Locally unmixed modules and linearly equivalent topologies

TL;DR

This paper characterizes locally unmixed modules over Noetherian rings through the linear equivalence of topologies defined by symbolic and adic powers of certain ideals, linking algebraic properties to topological behavior.

Contribution

It establishes a new equivalence between local unmixedness of modules and the linear equivalence of specific topologies, extending understanding of module and ideal interactions.

Findings

Locally unmixed modules are characterized by topology equivalence.

The topology defined by symbolic powers aligns with the $I$-adic topology under certain conditions.

The result provides a new criterion for local unmixedness in terms of topological properties.

Abstract

Let $R$ be a commutative Noetherian ring, and let $N$ be a non-zero finitely generated $R$-module. The purpose of this paper is to show that $N$ is locally unmixed if and only if, for any $N$-proper ideal $I$ of $R$ generated by $\Ht_N I$ elements, the topology defined by $(IN)^{(n)}$, $n \geq 0$, is linearly equivalent to the $I$-adic topology.

Equations

$$M=Q_1\cap \cdots \cap Q_r\cap Q_{r+1}\cap \cdots \cap Q_s$$

$$I^{n}N{:}_N⟨\mathfrak{m}⟩\subseteq D_{\mathfrak{m}}(I^{n}N).$$

$$D(I,N)\cong D_{\mathfrak{m}}({\oplus }_{n\ge 0}{I}^{n}N)\cong D_{\mathfrak{m}R[It]}({\oplus }_{n\ge 0}{I}^{n}N)),$$

$$l(I,N):=\dim N[It]/(\mathfrak{m},u)N[It],$$

$$l(I\widehat{R_{\mathfrak{q}}}+z/z\text{textbullet})<\dim \widehat{R_{\mathfrak{q}}}/z.(†)$$

$$k:=\max \{k_{\mathfrak{q}}:\mathfrak{q}\in A^{\ast }(I,N)\setminus {\mathrm{mAss}}_{R}N/IN\},$$

$$I^{s}N{:}_N⟨\mathfrak{m}⟩=(IN{)}^{(s)}$$

$$(IN{)}_{\mathfrak{q}}^{(n+k_{\mathfrak{q}})}\subseteq (I^{n}N{)}_{\mathfrak{q}},$$

$$(IN{)}_{\mathfrak{q}}^{(n+k)}\subseteq (I^{n}N{)}_{\mathfrak{q}}{:}_{N_{\mathfrak{q}}}⟨\mathfrak{m}{R}_{\mathfrak{q}}⟩,$$

$$(IN{)}^{(n+k)}\subseteq (I^{n}N){:}_N⟨\mathfrak{m}⟩.$$

$$I^{n+s}N{:}_N⟨\mathfrak{m}⟩\subseteq I^{n}N$$

$$(IN{)}^{(n+k+s)}\subseteq I^{n+s}N{:}_N⟨\mathfrak{m}⟩\subseteq I^{n}N,$$

Full text

2010 Mathematics Subject Classification: 13A30, 13E05.

∗Corresponding author: e-mail: naghipour@ipm.ir (Reza Naghipour)

1. Introduction

Let $R$ denote a commutative Noetherian ring, $I$ an ideal of $R$ and $N$ a non-zero finitely generated $R$-module. We denote by $R[It]$ (resp. $R[It,u]$) the graded ordinary (resp. extended) Rees ring $\oplus_{n\in\mathbb{N}_{0}}I^{n}t^{n}$ (resp. $\oplus_{n\in\mathbb{Z}}I^{n}t^{n}$) of $R$ with respect to $I$, where $t$ is an indeterminate and $u=t^{-1}$. Also, the graded ordinary Rees module $\oplus_{n\in\mathbb{N}_{0}}I^{n}N$ over $R[It]$ (resp. graded extended Rees module $\oplus_{n\in\mathbb{Z}}I^{n}N$ over $R[It,u]$) is denoted by $N[It]$ (resp. $N[It,u]$), which is finitely generated. For any multiplicatively closed subset $S$ of $R$, the $n$th $(S)$-symbolic power of $I$ with respect to $N$, denoted by $S(I^{n}N)$, is defined to be the union of $I^{n}N:_{N}s$ where $s$ varies in $S$. The $I$-adic filtration $\{I^{n}N\}_{n\geq 0}$ and the $(S)$-symbolic filtration $\{S(I^{n}N)\}_{n\geq 0}$ induce topologies on $N$ which are called the $I$-adic topology and the $(S)$-symbolic topology, respectively. These two topologies are said to be linearly equivalent if, there is an integer ${k\geq 0}$ such that $S(I^{n+k}N)\subseteq I^{n}N$ for all integers $n$. In particular, if $S=R\setminus\bigcup\{\mathfrak{p}\in\operatorname{mAss}_{R}N/IN\}$, where $\operatorname{mAss}_{R}N/IN$ denotes the set of minimal prime ideals of $\operatorname{Ass}_{R}N/IN$, the $n$th $(S)$-symbolic power of $I$ with respect to $N$, is denoted by $(IN)^{(n)}$, and the topology defined by the filtration $\{(IN)^{(n)}\}_{n\geq 0}$ is called the symbolic topology. The purpose of this paper is to show that $N$ is locally unmixed if and only if, for each $N$-proper ideal $I$ that is generated by $\operatorname{ht}_{N}I$ elements, the $I$-adic and the symbolic topologies are linearly equivalent.

