---
abstract: 'Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R$. It is shown that the sequences ${\operatorname{Ass}}_R R/(I^n)_a^{(N)}$, ${\operatorname{Ass}}_R (I^n)_a^{(N)}/ (I^{n+1})^{(N)}_a$ and ${\operatorname{Ass}}_R (I^n)_a^{(N)}/ (I^n)_a, n= 1,2, \dots,$ of associated prime ideals, are increasing and ultimately constant for large $n$. Moreover, it is shown that, if $S$ is a multiplicatively closed subset of $R$, then the topologies defined by $(I^n)_a^{(N)}$ and $S((I^n)_a^{(N)}), \,{n\geq1}$, are equivalent if and only if $S$ is disjoint from the quintasymptotic primes of $I$. By using this, we also show that, if $(R, \mathfrak{m})$ is local and $N$ is quasi-unmixed, then the local cohomology module $H^{\dim N}_I(N)$ vanishes if and only if there exists a multiplicatively closed subset $S$ of $R$ such that $\mathfrak{m} \cap S \neq \emptyset$ and that the topologies induced by $(I^n)_a^{(N)}$ and $S((I^n)_a^{(N)}), \, {n\geq1},$ are equivalent.'
address:
- 'Department of mathematics, University of Tabriz, Tabriz, Iran; and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746, Tehran, Iran.'
- 'Martin-lüther-universität halle-wittenberg, fachbereich mathematik and informatik, d-06099 halle (saale), germany'
author:
- 'Reza Naghipour$^*$ and Peter Schenzel'
title: 'Asymptotic behaviour of integral closures, quintasymptotic primes and ideal topologies'
---

[^1]

Introduction
============

The important concept of integral closure of an ideal of a commutative Noetherian ring (with identity), developed by D. G. Northcott and D. Rees in [@NR], is fundamental to a considerable body of recent and current research both in commutative algebra and algebraic geometry. Let $R$ be a commutative ring (with identity), $I$ an ideal of $R$. In the case when $R$ is Noetherian, we denote by $(I)_a$ the integral closure of $I$, i.e., $(I)_a$ is the ideal of $R$ consisting of all elements $x\in R$ which satisfy an equation $x^n+ r_1x^{n-1}+
\cdots + r_n= 0$, where $r_i\in I^i, i=1, \ldots, n$.

In [@R] L.J. Ratliff, Jr., has shown that (when $R$ is Noetherian), the sequence of associated prime ideals $${\operatorname{Ass}}_R R/(I^n)_a,  n= 1,2, \ldots ,$$ is increasing and ultimately constant; we use the notation $A^*_a(I)$ to denote ${\operatorname{Ass}}_R R/(I^n)_a$ for large $n$.

The notion of integral closures of ideals of $R$ relative to a Noetherian $R$-module $N$, was initiated by R.Y. Sharp et al., in [@STY]. An element $x\in R$ is said to be [*integrally dependent on $I$ relative to*]{} $N$ if there exists a positive integer $n$ such that $x^{n}N \subseteq \sum_{i=1}^n x^{n-i}I^iN.$ Then the set

$I^{(N)}_a=\{x\in R\,|\, x$ is integrally dependent on $I$ relative to $N$}

is an ideal of $R$, called the [*integral closure of $I$ relative to*]{} $N$, in the case $N=R$, $I^{(N)}_a$ is the classical integral closure $I_a$ of $I$. It is clear that $I\subseteq I^{(N)}_a.$ We say that $I$ is [*integrally closed*]{} relative to $N$ if $I= I^{(N)}_a.$

In the second section (among other things) we show that, when $R$ is a Noetherian ring and $N$ is a finitely generated $R$-module, the sequences $${\operatorname{Ass}}_R R/(I^n)^{(N)}_a, \,\,\,  {\operatorname{Ass}}_R (I^n)_a^{(N)}/ (I^{n+1})^{(N)}_a \text{ and }   {\operatorname{Ass}}_R (I^n)_a^{(N)}/ ((I+{\operatorname{Ann}}_R N)^n)_a, \,\, n= 1,2, \ldots,$$ of associated primes, are ultimately constant; we let $A^*_a(I, N):={\operatorname{Ass}}_R R/(I^n)^{(N)}_a$ and $C^*_a(I, N):={\operatorname{Ass}}_R (I^n)_a^{(N)}/ ((I+{\operatorname{Ann}}_R N)^n)_a$, for large $n$. Pursing this point of view further we shall show that $A^*_a(I+ {\operatorname{Ann}}_R N)\setminus C^*_a(I, N) \subseteq A^*_a(I, N).$

In [@Mc2], McAdam studied the following interesting set of prime ideals of $R$ associated with $I$, $$\bar{Q}^*(I)= \{\mathfrak{p} \in {\operatorname{Spec}}R  : \text{ there  exists
a } \mathfrak{q} \in {\operatorname{mAss}}\hat{R}_{\mathfrak{p}} \text{ such
that} {\operatorname{Rad}}(I\hat{R}_{\mathfrak{p}}+ \mathfrak{q})=
\mathfrak{p}\hat{R}_{\mathfrak{p}}\},$$ and he called $\bar{Q}^*(I)$ the set of [*quintasymptotic prime ideals*]{} of $I$.

On the other hand, Ahn in [@Ah] extended the notion of the quintasymptotic prime ideals to a finitely generated module over $R.$ More precisely, if $N$ is a finitely generated $R$-module then a prime ideal $\mathfrak{p}$ of $R$ is said to be a [*quintasymptotic prime ideal*]{} of $I$ with respect to $N$ whenever there exists a $\mathfrak{q}\in {\operatorname{mAss}}_{\hat{R}_\mathfrak{p}}\hat{N}_\mathfrak{p}$ such that ${\operatorname{Rad}}(I\hat{R}_{\mathfrak{p}}+ \mathfrak{q})=
\mathfrak{p}\hat{R}_{\mathfrak{p}}.$ The set of all [*quintasymptotic prime ideals*]{} of $I$ with respect to $N$ is denoted by $\bar{Q}^*(I,N).$

In the third section, for a multiplicatively closed subset $S$ of $R$, we examine the equivalence between the topologies defined by the filtrations $\{(I^n)_a^{(N)}\}_{n\geq1}$, $\{S((I^n)_a^{(N)})\}_{n\geq1}$, $\{S(((I+ {\operatorname{Ann}}_RN)^n)_a)\}_{n\geq 1}$ and $\{S((I+ {\operatorname{Ann}}_RN)^n)\}_{n\geq 1}$ by using the quintasymptotic prime ideals of $I$ with respect to $N$. Some of these results has been established, by Schenzel in [@Sc; @Sc1], McAdam in [@Mc2] and Mehrvarz et al., in [@MNS], in certain case when $N=R$.

A typical result in this direction is the following:

Let $N$ be a finitely generated module over a Noetherian ring $R$ and let $I$ be an ideal of $R$. Let $S$ be a multiplicatively closed subset of $R$. Then the topologies defined by $(I^n)_a^{(N)}$, $S((I^n)_a^{(N)})$, $S(((I+ {\operatorname{Ann}}_RN)^n)_a)$ and $S((I+ {\operatorname{Ann}}_RN)^n), \, {n\geq 1},$ are equivalent if and only if $S$ is disjoint from each of the quintasymptotic prime ideals of $I$ with respect to $N$.

The proof of Theorem 1.1 is given in Theorem 3.11. One of our tools for proving Theorem 1.1 is the following, which is a characterization of the quintasymptotic prime ideals of $I$ with respect to $N$. In the following, we use $I_a^{\langle N \rangle}$ to denote the union $I_a^{(N)}:_Rs$, where $s$ varies in $R\backslash \bigcup \{\frak p\in {\operatorname{mAss}}_RN/IN\}$; in particular, for every integer $k\geq1$ and every prime ideal $\mathfrak{p}$ of $R$, $$(\mathfrak{p}^k)^{\langle N \rangle}_a= \bigcup _{s\in R\backslash
\mathfrak{p}}((\mathfrak{p}^k)^{(N)}_a:_Rs).$$

\[3.1\] Let $R$ be a Noetherian ring and let $N$ be a finitely generated $R$-module. Let $I \subseteq \mathfrak{p}$ be ideals of $R$ such that $\mathfrak{p}\in {\operatorname{Supp}}(N).$ Then $\mathfrak{p}\in \bar{Q}^*(I, N)$ if and only if there exists an integer $k \geq 0$ such that for all integers $m \geq 0$ $$(I^m)^{(N)}_a :_R \langle \mathfrak{p}\rangle \nsubseteq
(\mathfrak{p}^k)^{\langle N \rangle}_a.$$

Finally in this section we derive the following consequence of Theorem 1.1.

\[3.3\] Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R.$ Then the following conditions are equivalent:

-   $\bar{Q}^*(I, N)= {\operatorname{mAss}}_R N/IN.$

-   The topologies defined by $\{(I^n)^{(N)}_a \}_{n \geq
    0}$ and $\{(I^n)^{\langle N \rangle}_a \}_{n \geq 0}$ are equivalent.

