Squarefree monomial ideals with maximal depth

TL;DR

This paper classifies squarefree monomial ideals with maximal depth, including edge ideals of cycle graphs, transversal polymatroidal ideals, and high powers of connected bipartite graph ideals, revealing their structural properties.

Contribution

It provides a classification of squarefree monomial ideals with maximal depth, expanding understanding of their algebraic and combinatorial characteristics.

Findings

Classified edge ideals of cycle graphs with maximal depth

Identified transversal polymatroidal ideals with this property

Analyzed high powers of connected bipartite graph ideals

Abstract

Let $(R,\mm)$ be a Noetherian local ring and $M$ a finitely generated $R$-module. We say $M$ has maximal depth if there is an associated prime $\pp$ of $M$ such that $\depth M=\dim R/\pp$. In this paper we study squarefree monomial ideals which have maximal depth. Edge ideals of cycle graphs, transversal polymatroidal ideals and high powers of connected bipartite graph with this property are classified.

Equations

$$\mathrm{depth}M\le \min \{\dim R/\mathfrak{p}:\mathfrak{p}\in \mathrm{Ass}(M)\}.$$

$$\mathrm{depth}M\le \min \{\dim R/\mathfrak{p}:\mathfrak{p}\in \mathrm{Ass}(M)\},$$

$$\mathrm{depth}M=\mathrm{mdepth}M.$$

$$\mathrm{pd}M+\mathrm{depth}M=\mathrm{depth}R.$$

$$\mathrm{reg}{I}^{\vee }$$

$$\mathrm{depth}S/I=n-\mathrm{pd}S/I=n-\mathrm{reg}{I}^{\vee }=n-\mathrm{bight}I=\mathrm{mdepth}S/I,$$

$$\operatorname{depth}L_{n}=\left\{\begin{array}[]{ccc}\frac{n}{3}&&\text{if \; $n\equiv 0(\operatorname{mod}3)$},\\[4.30554pt] \frac{n+2}{3}&&\text{if \; $n\equiv 1(\operatorname{mod}3)$,}\\[4.30554pt] \frac{n+1}{3}&&\text{if \; $n\equiv 2(\operatorname{mod}3)$.}\end{array}\right.$$

$$C=\{1,3,4,6,7,9,10,\dots ,n-3,n-2,n\}$$

$${\mathfrak{p}}_C=(x_1,x_3,x_4,x_6,x_7,x_9,x_{10},\dots ,x_{n-3},x_{n-2},x_n)$$

$$C=\{2,3,5,6,8,9,\dots ,n-2,n-1\}$$

$${\mathfrak{p}}_C=(x_2,x_3,x_5,x_6,x_8,x_9,\dots ,x_{n-2},x_{n-1})$$

$$C=\{2,3,5,6,8,9,\dots ,n-3,n-2,n\}$$

$${\mathfrak{p}}_C=(x_2,x_3,x_5,x_6,x_8,x_9,\dots ,x_{n-3},x_{n-2},x_n)$$

$$\operatorname{depth}C_{n}=\left\{\begin{array}[]{ccc}\frac{n}{3}&&\text{if \; $n\equiv 0(\operatorname{mod}3)$},\\[4.30554pt] \frac{n-1}{3}&&\text{if \; $n\equiv 1(\operatorname{mod}3)$,}\\[4.30554pt] \frac{n+1}{3}&&\text{if \; $n\equiv 2(\operatorname{mod}3)$.}\end{array}\right.$$

$$C=\{i,i+1,i+3,i+4,i+6,i+7,\dots ,n-i-1,n-i\}$$

$${\mathfrak{p}}_C=(x_i,x_{i+1},x_{i+3},x_{i+4},x_{i+6},x_{i+7},\dots ,x_{n-i-1},x_{n-i})$$

$$C=\{i,i+1,i+3,i+4,\dots ,n+i-5,n+i-4,n+i-2\}$$

$${\mathfrak{p}}_C=(x_i,x_{i+1},x_{i+3},x_{i+4},\dots ,x_{n+i-5},x_{n+i-4},x_{n+i-2})$$

$$\mathrm{mdepth}{C}_n=n-(2n-1)/3=(n+1)/3=\mathrm{depth}{C}_n.$$

$$C=\{i,i+2,i+3,i+5,i+6,\dots ,n+i-5,n+i-4,n+i-2\}$$

$${\mathfrak{p}}_C=(x_i,x_{i+2},x_{i+3},x_{i+5},x_{i+6},\dots ,x_{n+i-5},x_{n+i-4},x_{n+i-2})$$