P. Schenzel has characterized unmixed local rings [19, Theorem 7] in terms of comparison of the topologies defined by certain filtrations. Also, D. Katz [9, Theorem 3.5] and J. Verma [21, Theorem 5.2] have proved a characterization of locally unmixed rings in terms of $s$-ideals. Equivalence of $I$-adic topology and $(S)$-symbolic topology has been studied, in the case $N=R$, in [9, 15, 19, 18, 17], and has led to some interesting results.

Let $\mathfrak{p}\in\operatorname{Supp}(N)$. Then $N$-height of $\mathfrak{p}$, denoted by $\operatorname{ht}_{N}\mathfrak{p}$, is defined to be the supremum of lengths of chains of prime ideals of $\operatorname{Supp}(N)$ terminating with $\mathfrak{p}$. We have $\operatorname{ht}_{N}\mathfrak{p}=\dim_{R_{\mathfrak{p}}}N_{\mathfrak{p}}$. We shall say an ideal $I$ of $R$ is $N$-proper if $N/IN\neq 0$, and, when this is the case, we define the $N$-height of $I$ (written $\operatorname{ht}_{N}I$) to be

$\inf\{\operatorname{ht}_{N}\mathfrak{p}:\,\mathfrak{p}\in\operatorname{Supp}(N)\cap V(I)\}$

$(=\inf\{\operatorname{ht}_{N}\mathfrak{p}:\,\mathfrak{p}\in\operatorname{Ass}_{R}(N/IN)\}).$

If $(R,\mathfrak{m})$ is local, then $\widehat{R}$ (resp. $\widehat{N}$) denotes the completion of $R$ (resp. $N$) with respect to the $\mathfrak{m}$-adic topology. In particular, for any $\mathfrak{p}\in\operatorname{Spec}(R)$, we denote $\widehat{R_{\mathfrak{p}}}$ and $\widehat{{N_{\mathfrak{p}}}}$ the $\mathfrak{p}R_{\mathfrak{p}}$-adic completion of $R_{\mathfrak{p}}$ and $N_{\mathfrak{p}}$, respectively. Then $N$ is said to be an unmixed module if for any $\mathfrak{p}\in\operatorname{Ass}_{\widehat{R}}\widehat{N}$, $\dim\widehat{R}/\mathfrak{p}=\dim N$. More generally, if $R$ is not necessarily local and $N$ is non-zero finitely generated, $N$ is a locally unmixed module if for any $\mathfrak{p}\in\operatorname{Supp}(N)$, $N_{\mathfrak{p}}$ is an unmixed $R_{\mathfrak{p}}$-module.

As the main result of this paper we characterize the locally unmixed property of a non-zero finitely generated $R$-module $N$ in terms of the linearly equivalence of the topologies defined by $\{I^{n}N\}_{n\geq 0}$ and $\{(IN)^{(n)}\}_{n\geq 0}$, for certain $N$-proper ideals $I$ of $R$. More precisely we shall show that:

Theorem 1.1**.**

Let $R$ be a Noetherian ring and $N$ a non-zero finitely generated $R$-module. Then the following conditions are equivalent:

$\rm(i)$* $N$ is locally unmixed.*

$\rm(ii)$* For each $N$-proper ideal $I$ of $R$ that is generated by $\operatorname{ht}_{N}I$ elements, the topology given by $\{(IN)^{(n)}\}_{n\geq 0}$ is linearly equivalent to the $I$-adic topology on $N$.*