For any ideal $I$ of $R$ and any $R$-module $N$, the i-th [*local cohomology module of $N$ with respect to*]{} $I$ is defined by $$H^i_I(N):= \varinjlim {\operatorname{Ext}}^i_R(R/I^n,N).$$ We refer the reader to [@BS] for basic properties of local cohomology modules. The purpose of the fourth section is to characterize the equivalence between the topologies defined by $(I^n)_a^{(N)}$ and $S((I^n)_a^{(N)}), \, {n\geq1}$ in terms of the top local cohomology module $H^{\dim N}_I(N)$. This will generalize the main result of Marti-Farre [@MF], as an extension of the main results of Call [@Ca Corollary 1.4], Call-Sharp [@CS] and Schenzel [@Sc2 Corollary 4.3].

If $(R, \mathfrak{m})$ is a local (Noetherian) ring and $N$ a finitely generated quasi-unmixed $R$-module of dimension $d$, then $H^d_I(N)= 0$ if and only if there exists a multiplicatively closed subset $S$ of $R$ such that $\mathfrak{m} \cap S \neq \emptyset$ and the topologies induced by $(I^n)_a^{(N)}$ and $S((I^n)_a^{(N)}), \, {n\geq1}$ are equivalent.

The result in Theorem 1.4 is proved in Theorem 4.1. Pursing this point of view further we show that the support of the $(d-1)$-th local cohomology module of a finitely generated $R$-module $N$ is always finite ($d= \dim N$), which will be the strengthened and a generalized version of a corresponding one by Marley ([@M Corollaries 2.4 and 2.5]) and by Naghipour-Sedghi ([@NS Corollary 3.3]).

Assume that $R$ is a Noetherian ring. Let $N$ be a finitely generated $R$-module of dimension $d$ and $I$ an ideal of $R.$ Then ${\operatorname{Supp}}(H^d_I(N)) \subseteq \bar{Q}^*(I, N).$ Moreover, if $(R, \frak m)$ is local, then $${\operatorname{Supp}}(H^{d-1}_I(N)) \subseteq \bar{Q}^*(I,N) \cup \{\mathfrak{m}\}.$$

The proof of Theorem 1.5 is given in Corollaries 4.2 and 4.3.\
Throughout the paper, all rings are commutative, with identity, unless otherwise specified. We shall use $R$ to denote such a ring, $I$ an ideal of $R$, and $N$ a non-zero module over $R.$ If $(R, \mathfrak{m})$ is a Noetherian local ring and $N$ a finitely generated $R$-module, then $\hat{R}$ (resp. $\hat{N}$) denotes the completion of $R$ (resp. $N$) with respect to the $\mathfrak{m}$-adic topology. Then $N$ is said to be [*quasi-unmixed* ]{} if for every $\mathfrak{p}\in
{\operatorname{mAss}}_{\hat{R}}\hat{N}$, the condition $\dim\hat{R}/\mathfrak{p}=
\dim N$ is satisfied. For any ideal $J$ of $R$, the radical of $J$, denoted by ${\operatorname{Rad}}(J)$, is defined to be the set $\{ a\in R:\, a^n\in J \text{ for some }n \in \mathbb{N}\}.$ Moreover we use $V(J)$ to denote the set of prime ideals of $R$ containing $J.$ Finally, for any $R$-module $L$ we shall use ${\operatorname{mAss}}_RL$ to denote the set of minimal elements of ${\operatorname{Ass}}_RL.$ For any unexplained notation or terminology we refer the reader to [@Ma] and [@Na].

Asymptotic behaviour of integral closures of ideals
===================================================

The purpose of this section is to study the asymptotic behaviour of the integral closure of ideals with respect to a finitely generated module $N$ over a Noetherian ring $R$. More precisely we show that the sequences $$\{{\operatorname{Ass}}_R R/(I^n)^{(N)}_a\}_{n\geq1},\,\,   \{{\operatorname{Ass}}_R (I^n)_a^{(N)}/ (I^{n+1})^{(N)}_a\}_{n\geq1}, \,\,\,   \{{\operatorname{Ass}}_R (I^n)_a^{(N)}/ ((I+{\operatorname{Ann}}_RN)^n)_a\}_{n\geq1},$$ of associated prime ideals, are ultimately constant; and pursing this point of view further we show that $A^*_a(I+ {\operatorname{Ann}}_R N)\setminus C^*_a(I, N) \subseteq A^*_a(I, N).$

\[2.1\] Let $R$ be a ring (not necessarily Noetherian), and $N$ a Noetherian $R$-module. Then for any ideal $I$ of $R$, the following statements hold:

-   $I_a + {\rm Ann}_RN \subseteq I^{(N)}_a.$

-   $0^{(N)}_a= {\operatorname{Rad}}( {\operatorname{Ann}}_RN).$

-   $I^{(N)}_a/{\operatorname{Ann}}_RN = (I+ {\operatorname{Ann}}_RN / {\operatorname{Ann}}_RN)_a;$ so that ${\operatorname{Rad}}(I^{(N)}_a) = {\operatorname{Rad}}(I+ {\operatorname{Ann}}_RN).$

-   For any multiplicatively closed subset $S$ of $R$, $(S^{-1}I)^{(S^{-1}N)}_a = S^{-1}(I^{(N)}_a).$

-   $\bigcap_{n\geq1}(I^n)^{(N)}_a= \bigcap\{\mathfrak{p}\in
    {\operatorname{mAss}}_RN\,|\, \mathfrak{p}+ I\neq R\}.$ In particular, if $R$ is local then $$\bigcap_{n\geq1}(I^n)^{(N)}_a={\operatorname{Rad}}( {\operatorname{Ann}}_RN).$$

-   $(I^{(N)}_a J^{(N)}_a)^{(N)}_a = (IJ)^{(N)}_a$, where $J$ is a second ideal of $R$.

\(i) and (ii) follow from the definition. For the proof of (iii) see [@STY Remark 1.6]. (iv) follows from (iii) and [@S Lemma 2.3]. To show (v) use (iii) and [@Mc1 Lemma 3.11]. Finally, in order to prove (vi), use (iii) and the fact that $(KL)_a = (K_aL_a)_a$ for all ideals $K$ and $L$ of $R.$

The following corollary extends McAdam’s result [@Mc2 Lemma 1.4].

\[2.6\] Let $R$ and $N$ be as in Lemma [2.1]{}. Let $I$ be an integrally closed ideal with respect to $N.$ Then $I$ has a primary decomposition each primary component of which is integrally closed with respect to $N.$

The result follows from Lemma \[2.1\] and [@Mc2 Lemma 1.4].

\[2.2\] Let $R$ be a ring (not necessarily Noetherian), and $N$ a Noetherian $R$-module. If $I$ is an ideal of $R$ such that is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$, then $$\begin{aligned}
((I^n)^{(N)}_a :_R I^m )&= ((I^n)^{(N)}_a :_R (I^m)_a) \\
                        &=((I^n)^{(N)}_a :_R (I^m)^{(N)}_a)\\
                        & = (I^{n-m})^{(N)}_a\end{aligned}$$ for all integers $n\,\geq m\,\geq \,0.$

In view of Lemma \[2.1\] it is enough to show that $$((I^n)^{(N)}_a :_R I^m) \subseteq (I^{n-m})^{(N)}_a.$$ To this end, let $x\in R$ be such that $I^m x \subseteq
(I^n)^{(N)}_a.$ To simplify notation let us denote the Noetherian ring $R/{\rm Ann}_RN$ by $\widetilde{R};$ the natural image of $x$ in $\widetilde{R}$ is denoted by $\widetilde{x};$ and for each ideal $J$ of $R$ write $\widetilde{J}$ for the ideal $J+
{\operatorname{Ann}}_RN/{\operatorname{Ann}}_RN.$ Then by Lemma \[2.1\] $$\widetilde{x} \in ((\widetilde{I}^n)_a :
  _{\widetilde{R}} (\widetilde{I})^m).$$ Now, it follows from [@Mc1 Lemma 11.27] that $x\in
(I^{n-m})^{(N)}_a,$ as desired.

