On Linear Quotients of Squarefree Monomial Ideals

Abstract

Let $I\subset K[x_1,\ldots ,x_n]$ be a squarefree monomial ideal generated in one degree. Let $G_I$ be the graph whose nodes are the generators of I, and two vertices $u_i$ and $u_{ j}$ are adjacent if there exist variables x, y such that $xu_i = yu_j$. We show that if I is generated in degree $n-2$, then the following are equivalent:

• (i) $G_I$ is a connected graph;

• (ii) I has a $(n-2)$-linear resolution;

• (iii) I has linear quotients;

• (iv) I is a variable-decomposable ideal.

We also prove that if I has linear relations and $\overline{G_I}$ is chordal, then I has linear quotients.

1 Introduction

Let $S = K[x_{1},\ldots , x_{n}]$ be the polynomial ring in n variables over a field K and I be a squarefree monomial ideal in S Fröberg [5, Theorem 1] proved that the edge ideal of a finite graph G has a 2-linear resolution if and only if the complement graph $\overline{G}$ of G is a chordal graph. Emtander [4], Woodroofe [12], Van Tuyl and Villarreal [11] have partially generalized the Fröberg’s Theorem. It is known that monomial ideals with 2-linear resolution have linear quotients [7]. We denote by $\varDelta ^{\vee }$ the Alexander dual of $\varDelta$. Let I be the Stanley–Reisner ideal of $\varDelta ^{\vee }$ which is generated in degree 2 and has a linear resolution. By a result of Eagon–Reiner [3], $\varDelta$ is a Cohen–Macaulay of codimension 2. In [1], it is shown that if $\varDelta$ is a Cohen–Macaulay simplicial complex of codimension 2, then $\varDelta$ is vertex decomposable. Hence, by a result of [10], $I=I_{\varDelta ^{\vee }}$ is a variable-decomposable monomial ideal generated in degree 2. Therefore, if I is a monomial ideal generated in degree 2, then

$(*)\; \; I$ is a variable-decomposable ideal$\ \Leftrightarrow I$ has linear quotients $\Leftrightarrow I$ has a linear resolution.

In [8], some other classes of squarefree monomial ideals with property $(*)$ were introduced. These three concepts are not equivalent in general, but we have the following implications:

I is a variable-decomposable ideal $\Rightarrow I$ has linear quotients $\Rightarrow I$ has a linear resolution.

In the present paper, we consider squarefree monomial ideals generated in degree d and try to find some other classes of squarefree monomial ideals with $(*)$ property. We show that if $d=2$, $d=n-2$, $n\le 5$ or $\overline{G_I}$ is a chordal graph, then the property $(*)$ holds for I. We also study squarefree monomial ideals with linear quotients and show that if I has linear relations and $\overline{G_I}$ is chordal, then I has linear quotients.

The paper proceeds as follows. In Sect. 2, we recall some definitions and some known results which will be needed later. Let $I=I(G)$ be an edge ideal of a simple graph G. In Sect. 3, we present a different proof of the well-known fact that property $(*)$ holds for I. Also we show that the concepts of linear relations, linear resolution, linear quotients and variable–decomposability are equivalent when I is a squarefree monomial ideal generated in degree $n-2$. Consequently, if $I\subset K[x_1,\ldots ,x_5]$ is a squarefree monomial ideal, then I has property $(*)$, see Corollary 1. We consider monomial ideal I, where $\overline{G_I}$ is a chordal graph, in Sect. 4. If $\overline{G_I}$ is a chordal graph and I has linear relations, then I has linear quotients as will be proved in Theorem 3. Hence, if $\overline{G_I}$ is a chordal graph, then I has property $(*)$, see Corollary 2. In addition, another property of $\overline{G_I}$ which implies that I has linear quotients is given in Theorem 4.