$$\mathrm{depth}{C}_n=\mathrm{depth}{C}_n\cup W(x_1),$$

$$0\to R/(I:x_{n+1})\to R/I\to R/(I+(x_{n+1}))\to 0,$$

$$R/(I:x_{n+1})\cong K[x_2,\dots ,x_n][x_{n+1}]/(x_2{x}_3,x_3{x}_4,\dots ,x_{n-1}{x}_n),$$

$$R/(I+(x_{n+1}))\cong S/J.$$

$$\mathrm{depth}K[x_2,\dots ,x_n]/(x_2{x}_3,x_3{x}_4,\dots ,x_{n-1}{x}_n)=((n-1)+1)/3=n/3.$$

$$\mathrm{depth}R/I\ge \min \{n/3+1,n/3\}=n/3.$$

$$C=\{n+1,2,3,5,6,\dots ,n-4,n-3,n-1,n\}$$

$${\mathfrak{p}}_C=(x_{n+1},x_2,x_3,x_5,x_6,\dots ,x_{n-4},x_{n-3},x_{n-1},x_n)$$

$$\mathrm{mdepth}R/I=(n+1)-\mathrm{bight}{\mathfrak{p}}_C=(n+1)-(2n+3)/3=n/3.$$

Full text

Introduction

Let $K$ be a field and $(R,{\mathfrak{m}})$ be a Noetherian local ring, or a standard graded $K$-algebra with graded maximal ideal ${\mathfrak{m}}$. Let $M$ be a finitely generated $R$-module. A basic fact in commutative algebra says that

[TABLE]

We set $\operatorname{mdepth}_{R}M=\min\{\dim R/{\mathfrak{p}}:{\mathfrak{p}}\in\operatorname{Ass}(M)\}$. For simplicity, we write $\operatorname{mdepth}M$ instead of $\operatorname{mdepth}_{R}M$. We say $M$ has maximal depth if the equality holds, i.e., $\operatorname{depth}M=\operatorname{mdepth}M.$ In other words, there is an associated prime ${\mathfrak{p}}$ of $M$ such that $\operatorname{depth}M=\dim R/{\mathfrak{p}}$. In this paper, we study squarefree monomial ideals with maximal depth.

Let $I\subset S=K[x_{1},\ldots,x_{n}]$ be a squarefree monomial ideal. We say $I$ has maximal depth if $S/I$ has maximal depth. We observe that, $I$ has maximal depth is equivalent to say that $\operatorname{reg}I^{\vee}$ is the maximum degree of the generators of $I^{\vee}$. This fact motivates us to work on squarefree monomial ideals with maximal depth. Here $I^{\vee}$ is the Alexander dual of $I$ and $\operatorname{reg}M$ denotes the regularity of a finitely generated graded $S$-module $M$.

Several authors have been working on this topic and some known results in this regards are as follows: If $I\subset S$ is a generic monomial ideal, then it has maximal depth, see [11, Theorem 2.2]. If a monomial ideal $I$ has maximal depth, then so does its polarization, see [5]. Algebraic properties and some classifications of modules with maximal depth are given in [12].

In [8], the depth of the line graph $L_{n}$ is explicitly computed. In Section 2, we compute the depth of the line graph $L_{n}$ in a different way. Our proof relies on the fact that trees, and line graphs in particular, have maximal depth, see Proposition 2.1.

In [8], the depth of the cycle graph $C_{n}$ of length $n$ is also computed. This number is independent of the characteristic of the chosen field. By using this result, we classify all cycle graphs $C_{n}$ which have maximal depth. In fact, $C_{n}$ has maximal depth if and only if $n\equiv 0(\operatorname{mod}3)$ or $n\equiv 2(\operatorname{mod}3)$, see Proposition 2.3.

Adding a whisker to $C_{n}$ at a vertex $x_{1}$ means adding a new vertex $x_{n+1}$ and the edge $\{x_{1},x_{n+1}\}$ to $C_{n}$. We denote by $C_{n}\cup W(x_{1})$ the graph obtained from $C_{n}$ by adding a whisker at $x_{1}$. By using Proposition 2.1, we show that $C_{n}$ and $C_{n}\cup W(x_{1})$ have the same depth as well as $C_{n}\cup W(x_{1})$ has maximal depth.

In Section 3, we consider the transversal polymatroidal ideals. A transversal polymatroidal ideal is an ideal $I$ of the form $I={\mathfrak{p}}_{F_{1}}\dots\mathfrak{p}_{F_{r}}$ where $F_{1},\dots,F_{r}$ is a collection of non-empty subsets of $[n]$ with $r\geq 1$. Here for a non-empty subset $F$ of $[n]$, we denote by ${\mathfrak{p}}_{F}$ the monomial prime ideal $(\{x_{i}:i\in F\})$. The depth of a transversal polymatroidal ideal is explicitly given in [7]. By applying this result, we classify all transversal polymatroidal ideals which have maximal depth. In fact, we prove the following: Let $I\subset S$ be a transversal polymatroidal ideal. Then, $I$ has maximal depth if and only $I$ is a product of monomial prime ideals such that at most one of the factors is not principal. In the following, we also classify ideals of Veronese type which have maximal depth.

