Stability of Depth and Cohen-Macaulayness of Integral Closures of Powers of Monomial Ideals

TL;DR

This paper establishes an upper bound on the stability index for the depth of integral closures of powers of monomial ideals and classifies when these closures become Cohen-Macaulay for large powers.

Contribution

It provides a new upper bound on the depth stability index of integral closures of monomial ideals and characterizes ideals with eventually Cohen-Macaulay integral closures.

Findings

Upper bound on $ar{ ext{dstab}}(I)$ in terms of ring dimension and generating degree

Classification of monomial ideals with Cohen-Macaulay integral closures for large powers

Depth of $R/ar{I^n}$ stabilizes beyond a certain power

Abstract

Let $I$ be a monomial ideal $I$ in a polynomial ring $R = k[x_1,...,x_r]$. In this paper we give an upper bound on $\overline{\dstab} (I)$ in terms of $r$ and the maximal generating degree $d(I)$ of $I$ such that $\depth R/\overline{I^n}$ is constant for all $n\geqslant \overline{\dstab}(I)$. As an application, we classify the class of monomial ideals $I$ such that $\overline{I^n}$ is Cohen-Macaulay for some integer $n\gg 0$.

Equations

$$x^n+a_1{x}^{n-1}+\cdots +a_{n-1}x+a_n=0,$$

$$E(\overline{I})=NP(I)\cap {\mathbb{N}}^r=\{\mathbf{\alpha }\in {\mathbb{N}}^r\mid {\mathbf{x}}^{n\mathbf{\alpha }}\in I^n\text{ for some }n⩾1\}.$$

$$NP(I^n)=nNP(I)=n\mathrm{conv}\{E(I)\}+{\mathbb{R}}_{+}^r\text{for all}n⩾1.$$

$$d(I):=\max \{{\alpha }_1+\cdots +{\alpha }_r|{\mathbf{x}}^{\mathbf{\alpha }}\in G(I)\},$$

$$\{\mathbf{x}\in {\mathbb{R}}^r\mid ⟨{\mathbf{a}}_j,\mathbf{x}⟩⩾b_j,j=1,\dots ,q\},$$

$$a_{ji}⩽s_{j}d(I{)}^{s_j-1}\text{ for all }i=1,\dots ,r,$$

$$\left|\begin{array}[]{cccc}x_{1}&\cdots&x_{s}&1\\ \alpha_{11}&\cdots&\alpha_{1s}&1\\ \vdots&\vdots&\vdots&\vdots\\ \alpha_{s1}&\cdots&\alpha_{ss}&1\end{array}\right|=0.$$

$$a^{\prime}_{1}=\left|\begin{array}[]{cccc}\alpha_{12}&\cdots&\alpha_{1s}&1\\ \vdots&\vdots&\vdots&\vdots\\ \alpha_{s2}&\cdots&\alpha_{ss}&1\end{array}\right|,$$

$$(-1)^{s+1}a^{\prime}_{1}=\left|\begin{array}[]{ccc}\alpha_{22}&\cdots&\alpha_{2s}\\ \alpha_{32}&\cdots&\alpha_{3s}\\ \vdots&\vdots&\vdots\\ \alpha_{s2}&\cdots&\alpha_{ss}\end{array}\right|-\left|\begin{array}[]{ccc}\alpha_{12}&\cdots&\alpha_{1s}\\ \alpha_{32}&\cdots&\alpha_{3s}\\ \vdots&\vdots&\vdots\\ \alpha_{s2}&\cdots&\alpha_{ss}\end{array}\right|+\cdots+(-1)^{s-1}\left|\begin{array}[]{ccc}\alpha_{12}&\cdots&\alpha_{1s}\\ \alpha_{22}&\cdots&\alpha_{2s}\\ \vdots&\vdots&\vdots\\ \alpha_{s-1,2}&\cdots&\alpha_{s-1,s}\end{array}\right|.$$

$$(\det (c_{ij}){)}^2⩽{\Pi }_{i=1}^{s-1}({\Sigma }_{j=1}^{s-1}|c_{ij}{|}^2)⩽{\Pi }_{i=1}^{s-1}({\Sigma }_{j=1}^{s-1}|c_{ij}|{)}^2⩽d(I{)}^{2(s-1)}.$$

$$\mathfrak{m}\in \mathrm{Ass}R/\overline{I^n}\text{ for all }n⩾(r-1)rd(I{)}^{r-2}.$$

$$\{\mathbf{x}\in {\mathbb{R}}^r\mid ⟨{\mathbf{a}}_j,\mathbf{x}⟩=b_j,j=1,\dots ,q\},$$

$$a_{ji}⩽(r-1)d(J{)}^{r-2}\text{ for all }j⩽q-1\text{ and }i⩽r.$$

$$⟨{\mathbf{a}}_j,\mathbf{\alpha }⟩>b_j\text{ for all }j⩽q-1.$$

$$⟨{\mathbf{a}}_q,\mathbf{\beta }⟩=m⟨{\mathbf{a}}_q,\mathbf{\alpha }⟩-⟨{\mathbf{a}}_q,{\mathbf{e}}_r⟩=m{b}_q-a_{qr}<m{b}_q.$$

$$\mathbf{\beta }+{\mathbf{e}}_i\in NP(J^m)\text{ for all }i=1,\dots ,r.$$

$$\begin{array}[]{ll}\left<\mathbf{a}_{q},{\boldsymbol{\beta}}+\mathbf{e}_{i}\right>&=\left<\mathbf{a}_{q},m{\boldsymbol{\alpha}}-\mathbf{e}_{r}+\mathbf{e}_{i}\right>\\ &=m\left<\mathbf{a}_{q},{\boldsymbol{\alpha}}\right>-\left<\mathbf{a}_{q},\mathbf{e}_{r}\right>+\left<\mathbf{a}_{q},\mathbf{e}_{i}\right>\\ &=mb_{q}-a_{qr}+a_{qi}\\ &\geqslant mb_{q}\ \ (\text{since }a_{qr}=\min\{a_{q1},\ldots,a_{qr}\}).\end{array}$$

$$\begin{array}[]{ll}\left<\mathbf{a}_{j},{\boldsymbol{\beta}}+\mathbf{e}_{i}\right>&=\left<\mathbf{a}_{j},m{\boldsymbol{\alpha}}-\mathbf{e}_{r}+\mathbf{e}_{i}\right>\\ &=(r-1)d(I)^{r-2}\left<\mathbf{a}_{j},r{\boldsymbol{\alpha}}\right>-\left<\mathbf{a}_{j},\mathbf{e}_{r}\right>+\left<\mathbf{a}_{j},\mathbf{e}_{i}\right>\\ &\geqslant(r-1)d(I)^{r-2}(rb_{j}+1)-a_{jr}+a_{ji}\\ &=mb_{j}+((r-1)d(I)^{r-2}-a_{jr})+a_{ji}\\ &\geqslant mb_{j}\ \ \text{(by (\ref{ET1}))}.\end{array}$$

$$\Delta (I)=\{\{i_1,...,i_s\}\subseteq [r]|x_{i_1}\cdots x_{i_s}\notin \sqrt{I}\},$$

$${\Delta }_{\mathbf{\alpha }}(I)=\{F\subseteq [r]\setminus C{S}_{\mathbf{\alpha }}|{\mathbf{x}}^{\mathbf{\alpha }}\notin I{R}_{F\cup C{S}_{\mathbf{\alpha }}}\}.$$

$${\dim }_k{\tilde{H}}_{t-|C{S}_{\mathbf{\alpha }}|-1}({\Delta }_{\mathbf{\alpha }}(\overline{I});k)={\dim }_k{H}_{\mathfrak{m}}^t(R/\overline{I}{)}_{\mathbf{\alpha }}\ne 0.$$

$${\Delta }_{\mathbf{\alpha }}(\overline{I})=\{F\in \Delta \mid {\mathbf{x}}^{\mathbf{\alpha }}\notin \overline{I{R}_{F\cup C{S}_{\mathbf{\alpha }}}}\}=\{F\in \Delta \mid {\mathbf{x}}^{n\mathbf{\alpha }}\notin \overline{(I{R}_{F\cup C{S}_{\mathbf{\alpha }}}{)}^n}\}={\Delta }_{n\mathbf{\alpha }}(\overline{I^n}).$$