2 Preliminary

For a monomial ideal I in S, there is a minimal graded free resolution of the form

$$\begin{aligned} F_{\bullet }: \; 0 \longrightarrow F_p \longrightarrow \cdots \longrightarrow F_2 \longrightarrow F_1 \longrightarrow I \longrightarrow 0 \end{aligned}$$

where $F_i=\bigoplus _{j}S(-j)^{\beta _{i j}}$ and $S(-j)$ denotes the free S-module obtained by shifting the degrees of S by j. The numbers $\beta _{i j}=\beta _{i j} (I)$ are called the graded Betti numbers of I. Recall that I has a d-linear resolution over K if $\beta _{i j} (I) = 0$ for all $j\ne i + d$. It follows that I is generated in degree d. Let $\varphi : F_1 \longrightarrow I$ be the maps which send the basis element $e_i$s of $F_1$ to the generators $u_i$ of I. Recall that I has linear relations if the kernel of $\varphi$ is generated by linear forms. Note that if the ideal I has linear relations, then all of its generators have the same degree.

The unique minimal monomial set of generators of a monomial ideal I is denoted by G(I). Monomial ideal I has linear quotients if there exists an order $\sigma = u_1, \ldots , u_m$ of G(I) such that the colon ideal $\langle u_1, \ldots , u_{i-1}\rangle : u_i$ is generated by a subset of variables for $i = 2, \ldots , m$. Any order of the generators for which we have linear quotients will be called an admissible order. For a monomial $u\in S$, set $F(u_i) :={\,}$Supp$(u)=\{i\; :\; x_i\mid u\}$. It is easy to see that if I is a squarefree monomial ideal, then I has linear quotients if and only if for all i and all $j < i$ there exist $l \in F(u_j) {\setminus } F(u_i)$ and an integer $k < i$ such that $F(u_k) {\setminus } F(u_i) =\{l\}$ (see [6, Corollary 8.2.4]).

Let $u = x_1^{a_1} \ldots x_n^{a_n}$ be a monomial in S. For another monomial v, set $[u, v] = 1$ if $x_i^{a_i} \not \mid v$ for all $i \in F(u)$. Otherwise, set $[u, v] \ne 1$. For a monomial $u \in S$ and a monomial ideal $I \subseteq S$, set $I_u=\langle u_i \in G(I) :\ [u, u_i] = 1\rangle$ and $I^u = \langle u_j \in G(I) :\ [u, u_j] \ne 1\rangle>$.

In this paper, by the notation $u_j:u_i=x_l$ we mean $<u_j>:<u_i>=<x_l>.$

Definition 1

Let I be a monomial ideal with the minimal system of generators $\{u_1, \ldots , u_m \}$. A monomial $u =x_1^{a_1} \cdots x_n^{a_n}$ is called shedding if $I_u \ne 0$, and for each $u_i \in G(I_u)$ and $l \in F(u)$, there exists $u_j \in G(I^u)$ such that $u_j:u_i=x_l.$

Definition 2

Let I be a monomial ideal minimally generated by $\{u_1, \ldots , u_m \}$. Monomial ideal I is called a k-decomposable ideal if $m= 1$ or else has a shedding monomial u with $\mid$Supp$(u)\mid \le k + 1$ such that the ideals $I_u$ and $I^u$ are k-decomposable. (Note that since the number of minimal generators of I is finite, the recursion procedure will stop.)

A monomial ideal is decomposable if it is k-decomposable for some $k \ge 0$ . Also, a 0-decomposable ideal is called variable-decomposable. The concepts of shedding monomial and k-decomposable monomial ideals were first introduced by Rahmati and Yassemi in [10].

Let $I\subset S$ be a squarefree monomial ideal which is generated in one degree. We associate with I a simple graph $G_{I}$, called the first syzygies graph of I, whose vertices are the elements of G(I). Two vertices $u_i$ and $u_j$ are adjacent if there exist variables x and y such that $xu_i=yu_j$. Since the generators of a monomial ideal form a Gröbner basis, hence the S-polynomials of each pair of its generators would give the generators of the first syzygies of the monomial ideal, that is, $\frac{lcm(u_i,u_j)}{ u_i} e_i- \frac{lcm(u_i,u_j)}{ u_j} e_j$. So one can easily find nonlinear syzygies or, equivalently, linear syzygies.