We are now ready to state and prove the main results of this section. Namely, we prove that the sequences of associated primes $\{{\operatorname{Ass}}_R R/(I^n)^{(N)}_a\}_{n\geq1}$, $\{{\operatorname{Ass}}_R(I^n)^{(N)}_a/(I^{n+1})^{(N)}_a\}_{n\geq1}$ and $\{{\operatorname{Ass}}_R(I^n)^{(N)}_a/((I+{\operatorname{Ann}}_RN)^n)_a\}_{n\geq 1}$ are increasing and become eventually constant.

\[2.3\] Let $I$ denote an ideal of a ring $R$ (not necessarily Noetherian), and let $N$ be a Noetherian $R$-module. Then the sequence of associated primes $${\operatorname{Ass}}_R R/(I^n)^{(N)}_a ,  n= 1,2, \ldots,$$ is increasing and ultimately constant. Moreover, if $I$ is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$, then the sequence $${\operatorname{Ass}}_R (I^n)^{(N)}_a/
(I^{n+1})^{(N)}_a,  n= 1,2, \ldots,$$ is also increasing and eventually constant.

First assume that $n\geq 1$ is an integer and $\mathfrak{p}\in {\operatorname{Spec}}R.$ In order to simplify notation, we will use $\widetilde{R}$ to denote the commutative Noetherian ring $R/{\operatorname{Ann}}_R N$ and for each ideal $J$ of $R$ we will write $\widetilde{J}$ for the ideal $J+ {\operatorname{Ann}}_RN/{\operatorname{Ann}}_RN$ of $\widetilde{R}.$ Then by Lemma \[2.1\], it is easy to see that $\mathfrak{p}\in {\operatorname{Ass}}_R R/(I^n)^{(N)}_a$ if and only if $\widetilde{\mathfrak{p}}\in {\operatorname{Ass}}_{\widetilde{R}} \, \widetilde{R}
/(\widetilde{I}^n)_a.$ Hence it follows that $$\bigcup _{n\geq 1}{\operatorname{Ass}}_R R/ (I^n)^{(N)}_a= \{\mathfrak{q}\cap R\,
|\, \mathfrak{q} \in A^*_a(\widetilde{I})\}.$$ As, by Ratliff’s Theorem [@R], $A^*_a(\widetilde{I})$ is finite it follows that the set $ \bigcup _{n\geq 1}{\operatorname{Ass}}_R R/
(I^n)^{(N)}_a$ is a finite set. Moreover, since the sequence $\{{\operatorname{Ass}}_{\widetilde{R}} \, \widetilde{R} /(\widetilde{I}^n)_a\}_{n\geq 1}$ is increasing, it turns out that the sequence $\{{\operatorname{Ass}}_R R/(I^n)^{(N)}_a\}_{n\geq 1}$ is increasing, and therefore ultimately constant.

In order to prove the second part, we assume that $I$ is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$. Suppose that $\mathfrak{p}\in
{\operatorname{Ass}}_R R/(I^{n+1})^{(N)}_a.$ Then there exists an $x \in R
\setminus(I^{n+1})^{(N)}_a$ such that $\mathfrak{p}=
((I^{n+1})^{(N)}_a:_R x).$ Since $I \subseteq \mathfrak{p}$ it follows that $Ix \subseteq (I^{n+1})^{(N)}_a.$ Hence by Lemma \[2.2\], $x \in (I^n)^{(N)}_a.$ Whence we get $\mathfrak{p}\in
{\operatorname{Ass}}_R\,(I^n)^{(N)}_a/(I^{n+1})^{(N)}_a.$ Therefore, when $I$ is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$, it follows that $${\operatorname{Ass}}_R R/(I^{n+1})^{(N)}_a=
{\operatorname{Ass}}_R\,(I^n)^{(N)}_a/(I^{n+1})^{(N)}_a$$ for all integers $n\geq 0.$ This finally completes the proof.

\[2.5\] Suppose that $R$ is a Noetherian ring. Let $N$ denote a finitely generated $R$-module and let $I$ be an ideal of $R$ such that is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$. Then the sequence $$\{{\operatorname{Ass}}_R(I^n)^{(N)}_a/(I^n)_a\}_{n\geq 1}$$ is increasing and eventually constant.

Assume first that $\mathfrak{p}\in {\operatorname{Ass}}_R(I^n)^{(N)}_a/(I^n)_a$ for some integer $n \geq 1.$ Without loss of generality we may assume that $(R, \mathfrak{p})$ is a local ring. There exists an element $x\in
(I^n)^{(N)}_a$ such that $\mathfrak{p}= ((I^n)_a:_R x).$ Then, in view of Lemma \[2.2\], $((I^{n+1})_a:_R Ix)$ is a proper ideal of $R$ and $\mathfrak{p}\subseteq(I^{n+1})_a:_R Ix.$ Thus $\mathfrak{p}= ((I^{n+1})_a:_R Ix).$ Because of $Ix
\subseteq (I^{n+1})^{(N)}_a$ it follows that $\mathfrak{p}\in
{\operatorname{Ass}}_R(I^{n+1})^{(N)}_a/(I^{n+1})_a.$ Therefore the sequence $\{{\operatorname{Ass}}_R(I^n)^{(N)}_a/(I^n)_a\}_{n\geq 1}$ is increasing. As $$\bigcup_{n\geq1} {\operatorname{Ass}}_R(I^n)^{(N)}_a/(I^n)_a \subseteq
A^*_a(I),$$ and $A^*_a(I)$ is finite, we deduce that it is eventually constant for large $n.$

Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R$ such that is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$. Then the sequence $$\{{\operatorname{Ass}}_R (I^n)^{(N)}_a/((I+ {\operatorname{Ann}}_RN)^n)_a\}_{n\geq 1}$$ is increasing and ultimately constant.

This follows from Theorem \[2.5\], when $I$ is replaced by $I+{\operatorname{Ann}}_RN$, since the integral closure with respect to $N$ of their powers are the same.

\[4.4\] Suppose that $R$ is a Noetherian ring. Let $I$ be an ideal of $R.$ Let $N$ denote a finitely generated $R$-module. The eventual constant values of the sequences $$\{{\operatorname{Ass}}_R R/(I^n)^{(N)}_a \}_{n\geq1} \text{ and }
\{{\operatorname{Ass}}_R(I^n)^{(N)}_a/((I+ {\operatorname{Ann}}_R N)^n)_a \}_{n\geq1}$$ will be denoted by $A^*_a(I, N)$ and $C^*_a(I, N)$ respectively.

It is easy to see that $A^*_a(I, N)$ and $C^*_a(I, N)$ are stable under localization. Moreover, ${\operatorname{mAss}}_RN/IN\subseteq A^*_a(I, N)$ and $A^*_a(0, N)={\operatorname{mAss}}_R N.$

\[4.5\] Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $N$ a finitely generated $R$-module. Then $A^*_a(I+ {\operatorname{Ann}}_R N)\setminus C^*_a(I, N) \subseteq A^*_a(I, N).$

Let $\mathfrak{p}\in A^*_a(I+ {\operatorname{Ann}}_R N)\setminus C^*_a(I, N).$ Because $A^*_a(I+ {\operatorname{Ann}}_R N),$ $C^*_a(I, N)$ and $A^*_a(I, N)$ behave well under localization we may assume that $(R, \mathfrak{p})$ is a local ring. Let $\mathfrak{p} = ((I+ {\operatorname{Ann}}_R N)^n)_a:_R x)$ for some $x\in R$ and for large $n.$ Because $\mathfrak{p} \subseteq
((I^n)^{(N)}_a:_R x)$ and $\mathfrak{p} \not\in C^*_a(I, N)$ it follows that $\mathfrak{p}=( (I^n)^{(N)}_a:_R x).$ Hence $\mathfrak{p} \in A^*_a(I, N)$, as required.