$${\dim }_k{H}_{\mathfrak{m}}^t(R/\overline{I^n}{)}_{n\mathbf{\alpha }}={\dim }_k{\tilde{H}}_{t-|C{S}_{n\mathbf{\alpha }}|-1}({\Delta }_{n\mathbf{\alpha }}(\overline{I^n});k)={\dim }_k{\tilde{H}}_{t-|C{S}_{\mathbf{\alpha }}|-1}({\Delta }_{\mathbf{\alpha }}(\overline{I});k)\ne 0.$$

$$\underset{m\to \infty }{\lim }\mathrm{depth}R/J^m=\dim R-\ell (J).$$

$$\underset{n\to \infty }{\lim }\mathrm{depth}R/\overline{I^n}=\underset{m\to \infty }{\lim }\mathrm{depth}R/J^m=\dim R-\ell (J).$$

$$n_0(I):=\{\begin{array}{ll}1& \text{ if }\ell (I)⩽2,\\ \ell (I)(\ell (I)-1)d(I{)}^{\ell (I)-2}& \text{ if }\ell (I)>2.\end{array}$$

$$\overline{\mathrm{astab}}(I)=\min \{m|\mathrm{Ass}R/\overline{I^n}=\mathrm{Ass}R/\overline{I^m}\text{for all }n⩾m\}.$$

$${\Delta }_{\mathbf{\alpha }}(\overline{I^n})=⟨[r]\setminus \mathrm{supp}({\mathbf{a}}_j)\mid j\in \{1,\dots ,q\}\text{ and }⟨{\mathbf{a}}_j,\mathbf{\alpha }⟩<n{b}_j⟩.$$

$$\{\mathbf{x}\in {\mathbb{R}}^r\mid ⟨{\mathbf{a}}_j,\mathbf{x}⟩⩾n{b}_j,j=1,\dots ,q\}.$$

$$m>\max \{n{b}_j-⟨{\mathbf{a}}_j,\mathbf{\alpha }⟩\mid 1\le j⩽q\},$$

Full text

Introduction

Let $R=k[x_{1},\ldots,x_{r}]$ be a polynomial ring over a field $k$ and $\mathfrak{a}$ a homogeneous ideal in $R$. It was shown by Brodmann [2] that $\operatorname{depth}R/\mathfrak{a}^{n}$ is constant for $n\gg 0$. The smallest integer $m>0$ such that $\operatorname{depth}R/\mathfrak{a}^{n}=\operatorname{depth}R/\mathfrak{a}^{m}$ for all $n\geqslant m$ is called the index of depth stability and is denoted by $\operatorname{dstab}(\mathfrak{a})$. Since the behavior of depth function $\operatorname{depth}R/\mathfrak{a}^{n}$ is quite mysterious (see [7, 5]), it is of great interest to bound $\operatorname{dstab}(\mathfrak{a})$ in terms of $r$ and $\mathfrak{a}$. However, until now this problem is only solved for a few classes of monomial ideals (see, e.g., [7, 8, 20]). The bound obtained in [20] for ideals generated by square-free monomials of degree two is rather small and optimal. However, this problem is still open for a general square-free monomial ideal.

In this direction, it is also of interest to consider similar problems for other powers of $\mathfrak{a}$. In [10] together with Kimura and Terai we were able to solve the problem of bounding the index of depth stability for symbolic powers of square-free monomial ideals. In this paper we are interested in bounding the index of depth stability $\overline{\operatorname{dstab}}(\mathfrak{a})$ for integral closures, which is defined as the smallest integer $m>0$ such that $\operatorname{depth}R/\overline{\mathfrak{a}^{n}}=\operatorname{depth}R/\overline{\mathfrak{a}^{m}}$ for all $n\geqslant m$. Like in the case of ordinary powers, $\overline{\operatorname{dstab}}(\mathfrak{a})$ is well-defined. We only consider the problem for monomial ideals $I$. In this context one can use geometry and convex analysis to describe the integral closures of $I^{n}$ (see Definition 1.1 and some properties after it). Then one can use Takayama’s formula (see Lemma 1.4) to compute the local cohomology modules of $R/\overline{I^{n}}$. This approach was successfully applied in several papers (see, e.g., [10, 11, 19]). In particular, one can show that in the class of monomial ideals the behavior of the function $\operatorname{depth}R/\overline{I^{n}}$ is much better than that of $\operatorname{depth}R/I^{n}$: it is “quasi-decreasing” (see Lemma 1.5) while the function $\operatorname{depth}R/I^{n}$ can be any convergent non-negative numerical function (see [5]). Our main result is Theorem 2.3, where we can give an upper bound on $\overline{\operatorname{dstab}}(I)$ in terms of $r$ and the maximal generating degree $d(I)$ of $I$ for any monomial ideal $I$. Although our bound is very big, an example shows that an upper bound must depend on $d(I)$, and in the worst case must be an exponential function of $r$.

In order to bound $\overline{\operatorname{dstab}}(I)$ we have to study the index of stability for the associated primes on $R/\overline{I^{n}}$. This in some sense corresponds the zero depth case and was firstly done in [19]. In this paper we can improve the main result of [19] by giving an essentially better bound, see Theorem 1.7.

As an application we classify all monomial ideals such that $R/\overline{I^{n}}$ is a Cohen-Macaulay ring for all $n\geqslant 1$ (or for some fixed $n=n_{0}\gg 0$). It turns out that only equimultiple ideals have this property, see Theorem 3.1. In the case of square-free monomial ideals, we can then derive a criterion for the Cohen-Macaulayness of $R/\overline{I^{n}}$ for some fixed $n\geqslant 3$, see Theorem 3.7. This criterion is exactly the one for the Cohen-Macaulayness of $R/{I^{n}}$ given in [17, Theorem 1.2].

The paper is organized as follows. In Section 1 we study the stability of associated primes and give an upper bound on $\overline{\operatorname{astab}}(I)$ of a monomial ideal. In Section 2 we prove the main Theorem 2.3. The study of Cohen-Macaulay property of $R/\overline{I^{n}}$ is done in the last section.

1. Stability of associated primes

Let $R:=k[x_{1},\ldots,x_{r}]$ be a polynomial ring over a field $k$ with the maximal homogeneous ideal $\mathfrak{m}=(x_{1},\ldots,x_{r})$. Throughout this paper, let $I$ be a proper monomial ideal in $R$. Let $\mathbb{N},\ \mathbb{R},\ \mathbb{R}_{+}$ be the set of non-negative integers, real numbers and non-negative real numbers, respectively. For a vector ${\boldsymbol{\alpha}}=(\alpha_{1},\ldots,\alpha_{r})\in\mathbb{N}^{r}$, we denote by $\mathbf{x}^{{\boldsymbol{\alpha}}}=x_{1}^{\alpha_{1}}\cdots x_{r}^{\alpha_{r}}$.

The integral closure of an arbitrary ideal $\mathfrak{a}$ of $R$ is the set of elements $x$ in $R$ that satisfy an integral relation

[TABLE]

where $a_{i}\in\mathfrak{a}^{i}$ for $i=1,\ldots,n$. This is an ideal and is denoted by $\overline{\mathfrak{a}}$. The integral closure of a monomial ideal $I$ is a monomial ideal as well. We can geometrically describe $\overline{I}$ by using its Newton polyhedron.

Definition 1.1**.**

Let $I$ be a monomial ideal of $R$. We define

1. (1)

For a subset $A\subseteq R$, the exponent set of $A$ is $E(A):=\{{\boldsymbol{\alpha}}\mid\mathbf{x}^{{\boldsymbol{\alpha}}}\in A\}\subseteq\mathbb{N}^{r}$. 2. (2)

The Newton polyhedron of $I$ is $NP(I):=\operatorname{conv}\{E(I)\}$, the convex hull of the exponent set of $I$ in the space $\mathbb{R}^{r}$.