In the final section, we consider $G$ to be a connected bipartite graph and $I$ its edge ideal. We show $I^{k}$ has maximal depth for $k\gg 0$ if and only if $G$ is a star graph.

1. Preliminaries

Let $K$ be a field and $(R,{\mathfrak{m}})$ be a Noetherian local ring, or a standard graded $K$-algebra with graded maximal ideal ${\mathfrak{m}}$. It is a classical fact that if $M\neq 0$ is an $R$-module, then

[TABLE]

see [1]. We set $\operatorname{mdepth}_{R}M=\min\{\dim R/{\mathfrak{p}}:{\mathfrak{p}}\in\operatorname{Ass}(M)\}$. For simplicity, we write $\operatorname{mdepth}M$ instead of $\operatorname{mdepth}_{R}M$. Thus $\operatorname{depth}M\leq\operatorname{mdepth}M\leq\dim M$. Observe that $\operatorname{depth}(M)=0$ if and only if $\operatorname{mdepth}(M)=0$. Thus, if $\operatorname{mdepth}M=1$, then $\operatorname{depth}M=1$.

Definition 1.1**.**

We say $M$ has maximal depth if the equality holds, i.e.,

[TABLE]

In other words, there is an associated prime ${\mathfrak{p}}$ of $M$ such that $\operatorname{depth}M=\dim R/{\mathfrak{p}}$.**

Some examples of modules with maximal depth property are as follows:

• •

Cohen–Macaulay modules have maximal depth because $\operatorname{depth}M=\dim R/{\mathfrak{p}}$ for every associated prime of $M$, see [13, Proposition 2.3.13].

• •

Sequentially Cohen-Macaulay modules have maximal depth, see [12, Proposition 1.4], see also [13, Theorem 6.4.23] where the ring $R$ is a polynomial ring.

• •

If $M$ is unmixed, then $M$ has maximal depth if and only if $M$ is Cohen–Macaulay.

Let $I\subset S=K[x_{1},\ldots,x_{n}]$ be a squarefree monomial ideal. Then $I=\bigcap_{j=1}^{m}{\mathfrak{p}}_{j}$ where each of the ${\mathfrak{p}}_{j}$ is a monomial prime ideal of $I$. The ideal $I^{\vee}$ which is minimally generated by the monomials $u_{j}=\prod_{x_{i}\in{\mathfrak{p}}_{j}}x_{i}$ is called the Alexander dual of $I$. As usual we denote by $\operatorname{reg}M$ the regularity of a finitely generated graded $S$-module $M$. We quote the following facts which for example can be found in [6].

Theorem 1.2**.**

(Terai) $\operatorname{reg}I^{\vee}=\operatorname{pd}S/I$.

Theorem 1.3**.**

(Auslander-Buchsbam formula) Let $M$ be a finitely generated $R$-module with $\operatorname{pd}M<\infty$. Then

[TABLE]

The big height of an ideal $J\subset S$, denoted by $\operatorname{bight}J$, is the maximum height of the minimal primes of $J$. The following simple fact motivates us to work on squarefree monomial ideals with maximal depth. We say $I\subset S$ has maximal depth if $S/I$ has maximal depth.

Proposition 1.4**.**

Let $I\subset S$ be a squarefree monomial ideal. Then, $I$ has maximal depth if and only if $\operatorname{reg}I^{\vee}$ is the maximum degree of the generators of $I^{\vee}$.

Proof.

Suppose $I$ has maximal depth. Hence

[TABLE]

Theorem 1.2 explains the first step in this sequence. Theorem 1.3 provides the second step. Our assumption implies the third step. The forth step follows from that fact that when $I$ is squarefree, the associated primes are the same as minimal primes containing $I$. Notice that the $\operatorname{bight}I$ is the maximum degree of the generators of $I^{\vee}$. Therefore, the conclusion follows. Conversely, suppose $\operatorname{reg}I^{\vee}$ is the maximum degree of the generators of $I^{\vee}$. By the same reasons as above, we have

[TABLE]

as desired.

We recall the following fact from [13, Lemma 2.3.8].

Lemma 1.5**.**

(Depth Lemma) If $0\rightarrow N\rightarrow M\rightarrow L\rightarrow 0$ is a short exact sequence of $R$-modules, then

• (a)

If $\operatorname{depth}(M)<\operatorname{depth}(L)$, then $\operatorname{depth}(N)=\operatorname{depth}(M)$.