Remark 1

Let I be a squarefree monomial ideal . If $G_I$ is a complete graph, then I is a variable-decomposable ideal.

Let G be a finite graph on vertex set $V=\{1,\ldots , n\}$. The edge ideal of G is defined by $I(G) = \langle x_ix_j {:} \{i, j\} \text { is an edge in}\; G\rangle \subset S$. The induced subgraph $G_W$ is the graph whose vertex set is $W \subseteq V$ and the edge set consists of all the edges of G whose both endpoints are in W.

Definition 3

Let G be a graph on vertex set V and $u \in V$. The set of all neighborhoods of u is denoted by N(u).

Definition 4

A cut vertex is a vertex that when removed (with its boundary edges) from a graph creates more components than previously in the graph.

The following fact will be used later.

Proposition 1

[2, Corollary 5.6] Every non-trivial connected graph contains at least two vertices that are not cut-vertices.

A finite graph G is called chordal if each of whose cycles of length $> 3$ has a chord. It is clear that every induced subgraph of a chordal graph is again chordal. A perfect elimination ordering of G is an ordering $i_{n}, \ldots , i_{2}, i_{1}$ of the vertices $1, 2, \ldots , n$ of G such that, for each $1 < j \le n,$ the induced graph on $C_{i_{j}}=\{i_j\} \cup \lbrace i_{k} \in [n] : 1 \le k < j, \lbrace i_{k}, i_{j} \rbrace \in E(G) \rbrace$ is a clique of G. In 1961,  G. A. Dirac proved that a graph G is chordal if and only if it has a perfect elimination ordering.

Remark 2

Let $\overline{G}$ be the complementary graph of G and let the ordering $n, n-1, \ldots ,1$ be a perfect elimination ordering of $\overline{G}$. If $k> i,j$ and $\{ i , k \} , \{j , k \} \in E( \overline{G})$, then $\{ i , j \} \in E( \overline{G})$. In other words, if $\{ i, k \}, \{ j, k \} \notin E( G)$, then $\{ i, j \} \notin E(G)$. Hence, if $\{ i, j \} \in E(G)$ and $k> i,j$, then $\{ i, k\} \in E(G)$ or $\{j , k \} \in E(G)$.

3 Resolution of Monomial Ideals of Degree 2 and $n-2$

Let G be a finite graph and I be the edge ideal of G. Fröberg [5] showed that I has a linear resolution if and only if $\overline{G}$ is a chordal graph. As it is mentioned in the introduction, for monomial ideals generated in degree 2 the following conditions are equivalent:

• (a) I has a linear resolution;

• (b) I has linear quotients;

• (c) I is a variable-decomposable ideal.

We could not find a direct proof for equivalence of these three statements, but we conclude it from known results in [1, 3, 7, 10]. For convenience of reader, a direct proof of this fact is provided in the following.

Theorem 1

Let I be the edge ideal of a finite graph G. Then the following conditions are equivalent:

• (a) The complementary graph $\overline{G}$ of G has a perfect elimination ordering;

• (b) I is variable-decomposable ideal;

• (c) I has a linear resolution;

• (d) I has linear quotients.