The following results are well-known (see [14]):

[TABLE]

[TABLE]

Let $G(I)$ denote the minimal generating system of monomials of $I$ and

[TABLE]

the maximal generating degree of $I$. Let $\mathbf{e}_{1},...,\mathbf{e}_{r}$ be the canonical basis of $\mathbb{R}^{r}$. The first part of the following result is [19, Lemma 6]. It gives more precise information on the coefficients of defining equations of supporting hyperplanes of $NP(I)$.

Lemma 1.2**.**

The Newton polyhedron $NP(I)$ is the set of solutions of a system of inequalities of the form

[TABLE]

such that each hyperplane with the equation $\left<\mathbf{a}_{j},\mathbf{x}\right>=b_{j}$ defines a facet of $NP(I)$, which contains $s_{j}$ affinely independent points of $E(G(I))$ and is parallel to $r-s_{j}$ vectors of the canonical basis. Furthermore, we can choose $\mathbf{0}\neq\mathbf{a}_{j}\in\mathbb{N}^{r},b_{j}\in\mathbb{N}$ for all $j=1,...,q$; and if we write $\mathbf{a}_{j}=(a_{j1},\ldots,a_{jr})$, then

[TABLE]

where $s_{j}$ is the number of non-zero coordinates of $\mathbf{a}_{j}$.

Proof.

The first part of the lemma is [19, Lemma 6]. Moreover, it also claims that $\mathbf{a}_{j}\in\mathbb{R}_{+}^{r}$ and $b\in\mathbb{R}_{+}$. For the second part, let $H$ be a hyperplane which defines a facet of $NP(I)$. W.l.o.g, we may assume that $H$ is defined by $s$ affinely independent points ${\boldsymbol{\alpha}}_{1},...,{\boldsymbol{\alpha}}_{s}\in E(G(I))$ and is parallel to $r-s$ vectors $\mathbf{e}_{s+1},\ldots,\mathbf{e}_{r}$. Then the defining equation of $H$ can be written as

[TABLE]

Expanding this determinant in the first row, we get: $a^{\prime}_{1}x_{1}+\cdots+a^{\prime}_{s}x_{s}=b^{\prime}$, where $a^{\prime}_{i}$ are the $(1,i)$-cofactor for $i=1,\ldots,s$ and $b^{\prime}$ is the $(1,s+1)$-cofactor of this determinant. Clearly, $a^{\prime}_{1},\ldots,a^{\prime}_{s},b^{\prime}\in\mathbb{Z}$. Note that we may take $a_{i}=|a^{\prime}_{i}|$ and $b=|b^{\prime}|$. Expanding the determinant

[TABLE]

in the last column, we get

[TABLE]

Let $\det(c_{ij})$ be a determinant in the above sum. By Hadamard’s inequality, we have

[TABLE]

Hence $a_{1}=|a^{\prime}_{1}|\leqslant sd(I)^{s-1}$. Similarly, $a_{i}\leqslant sd(I)^{s-1}$ for $i=2,\ldots,r$, as required. ∎

The following lemma is a crucial result in the study of the stability of $\operatorname{Ass}(R/\overline{I^{n}})$.

Lemma 1.3**.**

Let $I$ be a monomial ideal in $R$ with $r>2$. If $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{s}}$ for some $s\geqslant 1$, then

[TABLE]

Proof.

Let $m:=(r-1)rd(I)^{r-2}$. Since the sequence $\{\operatorname{Ass}R/\overline{I^{n}}\}_{n\geqslant 1}$ is increasing by [6, Proposition 16.3], it suffices to show that $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{m}}$.

As $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{s}}$, by [19, Lemma 13], there is a supporting hyperplane of $NP(I)$, say $H$, of the form $\left<\mathbf{a},\mathbf{x}\right>=b$ such that all coordinates of $\mathbf{a}$ are positive. By Lemma 1.2, this hyperplane passes through $r$ affinely independent points of $E(G(I))$, say ${\boldsymbol{\alpha}}_{1},\ldots,{\boldsymbol{\alpha}}_{r}$. Let $J:=(\mathbf{x}^{{\boldsymbol{\alpha}}_{1}},\ldots,\mathbf{x}^{{\boldsymbol{\alpha}}_{r}})$. Clearly, $H$ is still a supporting plane of $NP(J)$. Again by Lemma 1.2, the Newton polyhedron $NP(J)$ can be represented by a system of inequalities

[TABLE]

where $\mathbf{0}\neq\mathbf{a}_{j}\in\mathbb{N}^{r}$ and $b_{j}\in\mathbb{N}$. Let $H_{j}=\{\mathbf{x}\in\mathbb{R}^{r}\mid\left<\mathbf{a}_{j},\mathbf{x}\right>=b_{j}\}$ for $j=1,\ldots,q$. We may assume that $q$ is minimal and $H_{q}=H$. Since $J$ is generated by exactly $r$ monomials and $q$ is taken to be minimal, by Lemma 1.2, each hyperplane $H_{j}$, where $j\leqslant q-1$, must be parallel to at least one of the vectors $\mathbf{e}_{1},\ldots,\mathbf{e}_{r}$. Hence, by the second statement of Lemma 1.2, we may assume that

[TABLE]

Consider the barycenter ${\boldsymbol{\alpha}}:=\frac{1}{r}({\boldsymbol{\alpha}}_{1}+\cdots+{\boldsymbol{\alpha}}_{r})$ of the simplex $[{\boldsymbol{\alpha}}_{1},\ldots,{\boldsymbol{\alpha}}_{r}]$. Then ${\boldsymbol{\alpha}}$ is a relative interior point of the facet $H_{q}\cap NP(J)$ of $NP(J)$. Therefore, ${\boldsymbol{\alpha}}$ does not lie in $H_{j}$ for all $j=1,\ldots,q-1$, and so

[TABLE]

Next, we may assume that $a_{qr}=\min\{a_{q1},\ldots,a_{qr}\}>0$. Let ${\boldsymbol{\beta}}:=m{\boldsymbol{\alpha}}-\mathbf{e}_{r}$. Then ${\boldsymbol{\beta}}=(r-1)d(I)^{r-2}({\boldsymbol{\alpha}}_{1}+\cdots+{\boldsymbol{\alpha}}_{r})-\mathbf{e}_{r}\in\mathbb{Z}^{r}$. Since ${\boldsymbol{\alpha}}_{1},...,{\boldsymbol{\alpha}}_{r}\in H_{q}$ are affinely independent and $a_{q1},...,a_{qr}>0$, there exists $j\leqslant r$ such that ${\boldsymbol{\alpha}}_{jr}>0$, whence ${\boldsymbol{\alpha}}_{jr}\geqslant 1$. Hence ${\boldsymbol{\beta}}\in\mathbb{N}^{r}$. Moreover,

[TABLE]

Therefore ${\boldsymbol{\beta}}\notin NP(J^{m})$ and also ${\boldsymbol{\beta}}\notin NP(I^{m})$ (recall that $H=H_{q}$).

On the other hand, we claim that

[TABLE]

Indeed, for $i=r$, ${\boldsymbol{\beta}}+\mathbf{e}_{r}=m{\boldsymbol{\alpha}}\in mNP(J)=NP(J^{m})$. For $i\leqslant r-1$, we have

[TABLE]

Let $j\leqslant q-1$. Since $r{\boldsymbol{\alpha}}={\boldsymbol{\alpha}}_{1}+\cdots+{\boldsymbol{\alpha}}_{r}\in NP(J^{r})\cap\mathbb{N}^{r}$, by (1.4), we have $\left<\mathbf{a}_{j},r{\boldsymbol{\alpha}}\right>>rb_{j}$, which implies $\left<\mathbf{a}_{j},r{\boldsymbol{\alpha}}\right>\geqslant rb_{j}+1$. Hence

[TABLE]

This completes the proof of (1.5).