• (b)

If $\operatorname{depth}(M)=\operatorname{depth}(L)$, then $\operatorname{depth}(N)\geq\operatorname{depth}(M)$.

• (c)

If $\operatorname{depth}(M)>\operatorname{depth}(L)$, then $\operatorname{depth}(N)=\operatorname{depth}(L)+1$.

2. Line and Cycle graphs

Let $G$ be a graph. The vertex set of $G$ will be denoted by $V(G)$ and will be the set $[n]=\{1,2,\ldots,n\}$. We denote the set of edges of $G$ by $E(G)$. We consider the edge ideal $I(G)$ which is generated by all monomials $x_{i}x_{j}$ with $\{i,j\}\in E(G)$. A subset $C\subset[n]$ is called a vertex cover of $G$ if $C\cap\{i,j\}\neq\emptyset$ for all edges $\{i,j\}$ of $G$. A vertex cover $C$ is called minimal if $C$ is a vertex of $G$, and no proper subset of $C$ is a vertex cover of $G$. A minimal vertex cover of $G$ is called maximum if it has maximum cardinality among the minimal vertex covers of $G$. Thus $\operatorname{bight}I(G)$ is the cardinality of the maximum minimal vertex covers of $G$.

It is well known that the minimal vertex covers of $G$ are the sets of generators of the minimal primes of $I(G)$. In fact, a subset $C=\{i_{1},\dots,i_{r}\}\subset[n]$ is a minimal vertex cover of $G$ if and only if ${\mathfrak{p}}_{C}=(x_{i_{1}},\dots,x_{i_{r}})$ is a minimal prime ideal of $I(G)$, see [6, Lemma 9.1.4].

The graph $G$ is called disconnected if $V(G)$ is the disjoint union of $W_{1}$ and $W_{2}$ and there is no edge $\{i,j\}$ of $G$ with $i\in W_{1}$ and $j\in W_{2}$. The graph $G$ is called connected if it is not disconnected. A graph which has no cycle and which is connected is called a tree.

For $n\geq 2$, we let $L_{n}$ denote the line graph on $n$ vertices. This is the graph with vertices $[n]$ and edges $\{j,j+1\}$ for all $j=1,\dots,n-1$. Hence, its edge ideal is $I(L_{n})=(x_{1}x_{2},x_{2}x_{3},\dots,x_{n-1}x_{n})$ in a polynomial ring with $n$ variables. In the following, we explicitly compute the depth of the line graph $L_{n}$. However, this is a known fact, see [8, Corollary 7.7.35] but here we prove it in a different way.

Notation: For any graph $G$, we write $\operatorname{depth}G$ for the depth of $S/I(G)$.

Proposition 2.1**.**

The depth of the line graph $L_{n}$ is independent of the characteristic of the chosen field and is

[TABLE]

Proof.

Notice that the line graph is a tree. Trees are sequentially Cohen-Macaulay, see [3]. As sequentially Cohen-Macaulay modules have maximal depth, all trees have maximal depth. In particular, $L_{n}$ has maximal depth for all $n$. Let $I=I(L_{n})$ be the edge ideal of $L_{n}$ in a polynomial ring $S$ with $n$ variables. We consider the following cases:

Case 1: $n\equiv 0(\operatorname{mod}3)$. We claim that the set

[TABLE]

is a maximum minimal vertex cover of $L_{n}$. A minimal vertex cover of $L_{n}$ cannot contain $3$ consecutive vertices because of minimality. This implies that if we divide the vertices of $L_{n}$ into blocks of $3$ vertices, then each block can have at most $2$ vertices in the cover. Therefore the cardinality of a minimal vertex cover can be at most $2n/3$. Thus

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=2n/3$. It follows that $\operatorname{mdepth}L_{n}=n-2n/3=n/3$ and hence $\operatorname{depth}L_{n}=n/3$.

Case 2: $n\equiv 1(\operatorname{mod}3)$. Hence $n-1\equiv 0(\operatorname{mod}3)$. We claim that the set

[TABLE]

is a maximum minimal vertex cover of $L_{n}$. In fact, the vertices of $L_{n}$ can be divided into blocks with $3$ vertices as well as one block with only $1$ vertex. Then each block of $3$ vertices can have at most $2$ vertices in the cover. The vertex in the block with one vertex need not to be in the cover. Therefore the cardinality of a minimal vertex cover can be at most $2(n-1)/3$. Hence

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=2(n-1)/3$. Consequently, $\operatorname{mdepth}L_{n}=n-2(n-1)/3=(n+2)/3$ and so $\operatorname{depth}L_{n}=(n+2)/3$.