Proof

$(a) \Rightarrow (b)$: We proceed by induction on $\mid V(\overline{G})\mid$. The assertion is trivial if $\mid V(\overline{G})\mid =2$. If $\mid V(\overline{G})\mid >2$, then without loss of generality, we may assume that $n, n-1, \ldots , 1$ is a perfect elimination ordering of $\overline{G}$ and $I_{x_n}=I(G_1)$. It is clear that $\overline{G_1}$ is an induced subgraph of $\overline{G}$, so $\overline{G_1}$ is a chordal graph. Since $\mid V(\overline{G_1})\mid <\mid V(\overline{G}) \mid$, by induction hypothesis, $I_{x_n}$ is a variable-decomposable ideal. It is not difficult to see that $G_{I^{x_n}}$ is a complete graph; hence, $I^{x_n}$ is a variable-decomposable ideal. So it is enough to show that $x_n$ is a shedding variable. If $u=x_ix_j \in I_{x_n}$, then by Remark 2, $x_ix_n \in G(I)$ or $x_jx_n \in G(I)$. Therefore, $x_n$ is a shedding monomial.

$(b) \Rightarrow (c)$: This proof is down by induction on $k=\mid G(I)\mid$. The assertion is obvious for $k = 1$. Now suppose $k>1$ and x is a shedding. Since $\mid G(I^x)\mid$ and $\mid G(I_x)\mid$ are less than $\mid G(I)\mid$, by induction hypothesis, $I^x$ and $I_x$ have linear resolution. The short exact sequence

$$\begin{aligned} 0\rightarrow I^x \rightarrow I=I^x\oplus I_x \rightarrow I_x \rightarrow 0 \end{aligned}$$

yields the long exact sequence

$$\begin{aligned} \cdots \rightarrow Tor_i^S(I^x,K)_{i+j} \rightarrow Tor_i^S(I,K)_{i+j} \rightarrow Tor_i^S(I_x,K)_{i+j} \rightarrow \cdots \end{aligned}$$

If $j \ne d$, then both ends in this exact sequence vanish. Hence, the middle term vanishes too.

$(c) \Rightarrow (a)$: It follows from Fröberg’s Theorem that $\overline{G}$ is chordal. It is well known that a graph is chordal if and only if it has a perfect elimination ordering.

$(b) \Rightarrow (d)$: Let I be a variable-decomposable ideal. Without loss of generality, assume that $x_1$ is a shedding where $I^{x_1}=\langle u_1, \ldots , u_{t-1}\rangle$ and $I_{x_1}=\langle u_t, \ldots , u_m\rangle$ are variable-decomposable. Arguing by induction on $\mid G(I)\mid$ yields that $I^{x_1}$ and $I_{x_1}$ have linear quotients. It can be assumed that $u_1, \ldots , u_{t-1}$ and $u_t, \ldots , u_m$ are admissible order for $I^{x_1}$ and $I_{x_1}$, respectively. It will be shown that $u_1, \ldots , u_{t-1}, u_t, \ldots , u_m$ is an admissible order for I. Let $u_i\in I_{x_1}$ and $u_j\in I^{x_1}$, then, by definition of shedding, there exists $u_k\in I^{x_1}$ such that $F(u_k) {\setminus } F(u_i)=\{1\}$. Since $u_j\in I^{x_1}$, one has $1 \in F(u_k){\setminus } F(u_i)$.

$(d) \Rightarrow (c)$: Follows from [6, Proposition 8.2.1]. $\square$

Now we consider squarefree monomial ideals generated in degree $n-2$.

Remark 3

Let $I \subset S$ be a squarefree monomial ideal generated in degree $n-2$. If there exists l, $1\le l\le n$, such that $x_l \not \mid u$ for all $u\in G(I)$, then $G_I$ is a complete graph.

Remark 4

Let $S=K[x_{1},\ldots ,x_{n}]$ and I be a squarefree monomial ideal generated in degree $n-2$. If $u_i, u_j \in G(I)$ are not adjacent in $G_I$, then $F(u_i) \cup F(u_j)=[n]$.

Theorem 2

Let $S=K[x_{1},\ldots ,x_{n}]$ and $I =\langle u_1,\ldots , u_m\rangle$ be a squarefree monomial ideal generated in degree $n-2$. The following are equivalent:

• (a) $G_{I}$ is a connected graph;

• (b) I has linear relations;

• (c) I has a $(n-2)$-linear resolution;

• (d) I has linear quotients;

• (e) I is a variable-decomposable ideal.