Since $NP(J^{m})=mNP(J)\subseteq mNP(I)=NP(I^{m})$, ${\boldsymbol{\beta}}+\mathbf{e}_{i}\in NP(I^{m})$, whence $\mathbf{x}^{{\boldsymbol{\beta}}}x_{i}\in\overline{I^{m}}$. As shown above, ${\boldsymbol{\beta}}\not\in NP(I^{m})$. Therefore, $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{m}}$, as required. ∎

A main tool in the study of the set of associated primes and the depth of rings is using local cohomology modules. In the setting of monomial ideals, one often uses a generalized version of a Hochster’s formula given by Takayama in [16]. Let us recall this formula here.

Since $R/I$ is an $\mathbb{N}^{r}$-graded algebra, $H_{\mathfrak{m}}^{i}(R/I)$ is an $\mathbb{Z}^{r}$-graded module over $R$. For every degree ${\boldsymbol{\alpha}}\in\mathbb{Z}^{r}$ we denote by $H_{\mathfrak{m}}^{i}(R/I)_{{\boldsymbol{\alpha}}}$ the ${\boldsymbol{\alpha}}$-component of $H_{\mathfrak{m}}^{i}(R/I)$.

Let $\Delta(I)$ denote the simplicial complex corresponding to the Stanley-Reisner ideal $\sqrt{I}$, i.e.

[TABLE]

where $[r]$ denotes the set $\{1,2,...,r\}$. For every ${\boldsymbol{\alpha}}=(\alpha_{1},\ldots,\alpha_{r})\in\mathbb{Z}^{r}$, we define its co-support to be the set $CS_{{\boldsymbol{\alpha}}}:=\{i\ |\ \alpha_{i}<0\}$. For a subset $F$ of $[r]$, let $R_{F}:=R[x_{i}^{-1}\ |\ i\in F]$. Set

[TABLE]

We set $\widetilde{H}_{i}(\emptyset;k)=0$ for all $i$, $\widetilde{H}_{i}(\{\emptyset\};k)=0$ for all $i\neq-1$, and $\widetilde{H}_{-1}(\{\emptyset\};k)=k$. Thanks to [4, Lemma 1.1] we may formulate Takayama’s formula as follows.

Lemma 1.4**.**

([16, Theorem 2.2])* $\dim_{k}H_{\mathfrak{m}}^{i}(R/I)_{{\boldsymbol{\alpha}}}=\dim_{k}\widetilde{H}_{i-|CS_{{\boldsymbol{\alpha}}}|-1}(\Delta_{{\boldsymbol{\alpha}}}(I);k).$*

As an immediate consequence of this result is the following“quasi-decreasing” property of the depth function $\operatorname{depth}R/\overline{I^{n}}$. We don’t know if this property holds for an arbitrary homogeneous ideal.

Lemma 1.5**.**

For any monomial ideal $I$ of $R$, we have

1. (1)

$\operatorname{depth}R/\overline{I^{m}}\geqslant\operatorname{depth}R/\overline{I^{mn}}\text{ for all }m,n\geqslant 1$. 2. (2)

$\lim_{n\rightarrow\infty}\operatorname{depth}R/\overline{I^{n}}=\dim R-\ell(I),$* where $\ell(I)$ denotes the analytic spread of $I$.*

Proof.

1. Replacing $I^{m}$ by $J$, it suffices to prove the statement for $m=1$. Let $t:=\operatorname{depth}R/\overline{I}$. Then we must have $H^{t}_{\mathfrak{m}}(R/\overline{I})_{{\boldsymbol{\alpha}}}\neq 0$ for some ${\boldsymbol{\alpha}}\in\mathbb{Z}^{r}$. By Lemma 1.4,

[TABLE]

For $n\geqslant 1$, we have $CS_{n{\boldsymbol{\alpha}}}=CS_{{\boldsymbol{\alpha}}}$ and

[TABLE]

The middle equality follows from (1.1). Together with Equation (1.7) and Lemma 1.4, this fact implies that

[TABLE]

This means $\operatorname{depth}R/\overline{I^{n}}\leqslant t$.

2. Let $J:=\overline{I^{r-1}}$. By [21, Theorem 7.29], $J$ is torsion-free. Thus, by [3, Proposition $3.3$], we have

[TABLE]

For each $m\geqslant 1$, by [21, Corollary 7.60], we have $J^{m}=\overline{I^{m(r-1)}}$. Hence

[TABLE]

Note that $\ell(J)=\ell(\overline{I^{r-1}})=\ell(I^{r-1})=\ell(I)$, so the desired equality follows. ∎

Let $F$ be a subset of $[r]$. Put $R[F]=k[x_{i}\mid i\notin F]$ and denote by $I[F]$ the ideal of $R[F]$ obtained from $I$ by setting $x_{i}=1$ for all $i\in F$. Then $I[F]R=IR_{F}\cap R$. If $F=\{i\}$ for some $i\in[r]$, then we write $R[i]$ and $I[i]$ instead of $R[\{i\}]$ and $I[\{i\}]$ respectively.

Remark 1.6**.**

Let $I$ be a monomial ideal in $R$. Then,

1. (1)

Using (1.1) it is easy to see that $\overline{I^{n}}[F]=\overline{I[F]^{n}}$ for any $n\geqslant 1$ (cf. [11, Lemma 4.6]). 2. (2)

If $I$ is Cohen-Macaulay, then $I[F]$ is Cohen-Macaulay.

We can now give an improvement of the main result, Theorem 16, in [19].

Theorem 1.7**.**

Let $I$ be a monomial ideal of $R$ and

[TABLE]

Then, $\operatorname{Ass}R/\overline{I^{n}}=\operatorname{Ass}R/\overline{I^{n_{0}(I)}}$ for all $n\geqslant n_{0}(I)$.

Proof.

Fix an index $i\leqslant r$. It is well known that the analytic spread of $I$ is equal to the minimal number of generators of a minimal reduction of $I$. Since $I[i]$ is obtained from $I$ by setting $x_{i}=1$, this implies that $\ell(I)\geqslant\ell(I[i])$ and $d(I[i])\leqslant d(I)$. Hence $n_{0}(I[i])\leqslant n_{0}(I)$. Using this remark, Remark 1.6(1) and Lemma 1.5, we can prove the theorem by induction on $r$. The proof is similar to that of [19, Theorem 16], so we omit details here. ∎

Remark. Set

[TABLE]

It can be called the index of stability for the associated primes of $R/\overline{I^{n}}$. An example given in [19, Proposition 17] shows that an upper bound on $\overline{\operatorname{astab}}(I)$ must be of the order $d(I)^{r-2}$, provided that $r$ is fixed. The coefficient of $d(I)^{r-2}$ in the upper bound given in [19, Theorem 16] is $r2^{r-1}$.

2. Stability of Depth

In this section we study the stability index of the depth function $\operatorname{depth}R/\overline{I^{n}}$. It is clear that a simplicial complex $\Delta$ is defined by the set of its maximal faces, say $F_{1},...,F_{s}$. In this case we write $\Delta=\left<F_{1},...,F_{s}\right>$. Keeping the notations in Lemma 1.2, we set $\operatorname{supp}(\mathbf{a}_{j}):=\{i\mid a_{ji}\neq 0\}$. We can describe $\Delta_{{\boldsymbol{\alpha}}}(\overline{I^{n}})$ as follows.

Lemma 2.1**.**

For any ${\boldsymbol{\alpha}}\in\mathbb{N}^{r}$ and $n\geqslant 1$, we have

[TABLE]

Proof.

Let $F\in\Delta_{{\boldsymbol{\alpha}}}(\overline{I^{n}})$. We may assume that $F=\{s+1,\ldots,r\}$ for some $0\leqslant s\leqslant r$. By Lemma 1.2 and (1.2), we can deduce that $NP(I^{n})$ is the set of solutions of the system

[TABLE]

Since $CS_{{\boldsymbol{\alpha}}}=\emptyset$, $\mathbf{x}^{{\boldsymbol{\alpha}}}\notin\overline{I^{n}}R_{F}$ if and only if $\mathbf{x}^{{\boldsymbol{\alpha}}}\mathbf{x}^{{\boldsymbol{\gamma}}}\notin\overline{I^{n}}$ for any monomial $\mathbf{x}^{{\boldsymbol{\gamma}}}\in k[x_{s+1},\ldots,x_{r}]$. Taking $\mathbf{x}^{{\boldsymbol{\gamma}}}=x_{s+1}^{m}\cdots x_{r}^{m}$, where

[TABLE]

is fixed, it implies that there is $1\leq p\leqslant q$ such that

[TABLE]

Assume that there is $i\geqslant s+1$ such that $a_{pi}>0$. Then

[TABLE]

a contradiction. Hence $a_{p(s+1)}=\cdots=a_{pr}=0$, whence $F\subseteq[r]\setminus\operatorname{supp}(\mathbf{a}_{p})$ . Then $\left<\mathbf{a}_{p},{\boldsymbol{\alpha}}\right>=\left<\mathbf{a}_{p},{\boldsymbol{\alpha}}+m(\mathbf{e}_{s+1}+\mathbf{e}_{r})\right><nb_{p}$.