Case 3: $n\equiv 2(\operatorname{mod}3)$. Hence, $n-2\equiv 0(\operatorname{mod}3)$. The set

[TABLE]

is a maximum minimal vertex cover of $L_{n}$. Indeed, the vertices of $L_{n}$ can be divided into blocks with $3$ vertices as well as only one block with $2$ vertex. Then each block of $3$ vertices can have at most $2$ vertices and the block of $2$ vertices can have at most $1$ vertex in the cover. Therefore the cardinality of a minimal vertex cover can be at most $2(n-2)/3+1=(2n-1)/3$. Hence

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=2(n-1)/3$. Consequently, $\operatorname{mdepth}L_{n}=n-(2n-1)/3=(n+1)/3$ and so $\operatorname{depth}L_{n}=(n+1)/3.$ We remark that the proof of proposition does not depend on the characteristic of the field $K$.

Let $C_{n}$ be a cycle graph of length $n$. We recall the following result from [8, Corollary 7.6.30].

Fact 2.2**.**

The depth of the cycle graph is independent of the characteristic of the chosen field and is

[TABLE]

In the following, we classify all cycle graphs which have maximal depth.

Proposition 2.3**.**

The cycle graph $C_{n}$ has maximal depth if and only if $n\equiv 0(\operatorname{mod}3)$ or $n\equiv 2(\operatorname{mod}3)$.

Proof.

Let $I=I(C_{n})$ be the edge ideal of $C_{n}$ in a polynomial ring $S$ with $n$ variables. We need to consider the following three cases.

Case 1: $n\equiv 0(\operatorname{mod}3)$. For the maximum minimal vertex covers of cycles one can use the line graphs. A similar argument as in the proof of Proposition 2.3 shows that, the cardinality of a minimal vertex cover of $C_{n}$ in this case can be at most $2n/3$. The set

[TABLE]

is a maximum minimal vertex cover of $C_{n}$ for all $i$. Thus,

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=2n/3$. Hence $\operatorname{mdepth}C_{n}=n-2n/3=n/3=\operatorname{depth}C_{n}.$ Fact 2.2 provides the last equality. Therefore, $C_{n}$ has maximal depth.

Case 2: $n\equiv 2(\operatorname{mod}3)$. A similar argument as in the proof of Proposition 2.3 shows that, the cardinality of a minimal vertex cover of $C_{n}$ in this case can be at most $2(n-2)/3+1=(2n-1)/3$. One observes that the set

[TABLE]

is a maximum minimal vertex cover of $C_{n}$ for all $i$. Hence

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=(2n-1)/3$. Consequently,

[TABLE]

Fact 2.2 explains the last equality.

Case 3: $n\equiv 1(\operatorname{mod}3)$. In this case, one has that, the cardinality of a minimal vertex cover of $C_{n}$ can be at most $2(n-1)/3$ and the set

[TABLE]

is a maximum minimal vertex cover of $C_{n}$ for all $i$. Hence

[TABLE]

is a minimal prime ideal of $I$ with maximum height and so $\operatorname{bight}I=2(n-1)/3$. Thus $\operatorname{mdepth}C_{n}=n-2(n-1)/3=(n+2)/3$. Fact 2.2 provides $\operatorname{depth}C_{n}=(n-1)/3$. Thus, $C_{n}$ has no maximal depth in this case.

Adding a whisker to $C_{n}$ at a vertex $x_{1}$ means adding a new vertex $x_{n+1}$ and the edge $\{x_{1},x_{n+1}\}$ to $C_{n}$. We denote by $C_{n}\cup W(x_{1})$ the graph obtained from $C_{n}$ by adding a whisker at $x_{1}$. Thus $I(C_{n}\cup W(x_{1}))=I(C_{n})+(x_{1}x_{n+1})$. In the following, by using Proposition 2.1, we show that $C_{n}$ and $C_{n}\cup W(x_{1})$ have the same depth as well as $C_{n}\cup W(x_{1})$ has maximal depth.

Proposition 2.4**.**

The following statements hold.

[TABLE]

and $C_{n}\cup W(x_{1})$ has maximal depth.

Proof.