Proof

$(a) \Rightarrow (d)$: We order the vertices of $G_I$ in a way that the induced subgraph on $\{ u_1, \ldots , u_i \}$ is a connected graph for $2\le i\le m$. It will be shown that $u_1, \ldots , u_m$ is an admissible order. If $j < i$ and $u_i, u_j$ are adjacent, take $u_k=u_j$. Otherwise, by connectivity of the induced subgraph $\{ u_1, \ldots , u_i \}$, there exist a vertex $u_k$ such that $u_i$ and $u_k$ are adjacent. Set $\{l\}=F(u_k) {\setminus } F(u_i)$. Then by Remark 4 one has $l \in F(u_j) {\setminus } F(u_i)$.

$(d) \Rightarrow (c)$: It follows from [6, Proposition 8.2.1].

$(c) \Rightarrow (b)$: The implication is trivial.

$(b) \Rightarrow (a)$: See [8, Lemma 1.8].

$(a)\Rightarrow (e)$: We proceed by induction on $\mid G(I)\mid$. If $\mid G(I) \mid =1$ the assertion is trivial; hence, assume that $\mid G(I)\mid \ge 2$. Let $x_i$ be an arbitrary variable. By Remark 3, all elements in $G_{I_{x_{i}}}$ are adjacent. If $x_i$ is not a shedding, then there exist $u \in I_{x_i}$ such that $v: u \ne x_i$ for all $v \in I^{x_i}$. Hence, u and v are not adjacent for all $v \in I^{x_i}$. Assume that $F(u)=[n] {\setminus } \{ i,j\}$. Now, we show that $x_j$ is a shedding. If $w \in I_{x_j}$, then u and w are adjacent which implies $w \in I_{x_i}$, $F(w)=[n] {\setminus } \{ i,j\}=F(u)$ and $I_{x_j}=\langle u\rangle$. Since $G_I$ is a connected graph, there exists $v \in I^{x_j}$ such that u and v are adjacent and $v:u=x_j$.

Now, it is enough to show that $G_{I^{x_j}}\cong G_I {\setminus } \{u\}$ is a connected graph. If $v ,w \in G(I){\setminus } \{u\}$, then there exists a path between v and w in $G_I$. If u is not a vertex in this path, then v and w are connected in $G_{I^{x_j}}$. Otherwise, we have a path $v, \ldots , z, u, q, \ldots , w$. We know z, q are adjacent to u; hence, $z, q \in I_{x_i}$ and z and q are adjacent. Therefore, v and w are connected in $G_{I^{x_j}}$.

Now, we consider a case that every variable is a shedding for I. By Remark 3, it is enough to show that $G_{I^{x_i}}$ is a connected graph for some $1 \le i \le n$. Assume $G_{I^{x_1}}$ is not a connected graph and $C_1, \ldots , C_p$ are its connected components. Since $G_I$ is a connected graph, $C_i$ is connected to $G_{ I_{x_1}}$, for $i=1,\ldots ,p$. If $u_1 \in C_1$ and $l \notin F(u_1)$, then $u_1 \in I_{x_l}$. Since $G_{ I_{x_l}}$ is a complete graph and $u \in C_i$, $1 < i \le p$, is not adjacent to $u_1$, we conclude that $V(C_i)\subseteq G( I^{x_l})$. We claim that if $u_t \in I_{x_1}$ and $u_t$ is adjacent to $u \in C_i$, for some $1 < i \le p$, then $u_t \in I^{x_l}$. Suppose on the contrary that $u_t \in I_{x_l}$, then $F(u_t)=[n] {\setminus } \{ 1, l\}$. On the other hand, $u \in I^{x_1}$ and $u \in I^{x_l}$. Hence, $\{1, l\} \subseteq F(u)$; therefore, $u_t$ and u are not adjacent, a contradiction.