Conversely, assume that there is $j\leqslant q$ such that $F\subseteq[r]\setminus\operatorname{supp}(\mathbf{a}_{j})$, i.e. $a_{j(s+1)}=\cdots=a_{jr}=0$, and $\left<\mathbf{a}_{j},{\boldsymbol{\alpha}}\right><nb_{j}$. Then for all monomials $\mathbf{x}^{{\boldsymbol{\gamma}}}\in k[x_{s+1},\ldots,x_{r}]$, we have $\left<\mathbf{a}_{j},{\boldsymbol{\alpha}}+{\boldsymbol{\gamma}}\right>=\left<\mathbf{a}_{j},{\boldsymbol{\alpha}}\right><nb_{j}.$ By (1.1) and Lemma 1.2, this implies $\mathbf{x}^{{\boldsymbol{\alpha}}}\mathbf{x}^{{\boldsymbol{\gamma}}}\notin\overline{I^{n}}$. From (1.6), we get that $F\in\Delta_{{\boldsymbol{\alpha}}}(\overline{I^{n}})$. This completes the proof of the lemma. ∎

The following lemma is the main step in the proof of Theorem 2.3

Lemma 2.2**.**

Let $m\geqslant 1$ and $t:=\operatorname{depth}R/\overline{I^{m}}$. Assume that $H_{\mathfrak{m}}^{t}(R/\overline{I^{m}})_{{\boldsymbol{\beta}}}\neq 0$ for some ${\boldsymbol{\beta}}\in\mathbb{N}^{r}$. If $r\geqslant 3$, then

[TABLE]

Proof.

For simplicity, set $n^{*}:=r(r^{2}-1)r^{r/2}(r-1)^{r}d(I)^{(r-2)(r+1)}$. We keep the notations in Lemma 1.2.

Assume that $\operatorname{supp}(\mathbf{a}_{j})=[r]$ for some $1\leqslant j\leqslant q$. By [19, Lemma 14] we have $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{n}}$ for all $n\gg 0$. By Lemma 1.3, it yields $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{n}}$ for all $n\geqslant n^{*}$. Thus, $\operatorname{depth}R/\overline{I^{n}}=0$ for $n\geqslant n^{*}$, and the lemma holds in this case.

We now assume that $\operatorname{supp}(\mathbf{a}_{j})\neq[r]$ for all $j=1,\ldots,q$, i.e. the number of non-zero coordinates of $\mathbf{a}_{j}$ is strictly less than $r$. By Lemma 1.2, we have

[TABLE]

Assume that

[TABLE]

and

[TABLE]

for some $0\leqslant p\leqslant q$. Then, by Lemma 2.1,

[TABLE]

By Lemma 1.4, we have

[TABLE]

Hence $\Delta_{{\boldsymbol{\beta}}}(\overline{I^{m}})$ is not acyclic. In particular, $p\geqslant 1$. For each $n\geqslant 1$, put

[TABLE]

and

[TABLE]

It is clear that $C_{n}=nC_{1}$. By Lemma 2.1, $C_{n}\cap\mathbb{N}^{r}\subseteq\Gamma(\overline{I^{n}})$.

Assume that $C_{n}\cap\mathbb{N}^{r}\neq\emptyset$. Then for any ${\boldsymbol{\alpha}}\in C_{n}\cap\mathbb{N}^{r}$, by Lemma 1.4, we have

[TABLE]

whence $\operatorname{depth}R/\overline{I^{n}}\leqslant t$.

Thus, in order to complete the proof of the lemma, it remains to show that $C_{n}\cap\mathbb{N}^{r}\neq\emptyset$ for any $n\geqslant n^{*}$. Fix such an integer $n$.

Since ${\boldsymbol{\beta}}\in C_{m}=mC_{1}$, $C_{1}\neq\emptyset$. First, we prove that $C_{1}$ is bounded in $\mathbb{R}^{r}$. Assume that $a_{ji}=0$ for some $1\leqslant i\leqslant r$ and for all $j=1,\ldots,p$. Then, for any $s\gg 0$, by Formula (2.2) we get that ${\boldsymbol{\beta}}+s\mathbf{e}_{i}\in C_{m}$, which implies $\Delta_{{\boldsymbol{\beta}}+s\mathbf{e}_{i}}(\overline{I^{m}})=\Delta_{{\boldsymbol{\beta}}}(\overline{I^{m}})$. Again by Lemma 1.4, we have

[TABLE]

This contradicts the Artiness of $H_{\mathfrak{m}}^{t}(R/\overline{I^{m}})$. Hence, for each $i\leqslant r$, there is $j_{i}\leqslant p$ such that $a_{j_{i}i}\geqslant 1$. Let $\mathbf{y}=(y_{1},\ldots,y_{r})$ be an arbitrary point of $C_{1}\subseteq\mathbb{R}_{+}^{r}$ . Then for each $i\leqslant r$, we have $y_{i}\leqslant a_{j_{i}i}y_{i}\leqslant\left<\mathbf{a}_{j_{i}},\mathbf{y}\right><b_{j_{i}}$. This implies that $C_{1}$ is bounded and so is $C_{n}$.

Let $\overline{C}_{n}$ be the closure of $C_{n}$ in $\mathbb{R}^{r}$ with respect to the usual Euclidean topology. Then $\overline{C_{n}}$ is bounded as well. Moreover,

[TABLE]

and hence $\overline{C}_{n}$ is a polytope.

We next claim that $\overline{C}_{1}$ is full dimensional. Indeed, for any $\mathbf{y}\in C_{1}$, by Formula (2.2) we can choose a real number $\varepsilon>0$ such that for all real numbers $\varepsilon_{1},\ldots,\varepsilon_{r}$ with $0\leqslant\varepsilon_{1},\ldots,\varepsilon_{r}\leqslant\varepsilon$, we have $\mathbf{y}+\varepsilon_{1}\mathbf{e}_{1}+\cdots+\varepsilon_{r}\mathbf{e}_{r}\in C_{1}$. This means that the parallelotope $[y_{1},y_{1}+\varepsilon]\times\cdots\times[y_{r},y_{r}+\varepsilon]\subseteq C_{1}$ , and thus $\overline{C}_{1}$ is full dimensional in $\mathbb{R}^{r}$, as claimed.

Since the polytope $\overline{C}_{1}$ is full dimensional, by the Decomposition Theorem for polyhedra (see [15, Corollary 7.1.b]), we can find $r+1$ vertices, say ${\boldsymbol{\alpha}}_{0},\ldots,{\boldsymbol{\alpha}}_{r}$, of the polytope $\overline{C}_{1}$, which are affinely independent. Let ${\boldsymbol{\alpha}}=\frac{1}{r+1}({\boldsymbol{\alpha}}_{0}+\cdots+{\boldsymbol{\alpha}}_{r})$ be the barycenter of the $r$-simplex $[{\boldsymbol{\alpha}}_{0},{\boldsymbol{\alpha}}_{1},\ldots,{\boldsymbol{\alpha}}_{r}]\subseteq\overline{C_{1}}$.