We set $I(C_{n})=J$ and $I(C_{n}\cup W(x_{1}))=I$. Consider the exact sequence

[TABLE]

where $R=S[x_{n+1}]$. One has

[TABLE]

and

[TABLE]

We consider the following three cases:

Case 1: $n\equiv 0(\operatorname{mod}3)$. Thus $n-1\equiv 2(\operatorname{mod}3)$. By Proposition 2.1

[TABLE]

Hence $\operatorname{depth}R/(I:x_{n+1})=n/3+1$. Fact 2.2 provides $\operatorname{depth}S/J=n/3$. Thus by using (1) we have

[TABLE]

For computing $\operatorname{mdepth}R/I$ in this case, a similar argument as in the proof of Proposition 2.3 shows that, the cardinality of a minimal vertex cover of $C_{n}\cup W(x_{1})$ can be at most $(2n+3)/3$. One observes that the set

[TABLE]

is a maximum minimal vertex cover of $C_{n}\cup W(x_{1})$. Thus

[TABLE]

is a minimal prime ideal of $I$ with $\operatorname{bight}{\mathfrak{p}}_{C}=(2n+3)/3$. Hence

[TABLE]

Consequently,

[TABLE]

Thus, the results follow in this case.

Case 2: $n\equiv 1(\operatorname{mod}3)$. Thus $n-1\equiv 0(\operatorname{mod}3)$. By Proposition 2.1

[TABLE]

Hence $\operatorname{depth}R/(I:x_{n+1})=(n+2)/3$. Fact 2.2 explains $\operatorname{depth}S/J=(n-1)/3$. Thus by using (1) we have

[TABLE]

One observes that the cardinality of a minimal vertex cover of $C_{n}\cup W(x_{1})$ in this case can be at most $2(n+2)/3$ and the set

[TABLE]

is a maximum minimal vertex cover of $C_{n}\cup W(x_{1})$. Thus

[TABLE]

is a minimal prime ideal of $I$ with $\operatorname{bight}{\mathfrak{p}}_{C}=2(n+2)/3$. Hence

[TABLE]

We conclude that

[TABLE]

Therefore, the desired conclusions follow in this case.

Case 3: $n\equiv 2(\operatorname{mod}3)$. Thus $n-1\equiv 1(\operatorname{mod}3)$. By Proposition 2.1

[TABLE]

Hence $\operatorname{depth}R/(I:x_{n+1})=(n+4)/3$. Fact 2.2 provides $\operatorname{depth}S/J=(n+1)/3$. Thus by using (1) we have

[TABLE]

One observes that the cardinality of a minimal vertex cover of $C_{n}\cup W(x_{1})$ can be at most $2(n+1)/3$ and the set

[TABLE]

is a maximum minimal vertex cover of $C_{n}\cup W(x_{1})$. Thus

[TABLE]

is a minimal prime ideal of $I$ with $\operatorname{bight}{\mathfrak{p}}_{C}=2(n+1)/3$. Hence

[TABLE]

Consequently,

[TABLE]

Therefore, the desired conclusions follow in this case too.

We remark that the second part of Proposition 2.4 also follows from [4, Corollary 3.4] in a different way.

3. Transversal polymatroids and Ideals of Veronese type

In this section, we classify all transversal polymatroidal ideals and all ideals of Veronese type which have maximal depth. Let $F$ be a non-empty subset of $[n]$. We denote by ${\mathfrak{p}}_{F}$ the monomial prime ideal $(\{x_{i}:i\in F\})$. A transversal polymatroidal ideal is an ideal $I$ of the form $I={\mathfrak{p}}_{F_{1}}\dots\mathfrak{p}_{F_{r}}$ where $F_{1},\dots,F_{r}$ is a collection of non-empty subsets of $[n]$ with $r\geq 1$. Let $G_{I}$ be the graph with vertex set $\{1,\dots,r\}$ and for which $\{i,j\}$ is an edge of $G_{I}$ if and only if $F_{i}\cap F_{j}\neq\emptyset$. We recall the following fact from [7, Theorem 4.12].

Fact 3.1**.**

Let $I={\mathfrak{p}}_{F_{1}}\dots\mathfrak{p}_{F_{r}}\subset S$ be a transversal polymatroidal ideal. Then

[TABLE]

where by $c(G_{I})$ we denote the number of connected components of the graph $G_{I}$.**

Let $\mathcal{H}$ be a subgraph of $G_{I}$. We associate the prime ideal ${\mathfrak{p}}_{\mathcal{H}}=\sum_{i\in\mathcal{V}(\mathcal{H})}{\mathfrak{p}}_{F_{i}}$. We denote by $\operatorname{Ass}(I)$ the set of associated prime ideals of $R/I$. The set associated primes of $R/I$ is explicitly described in [7, Theorem 4.7] as follows.

Fact 3.2**.**

Let $I\subset S$ be a transversal polymatroidal ideal. Then

[TABLE]

In the following we characterize all transversal polymatroidal ideals which have maximal depth.

Proposition 3.3**.**

Let $I={\mathfrak{p}}_{F_{1}}\dots\mathfrak{p}_{F_{r}}\subset S$ be a transversal polymatroidal ideal. The following conditions are equivalent:

• (a)

$I$* has maximal depth;*

• (b)

$I$* is a product of monomial prime ideals such that at most one of the factors is not principal.*

Proof.