\[4.6\] Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R$ such that is not contained in any of the minimal prime ideals of ${\operatorname{Ann}}_RN$ and $(I^nR_\mathfrak{p})^{(N_\mathfrak{p})}_a = ((IR_\mathfrak{p}+{\operatorname{Ann}}_{R_\mathfrak{p}}N_\mathfrak{p})^n)_a$, for large $n$ and for all $\mathfrak{p} \in C^*_a(I, N).$ Then $(I^n)^{(N)}_a = ((I+{\operatorname{Ann}}_RN)^n)_a$ for large $n$. For this, let $C^*_a(I, N)= \{\mathfrak{p}_1, \ldots, \mathfrak{p}_t\}.$ Choose an integer $k$ such that $$C^*_a(I, N)= {\operatorname{Ass}}_R (I^n)^{(N)}_a /((I+ {\operatorname{Ann}}_RN)^n)_a$$ for all $n \geq k$, and let $(I^nR_{\mathfrak{p}_i})^{(N_{\mathfrak{p}_i})}_a=
((IR_{\mathfrak{p}_i}+ {\operatorname{Ann}}_{R_\mathfrak{p_i}}
N_\mathfrak{p_i})^n)_a$ for all $i= 1, \ldots, t.$ Now, if $(I^n)^{(N)}_a\neq ((I+ {\operatorname{Ann}}_RN)^n)_a,$ then there exists $x\in (I^n)^{(N)}_a\setminus ((I+ {\operatorname{Ann}}_RN)^n)_a,$ and so $((I+ {\operatorname{Ann}}_RN)^n)_a:_R x$ is a proper ideal of $R$. As $R$ is Noetherian, there is $r \in R$ such that $\mathfrak{p}:=((I+ {\operatorname{Ann}}_RN)^n)_a:_R rx$ is a prime ideal of $R,$ and hence $\mathfrak{p}\in
C^*_a(I, N).$ Thus $(I^nR_\mathfrak{p})^{(N_\mathfrak{p})}_a=
((IR_\mathfrak{p}+ {\operatorname{Ann}}_{R_\mathfrak{p}}N_\mathfrak{p})^n)_a$; so that $x/1 \in ((I+ {\operatorname{Ann}}_RN)^n)_aR_\mathfrak{p}.$ Hence there is $s
\in R \setminus \mathfrak{p}$ such that $sx \in ((I+
{\operatorname{Ann}}_RN)^n)_a.$ That is $s \in ((I+ {\operatorname{Ann}}_RN)^n)_a:_R x \subseteq
\mathfrak{p}$, which is a contradiction.

Quintasymptotic primes and ideal topologies
===========================================

In this section we study the equivalence of the topologies defined by $(I^n)_a^{(N)}$, $S((I^n)_a^{(N)})$, $S(((I+ {\operatorname{Ann}}_RN)^n)_a)$ and $S((I+ {\operatorname{Ann}}_RN)^n), \, {n\geq 1},$ by using the quintasymptotic prime ideals of $I$ with respect to $N$. The main results are Proposition 3.10 and Theorem 3.11. As a consequence we show that $\bar{Q}^*(I, N)= {\operatorname{mAss}}_R N/IN$ if and only if the topologies $(I^n)^{(N)}_a$ and $(I^n)^{\langle N \rangle}_a, \, {n \geq 1},$ are equivalent. We begin with the following elementary result.

\[2.7\] Let $R$ be a Noetherian ring and $N$ a finitely generated $R$-module. Let $T$ be a faithfully flat Noetherian ring extension of $R.$ Then, for any ideal $I$ of $R,$ $$(IT)^{(N\otimes_RT)}_a \cap R= I^{(N)}_a.$$

Let $x\in (IT)^{(N\otimes_RT)}_a\cap R.$ Then, in view of [@STY Corollary 1.5], there is an integer $n\geq1$ such that $$(IT+Tx)^{n+1}(N\otimes_RT)= IT(IT+Tx)^n(N\otimes_RT).$$ Hence $$(I+Rx)^{n+1}(N\otimes_RT)= I(I+Rx)^n(N\otimes_RT).$$ Therefore $$(I+Rx)^{n+1}N \otimes_RT= I(I+Rx)^nN \otimes_RT.$$ Now, by faithfully flatness, we deduce that $(I+Rx)^{n+1}N= I(I+Rx)^nN$; hence $x\in I^{(N)}_a$, by [@STY Corollary 1.5]. Therefore the conclusion follows, since the opposite inclusion is clear by the faithfully flatness of $T$ over $R$.

\[2.8\] Before continuing, let us fix two notations, which are employed by Schenzel in [@Sc3] and McAdam in [@Mc2] in the case $N= R$, respectively.

For any multiplicatively closed subset $S$ of $R$ and for each ideal $J$ of $R$ we use $S(J)$ to denote the ideal $\cup_{s\in
S}(J:_Rs).$ Note that $${\operatorname{Ass}}_RR/S(J)= \{ \mathfrak{p}\in {\operatorname{Ass}}_R R/J \,:\,\mathfrak{p}\cap
S=\emptyset\}.$$ In the case $N$ is a finitely generated $R$-module and $S=R\backslash \bigcup \{\frak p\in {\operatorname{mAss}}_RN/JN\}$, we use $J_a^{\langle N \rangle}$ to denote the ideal $S(J_a^{(N)})$. In particular, for every integer $k\geq1$ and every prime ideal $\mathfrak{p}$ of $R$, we have $$(\mathfrak{p}^k)^{\langle N \rangle}_a= \bigcup _{s\in R\backslash
\mathfrak{p}}((\mathfrak{p}^k)^{(N)}_a:_Rs).$$

\[2.9\] Let $R$ be a Noetherian ring and let $N$ be a finitely generated $R$-module.

-   If $(R, \mathfrak{m})$ is local and $\mathfrak{p}\in {\operatorname{mAss}}_RN,$ then there exists an element $x\in R$ not in $\mathfrak{p}$ such that for every ideal $J$ of $R$ with $\mathfrak{m}$ is minimal over $J+ \mathfrak{p}$, $x\in
    J+ {\operatorname{Ann}}_RN$ or $\mathfrak{m}\in {\operatorname{Ass}}_RR/J+ {\operatorname{Ann}}_RN$.

-   If $\mathfrak{p}\in {\operatorname{Spec}}R$ and $\mathfrak{q}\in
    {\operatorname{mAss}}_RN$ with $\mathfrak{q}\subseteq\mathfrak{p},$ then there is an integer $k\geq1$ such that $\mathfrak{p}\in
    {\operatorname{Ass}}_RR/J+ {\operatorname{Ann}}_RN$ for any ideal $J$ of $R$ with $J\subseteq
    (\mathfrak{p}^k)^{\langle N \rangle}_a$ and $\mathfrak{p}\in {\operatorname{mAss}}_RR/J+
    \mathfrak{q}.$

In order to show (i), let $$\frak q_1\cap \dots \cap \frak q_t= {\operatorname{Ann}}_RN,$$ be an irredundant primary decomposition of the ideal ${\operatorname{Ann}}_RN$ with $\frak q_1$ $\frak p$-primary. (Note that $\mathfrak{p}\in {\operatorname{Ass}}_RR/{\operatorname{Ann}}_RN$.) It follows from $\mathfrak{p}\in {\operatorname{mAss}}_RN$ that $\bigcap _{i=2}^t\frak q_i\nsubseteq \frak p$. Hence there exists $x\in \bigcap _{i=2}^t\frak q_i$ such that $x\not\in \frak p$. Now, let $J$ be any ideal of $R$ such that ${\rm Rad}(J+ \mathfrak{p})= \mathfrak{m}$ and that $\mathfrak{m}\not\in {\operatorname{Ass}}_RR/J+ {\operatorname{Ann}}_RN$. It is enough for us to show that $x\in J+ {\operatorname{Ann}}_RN$. To this end, let $$Q_1\cap \dots \cap Q_l=J+{\operatorname{Ann}}_RN,$$ be an irredundant primary decomposition of the ideal $J+{\operatorname{Ann}}_RN$, with $Q_i$ is $\frak p_i$-primary ideal, for all $i=1, \dots, l$. Then $\frak m\neq \frak p_i$, for all $i=1, \dots, l$, and so it follows from ${\rm Rad}(J+ \mathfrak{p})= \mathfrak{m}$ that $\frak p\nsubseteq \frak p_i$. Hence $\frak p^v\nsubseteq \frak p_i$, where $v\geq1$ is an integer such that $\frak p^v\subseteq \frak q_1$. Therefore, because of $x\frak q_1\subseteq {\operatorname{Ann}}_RN$, it follows that $x\frak p^v\subseteq  Q_i$, for all $i=1, \dots, l$. Consequently $x\in Q_1\cap \dots \cap Q_l$, so that $x\in J+{\operatorname{Ann}}_RN,$ as required.