For each $i\leqslant r$, set $\lambda_{i}=\lceil\alpha_{i}\rceil-\alpha_{i}\geqslant 0$, where $\lceil\alpha_{i}\rceil$ is the least integer which is bigger than or equal $\alpha_{i}$. Then $\lambda_{1}+\cdots+\lambda_{r}<r$ and ${\boldsymbol{\gamma}}:=n{\boldsymbol{\alpha}}+\lambda_{1}\mathbf{e}_{1}+\cdots+\lambda_{r}\mathbf{e}_{r}\in\mathbb{N}^{r}$. In order to show $C_{n}\cap\mathbb{N}^{r}\neq\emptyset$, it suffices to show that ${\boldsymbol{\gamma}}\in C_{n}.$

Since ${\boldsymbol{\alpha}}\in\overline{C_{1}}$, by Formula (2.3), we have

[TABLE]

for all $l=p+1,\ldots,q$.

Now, fix an index $j\leqslant p$. Since ${\boldsymbol{\alpha}}_{0},{\boldsymbol{\alpha}}_{1},\ldots,{\boldsymbol{\alpha}}_{r}$ are affinely independent in $\mathbb{R}^{r}$, there is at least one point not lying in the hyperplane $\left<\mathbf{a}_{j},\mathbf{x}\right>=b_{j}$. We may assume that ${\boldsymbol{\alpha}}_{0}$ is such a point. From Formula (2.3) we then have

[TABLE]

Since ${\boldsymbol{\alpha}}_{0}$ is a vertex of the polytope $\overline{C}_{1}$, by [15, Formula 23 in Page 104], ${\boldsymbol{\alpha}}_{0}$ can be represented as the unique solution of a system of linear equations of the form:

[TABLE]

where $|S|=r$. By Cramer’s rule we have $\alpha_{0i}=\delta_{i}/{\delta}$ for all $i\leqslant r$, where $\delta,\ \delta_{1},\ldots,\ \delta_{r}\in\mathbb{N}$ and $\delta$ is the absolute value of the determinant of this system of linear equations. In particular, $\delta{\boldsymbol{\alpha}}_{0}\in\mathbb{N}^{r}$. Using the inequalities (2.1) and Hadamard’s inequality applied to $\delta$, we get

[TABLE]

By (2.4) we have $\left<\mathbf{a}_{j},\delta{\boldsymbol{\alpha}}_{0}\right><\delta b_{j}$, whence $\left<\mathbf{a}_{j},\delta{\boldsymbol{\alpha}}_{0}\right>\leqslant\delta b_{j}-1$ because $\mathbf{a}_{j},\ \delta{\boldsymbol{\alpha}}_{0}\in\mathbb{N}^{r}$. Let $c=n/(r+1)\delta$, then by (2.5), $c\geqslant r(r-1)d(I)^{r-2}$. We then have

[TABLE]

Hence

[TABLE]

So $\left<\mathbf{a}_{j},{\boldsymbol{\gamma}}\right>\leqslant nb_{j}-1$, for all $j\leqslant p$. This means that ${\boldsymbol{\gamma}}\in C_{n}$, as required. ∎

We are now in position to prove the main result of this section.

Theorem 2.3**.**

Let $I$ be a monomial ideal of $R$. Let

[TABLE]

Then, $\operatorname{depth}R/\overline{I^{n}}=\dim R-\ell(I)$ for all $n\geqslant n_{1}(I)$.

Proof.

We prove the theorem by induction on $r$. If $r\leqslant 2$, and $I\neq 0$, then $\operatorname{depth}R/\overline{I^{n}}=0,1$, and $\operatorname{depth}R/\overline{I^{n}}=0$ if and only if $\mathfrak{m}\in\operatorname{Ass}(R/\overline{I^{n}})$. Since $\operatorname{Ass}(R/\overline{I^{n}})$ is constant for all $n\geqslant 1$ in this case (by [12, Proposition 16]), we get

[TABLE]

and the theorem follows from Lemma 1.5(2).

Assume that $r\geqslant 3$. By virtue of Lemma 1.5(2) and symmetry, it suffices to show that

[TABLE]

for any $m,n\geqslant n_{1}(I)$. Let $t:=\operatorname{depth}R/\overline{I^{m}}$. As $H^{t}_{\mathfrak{m}}(R/\overline{I^{m}})\neq 0$, by Lemma 1.4, there is ${\boldsymbol{\beta}}\in\mathbb{Z}^{r}$ such that

[TABLE]

In particular, $\Delta_{{\boldsymbol{\beta}}}(\overline{I^{m}})\neq\emptyset$. If $CS_{{\boldsymbol{\beta}}}=\emptyset$, i.e., ${\boldsymbol{\beta}}\in\mathbb{N}^{r}$, then (2.6) follows from Lemma 2.2.

We now assume that $CS_{{\boldsymbol{\beta}}}\neq\emptyset$. Without loss of generality, we may assume that $CS_{{\boldsymbol{\beta}}}=\{s+1,\ldots,r\}$ for some integer $0\leqslant s\leqslant r$. If $s=0$, i.e. $CS_{{\boldsymbol{\beta}}}=[r]$, then $\Delta_{{\boldsymbol{\beta}}}(\overline{I^{m}})=\{\emptyset\}$. By Lemma 1.4, it follows that $H^{0}_{\mathfrak{m}}(R/\overline{I^{m}})=0$, which is equivalent to $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{m}}$. Since $r\geqslant 3$, $n\geqslant n_{1}(I)>(r-1)rd(I)^{r-2}$, $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{n}}$ by Lemma 1.3. Hence, $\operatorname{depth}R/\overline{I^{n}}=0=\operatorname{depth}R/\overline{I^{m}}$ in this case.

Assume that $s\geqslant 1$. Let $R^{\prime}:=k[x_{1},\ldots,x_{s}]=R[\{s+1,...,r\}]$ (in the notation before Remark 1.6) and $I^{\prime}:=I[\{s+1,...,r\}]\subseteq R^{\prime}$. Let ${\boldsymbol{\beta}}=(\beta_{1},\ldots,\beta_{r})$ and ${\boldsymbol{\beta}}^{\prime}:=(\beta_{1},\ldots,\beta_{s})\in\mathbb{N}^{s}$. Then, by Formula (1.6), $\Delta_{{\boldsymbol{\beta}}^{\prime}}(\overline{I^{\prime m}})=\Delta_{{\boldsymbol{\beta}}}(\overline{I^{m}})$. Let $\mathfrak{n}:=(x_{1},\ldots,x_{s})$ be the maximal homogeneous ideal of $R^{\prime}$. Using (2.7) and Lemma 1.4 we obtain

[TABLE]

Hence $I^{\prime}\neq R^{\prime}$ and $\operatorname{depth}R^{\prime}/\overline{I^{\prime m}}\leqslant t-|CS_{{\boldsymbol{\beta}}}|$, or equivalently $\operatorname{depth}R/\overline{I^{m}}\geqslant|CS_{{\boldsymbol{\beta}}}|+\operatorname{depth}R^{\prime}/\overline{I^{\prime m}}$. On the other hand, by [10, Lemma 1.3], we have $\operatorname{depth}R/\overline{I^{m}}\leqslant|CS_{{\boldsymbol{\beta}}}|+\operatorname{depth}R^{\prime}/\overline{I^{\prime m}}$ and $\operatorname{depth}R/\overline{I^{n}}\leqslant|CS_{{\boldsymbol{\beta}}}|+\operatorname{depth}R^{\prime}/\overline{I^{\prime n}}$. Hence

[TABLE]

and (noticing that $n_{1}(I^{\prime})\leqslant n_{1}(I)$, since $d(I^{\prime})\leqslant d(I)$ and $s\leqslant r$)

[TABLE]

∎

Remark. Set

[TABLE]

One can call it the index of depth stability for integral closures. Then Theorem 2.3 says that $\overline{\operatorname{dstab}}(I)\leq n_{1}(I)$. It seems that this bound is too big. However, an example given in [19, Proposition 17] shows that an upper bound on $\overline{\operatorname{dstab}}(I)$ must be at least of the order $d(I)^{r-2}$.

3. Cohen-Macaulay property

In this section we apply results in previous sections to study the Cohen-Macaulayness of integeral closures of powers of monomial ideals. We say that $I$ is equimultiple if $\ell(I)=\operatorname{ht}(I)$. Note that, by [1, Theorem 2.3], we can compute $\ell(I)$ in terms of geometry of $NP(I)$.