(a)${}\Rightarrow{}$(b): We may assume that $\cup_{i=1}^{r}F_{i}=[n]$. Let $k=c(G_{I})$ and $G_{1},\dots,G_{k}$ be the connected components of $G_{I}$. Fact 3.1 provides $\operatorname{depth}S/I=k-1$. We denote by $I_{1},\dots,I_{k}$ the transversal polymatroidal ideals for which the associated graphs are the connected components of $G_{I}$. Hence $I=I_{1}\dots I_{k}=I_{1}\cap\dots\cap I_{k}$, since the ideals $I_{j}$ are generated in pairwise disjoint sets of variables. Thus $\operatorname{Ass}(I)=\bigcup_{i=1}^{k}\operatorname{Ass}(I_{i})$. We may assume that $1\leq l_{1}\leq\dots\leq l_{k}$ where for all $j$ we have $l_{j}=|\bigcup_{i\in\mathcal{V}(G_{j})}F_{i}|$. Note that $l_{1}+\dots+\l_{k}=n$. In view of Fact 3.2, we have $\operatorname{mdepth}S/I=n-l_{k}$. Since $I$ has maximal depth, it follows that $n-l_{k}=k-1$, and hence $l_{1}+\dots+l_{k-1}=k-1$. Consequently, $I=(x_{1})\dots(x_{k-1})(x_{k},\dots,x_{n})$, as desired.

(b)${}\Rightarrow{}$(a): If $I$ is a product of monomial prime ideals such that all the factors are principal, then $S/I$ is Cohen–Macaulay and hence $I$ has maximal depth. Thus, we may assume that $I=(x_{1})\dots(x_{k-1})(x_{k},\dots,x_{n})$. As

[TABLE]

we have $\operatorname{mdepth}S/I=n-(n-k+1)=k-1$. The ideal $I$ is a transversal polymatroidal ideal. It follows from Theorem 3.1 that $\operatorname{depth}S/I=k-1$. Here $c(G_{I})=k$ and $|\cup_{i=1}^{r}F_{i}\mid=n$. Therefore, $I$ has maximal depth.

As a consequence one has

Corollary 3.4**.**

Let $I\subset S$ be the intersection of monomial prime ideals in pairwise disjoint sets of variables. Then $I$ has maximal depth if and only if $I$ is a product of monomial prime ideals such that at most one of the factors is not principal.

One of the most distinguished polymatroidal ideals is the ideal of Veronese type. Let $S=K[x_{1},\dots,x_{n}]$ and fix positive integers $d$ and $a_{1},\dots,a_{n}$ with $1\leq a_{1}\leq\dots\leq a_{n}\leq d$. The ideal of Veronese type of $S$ indexed by $d$ and $(a_{1},\dots,a_{n})$ is the ideal $I_{d;a_{1},\dots,a_{n}}$ which is generated by those monomials $u=x_{1}^{u_{1}}\dots x_{n}^{u_{n}}$ of $S$ of degree $d$ with $u_{i}\leq a_{i}$ for each $1\leq i\leq n$.

The set of associated prime ideals and depth of the ideal of Veronese type are described in [7, Proposition 5.2 and Corollary 5.7] as follows

[TABLE]

and

[TABLE]

Proposition 3.5**.**

The ideal of Veronese type has maximal depth if and only if there exists a ${\mathfrak{p}}_{F}\in\operatorname{Ass}(S/I)$ where $|F|=\sum_{i=1}^{n}a_{i}-(d-1)$.

Proof.

In view of (2), ${\mathfrak{p}}_{F}$ has the maximum height if $|F|=\sum_{i=1}^{n}a_{i}-(d-1)$. Thus $\operatorname{mdepth}S/I=n-|F|=n+d-1-\sum_{i=1}^{n}a_{i}$ which is the same as $\operatorname{depth}S/I$ by (3). Therefore, the conclusion follows.

Here is an example

Example 3.6**.**

Consider $I=I_{5;1,2,3}\subset S=K[x_{1},x_{2},x_{3}].$ Then $I=(x_{1}^{2}x_{2}^{2}x_{3},x_{1}^{3}x_{2}x_{3},x_{1}^{3}x_{2}^{2})$. Formula (2) yields

[TABLE]

As ${\mathfrak{p}}_{F}\in\operatorname{Ass}(S/I)$ with $|F|=2$, $I$ has maximal depth and $\operatorname{depth}S/I=\operatorname{mdepth}S/I=1$. **

4. Powers of ideals

A subset $D\subset[n]$ is called an independent set of $G$ if $D$ contains no set $\{i,j\}$ which is an edge of $G$. The graph $G$ is called bipartite if $V(G)$ is the disjoint union of $V_{1}$ and $V_{2}$ such that $V_{1}$ and $V_{2}$ are independent sets. The bipartite graph $G$ is called a complete bipartite graph if $\{i,j\}\in E(G)$ for all $i\in V_{1}$ and $j\in V_{2}$.