One of our tools for proving Theorem 1.1 is the following, which plays a key role in this paper. Recall that a prime ideal $\mathfrak{p}$ of $R$ is called a quitessential prime ideal of $I$ with respect to $N$ precisely when there exists $\mathfrak{q}\in\operatorname{Ass}_{\widehat{R}_{\mathfrak{p}}}\widehat{N}_{\mathfrak{p}}$ such that $\operatorname{Rad}(I\widehat{R}_{\mathfrak{p}}+\mathfrak{q})=\mathfrak{p}\widehat{R}_{\mathfrak{p}}$. The set of quitessential primes of $I$ is denoted by $Q(I,N)$. Then, the set of essential primes of $I$ with respect to $N$, denoted by $E(I,N)$, is defined to be the set $\{\mathfrak{q}\cap R\mid\,\mathfrak{q}\in Q(uR[It,u],N[It,u])\}$.

Theorem 1.2**.**

Let $R$ denote a Noetherian ring, $N$ a non-zero finitely generated $R$-module and $I$ a $N$-proper ideal of $R$ such that $E(I,N)=\operatorname{mAss}_{R}N/IN$. Then, the $I$-adic topology $\{I^{n}N\}_{n\geq 0}$ and the topology defined by $\{(I^{n}N)^{(n)}\}_{n\geq 0}$ are linearly equivalent.

The proof of Theorem 1.2 is given in 1.13.

Throughout this paper, $R$ will always be a commutative Noetherian ring with non-zero identity, $N$ will be a non-zero finitely generated $R$-module, and $I$ will be an $N$-proper ideal of $R$, i.e., $N/IN\neq 0$. For each $R$-module $L$, we denote by $\operatorname{mAss}_{R}L$ the set of minimal primes of $\operatorname{Ass}_{R}L$. For any ideal $J$ of $R$, the radical of $J$, denoted by $\operatorname{Rad}(J)$, is defined to be the set $\{x\in R\,:\,x^{n}\in J$ for some $n\in\mathbb{N}\}$. For any unexplained notation and terminology we refer the reader to [6] or [12].

2. The Results

The main result of this section is to show that a non-zero finitely generated module $N$ over a Noetherian ring $R$ is locally unmixed if and only if, for any $N$-proper ideal $I$ of $R$ that can be generated by $\operatorname{ht}_{N}I$ elements, the topologies defined by $\{I^{n}N\}_{n\geq 0}$ and $\{(IN)^{(n)}\}_{n\geq 0}$, on $N$, are linearly equivalent. We begin with the following remark.

Remark 2.1*.*

Let $R$ be a Noetherian ring and $N$ a finitely generated $R$-module. For a submodule $M$ of $N$ and an ideal $I$ of $R$, the increasing sequence of submodules

$M\subseteq M:_{N}I\subseteq M:_{N}I^{2}\subseteq\cdots\subseteq M:_{N}I^{n}\subseteq\cdots$

becomes stationary. Denote its ultimate constant value by $M:_{N}\langle I\rangle$. Note that $M:_{N}\langle I\rangle=M:_{N}I^{n}$ for all large $n$. Let

[TABLE]

be an irredundant primary decomposition of $M$, with $I\subseteq\operatorname{Rad}(Q_{i}:_{R}N)$, exclusively for $r+1\leq i\leq s$. Then, from the definition, it easily follows that $M:_{N}\langle I\rangle=Q_{1}\cap\dots\cap Q_{r}$. Therefore

$\operatorname{Ass}_{R}{N}/{(M:_{N}\langle I\rangle)}=\{\mathfrak{p}\in\operatorname{Ass}_{R}{N}/{M\textbullet}:I\nsubseteq\mathfrak{p}\}=\operatorname{Ass}_{R}({N}/{M\textbullet})\setminus V(I)$.

Now we can state and prove the following lemma. Here $D_{I}(L)$ denotes the ideal transform of the $R$-module $L$ with respect to an ideal $I$ of $R$ (see [5, 2.2.1]).

Lemma 2.2**.**

Let $(R,\mathfrak{m})$ be local (Noetherian) ring, $I$ an ideal of $R$ and $N$ a non-zero finitely generated $R$-module such that $0ptN>0$. Then, for all integers $n\geq 0$, we have

[TABLE]

Proof.

The assertion follows from [5, Corollary 2.2.18] and the fact that $0ptI^{n}N>0$ for all integers $n\geq 0$. ∎

The next result concerns the associated prime ideals of the Rees module $N[It]$ for a non-zero finitely generated module $N$ over a Noetherian ring $R$ and an ideal $I$ in $R$.