[TABLE]

Therefore, the condition $I$ being equimultiple is independent on the characteristic of the base field $k$.

Theorem 3.1**.**

Let $I$ be a monomial ideal of $R$. The following conditions are equivalent

1. (1)

$R/\overline{I^{n}}$* is a Cohen-Macaulay ring for all $n\geqslant 1$,* 2. (2)

$R/\overline{I^{n}}$* is a Cohen-Macaulay ring for some $n\geqslant n_{1}(I)$, where $n_{1}(I)$ is defined in Theorem 2.3,* 3. (3)

$I$* is an equimultiple ideal of $R$.*

Proof.

If $R/\overline{I^{n}}$ is a Cohen-Macaulay ring for some $n\geq n_{1}(I)$, then by Theorem 2.3 we have $\dim R/\overline{I^{n}}=\operatorname{depth}R/\overline{I^{n}}=\dim R-\ell(I).$ On the other hand, $\dim R/\overline{I^{n}}=\dim R/I=\dim R-\operatorname{ht}(I)$. Hence, $\ell(I)=\operatorname{ht}(I)$.

Conversely, assume that $\ell(I)=\operatorname{ht}(I)$. Then,

[TABLE]

For all $n\geqslant 1$, by Lemma 1.5 applied to $m\gg 0$, we have

[TABLE]

Hence,

[TABLE]

which means that $R/\overline{I^{n}}$ is a Cohen-Macaulay ring. ∎

In the rest of this section we will improve the above theorem for the class of square-free monomial ideals. We need some auxiliary results.

Lemma 3.2**.**

Let $I$ be an unmixed monomial ideal and $I=Q_{1}\cap\cdots\cap Q_{s}$ be an irredundant primary decomposition of $I$. Assume that $\overline{I^{n}}$ is unmixed for some $n\geqslant 1$. Then

[TABLE]

Proof.

We prove by induction on $s$ and on $r$. If $s=1$ (which also includes the case $r=1$) there is nothing to prove. Assume that $s\geqslant 2$ and $r\geqslant 2$. Since $I$ is unmixed, $\sqrt{Q_{j}}\neq\mathfrak{m}$ for all $j\leqslant s$. For each $i\leqslant r$, let

[TABLE]

If there is $j_{0}\leq s$ such that $j_{0}\not\in\cup_{i=1}^{r}A_{i}$, then $x_{1},...,x_{r}\in\sqrt{Q_{j_{0}}}$, a contradiction. Hence $\cup_{i=1}^{r}A_{i}=[s]$ and $I=\bigcap_{i=1}^{r}(\cap_{j\in A_{i}}Q_{j}))$, where we set $\cap_{j\in A_{i}}Q_{j}=R$ if $A_{i}=\emptyset$. Moreover, using Remark 1.6(1) and the induction hypothesis on $r$, we may assume that $|A_{i}|<s$ for all $i\leqslant r$.

It is well-known that one can get a primary decomposition of a monomial ideal $\mathfrak{a}\subset R$ by repeated application of the formula $(B,uv)=(B,u)\cap(B,v)$, where $B$ is a set of monomials and $u,v$ are monomials having no common variable. Based on this fact, it is immediate to see that one can get a primary decomposition of $\mathfrak{a}[i]R$ from that of $\mathfrak{a}$ by deleting those primary components whose associated prime ideals contain $x_{i}$ (recall that $\mathfrak{a}[i]$ is obtained from $G(I)$ by setting $x_{i}=1$).

Using this remark we see that $I[i]$ is an unmixed ideal for all $i\leqslant r$. Moreover $I[i]=\cap_{j\in A_{i}}Q_{j}$ and $I=\cap_{i=1}^{r}I[i]$.

By Remark 1.6(1) $\overline{I[i]^{n}R}=(\overline{I^{n}})[i]R$ and $\overline{I^{n}}=\cap_{i=1}^{r}\overline{I[i]^{n}R}$. Since $\overline{I^{n}}$ is unmixed, by the above remark, all $\overline{I[i]^{n}R}$ are unmixed ideals. Since $|A_{i}|<s$, by the induction hypothesis on $s$, we also get $\overline{I[i]^{n}R}=\cap_{j\in A_{i}}\overline{Q_{j}^{n}}$.

Let $J:=\overline{Q_{1}^{n}}\cap\cdots\cap\overline{Q_{s}^{n}}$. This is an unmixed ideal. Hence, as shown above, $J=\bigcap_{i=1}^{r}J[i]R$ and $J[i]R=\cap_{j\in A_{i}}\overline{Q_{j}^{n}}=\overline{I[i]^{n}R}$ . Then $J=\cap_{i=1}^{r}\overline{I[i]^{n}R}=\overline{I^{n}}$, as required. ∎

Lemma 3.3**.**

Let $y$ be a new variable and $S=R[y]=k[x_{1},...,x_{r},y]$. Then, for every $n\geqslant 1$ we have

1. (1)

$\overline{(I,y)^{n}}=\sum_{i=0}^{n}y^{i}\overline{I^{n-i}}S$. 2. (2)

$\overline{(I,y)^{n}}$* is Cohen-Macaulay if and only if $\overline{I^{i}}$ is Cohen-Macaulay for all $i\leqslant n$.*

Proof.

(1) The inclusion $\sum_{i=0}^{n}y^{i}\overline{I^{n-i}}S\subseteq\overline{(I,y)^{n}}$ follows from the fact that $\overline{I_{1}}\cdot\overline{I_{2}}\subseteq\overline{I_{1}I_{2}}$ for all ideals $I_{1}$ and $I_{2}$ in $S$.

In order to prove the reverse inclusion, let $G(I)=\{\mathbf{x}^{{\boldsymbol{\alpha}}_{1}},\ldots,\mathbf{x}^{{\boldsymbol{\alpha}}_{s}}\}$. Set ${\boldsymbol{\alpha}}^{*}_{j}=({\boldsymbol{\alpha}}_{j},0)\in\mathbb{N}^{r+1}$ and let $\mathbf{e}_{r+1}$ be the $(r+1)$-th unit vector of $\mathbb{R}^{r+1}$. Assume that $({\boldsymbol{\alpha}},\beta)\in NP((I,y)^{n})\cap\mathbb{N}^{r+1}$. From (1.2) we see that there are non-negative numbers $a_{1},...,a_{s},b$ such that $\sum_{j=1}^{r}a_{j}+b\geqslant n$ and $({\boldsymbol{\alpha}},\beta)=\sum_{j=1}^{s}a_{j}{\boldsymbol{\alpha}}_{j}^{*}+b\mathbf{e}_{r+1}.$ Then $b=\beta\in\mathbb{N}$. If $b\geqslant n$, then $\mathbf{x}^{{\boldsymbol{\alpha}}}y^{\beta}\in y^{n}S$. Assume that $b<n$. Then $\sum_{j=1}^{r}a_{j}\geqslant n-b>0$ and ${\boldsymbol{\alpha}}=\sum_{j=1}^{s}a_{j}{\boldsymbol{\alpha}}_{j}\in NP(I^{n-b})$. Hence, by (1.1), $\mathbf{x}^{\boldsymbol{\alpha}}y^{\beta}\in y^{m}\overline{I^{n-m}}S$. In both cases, $\mathbf{x}^{\boldsymbol{\alpha}}y^{\beta}\in\sum_{i=0}^{n}y^{i}\overline{I^{n-i}}S$, i.e., $\overline{(I,y)^{n}}\subseteq\sum_{i=0}^{n}y^{i}\overline{I^{n-i}}S$.

(2) From (1) we deduce that

[TABLE]

as $R$-modules, and the conclusion follows. ∎

From now on, let $I$ be an ideal generated by square-free monomials. Such an ideal is often called a Stanley-Reisner ideal and is associated to the simplicial complex $\Delta:=\Delta(I)$. In this case we also denote $I$ by $I_{\Delta}$. Note that we do not require that $\Delta$ contains all vertices $\{i\}$, $i\leqslant r$. Recall that for a face $F\in\Delta$, the link of $F$ is defined by

[TABLE]

We simply write $\operatorname{lk}_{\Delta}i$ for $\operatorname{lk}_{\Delta}\{i\}$.