Now we consider the following two cases:

Case 1 If for all $u \in C_1$, then there exists $u_t \in I_{x_1}$ such that $u_t$ is adjacent to u, then $G_{I^{x_l}}$ is a connected graph.

Case 2 There is a $u \in C_1$ such that u is not adjacent to any $u_t \in I_{x_1}$. Assume $F(u)=[n] {\setminus } \{ 1, r\}$. If $G_{I^{x_r}}$ is a connected graph, we are done. Otherwise, since u is not adjacent to any $u_t \in I_{x_1}$ and $G_{I_{x_r}}$ is a complete graph, one has $G( I_{x_1}) \subseteq G(I^{x_r})$. It is clear that $G_{I_{x_1}}$ and $C_i$, for all $2 \le i \le p$, form a connected component of $G_{I^{x_r}}$. Hence, $G_{I^{x_r}}$ has two connected components and since $u \notin C_1^{'}$, its second connected component is $C_1^{'}\subsetneq C_1$. Now repeat the above procedure for $C_1^{'}$. Since the graph $G_I$ is finite, after a finite number of steps we find ${x_r}$ for which $G_{I^{x_r}}$ is a connected graph.

$(e) \Rightarrow (a)$: We proceed by induction on $\mid G(I)\mid$. Since I is variable-decomposable, there exists a shedding x such that $I_x$ and $I^x$ are variable-decomposable. By induction hypothesis, $G_{I_x}$ and $G_{I^x}$ are connected graphs. Since x is a shedding, one has every vertex in $G_{I_x}$ is adjacent to some vertex in $G_{I^x}$. $\square$

Corollary 1

Let $I \subseteq K[x_1,\ldots , x_n]$ be a squarefree monomial ideal. If $n\le 5$, then the following are equivalent:

• (a) I has a linear resolution;

• (b) I has linear quotients;

• (c) I is variable-decomposable ideal.

Proof

If I is generated in degree 2, then the assertion follows from Theorem 1. If I is generated in degree 3, then the assertion follows from Theorem 2. The assertion is trivial when I is generated in degree 1 or 4. $\square$

Notice that the following examples show that the condition $n \le 5$ is the best possible.

Example 1

Consider monomial ideal $I=\langle x_1x_2x_3, x_1x_3x_5, x_1x_5x_6, x_2x_5x_6, x_2x_3x_6,$$x_3x_4x_6, x_1x_4x_6, x_1x_2x_4, x_2x_4x_5, x_3x_4x_5\rangle$. This ideal is corresponding to triangulation of the projective plane and has a linear resolution in characteristics zero but has not a linear quotients (see [13]).

Example 2

Consider monomial ideal $I=\langle x_1x_2x_3, x_1x_2x_6, x_1x_2x_5, x_2x_4x_5, x_3x_4x_5,$$x_1x_3x_4, x_1x_3x_6, x_2x_4x_6, x_3x_5x_6, x_4x_5x_6\rangle$. This ideal has linear quotients and is not a variable-decomposable (see [9]).

4 Linear Quotients of Monomial Ideals

It is shown in [8] that from any path between $u_i$ and $u_j$ in $G_I$, one can find $w_i$ and $w_j$ in Mon(S) such that $w_i e_i-w_je_j \in \ker (\varphi )$.

Remark 5

If there exist two paths $u_i, v, u_j$ and $u_i, w, u_j$ in $G_I$ between $u_i$ and $u_j$, then the coefficients of $e_i$ which comes from the first and the second paths are equal.

Lemma 1

Let $I=\langle u_1, \ldots , u_m\rangle$ be a squarefree monomial ideal which has linear relations. If $\overline{G_I}$ is a chordal graph, then there exists a t such that the monomial ideal J with $G(J)=G(I){\setminus } \{u_t\}$ has linear relations.