Proposition 2.3**.**

Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $N$ a non-zero finitely generated $R$-module. Then

$\operatorname{Ass}_{R[It]}N[It]=\{\oplus_{n\geq 0}(I^{n}\cap\mathfrak{p}):\,\mathfrak{p}\in\operatorname{Ass}_{R}N\}$.

Proof.

Let $\mathfrak{q}\in\operatorname{Ass}_{R[It]}N[It]$. Then in view of [6, Lemma 1.5.6] there exists a homogenous element $x$ of $N[It]$ such that $\mathfrak{q}=\operatorname{Ann}_{R[It]}x$. Suppose that $x\in I^{v}N$ for some integer $v\geq 0$. Then we have

$\mathfrak{q}=(0:_{R[It]}x)=\oplus_{n\geq 0}(0:_{R}x)\cap I^{n}$.

Now, it is easy to see that $\mathfrak{p}:=(0:_{R}x)$ is a prime ideal of $R$ and so $\mathfrak{p}\in\operatorname{Ass}_{R}N$. Hence $\mathfrak{q}=\oplus_{n\geq 0}(I^{n}\cap\mathfrak{p})$ for some $\mathfrak{p}\in\operatorname{Ass}_{R}N$. Conversely, let $\mathfrak{p}\in\operatorname{Ass}_{R}N$ and $\mathfrak{p}=(0:_{R}x)$ for an element $x\in N$. Then

$\mathfrak{q}:=(0:_{R[It]}x)=\oplus_{n\geq 0}(I^{n}\cap\mathfrak{p})$

is a prime ideal of $\operatorname{Ass}_{R[It]}N[It]$, because $R[It]/\mathfrak{q}\cong R/{\mathfrak{p}}[(I+\mathfrak{p}/\mathfrak{p})t]$ is a domain. ∎

Definition 2.4**.**

Let $R$ be a Noetherian ring and $N$ an $R$-module. A decreasing sequence $\{N_{n}\}_{n\geq 0}$ of submodules of $N$ is called a filtration of $N$. If $I$ is an ideal of $R$, then the filtration $\{N_{n}\}_{n\geq 0}$ is called $I$-filtration whenever $IN_{n}\subseteq N_{n+1}$ for all integers $n\geq 0$.

Lemma 2.5**.**

Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $N$ an $R$-module. Let $\{N_{n}\}_{n\geq 0}$ be an $I$-filtration of submodules of $N$ such that the ordinary Rees module $N[It]$ is finitely generated over $R[It]$. Then there exists an integer $k$ such that $N_{n+k}=I^{n}N_{k}$, for all integers $n\geq 0$.

Proof.

The result follows easily from [7, Lemma 2.5.4]. ∎

Corollary 2.6**.**

Let $(R,\mathfrak{m})$ be a local (Noetherian) ring and $I$ an ideal of $R$. Let $N$ be an $R$-module and set $N_{n}=I^{n}N:_{N}\langle\mathfrak{m}\rangle$ for each integer $n\geq 0$. Suppose that the module $\oplus_{n\geq 0}N_{n}$ is finitely generated over the ordinary Rees ring $R[It]$. Then there is an integer $k$ such that $I^{n+k}N:_{N}\langle\mathfrak{m}\rangle\subseteq I^{n}N$, for all integer $n\geq 0$.

Proof.

As $I(I^{n}N:_{N}\langle\mathfrak{m}\rangle)\subseteq I^{n+1}N:_{N}\langle\mathfrak{m}\rangle$, for all integers $n\geq 0$, the claim follows from Lemma 2.5. ∎

Definition 2.7**.**

Let $(R,\mathfrak{m})$ be a local (Noetherian) ring, $I$ an ideal of $R$ and $N$ an $R$-module. We define the $R$-module $\operatorname{D}(I,N)$ as the following:

$\operatorname{D}(I,N):=\bigoplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}(I^{n}N)$.

As $\operatorname{D}_{\mathfrak{m}}(.)$ is an $R$-linear and left exact functor, it follows that $\{\operatorname{D}_{\mathfrak{m}}(I^{n}N)\}_{n\geq 0}$ is a decreasing sequence and $I\operatorname{D}_{\mathfrak{m}}(I^{n}N)\subseteq\operatorname{D}_{\mathfrak{m}}(I^{n+1}N)$ for all integers $n\geq 0$. Hence $\operatorname{D}(I,N)$ is an $R[It]$-module, by Lemma 2.5.