Proposition 4.1**.**

Let $G$ be a complete bipartite graph on the vertex set $V$ with bipartition $V=V_{1}\cup V_{2}$ where $V_{1}=\{v_{1},\ldots,v_{n}\}$ and $V_{2}=\{w_{1},\ldots,w_{m}\}$ with $1\leq n\leq m$. Then $G$ has maximal depth if and only if $n=1$, i.e., $G$ is a star graph.

Proof.

The edge ideal of $G$ is ${\mathfrak{p}}_{1}\cap{\mathfrak{p}}_{2}$ where ${\mathfrak{p}}_{1}=(x_{1},\dots,x_{n})$ and ${\mathfrak{p}}_{2}=(y_{1},\dots,y_{m})$. We set $S=K[x_{1},\dots,x_{n},y_{1},\dots,y_{m}]$ and $R=S/({\mathfrak{p}}_{1}\cap{\mathfrak{p}}_{2})$. Consider the exact sequence $0\rightarrow S/({\mathfrak{p}}_{1}\cap{\mathfrak{p}}_{2})\rightarrow S/{\mathfrak{p}}_{1}\oplus S/{\mathfrak{p}}_{2}\rightarrow S/({\mathfrak{p}}_{1}+{\mathfrak{p}}_{2})\rightarrow 0.$ Since $\operatorname{depth}S/({\mathfrak{p}}_{1}+{\mathfrak{p}}_{2})=0$, it follows from Lemma 1.5(Depth lemma) that $\operatorname{depth}R=1$. On the other hand, $\operatorname{Ass}(R)=\{{\mathfrak{p}}_{1},{\mathfrak{p}}_{2}\}$. It follows that $\operatorname{mdepth}R=n$. Consequently, $\operatorname{mdepth}R-\operatorname{depth}R=n-1$. Therefore, the conclusion follows.

Remark 4.2**.**

In Proposition 4.1, we showed $\operatorname{mdepth}R-\operatorname{depth}R=n-1$. Thus, the difference between $\operatorname{depth}$ and $\operatorname{mdepth}$ can be any number. **

In the following we classify all connected bipartite graph such that $I^{k}$ has maximal depth for all $k\gg 0$.

Proposition 4.3**.**

Let $G$ be a connected bipartite graph and $I=I(G)$ its edge ideal. Then $I^{k}$ has maximal depth for $k\gg 0$ if and only if $G$ is a star graph.

Proof.

Suppose $I^{k}$ has maximal depth for $k\gg 0$. By [6, Corollary 10.3.18], we have $\operatorname{depth}S/I^{k}=1$ for $k\gg 0$. Hence $\operatorname{mdepth}S/I^{k}=1$ for $k\gg 0$. As $G$ is bipartite, we have $\operatorname{Ass}(I)=\operatorname{Ass}(I^{k})$ for all $k$, see [9]. It follows that $\operatorname{mdepth}S/I=1$ and hence $\operatorname{depth}S/I=1.$ Thus there exists a minimal vertex cover $F$ of $G$ such that $|[n]\setminus F|=1$. Therefore, $G$ is a star graph.

Now, suppose $G$ is a star graph and $J$ is its edge ideal. By Proposition 4.1 we have $\operatorname{depth}S/J=\operatorname{mdepth}S/J=1$. As $G$ is bipartite, we have $\operatorname{mdepth}S/J^{k}=1$ for all $k$. It follows that $\operatorname{depth}S/J^{k}=1$ for all $k$ and hence $J^{k}$ has maximal depth for $k\gg 0$.

Remark 4.4**.**

Let $I$ be an ideal in a Noetherian ring $R$. Brodmann [3] showed that $\operatorname{Ass}(I^{k})=\operatorname{Ass}(I^{k+1})$ for all $k\gg 0$. The ideal $I$ for which $\operatorname{Ass}(I^{k})\subset\operatorname{Ass}(I^{k+1})$ for all $k\geq 1$, is said to satisfy the persistence property. Edge ideals of graphs and polymatroidal ideals have persistence property, see [7], [10]. In this case, we have $\operatorname{mdepth}S/I^{k+1}\leq\operatorname{mdepth}S/I^{k}$ for all $k$ and say the ideal $I$ has non-increasing mdepth.**

Acknowledgment

I would like to thank Jürgen Herzog for helpful discussions on this work. I would also like to thank the referee for helpful comments on this article.