Corollary 3.4**.**

If $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay, then so is $\overline{I_{\operatorname{lk}_{\Delta}i}^{n}}$ for every vertex $i$ of $\Delta$.

Proof.

Let $S=k[x_{j}\mid j\neq i]$. Then $R=S[x_{i}]$. Let $J=I_{\Delta}R[x_{i}^{-1}]\cap S$. We have $\overline{J^{n}}=\overline{I_{\Delta}^{n}}R[x_{i}^{-1}]\cap S$, whence $\overline{J^{n}}$ is Cohen-Macaulay.

Denote by $V(\Delta)$ the set of vertices of a simplicial complex $\Delta$. Let $Y=\{x_{j}\mid j\notin V(\operatorname{lk}_{\Delta}i)\text{ and }j\neq i\}$. By [17, Lemma 2.1] we have

[TABLE]

It follows that $J=(I_{\operatorname{lk}_{\Delta}i},Y)$ as ideals in $S$. By Lemma 3.3 we conclude that $\overline{I_{\operatorname{lk}_{\Delta}i}^{n}}$ is Cohen-Macaulay. ∎

Note that $I_{\Delta}$ is a complete intersection if and only if any two of its minimal monomial generators have no common variable. Recall that $\dim\Delta=\max\{|F||\ F\in\Delta\}-1$.

Lemma 3.5**.**

Assume that $\dim\Delta=0$ and that $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay for some $n\geqslant 2$. Then, $I_{\Delta}$ is a complete intersection. Moreover, $\Delta$ has at most two vertices.

Proof.

Since $\dim\Delta=0$, we may assume that $\Delta=\left<\{1\},\ldots,\{s\}\right>$ for some $s\leqslant r$. Then

[TABLE]

If $s\leqslant 2$, then $I_{\Delta}$ is a complete intersection. It remains to show that $s\leqslant 2$. Assume on the contrary that $s\geqslant 3$. Since

[TABLE]

we can check that $\mathfrak{m}=\overline{I^{2}}\colon(x_{1}x_{2}x_{3})$, but $x_{1}x_{2}x_{3}\not\in I_{\Delta}^{2}$. Hence $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{2}}$. Using the non-decreasing property of the sets associated prime ideals of integral closures of powers of an ideal (see [6, Proposition 16.3]), we have $\mathfrak{m}\in\operatorname{Ass}R/\overline{I^{n}}$. Hence $\overline{I_{\Delta}^{n}}$ is not Cohen-Macaulay, a contradiction. ∎

Lemma 3.6**.**

Assume that $\dim\Delta=1$ and that $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay for some $n\geqslant 3$. Then, $I_{\Delta}$ is a complete intersection.

Proof.

Since $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay and $\sqrt{\overline{I_{\Delta}^{n}}}=I_{\Delta}$, $\Delta$ is Cohen-Macaulay by [9, Theorem 2.6]. Since $\dim\Delta=1$, this property implies that $\Delta$ is connected. In particular, every facet of $\Delta$ has exactly two vertices, and we can regard $\Delta$ as a connected graph without isolated vertices. We may assume that $V(\Delta)=[s]$ for some $s\leqslant r$.

If $s=2$, then $\Delta$ is just an edge, and $I_{\Delta}=(x_{3},...,x_{r})$. Assume that $s\geqslant 3$. For each vertex $i$ of $\Delta$, by Corollary 3.4 and Lemma 3.5, $\operatorname{lk}_{\Delta}(i)$ is either one vertex or consists of exactly two vertices. Consequently, $\Delta$ is either a path or a cycle.

For each edge $\{i,j\}$ of $\Delta$ , set $P_{ij}=(x_{s+1},...,x_{r};x_{l}\mid 1\leq l\leqslant s;\ l\neq i\text{ and }l\neq j)$. Then, $I_{\Delta}=\cap_{\{i,j\}\in\Delta}P_{ij}.$ Since $\overline{I_{\Delta}^{n}}$ is unmixed, by Lemma 3.2 we have

[TABLE]

where $I_{\Delta}^{(n)}$ is the $n$-th symbolic power of $I_{\Delta}$. This implies that $I_{\Delta}^{(n)}$ is Cohen-Macaulay. By [13, Theorem 2.4], every pair of disjoint edges of $\Delta$ is contained in a cycle of length $4$. Since $\Delta$ is either a path or a cycle, we conclude that $\Delta$ is either a path of length two, or a cycle of length 3 or 4. Hence, either $I_{\Delta}=(x_{4},...,x_{r};x_{1}x_{3}),\ (x_{4},...,x_{r};x_{1}x_{2}x_{3})$ or $I_{\Delta}=(x_{5},...,x_{r};x_{1}x_{3},x_{2}x_{4})$ - all are complete intersections. ∎

We can now improve Theorem 3.1 for square-free monomial ideals by giving an exact description of all square-free monomial ideals $I_{\Delta}$ such that $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay for some $n\geqslant 3$.

Theorem 3.7**.**

Let $\Delta$ be a simplicial complex. Then the following conditions are equivalent:

1. (1)

$\overline{I_{\Delta}^{n}}$* is Cohen-Macaulay for every $n\geqslant 1$;* 2. (2)

$\overline{I_{\Delta}^{n}}$* is Cohen-Macaulay for some $n\geqslant 3$;* 3. (3)

$I_{\Delta}$* is a complete intersection;* 4. (4)

$I_{\Delta}$* is an equimultipe ideal.*

Proof.

$(1)\Rightarrow(2)$ and $(3)\Rightarrow(4)$ are clear. $(4)\Rightarrow(1)$ follows from Theorem 3.1.

It remains to prove that $(2)\Rightarrow(3)$. The following proof is similar to that of [17, Theorem 4.3]. We prove the implication by induction on $\dim\Delta$. The case $\dim\Delta\leqslant 1$ follows from Lemmas 3.5 and 3.6. Assume that $\dim\Delta\geqslant 2$. Since $\overline{I_{\Delta}^{n}}$ is Cohen-Macaulay and $\sqrt{\overline{I_{\Delta}^{n}}}=I_{\Delta}$, $I_{\Delta}$ is Cohen-Macaulay by [9, Theorem 2.6]. In particular, $\Delta$ is connected. On the other hand, by Corollary 3.4, $\overline{I_{\operatorname{lk}_{\Delta}i}^{n}}$ is Cohen-Macaulay for all $i\leqslant s$, where w.l.o.g. we assume $V(\Delta)=[s]$. By the induction hypothesis, $I_{\operatorname{lk}_{\Delta}}i$ is a complete intersection. Since $\Delta$ is connected, this implies that $I_{\Delta}$ is a complete intersection complex by [18, Theorem 1.5]. ∎

Note that [17, Theorem 1.2] states that the last condition in Theorem 3.7 is also equivalent to the Cohen-Macaulayness of $R/I^{n}$ for all $n\geqslant 1$ (or for some fixed $n\geqslant 3$).

The property that the Cohen-Macaulay property of $\overline{I^{n}}$ for some $n\geqslant 3$ forces that for all $n$ is very specific for square-free monomial ideals. For an arbitrary monomial ideal, the picture is much more complicate, as shown by the following example.

Example 3.8**.**

Let $d\geqslant 3$ and $I=(x^{d},xy^{d-2}z,y^{d-1}z)\subset R=k[x,y,z]$. Then $(x,y,z)\in\operatorname{Ass}(R/\overline{I^{n}})$ if and only if $n\geqslant d$ (see the example in [19, Page 54]). Since $\dim R/I=1$, it follows that:

1. (1)

$R/\overline{I^{n}}$ is Cohen-Macaulay for each $n=1,\ldots,d-1$; 2. (2)

$R/\overline{I^{n}}$ is not Cohen-Macaulay for any $n\geqslant d$.

Note that $\operatorname{ht}(I)=2$ and $\ell(I)=3$ in this case.

Acknowledgment

This work is partially supported by NAFOSTED (Vietnam) under the grant number 101.04-2015.02.