In order to prove (ii), without loss of generality, we may assume that $(R, \mathfrak{p})$ is local. Then $(\mathfrak{p}^k)^{\langle N \rangle}_a= (\mathfrak{p}^k)^{(N)}_a.$ Now, let $x$ be as in (i). Then, in view of Lemma \[2.1\](v), there exists an integer $k\geq1$ such that $x \not\in (\mathfrak{p}^k)^{(N)}_a.$ Therefore if $J$ is an ideal of $R$ such that $J\subseteq
(\mathfrak{p}^k)^{\langle N \rangle}_a$ and $\mathfrak{p}\in {\operatorname{mAss}}_RR/J+
\mathfrak{q}$, then $x\not\in J+{\operatorname{Ann}}_RN,$ and so it follows from (i) that $\mathfrak{p}\in {\operatorname{Ass}}_RR/J+ {\operatorname{Ann}}_RN$.

\[2.10\] Let $I$ be an ideal of a Noetherian ring $R$ and $S$ a multiplicatively closed subset of $R.$ Then for any finitely generated $R$-module $N$, $$\bigcap_{n\geq1}S((I^n)^{(N)}_a)= \bigcap \{\mathfrak{p}\in
{\operatorname{mAss}}_RN\,\mid\,(I+ \mathfrak{p})\cap S= \emptyset \}.$$

Let $x\in \bigcap_{n\geq1}S((I^n)^{(N)}_a).$ Then, for all $n\geq1$, there exists $s\in S$ such that $sx \in (I^n)^{(N)}_a.$ Now let $\mathfrak{p}\in {\operatorname{mAss}}_R N$ be such that $(\mathfrak{p}+ I)\cap S=
\emptyset.$ Then, it follows from Lemma \[2.1\](v) that $x\in \mathfrak{p}.$

Conversely, suppose that $x\in \mathfrak{p}$ for all $\mathfrak{p}\in {\operatorname{mAss}}_RN$ with $(\mathfrak{p}+ I)\cap S=
\emptyset.$ Then, by virtue of Lemma \[2.1\](v), $x/1\in
(S^{-1}I^n)^{(S^{-1}N)}_a$ for all $n\geq1.$ Hence, in view of Lemma \[2.1\](iv), $x/1\in
S^{-1}((I^n)^{(N)}_a)$, and so $sx \in (I^n)^{(N)}_a$ for some $s\in S.$ Consequently, we have $x\in S((I^n)^{(N)}_a)$, as required.

\[2.11\] Let $R$ be a Noetherian ring and let $N$ be a finitely generated $R$-module. Let $I$ and $J$ be ideals of $R.$ Then $$\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle J\rangle)=
\bigcap\{\mathfrak{p}\in {\operatorname{mAss}}_RN\,\mid\,J\nsubseteq {\rm Rad}(I+
\mathfrak{p})\}.$$

In view of Theorem \[2.3\] the set $A^*_a(I,N):= \bigcup _{n\geq1}
{\operatorname{Ass}}_RR/(I^n)^{(N)}_a$ is finite. Let $A^*_a(I,N)=
\{\mathfrak{p}_1,\ldots, \mathfrak{p}_t\}.$ Let $r$ be an integer such that $0\leq r\leq t$ and $J\nsubseteq\bigcup_{i=1}^r
\mathfrak{p}_i$ but $J\subseteq \bigcap_{i=r+1}^t \mathfrak{p}_j.$ Then there exists an element $s\in J$ such that $s\not\in\bigcup_{i=1}^r \mathfrak{p}_i.$ Suppose $S= \{ s^{i}
\,\mid\, i\geq0 \}.$ Then it easily seen that $((I^n)^{(N)}_a:_R \langle J\rangle)= S((I^n)^{(N)}_a)$ for each integer $n\geq1.$ Now in view of Proposition \[2.10\] it is enough to show that $J\subseteq {\rm Rad}(I+ \mathfrak{p})$ if and only if $s\in {\rm Rad}(I+ \mathfrak{p})$ for each $\mathfrak{p}\in {\operatorname{mAss}}_R N.$ To do this, as $s\in J$ one direction is clear. For other direction, let $\mathfrak{q}$ be a minimal prime ideal over $I+ \mathfrak{p}.$ Then, as $s\in {\rm Rad}(I+ \mathfrak{p})$ and $I+ \mathfrak{p}\subseteq \frak q$, we have $s\in\mathfrak{q}$, and hence in view of the choice of $s$, it suffices to show that $\mathfrak{q}\in A^*_a(I,N).$ By virtue of Lemma \[2.1\], we may assume that $R$ is local with maximal ideal $\mathfrak{q}.$ Let $x$ be as in the Proposition \[2.9\]. Then by Lemma \[2.1\], there is an integer $n\geq1$ such that $x\not\in (I^n)^{(N)}_a.$ Now it is easy to see that $\mathfrak{q}$ is minimal over $
(I^n)^{(N)}_a+\mathfrak{p}.$ Therefore, it follows from Proposition \[2.9\] that $\mathfrak{q}\in {\operatorname{Ass}}_R R/(I^n)^{(N)}_a,$ and so $\mathfrak{q}\in A^*_a(I,N)$, as required.\

Let $R$ be a Noetherian ring and $I$ an ideal of $R$. Let $N$ be a finitely generated $R$-module and $\mathfrak{p}\in {\operatorname{mAss}}_RN$. Then ${\operatorname{mAss}}_RR/(I+\frak p) \subseteq A^*_a(I,N).$

The assertion follows from the last argument in the proof of Theorem \[2.11\].

\[3.7\] Let $(R,\mathfrak{m})$ be a local (Noetherian) ring and let $N$ be a finitely generated $R$-module. Then, for any proper ideal $I$ of $R$,

$$\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle \mathfrak{m}\rangle)=
\bigcap\{\mathfrak{p}\in {\operatorname{mAss}}_RN\,\mid\,{\rm Rad}(I+
\mathfrak{p})\subsetneqq \mathfrak{m}\}.$$

The assertion follows from Theorem \[2.11\].\

\[2.13\] Let $(R,\mathfrak{m})$ be a local (Noetherian) ring, $I$ a proper ideal of $R$ and $N$ a finitely generated $R$-module. Then the following conditions are equivalent:

-   For all $\mathfrak{p}\in {\operatorname{mAss}}_RN,
    {\operatorname{Rad}}(I+\mathfrak{p})\neq\mathfrak{m}.$

-   $\bigcap_{n\geq1}(((I+{\operatorname{Ann}}_RN)^n)_a:_R\langle
    \mathfrak{m}\rangle)\subseteq {\rm Rad}({\operatorname{Ann}}_RN).$

-   $\bigcap_{n\geq1}((I+{\operatorname{Ann}}_RN)^n:_R\langle
    \mathfrak{m}\rangle)\subseteq {\rm Rad}({\operatorname{Ann}}_RN).$

-   $\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle
    \mathfrak{m}\rangle)= {\operatorname{Rad}}({\operatorname{Ann}}_RN).$

\(i) $\Longrightarrow$ (ii): In view of Corollary \[3.7\], $$\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle\mathfrak{m}\rangle)= {\operatorname{Rad}}({\operatorname{Ann}}_RN).$$ Hence as $$\bigcap_{n\geq1}(((I+{\operatorname{Ann}}_RN)^n)_a:_R\langle \mathfrak{m}\rangle)
\subseteq \bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle
\mathfrak{m}\rangle),$$ it follows that (ii) holds. The implication (ii) $\Longrightarrow$ (iii) is trivial.

In order to show that (iii) $\Longrightarrow$ (iv), suppose the contrary, that is (iv) is not true. Then $${\operatorname{Rad}}({\operatorname{Ann}}_RN) \subsetneqq
\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle \mathfrak{m}\rangle).$$ Hence, according to Corollary 3.7, there exists $\mathfrak{p}\in
{\operatorname{mAss}}_RN$ such that ${\operatorname{Rad}}(I+ \mathfrak{p})= \mathfrak{m}.$ Moreover, applying the assumption it is easily seen that ${\operatorname{Rad}}(I+ {\operatorname{Ann}}_RN) \neq
\mathfrak{m}.$ Therefore $${\operatorname{Rad}}((I+ {\operatorname{Ann}}_RN)^n:_R\langle \mathfrak{m} \rangle + \mathfrak{p})=
\mathfrak{m},$$ for each integer $n\geq1$.

