5. The lower bound for ${\text {reg}}(R/I_3(G))$ follows from Corollary 3.2. So, here we only establish the upper bound. Consider the short exact sequence (1). By Lemma 4.1(1), $I_3(G):uv=\langle N_G(e)\rangle +I_3(G\setminus N_G[e])$, where $e=\{u,v\}$. Since $G{\setminus } N_G[e]$ is an induced subgraph of G, by Remark 2.2, $\nu _3(G{\setminus } N_G[e])\le \nu _3(G)$. Note that $G\setminus N_G[e]$ is a tree. Thus, it follows from Theorem 4.4 that

$$\begin{aligned} {\text {reg}}(R/(I_3(G):uv))=2\nu _3(G\setminus N_G[e])\le 2\nu _3(G). \end{aligned}$$

Now, set $J=\langle uv,I_3(G)\rangle$ and we consider the short exact sequence (2), where $J:u=\langle v\rangle +I_2(H) +I_3(G{\setminus } \{N_G[u]\}),$ where H is the union of $N_G^{\text{ e }dge}(u)$ and the complete graph on the vertex set $N_G(u){\setminus } \{v\}$. Also, $\langle u,J\rangle =\langle u,I_3(G{\setminus } \{u\})\rangle$. Since $G{\setminus } \{u\}$ is a tree, by Theorem 4.4, we have

$$\begin{aligned} {\text {reg}}(R/\langle u,J\rangle )=2\nu _3(G\setminus \{u\})\le 2\nu _3(G). \end{aligned}$$

Set $J_1=J:u$ and $J_i=J_{1}+\langle u_2,\dots ,u_{i}\rangle$, where $N_G(u)=\{v,u_2,\dots ,u_t\}$ for $2\le i\le t$ and consider short exact sequences (3).

It can be observed that $J_{i-1}:u_i=\langle u_j:1\le j\le t, \ j\ne i\rangle +\langle w: w\in N_G(u_i){\setminus } \{u\}\rangle +I_3(G{\setminus } \{N_G[u]\cup N_G[u_i]\})$ for $2\le i\le t$ and $J_{t}=\langle v,u_2,\dots ,u_t\rangle +I_3(G\setminus \{N_G[u]\})$. Now it follows from [15, Corollary 18.7] applying to the short exact sequence (3) that

$$\begin{aligned} {\text {reg}}(R/J_1)\le \max \{{\text {reg}}(R/(J_{i-1}:u_i))+1, {\text {reg}}(R/J_{t}):2\le i\le t \}. \end{aligned}$$

Now, $\nu _3(G\setminus \{N_G[u]\cup N_G[w_i]\})\le \nu _3(G)$ and $\nu _3(G\setminus \{N_G[u]\})\le \nu _3(G)$ follow from Remark 2.2. Since $G{\setminus } \{N_G[u]\cup N_G[w_i]\}$ and $G{\setminus } \{N_G[u]\}$ are trees, by Theorem 4.4, ${\text {reg}}(R/(J_i:w_i))=2\nu _3(G{\setminus } \{N_G[u]\cup N_G[w_i]\})\le 2\nu _3(G)$ and ${\text {reg}}(R/J_{t})=2\nu _3(G\setminus \{N_G[u]\})\le 2\nu _3(G)$. Therefore, it follows from applying [15, Corollary 18.7] to short exact sequences 1, 2 and 3 that ${\text {reg}}(R/I_3(G))\le 2\nu _3(G)+2$. $\square$

We now show by examples that all the three possibilities for the regularity of $R/I_3(G)$, namely $2\nu _3(G)$, $2\nu _3(G)+1$ and $2\nu _3(G)+2$, indeed occur for unicyclic graphs.

Example 4.7

Consider graphs $G_1$, $G_2$ and $G_3$ as in Fig. 2. Then, using Macaulay 2 ( [9]), it can be computed that ${\text {reg}}(R/I_3(G_1))=2$, ${\text {reg}}(R/I_3(G_2))=3$ and ${\text {reg}}(R/I_3(G_3))=6$. Note that $\nu _3(G_1)=1=\nu _3(G_2)$ and $\nu _3(G_3)=2$.

Fig. 2

Unicyclic graphs