Lemma 2.8**.**

Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $N$ a finitely generated $R$-module. Then the following conditions are equivalent:

$\rm(i)$* $D_{I}(N)$ is a finitely generated $R$-module.*

$\rm(ii)$* For all $\mathfrak{p}\in\operatorname{Ass}_{R}N$, the $R/\mathfrak{p}$-module $\operatorname{D}_{I(R/\mathfrak{p})}(R/\mathfrak{p})$ is finitely generated.*

Proof.

See [4, Lemma 3.3]. ∎

Proposition 2.9**.**

Let $(R,\mathfrak{m})$ be a local (Noetherian) ring, $I$ an ideal of $R$ and $N$ a finitely generated $R$-module. Then the following conditions are equivalent:

$\rm(i)$* $\operatorname{D}(I,N)$ is a finitely generated $R[It]$-module.*

$\rm(ii)$* For all $\mathfrak{p}\in\operatorname{Ass}_{R}N$, the module $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}(I^{n}+\mathfrak{p}/\mathfrak{p})$ is finitely generated over the Rees ring $R/\mathfrak{p}[(I+\mathfrak{p}/\mathfrak{p})t]$.*

Proof.

In order to prove the implication $\rm(i)$ $\Longrightarrow$ $\rm(ii)$, suppose that $\mathfrak{p}\in\operatorname{Ass}_{R}N$. Then in view of Proposition 2.3, there exists $\mathfrak{q}\in\operatorname{Ass}_{R[It]}(\oplus_{n\geq 0}I^{n}N)$ such that $\mathfrak{q}=\oplus_{n\geq 0}(I^{n}\cap\mathfrak{p})$. Since

[TABLE]

is a finitely generated $R[It]$-module, it follows from Lemma 2.8 that the $R[It]/\mathfrak{q}$-module $\operatorname{D}_{\mathfrak{m}({R[It]}/{\mathfrak{q}})}(R[It]/\mathfrak{q})$ is finitely generated. Now, as

$R[It]/\mathfrak{q}\cong R/\mathfrak{p}[(I+\mathfrak{p}/\mathfrak{p})t],$ and $\operatorname{D}_{\mathfrak{m}({R[It]}/{\mathfrak{q}})}({R[It]}/{\textbullet\mathfrak{q}})\cong\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}({I^{n}+\mathfrak{p}}/{\mathfrak{p}\textbullet}),$

we deduce that the $R/\mathfrak{p}[({I+\mathfrak{p}}/{\mathfrak{p}\textbullet})t]$-module $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}({I^{n}+\mathfrak{p}}/{\mathfrak{p}\textbullet})$ is finitely generated.

Now, we show the conclusion $\rm(ii)$ $\Longrightarrow$ $\rm(i)$. To do this end, let $\mathfrak{q}\in\operatorname{Ass}_{R[It]}N[It]$. Then, by virtue of Proposition 2.3, there exists $\mathfrak{p}\in\operatorname{Ass}_{R}N$ such that $\mathfrak{q}=\oplus_{n\geq 0}(I^{n}\cap\mathfrak{p})$. Since

${R[It]}/{\mathfrak{q}\textbullet}\cong R/\mathfrak{p}[({I+\mathfrak{p}}/{\mathfrak{p}\textbullet})t]$ and $\operatorname{D}_{\mathfrak{m}({R[It]}/{\mathfrak{q}})}({R[It]}/{\mathfrak{q}\textbullet})\cong\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}({I^{n}+\mathfrak{p}}/{\mathfrak{p}\textbullet})$,

it follows from Lemma 2.8 that the $R[It]$-module $\operatorname{D}_{\mathfrak{m}R[It]}(N[It])$ is finitely generated, and so the $R[It]$-module $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}(I^{n}N)$ is finitely generated, as required. ∎

The next proposition gives us a criterion for the finiteness of $R[It]$-module $D_{\mathfrak{m}R[It]}(N[It])$, whenever $(R,\mathfrak{m})$ is a local ring and $N$ is a finitely generated module over $R$. To this end, let us, firstly, recall the important notion analytic spread of $I$ with respect to $N$, over a local ring $(R,\mathfrak{m}$), introduced by Brodmann in [3]:

[TABLE]

in the case $N=R$, $l(I,N)$ is the classical analytic spread $l(I)$ of $I$, introduced by Northcott and Rees (see [14]).