Now, let $x$ be as in the Proposition \[2.9\]. Since $\mathfrak{m}\not\in {\operatorname{Ass}}_R R/((I+{\operatorname{Ann}}_RN)^n:_R\langle \mathfrak{m}
\rangle)$ it follows that $$x \in \bigcap_{n\geq1}((I+{\operatorname{Ann}}_RN)^n:_R\langle \mathfrak{m}\rangle).$$ Thus $x\in {\rm {\operatorname{Rad}}}({\operatorname{Ann}}_RN)$, i.e., $x\in \mathfrak{p}$ which is a contradiction.

Finally, the implication (iv) $\Longrightarrow$ (i) follows from Corollary 3.7.

\[2.14\] Let $(R,\mathfrak{m})$ be a local (Noetherian) ring, let $N$ be a finitely generated $R$-module, and let $I$ be an ideal of $R.$ Then the following conditions are equivalent:

-   $\bigcap_{n\geq1}((I^n \hat{R})^{(\hat{N})}_a:_{\hat{R}}
    \langle \mathfrak{m} \hat{R} \rangle)= {\operatorname{Rad}}({\operatorname{Ann}}_ {\hat{R}}
    \hat{N}).$

-   For all integers $n\geq1$ there exists an integer $k\geq1$ such that $$(I+{\operatorname{Ann}}_RN)^k:_R\langle \mathfrak{m}\rangle \subseteq
    (\mathfrak{m}^n)^{(N)}_a.$$

-   For all integers $n\geq1$ there exists an integer $k\geq1$ such that $$((I+{\operatorname{Ann}}_RN)^k)_a:_R\langle \mathfrak{m}\rangle \subseteq
    (\mathfrak{m}^n)^{(N)}_a.$$

-   For all integers $n\geq1$ there exists an integer $k\geq1$ such that $$(I^k)^{(N)}_a:_R\langle \mathfrak{m} \rangle \subseteq
    (\mathfrak{m}^n)^{(N)}_a.$$

Without loss of generality we may assume that $(R,\mathfrak{m})$ is a complete local ring as follows by virtue of the faithfully flatness of $\hat{R}$. Now suppose that (i) is satisfied, then $$\bigcap_{n\geq1}((I^n)^{(N)}_a:_R \langle \mathfrak{m} \rangle/
{\operatorname{Rad}}({\operatorname{Ann}}_AN))= 0.$$ As $R/{\operatorname{Ann}}_RN$ is a complete local ring, Chevalley’s Theorem (see [@Na Theorem 30.1]) says that for all $n\geq1$ there exists an integer $k\geq1$ such that $$((I^k)^{(N)}_a:_R \langle \mathfrak{m}
\rangle)/{\operatorname{Rad}}({\operatorname{Ann}}_AN)\subseteq (\mathfrak{m}/{\operatorname{Rad}}({\operatorname{Ann}}_RN))^n.$$ Therefore $$((I^k)^{(N)}_a:_R \langle \mathfrak{m}\rangle) \subseteq
\mathfrak{m}^n+ {\operatorname{Rad}}({\operatorname{Ann}}_RN) \subseteq
(\mathfrak{m}^n)^{(N)}_a,$$ and so the statement (iv) is shown to be true.

The conclusions (iv) $\Longrightarrow$ (iii) and (iii) $ \Longrightarrow$ (ii) are obviously true. So, in order to complete the proof, it is enough to show that (ii) $\Longrightarrow$ (i). To this end, suppose that for all $n\geq1$ there exists an integer $k\geq1$ such that $$(I+{\operatorname{Ann}}_RN)^k:_R\langle \mathfrak{m}\rangle \subseteq
(\mathfrak{m}^n)^{(N)}_a.$$ Then in view of Lemma \[2.1\] we have $$\bigcap_{k\geq1}((I+{\operatorname{Ann}}_RN)^k:_R \langle \mathfrak{m}\rangle)
\subseteq {\operatorname{Rad}}({\operatorname{Ann}}_RN).$$ Now use Proposition 3.8 to complete the proof.

We are now ready to prove the first main result of this section. In fact there is a characterization of the quintasymptotic prime ideals of $I$ with respect to $N$, which is a generalization of [@Mc2 Proposition 3.5].

\[3.1\] Let $R$ be a Noetherian ring and let $N$ be a finitely generated $R$-module. Let $I \subseteq \mathfrak{p}$ be ideals of $R$ such that $\mathfrak{p}\in {\operatorname{Supp}}(N).$ Then the following conditions are equivalent:

-   $\mathfrak{p}\in \bar{Q}^*(I, N).$

-   There exists an integer $k\geq0$ such that $\mathfrak{p}\in {\operatorname{Ass}}_R R/J+{\operatorname{Ann}}_RN$ for any ideal $J$ of $R$ with $I\subseteq {\operatorname{Rad}}(J)$ and $J\subseteq (\mathfrak{p}^k)^{\langle N \rangle}_a.$

-   There exists an integer $k \geq 0$ such that for all integers $m \geq 0$ $$(I+ {\operatorname{Ann}}_RN)^m :_R \langle \mathfrak{p}\rangle \nsubseteq
    (\mathfrak{p}^k)^{\langle N \rangle}_a.$$

-   There exists an integer $k \geq 0$ such that for all integers $m \geq 0$ $$((I+ {\operatorname{Ann}}_RN)^m)_a :_R \langle \mathfrak{p}\rangle \nsubseteq
    (\mathfrak{p}^k)^{\langle N \rangle}_a.$$

-   There exists an integer $k \geq 0$ such that for all integers $m \geq 0$ $$(I^m)^{(N)}_a :_R \langle \mathfrak{p}\rangle \nsubseteq
    (\mathfrak{p}^k)^{\langle N \rangle}_a.$$

\(i) $\Longrightarrow$ (ii): Let $\mathfrak{p}\in
\bar{Q}^*(I, N).$ Then there exists a prime ideal $\mathfrak{q}\in
{\operatorname{mAss}}_{\hat{R_{\mathfrak{p}}}}\hat{N_{\mathfrak{p}}}$ such that ${\operatorname{Rad}}(I\hat{R_{\mathfrak{p}}}+ \mathfrak{q})=
\mathfrak{p}\hat{R_{\mathfrak{p}}}.$ Now, let $k$ be as in the Proposition \[2.9\](ii), applied to $\mathfrak{q}\in
{\operatorname{mAss}}_{\hat{R_{\mathfrak{p}}}}\hat{N_{\mathfrak{p}}}.$ Let $J$ be any ideal of $R$ such that $I\subseteq {\operatorname{Rad}}(J)$ and $J\subseteq
(\mathfrak{p}^k)^{\langle N \rangle}_a.$ Then $I\hat{R_{\mathfrak{p}}} \subseteq
{\operatorname{Rad}}(J\hat{R_{\mathfrak{p}}})$ and $J\hat{R_{\mathfrak{p}}}\subseteq (\mathfrak{p}^k
\hat{R_{\mathfrak{p}}})^{(\hat{N_{\mathfrak{p}}})}_a$ by virtue of Lemma \[2.7\]. Since $\mathfrak{p}\hat{R_{\mathfrak{p}}}$ is the maximal ideal of $\hat{R_{\mathfrak{p}}}$, it follows that $(\mathfrak{p}\hat{R_{\mathfrak{p}}})^{(\hat{N_{\mathfrak{p}}})}_a=
\mathfrak{p}\hat{R_{\mathfrak{p}}}$, and so $J\hat{R_{\mathfrak{p}}}$ is a proper ideal of $\hat{R_{\mathfrak{p}}}.$ Thus ${\operatorname{Rad}}(J\hat{R_{\mathfrak{p}}}+
\mathfrak{q})= \mathfrak{p}\hat{R_{\mathfrak{p}}}$. Hence Proposition \[2.9\] shows that $\mathfrak{p}\hat{R_{\mathfrak{p}}}\in {\operatorname{Ass}}_{\hat {R_{\mathfrak{p}}}} \hat {R_{\mathfrak{p}}}/J\hat {R_{\mathfrak{p}}}+{\operatorname{Ann}}_{\hat {R_{\mathfrak{p}}}}\hat {N_{\mathfrak{p}}}$, and so by [@Ma Theorem 23.2] we have $\mathfrak{p}\in {\operatorname{Ass}}_R R/J+{\operatorname{Ann}}_RN$. That is (ii) holds.