Proof

By [8, Lemma 1.8], $G_I$ is a connected graph; thus, the existence of non-cut-vertex $u_t$ follows from Proposition 1. Let $J=\langle u_1, \ldots ,u_{t-1}, u_{t+1}, \ldots , u_m\rangle$ and $u_i$ and $u_j$ be two arbitrary elements in $G(J)= G(I){\setminus }\{u_t\}$. We have

$$\begin{aligned} X_{F(u_j) {\setminus } F(u_i)}e_i-X_{F(u_i) {\setminus } F(u_j)}e_j\in \ker (F^{'}_1\longrightarrow J)\subset \ker (\varphi ). \end{aligned}$$

Since I has linear relations, there exists a path $L_1$ between $u_i$ and $u_j$ in $G_I$ such that $X_{F(u_j) {\setminus } F(u_i)}e_i-X_{F(u_i) {\setminus } F(u_j)}e_j$ is a linear combination of linear forms which comes from $L_1$. If $L_1$ is a subgraph of $G_J$, then we are done. Otherwise, there exists a path $L_2$ of minimum length between $u_i$ and $u_j$ in $G_J$. Since $\overline{G_I}$ is chordal, it is easy to see that $G_I$ has no cycle of length greater than 4 without chord, and so by Remark 5 the linear combinations of linear forms which come from $L_2$ are equal to $X_{F(u_j) {\setminus } F(u_i)}e_i-X_{F(u_i) {\setminus } F(u_j)}e_j$. $\square$

Theorem 3

Let $I=<u_1, \ldots , u_m>$ be a squarefree monomial ideal. If I has linear relations and $\overline{G_I}$ is a chordal graph, then I has linear quotients.

Proof

We proceed by induction on $|G(I)|=m$. The assertion is trivial for $m=1$ and $m=2$. If $m>2$, then by Lemma 1 there exists a t (assume $t=m$) such that $J=\langle u_1, \ldots ,u_{m-1} \rangle$ has linear relations. Induction hypothesis implies that J has linear quotients. Assume $u_1, \ldots , u_{m-1}$ is an admissible order for J. We show $u_1, \ldots , u_{m-1}, u_m$ is an admissible order for I. Let $j<m$, then by [8, Proposition 1.22] there exists a path $u_m=v_1, v_2, \ldots , v_r=u_j$ in $G_I$ such that $F(v_i) \subseteq F(u_m) \cup F(u_j)$ for all $1 \le i \le r$. If $\{l\}=F(v_2) {\setminus } F(u_m)$, then $l \in F(u_j) {\setminus } F(u_m)$. $\square$

As an immediate consequence of Theorem 3, we have:

Corollary 2

Let $I=\langle u_1, \ldots , u_m \rangle$ be a squarefree monomial ideal. If $\overline{G_I}$ is a chordal graph, then the following are equivalent:

• (a) I has linear relations;

• (b) I has a linear resolution;

• (c) I has linear quotients.

The following simple lemma is crucial for the proof of Theorem 4.

Lemma 2

Suppose I is a squarefree monomial ideal and $\{u_{i},u_{r} \}$ and $\{ u_{r},u_{j} \}$ are edges in $G_{I}$. If $\{u_{i},u_{j} \} \notin E(G_{I})$ and $F(u_{r}) {\setminus } F(u_{i}) =\{l\}$, then $l \in F(u_{j}) {\setminus } F(u_{i})$.

Proof

Since $\{u_{i},u_{r} \} \in E(G_{I})$, one has $F(u_{r}) {\setminus } F(u_{i}) =\{l\}$. Assume $F(u_{i})=A \cup \{t\}$ and $F(u_{r})=A \cup \{l\}$. If $l \notin F(u_{j})$, then $A \subseteq F(u_{j})$. Hence, $\{u_{i},u_{j} \} \in E(G_{I})$, a contradiction. $\square$

Theorem 4

Assume that the ordering $u_{m}, u_{m-1}, \ldots ,u_{1}$ is a perfect elimination ordering of $\overline{G_{I}}$. Let $u_{g_t}, u_{g_{t-1}},\ldots , u_{g_1}$ be the neighborhoods of $u_{m}$ in $\overline{G_I}$ and $t>1$. Set $V=\{u_{g_2-1},\ldots ,u_{g_1}, \ldots , u_{1} \}$. If $N_{G^{'}}(u_{g_1}) \ne \emptyset$, where $G^{'}$ is the induced subgraph of $G_I$ on V, then I has linear quotients.