Proposition 2.10**.**

Let $(R,\mathfrak{m})$ be a local (Noetherian) ring and $I$ an ideal of $R$. Let $N$ be a finitely generated $R$-module such that $l({I\widehat{R}+\mathfrak{p}}/{\mathfrak{p}})<\dim{\widehat{R}}/{\mathfrak{p}\textbullet}$ for all $\mathfrak{p}\in\operatorname{Ass}_{\widehat{R}}\widehat{N}$. Then the $R[It]$-module $D_{\mathfrak{m}R[It]}(N[It])$ is finitely generated, and $0ptN>0$.

Proof.

It is easy to see that

$D_{\mathfrak{m}R[It]}(N[It])\otimes_{R[It]}\widehat{R}[(I\widehat{R})t]\cong D_{\mathfrak{m}\widehat{R}[(I\widehat{R})t]}(\oplus_{n\geq 0}I^{n}\widehat{N})$,

and so by faithfully flatness of $\widehat{R}[(I\widehat{R})t]$ over $R[It]$, it is enough for us to show that the $\widehat{R}[(I\widehat{R})t]$-module $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}\widehat{R}[(I\widehat{R})t]}(I^{n}\widehat{N})$ is finitely generated. In order to do this, in view of Proposition 2.9, it is enough to show that $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}(I^{n}\widehat{R}+\mathfrak{p}/\mathfrak{p})$ is finitely generated over $\widehat{R}/\mathfrak{p}[(I\widehat{R}+\mathfrak{p}/\mathfrak{p})t]$ for all $\mathfrak{p}\in\operatorname{Ass}_{\widehat{R}}\widehat{N}$. But this follows easily from [19, Proposition] and the assumption $l(I\widehat{R}+\mathfrak{p}/\mathfrak{p})<\dim\widehat{R}/\mathfrak{p}$. ∎

Remark 2.11*.*

Before bringing the next result we fix a notation, which is employed by P. Schenzel in [18] in the case $N=R$. Let $S$ be a multiplicatively closed subset of a Noetherian ring $R$. For a submodule $M$ of a finitely generated $R$-module $N$, we use $S(M)$ to denote the submodule $\bigcup_{s\in S}(M:_{N}s)$. Note that the primary decomposition of $S(M)$ consists of the intersection of all primary components of $M$ whose associated prime ideals do not meet $S$. In other words

$\operatorname{Ass}_{R}{N}/{S(M)}=\{\mathfrak{p}\in\operatorname{Ass}_{R}N/M:\mathfrak{p}\cap S=\emptyset\}$.

In particular, if $S=R\setminus\bigcup\{\mathfrak{p}\in\operatorname{mAss}_{R}{N}/{IN}\}$, then for any $n\in\mathbb{N}$, $S(I^{n}N)$ is denoted by $(IN)^{(n)}$, where $I$ is an ideal of $R$.

The following lemma is needed in the proof of Theorem 2.13.

Lemma 2.12**.**

Let $R$ be a Noetherian ring and $N$ an $R$-module. Let $M$ and $L$ be two submodules of $R$ such that $M_{\mathfrak{p}}\subseteq L_{\mathfrak{p}}$ for all $\mathfrak{p}\in\operatorname{Ass}_{R}{N}/{L\textbullet}$. Then $M\subseteq L$.

Proof.

The assertion follows from the fact that $\operatorname{Ass}_{R}({M+L}/{L\textbullet})\subseteq\operatorname{Ass}_{R}{N}/{L\textbullet}$. ∎

Following, we investigate a fundamental characterization for linearly equivalence between the $I$-adic and symbolic topologies on a finitely generated $R$-module $N$, for certain ideal $I$ of $R$. This result plays a key role in the proof of the main theorem.

To this end, recall that, in [16], L.J. Ratliff, Jr., (resp. in [2] Brodmann) introduced the interesting set of associated primes $\bar{A^{*}}(I):=\operatorname{Ass}_{R}R/(I^{n})_{a}$ (resp. $A^{*}(I,N):=\operatorname{Ass}_{R}{N}/{I^{n}N}$), for large $n$. Here $I_{a}$ denotes the integral closure of $I$ in $R$, i.e., $I_{a}$ is the ideal of $R$ consisting of all elements $x\in R$ which satisfy an equation $x^{n}+r_{1}x^{n-1}+\cdots+r_{n}=0$, where $r_{i}\in I^{i},i=1,\ldots,n$.

Moreover, recall that a local ring $(R,\mathfrak{m})$ is said to be a quasi-unmixed ring if for every $\mathfrak{p}\in\operatorname{mAss}\widehat{R}$, the condition $\dim\widehat{R}/\mathfrak{p}=\dim R$ is satisfied.