The implication (ii) $\Longrightarrow$ (iii) follows easily from the fact that $$\mathfrak{p}\not\in {\operatorname{Ass}}_R R/((I+ {\operatorname{Ann}}_RN)^n :_R\langle
\mathfrak{p}\rangle),$$ for all integers $n \geq 0.$

The conclusions (iii) $\Longrightarrow$ (iv) and (iv) $\Longrightarrow$ (v) are obviously true.

Finally, in order to complete the proof, we have to show the implication (v) $\Longrightarrow$ (i). To this end, suppose that there is an integer $k \geq 0$ such that $((I^m)^{(N)}_a :_R \langle \mathfrak{p}\rangle) \nsubseteq
(\mathfrak{p}^k)^{\langle N \rangle}_a$, for all integers $m \geq 0$. Then by Lemma \[2.7\], $$((I^m\hat{R_{\mathfrak{p}}})^{(\hat{N_{\mathfrak{p}}})}_a :_
{\hat{R_{\mathfrak{p}}}} \langle \mathfrak{p}\hat{R_{\mathfrak{p}}}\rangle) \nsubseteq
(\mathfrak{p}^k\hat{R_{\mathfrak{p}}})^{(\hat{N_{\mathfrak{p}}})}_a.$$ Hence, in view of Proposition \[2.13\], there is a $\mathfrak{q}\in {\operatorname{mAss}}_{\hat{R_{\mathfrak{p}}}}
\hat{N_{\mathfrak{p}}}$ such that ${\operatorname{Rad}}(I\hat{R_{\mathfrak{p}}}+
\mathfrak{q})= \mathfrak{p}\hat{R_{\mathfrak{p}}}$, and so $\mathfrak{p}\in \bar{Q}^*(I, N)$, as required.

We are now ready to state and prove the second main theorem of this section, which is a characterization of the equivalence between the topologies $\{(I^n)_a^{(N)}\}_{n\geq1}$, $\{S((I^n)_a^{(N)})\}_{n\geq1}$, $\{S(((I+ {\operatorname{Ann}}_RN)^n)_a)\}_{n\geq 1}$ and $\{S((I+ {\operatorname{Ann}}_RN)^n)\}_{n\geq 1}$ in terms of the quintasymptotic primes of $I$ with respect to $N$. This will generalize the main result of McAdam [@Mc2].

\[3.2\] Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R.$ Then for any multiplicatively closed subset $S$ of $R$ the following are equivalent:

-   $S \subseteq R\setminus\bigcup\{\mathfrak{p}\in
    \bar{Q}^*(I, N)\}.$

-   The topologies defined by $\{S((I^n)^{(N)}_a)\}_{n
    \geq 0}$ and $\{(I^n)^{(N)}_a \}_{n \geq 0}$ are equivalent.

-   The topology defined by $\{S(((I+ {\operatorname{Ann}}_RN)^n)_a)\}_{n
    \geq 0}$ is finer than the topology defined by $\{(I^n)^{(N)}_a
    \}_{n \geq 0}.$

-   The topology defined by $\{S((I+ {\operatorname{Ann}}_R N)^n) \}_{n
    \geq 0}$ is finer than the topology defined by $\{(I^n)^{(N)}_a
    \}_{n\geq0}.$

-   For all $\mathfrak{p}\in {\operatorname{Supp}}(N) \cap V(I)$ the topology defined by $\{S((I^n)^{(N)}_a)\}_{n\geq0}$ is finer than the topology defined by $\{(\mathfrak{p}^n)^{\langle N \rangle}_a)\}_{n \geq
    0}.$

-   For all $\mathfrak{p}\in {\operatorname{Supp}}(N) \cap V(I)$ the topology defined by $\{S(((I+ {\operatorname{Ann}}_RN)^n)_a)\}_{n \geq 0}$ is finer than the topology defined by $\{(\mathfrak{p}^n)^{\langle N \rangle}_a)\}_{n \geq 0}.$

-   For all $\mathfrak{p}\in {\operatorname{Supp}}(N) \cap V(I)$ the topology defined by $\{S((I+{\operatorname{Ann}}_R N)^n) \}_{n \geq 0}$ is finer than the topology defined by $\{(\mathfrak{p}^n)^{\langle N \rangle}_a)\}_{n \geq 0}.$

In order to show (i) $\Longrightarrow$ (ii), let $\mathfrak{p}\in
{\operatorname{Spec}}R$ with $I+ {\operatorname{Ann}}_RN \subseteq \mathfrak{p}$ and let $l \geq
1.$ We first show that there exists an integer $m \geq 1$ such that $S((I^m)^{(N)}_a) \subseteq (\mathfrak{p}^l)^{\langle N \rangle}_a.$ To do this, let $S'$ be the natural image of $S$ in $R_\mathfrak{p}.$ Since $\bar{Q}^*(I, N)$ behaves well under localization, we have $S' \subseteq R_\mathfrak{p}\setminus\cup\{\mathfrak{q}\in
\bar{Q}^*(IR_\mathfrak{p}, N_\mathfrak{p})\}.$ Moreover, it is easy to see that $S'((I^mR_\mathfrak{p})^{(N_\mathfrak{p})}_a)
\subseteq (\mathfrak{p}^lR_\mathfrak{p})^{\langle N_\mathfrak{p}\rangle }_a$ implies $S((I^n)^{(N)}_a) \subseteq (\mathfrak{p}^n)^{\langle N \rangle}_a.$ Therefore we may assume that $R$ is local with maximal ideal $\mathfrak{p}.$ Then $(\mathfrak{p}^n)^{\langle N \rangle}_a=
(\mathfrak{p}^n)^{(N)}_a.$ By view of Lemma \[2.7\] and [@Ah Proposition 3.8] we may assume in addition that $R$ is complete. Whence, in view of [@Ah Lemma 3.5], for any $\mathfrak{q} \in {\operatorname{mAss}}_R N$, $S$ is disjoint from $I+ \mathfrak{q}.$ Therefore, by Proposition \[2.10\] we have $
\bigcap_{n \geq 1} S((I^n)^{(N)}_a) = {\operatorname{Rad}}({\operatorname{Ann}}_RN).$ Consequently $$\bigcap_{n\geq1}S((I^n)^{(N)}_a)/ {\operatorname{Rad}}({\operatorname{Ann}}_R N)= 0.$$ As the ring $R/{\operatorname{Rad}}({\operatorname{Ann}}_RN)$ is complete, Chevalley’s Theorem (see [@Na Theorem 30.1]) implies the existence of an integer $m \geq 1$ such that $$S((I^m)^{(N)}_a)/{\operatorname{Rad}}({\operatorname{Ann}}_RN) \subseteq
(\mathfrak{p}/{\operatorname{Rad}}({\operatorname{Ann}}_RN))^l.$$ Hence $$S((I^m)^{(N)}_a) \subseteq \mathfrak{p}^l+ {\operatorname{Rad}}({\operatorname{Ann}}_RN) \subseteq
(\mathfrak{p}^l)^{(N)}_a.$$ Now, in view of Corollary \[2.6\], we can consider $\mathfrak{q}_1\cap \dots \cap \mathfrak{q}_n$ a minimal primary decomposition of $(I^l)^{(N)}_a$ where $\mathfrak{q}_i$ is $\mathfrak{p}_i$-primary and integrally closed with respect to $N$ for every $i= 1, \ldots, n.$ Then there exists an integer $l_i$ such that $\mathfrak{p}_i^{l_i} \subseteq
\mathfrak{q}_i$ for $i= 1, \ldots, n$, and moreover for some $m_i$ we have $S((I^{m_i})^{(N)}_a)\subseteq
(\mathfrak{p}_i^{l_i})^{\langle N \rangle}_a.$ Let $m=
\max\{m_1,\ldots, m_n\}.$ Then we deduce that $S((I^m)^{(N)}_a)\subseteq (\mathfrak{p}_i^{l_i})^{\langle N \rangle}_a$ for each $1\leq i\leq n.$ On the other hand, we have $$(\mathfrak{p}_i^{l_i})^{\langle N \rangle}_a \subseteq \bigcup _{s\in
R\setminus \mathfrak{p}_i}((\mathfrak{q}_i)^{(N)}_a:_Rs)=
\mathfrak{q}_i,$$ and therefore $S((I^m)^{(N)}_a)\subseteq
\bigcap_{i=1}^n \mathfrak{q}_i.$ This completes the proof of (ii).