Proof

Let u be a neighborhood of $u_{g_1}$. By Remark 2, u is a neighborhood of $u_{g_i}$ for $2 \le i \le t$. We order the elements of G(I) as follows: $u_{g_1}$, neighborhoods of $u_{g_1}$, $u_{g_2}$, neighborhoods of $u_{g_2}$ which are not appeared before, ..., $u_{g_t}$, neighborhoods of $u_{g_t}$ which are not appeared before, $u_{m}$, neighborhoods of $u_{m}$ which are not appeared before.

Assume that $v_{1}, \ldots , v_{m}$ come from the above ordering. In what follows, we show that this order is an admissible order. For $j<i$, we must find $k<i$ such that $F(v_k){\setminus } F(v_i)=\{l\}$ and $l\in F(v_j)$. Consider the following four cases:

• (i) $v_{i}=u_{g_{s}}$ and $v_{j}=u_{g_{l}}$ with $l<s$. Since u is a neighborhood of $u_{g_s}$ and $u_{g_l}$, if we set $v_k=u$, then we are done by Lemma 2.

• (ii) $v_{i}=u_{g_{s}}$ and $v_{j}$ is a neighborhood of $u_{g_{l}}$ with $l<s$. Remark 2 implies that $v_{j}$ is a neighborhood of $v_{i}$. Hence, take $v_k=v_j$.

• (iii) $v_{i}$ is a neighborhood of $u_{g_{s}}$ and $v_{j}$ is a neighborhood of $u_{g_{l}}$ with $l \le s$. Remark 2 implies that $v_{j}$ is a neighborhood of $u_{g_{s}}$. If $\{v_{i},v_{j} \} \in E(G_{I})$, then we take $v_k=v_j$. Otherwise, we take $v_k= u_{g_{s}}$ and the assertion follows from Lemma 2.

• (iv) $v_{i}$ is a neighborhood of $u_{g_{s}}$ and $v_{j}= u_{g_{l}}$ with $l \le s$. If $l=s$, then take $v_k=v_{j}$. So assume that $l<s$. If $\{v_i, u \} \in E(G_I)$, then take $v_k=u$ and the assertion follows from Lemma 2. If $\{v_i, u \} \notin E(G_I)$, then $F(v_{i})= A \cup \{r_1, r_2\}$ and $F(u)= A^{'} \cup \{r_3, r_4\}$. Remark 2 implies that $v_{i}$ and u are neighborhoods of $u_{m}$. It is easy to see that $A=A^{'} \subseteq F(u_{m})$. Hence, $F(u_{m})= A\cup \{p,q\}$ such that $p \in \{r_1, r_2\}$ and $q \in \{r_3, r_4\}$. Without loss of generality, we may assume $F(u_{m})= A \cup \{r_1, r_3\}$. Since $v_{i}$ and u are neighborhoods of $u_{g_{s}}$ and $\{u_{m},u_{g_{s}} \} \notin E(G_{I})$, one has $F(u_{g_{s}})= A \cup \{r_2, r_4\}$. Take $v_{k} =u_{g_{s}}$, then $F(v_{k}){\setminus } F(v_{i}) =\{r_4\}$. If $r_4 \notin F(v_{j})$, since $\lbrace u, v_{j} \rbrace \in E(G_{I})$ , one has $A \cup \lbrace r_3 \rbrace \subseteq F(v_{j})$. Therefore, $\lbrace u_{m}, v_{j} \rbrace \in E(G_{I})$, a contradiction.

$\square$