Theorem 2.13**.**

Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and let $N$ be a finitely generated $R$-module such that $E(I,N)=\operatorname{mAss}_{R}{N}/{IN\textbullet}$. Then, the $I$-adic topology, $\{I^{n}N\}_{n\geq 0}$ and the topology defined by the filtration $\{(IN)^{(n)}\}_{n\geq 0}$ are linearly equivalent.

Proof.

Let $\mathfrak{q}\in A^{*}(I,N)\setminus\operatorname{mAss}_{R}N/IN$ and let $z\in\operatorname{Ass}_{\widehat{R_{\mathfrak{q}}}}\widehat{N_{\mathfrak{q}}}$. Then, by assumption, $\mathfrak{q}\notin E(I,N)$. Hence, in view of [1, Lemma 3.2], $\mathfrak{q}R_{\mathfrak{q}}\notin E(IR_{\mathfrak{q}},N_{\mathfrak{q}})$, and so it follows from [1, Proposition 3.6] that $\mathfrak{q}\widehat{R_{\mathfrak{q}}}/z\notin E(I\widehat{R_{\mathfrak{q}}}+z/z)$. Thus by virtue of [11, Lemma 2.1], $\mathfrak{q}\widehat{R_{\mathfrak{q}}}/z\notin\bar{A^{*}}(I\widehat{R_{\mathfrak{q}}}+z/z)$. As $\widehat{R_{\mathfrak{q}}}/z$ is quasi-unmixed, it follows from McAdam’s result [10, Proposition 4.1] that

[TABLE]

Now, we show that there exists a non-negative integer $k$ such that $(IN)^{(n+k)}\subseteq I^{n}N$ for all integers $n\geq 0$. To do this, it is easy to see that, $(IN)^{(s)}_{\mathfrak{p}}\subseteq(I^{s}N)_{\mathfrak{p}}$ for all $\mathfrak{p}\in\operatorname{mAss}_{R}{N}/{IN}$ and for all integers $s\geq 0$. Moreover, if for every $\mathfrak{q}\in A^{*}(I,N)\setminus\operatorname{mAss}_{R}{N}/{IN}$ there exists an integer $k_{\mathfrak{q}}$ such that

$(IN)^{(n+k_{\mathfrak{q}})}_{\mathfrak{q}}\subseteq(I^{n}N)_{\mathfrak{q}}$,

then by considering

[TABLE]

one easily sees that $(IN)^{(n+k)}\subseteq I^{n}N$. Since both $A^{*}(I,N)$ and $\operatorname{mAss}_{R}{N}/{IN}$ behave well under localization, we may assume by localizing at $\mathfrak{q}$ that $(R,\mathfrak{m})$ is a local ring.

Now, we use induction on $\dim{N}/{IN\textbullet}:=d$. It is clear that $d\geq 1$. Now, if $d=1$, then, as $\operatorname{Ass}_{R}{N}/{IN}\subseteq\operatorname{Supp}{N}/{IN\textbullet}$ and $\mathfrak{m}\in\operatorname{Supp}{N}/{IN\textbullet}$ it follows that the only possible embedded prime of $\operatorname{Ass}_{R}{N}/{IN}$ is $\mathfrak{m}$, and so in view of Remark 2.1 we have

[TABLE]

for all integers $s\geq 0$. Next, it follows from $({\dagger})$ and Proposition 2.10 that the $R[It]$-module $\oplus_{n\geq 0}\operatorname{D}_{\mathfrak{m}}(I^{n}N)$ is finitely generated and $0ptN>0$. Hence in view of Lemma 2.2, the module $\oplus_{n\geq 0}(I^{n}N:_{N}\langle\mathfrak{m}\rangle)$ is finitely generated over the Rees ring $R[It]$, and so by virtue of Corollary 2.6, there exists an integer $t$ such that $I^{n+t}N:_{N}\langle\mathfrak{m}\rangle\subseteq I^{n}N$ for all integers $n\geq 0$. Therefore $(IN)^{(n+k)}\subseteq I^{n}N$, and so the result holds for $d=1$.

We therefore assume, inductively, that $d>1$ and the result has been proved for smaller values of $d$. If $\mathfrak{q}\neq\mathfrak{m}$ and $\mathfrak{q}\in A^{*}(I,N)$, then

$\dim{N_{\mathfrak{q}}}/{IN_{\mathfrak{q}}}=\operatorname{ht}_{{N}/{IN\textbullet}}\mathfrak{q}<\operatorname{ht}_{{N}/{IN\textbullet}}\mathfrak{m}=\dim{N}/{IN\textbullet}=d$.

Hence by induction hypothesis, there exists a non-negative integer $k_{\mathfrak{q}}$ such that

[TABLE]

for all integers $n\geq 0$. Now, in view of Remark 2.1,

$\operatorname{Ass}_{R}{N}/{(I^{n}N:_{N}\langle\mathfrak{m}\rangle)}=\operatorname{Ass}_{R}{N}/{I^{n}N}\setminus V(\mathfrak{m})$,

it follows that for all $\mathfrak{q}\in\operatorname{Ass}_{R}{N}/{(I^{n}N:_{N}\langle\mathfrak{m}\rangle)}$, there exists a non-negative integer $k_{\mathfrak{q}}$ such that

$(IN)^{(n+k_{\mathfrak{q}})}_{\mathfrak{q}}\subseteq(I^{n}N)_{\mathfrak{q}}\subseteq((I^{n}N)_{\mathfrak{q}}:_{N_{\mathfrak{q}}}\langle\mathfrak{m}R_{\mathfrak{q}}\rangle)$,

for all integers $n\geq 0$. Hence by considering

$k:=\max\{k_{\mathfrak{q}}:\mathfrak{q}\in\operatorname{Ass}_{R}{N}/({I^{n}N:_{N}\langle\mathfrak{m}\rangle})\}$,

we get

[TABLE]

for all $\mathfrak{q}\in\operatorname{Ass}_{R}{N}/({I^{n}N:_{N}\langle\mathfrak{m}\rangle})$ and all integers $n\geq 0$. Therefore, by virtue of the Lemma 2.12, we have

[TABLE]

On the other hand, in view of Corollary 2.6, there exists an integer $s\geq 0$ such that

[TABLE]

for all integers $n\geq 0$. Consequently

[TABLE]

for all integers $n\geq 0$, and thus the topologies defined by the filtrations $\{I^{n}N\}_{n\geq 0}$ and $\{(IN)^{(n)}\}_{n\geq 0}$ are linearly equivalent. ∎

We are now ready to state and prove the main theorem of this paper, which is a new characterization of locally unmixed modules in terms of comparison of the topologies defined be certain decreasing families of a submodules of finitely generated modules over a commutative Noetherian ring. One of the implications in the proof of this theorem follows from [13, Theorem 3.2].

Theorem 2.14**.**

Let $R$ be a Noetherian ring and $N$ a non-zero finitely generated $R$-module. Then the following conditions are equivalent:

$\rm(i)$* $N$ is locally unmixed.*

$\rm(ii)$* For any $N$-proper ideal $I$ of $R$ generated by $\operatorname{ht}_{N}I$ elements, the $I$-adic topology is linearly equivalent to the symbolic topology.*

Proof.

The implication $\rm(ii)$ $\Longrightarrow$ $\rm(i)$ follows easily from [13, Theorem 3.2]. In order to prove the conclusion $\rm(i)$ $\Longrightarrow$ $\rm(ii)$, let $I$ be an $N$-proper ideal of $R$ which is generated by $\operatorname{ht}_{N}I$ elements. Then, in view of Theorem 2.13 it is enough for us to show that $E(I,N)=\operatorname{mAss}_{R}{N}/{IN}$. Suppose that $\mathfrak{p}\in E(I,N)$, and we show that $\mathfrak{p}\in\operatorname{mAss}_{R}{N}/{IN}$. Let $\operatorname{ht}_{N}I:=n$. Then by [13, Theorem 2.1], there exist the elements $x_{1},\dots,x_{n}$ in $I$ such that $\operatorname{ht}_{N}(x_{1},\dots,x_{i})=i$ for all $1\leq i\leq n$. As, in view of [13, Corollary 3.11], $x_{1},\dots,x_{n}$ is an essential sequence on $N$, and the fact that ${\rm egrade}(I,N)\leq\operatorname{ht}_{N}I$, it follows that ${\rm egrade}(I,N)=n$. Now, analogous to the proof of [8, Theorem 125], it is easy to see that $I$ can be generated by an essential sequence of length n. Therefore by [13, Lemma 3.8], we have $\mathfrak{p}\in\operatorname{mAss}_{R}{N}/{IN}$, and so $E(I,N)\subseteq\operatorname{mAss}_{R}{N}/{IN}$. As the opposite inclusion is obvious, the result follows. ∎

Acknowledgments

The authors are deeply grateful to the referee for his/her careful reading of the paper and valuable suggestions. Also, we would like to thank Professors M.P. Brodmann and S. Goto for their useful comments on Theorem 2.13.