The implications (ii) $\Longrightarrow$ (iii) $\Longrightarrow$ (iv) are obviously true. In order to prove the implication (iv)$\Longrightarrow$ (v), it is enough to show that $$S \subseteq R\setminus \bigcup \{\mathfrak{p}\in \bar{Q}^*(I,
N)\}.$$ To do this, let $\mathfrak{p}\in \bar{Q}^*(I, N).$ Then, by Proposition \[3.1\] there exists an integer $k \geq 0$ such that $((I+
{\operatorname{Ann}}_R N)^m :_R \langle \mathfrak{p}\rangle) \nsubseteq
(\mathfrak{p}^k)^{\langle N \rangle}_a$ for all integers $m \geq 0.$ On the other hand, by the assumption there is an integer $l \geq 0$ such that $S((I+ {\operatorname{Ann}}_RN)^l) \subseteq (I^k)^{(N)}_a.$ Therefore, $$(I+ {\operatorname{Ann}}_RN)^l :_R \langle \mathfrak{p}\rangle \nsubseteq
S((I+ {\operatorname{Ann}}_R N)^l).$$ Then it is readily to see that $\mathfrak{p}
\cap S = \emptyset$, as required.

The conclusions (v) $\Longrightarrow$ (vi) $\Longrightarrow$ (vii) are trivial. Finally, an argument similar to that used in the proof of the implication (iv) $\Longrightarrow$ (v) shows that (vii) $\Longrightarrow$ (ii) holds.

An immediate consequence of the Theorem \[3.2\] is the following corollary.

\[3.3\] Let $R$ be a Noetherian ring, $N$ a finitely generated $R$-module and $I$ an ideal of $R.$ Then the following conditions are equivalent:

-   $\bar{Q}^*(I, N)= {\operatorname{mAss}}_R N/IN.$

-   The topologies defined by $\{(I^n)^{(N)}_a \}_{n \geq
    0}$ and $\{(I^n)^{\langle N \rangle}_a \}_{n \geq 0}$ are equivalent.

Let $S=R\setminus\bigcup\{\mathfrak{p}\in {\operatorname{mAss}}_RN/IN\}.$ Then $S((I^n)^{(N)}_a )=(I^n)^{\langle N\rangle}_a $. Now, if $\bar{Q}^*(I, N)= {\operatorname{mAss}}_R N/IN,$ then $S=R\setminus\bigcup\{\mathfrak{p}\in \bar{Q}^*(I, N)\}.$ Hence Theorem \[3.2\] implies that the topologies defined by $\{(I^n)^{(N)}_a \}_{n \geq 0}$ and $\{(I^n)^{\langle N \rangle}_a \}_{n \geq 0}$ are equivalent. Conversely, if these topologies are equivalent, then it follows from Theorem \[3.2\] that $S\subseteq R\setminus\bigcup\{\mathfrak{p}\in \bar{Q}^*(I, N)\}$, and so $\bar{Q}^*(I, N)\subseteq{\operatorname{mAss}}_R N/IN$. On the other hand, using [@Ah Lemma 3.5], one easily sees that ${\operatorname{mAss}}_R N/IN\subseteq \bar{Q}^*(I, N)$; and so $\bar{Q}^*(I, N)= {\operatorname{mAss}}_R N/IN.$ This completes the proof.

Local cohomology and ideal topologies
=====================================

The purpose of this section is to establish the equivalence between the topologies defined by $\{(I^n)_a^{(N)}\}_{n\geq1}$ and $\{S((I^n)_a^{(N)})\}_{n\geq1}$ in terms of the vanishing of the top local cohomology module $H^{\dim N}_I(N)$. This will generalize the main result of Marti-Farre [@MF], as an extension of the main results of Call [@Ca Corollary 1.4], Call-Sharp [@CS] and Schenzel [@Sc2 Corollary 4.3].

\[4.1\] Let $(R, \mathfrak{m})$ be a local (Noetherian) ring, $N$ a finitely generated $R$-module of dimension $d$ and $I$ an ideal of $R.$ Consider the following conditions:

-   There exists a multiplicatively closed subset $S$ of $R$ such that $ \mathfrak{m} \cap S \neq\emptyset$ and that the topologies defined by $\{S((I^n)^{(N)}_a)\}_{n\geq0}$ and $\{(I^n)^{(N)}_a \}_{n\geq0}$ are equivalent.

-   $H^d_I(N)= 0.$

Then $\Longrightarrow$ ; and these conditions are equivalent, whenever $N$ is quasi-unmixed.

We start with the proof of the implication (i) $\Longrightarrow$ (ii). By Theorem \[3.2\] we have $S
\subseteq R \setminus \bigcup \{\mathfrak{p}\in \bar{Q}^*(I, N)\}.$ Then $\mathfrak{m} \not\in \bar{Q}^*(I, N).$ Therefore for all $\mathfrak{q}\in {\operatorname{mAss}}_{\hat{R}}\hat{N}$ we have $\dim \hat{R}/I\hat{R}+ \mathfrak{q} >0.$ By the Lichtenbaum-Hartshorne Theorem (see [@DS Corollary 3.4]), it follows that $H^d_I(N)= 0.$

Now, assume that $N$ is quasi-unmixed and that (ii) holds. We show that (i) is true. To this end, let $S= R \setminus \bigcup
\{\mathfrak{p} \in \bar{Q}^*(I, N)\}.$ Then, in view of Theorem \[3.2\], the topologies defined by $\{S((I^n)^{(N)}_a)\}_{n\geq0}$ and $\{(I^n)^{(N)}_a \}_{n\geq0}$ are equivalent. Hence, it is enough to show that $ \mathfrak{m}
\cap S \neq\emptyset.$ Suppose, the contrary, namely $
\mathfrak{m} \cap S = \emptyset.$ Then $\mathfrak{m}\in
\bar{Q}^*(I, N).$ So there exists $ \mathfrak{q}\in
{\operatorname{mAss}}_{\hat{R}}\hat{N}$ such that $\mathfrak{m}\hat{R}=
{\operatorname{Rad}}(I\hat{R}+ \mathfrak{q}).$ As $N$ is quasi-unmixed it follows that $\dim \hat{R}/I\hat{R}+ \mathfrak{q}= 0$ for some $
\mathfrak{q} \in {\operatorname{mAss}}_{\hat{R}}\hat{N}$ such that $\dim
\hat{R}/\mathfrak{q}= d.$ Now, use [@DS Corollary 3.4] to see that $H^d_I(N) \neq 0$, which is a contradiction.

The final results will be the strengthened and a generalized version of a corresponding one by Marley ([@M Corollaries 2.4 and 2.5]) and Naghipour-Sedghi ([@NS Corollary 3.3]).

\[3.5\] Assume that $R$ is a Noetherian ring. Let $N$ be a finitely generated $R$-module of dimension $d$ and $I$ an ideal of $R.$ Then ${\operatorname{Supp}}(H^d_I(N)) \subseteq \bar{Q}^*(I, N).$ Moreover the equality holds, whenever $N$ is Cohen-Macaulay.

Let $\mathfrak{p}\in {\operatorname{Supp}}(H^d_I(N)).$ Then $H^d_{IR_\mathfrak{p}}(N_{\mathfrak{p}})\neq 0,$ and so $\dim
N_{\mathfrak{p}} = d.$ Hence, in view of the Lichtenbaum-Hartshorne Theorem (cf. [@DS Corollary 3.4]) there exists $ \mathfrak{q}\in {\operatorname{mAss}}_{\hat{R_{\mathfrak{p}}
}}\hat{N_{\mathfrak{p}}}$ such that $\mathfrak{p}\hat{R_{\mathfrak{p}}}= {\operatorname{Rad}}(I\hat{R_{\mathfrak{p}}}+
\mathfrak{q}).$ Thus $\mathfrak{p}\in
\bar{Q}^*(I, N)$, and so ${\operatorname{Supp}}(H^d_I(N))\,\subseteq\,\bar{Q}^*(I,N).$

In order to prove the second assertion, let $\mathfrak{p}\in\bar{Q}^*(I, N)$. Then there exists $ \mathfrak{q}\in {\operatorname{mAss}}_{\hat{R_{\mathfrak{p}} }}\hat{N_{\mathfrak{p}}}$ such that $\mathfrak{p}\hat{R_{\mathfrak{p}}}= {\operatorname{Rad}}(I\hat{R_{\mathfrak{p}}}+\mathfrak{q}).$ Now, since $N$ is Cohen-Macaulay, we deduce that $\dim \hat{R_{\mathfrak{p}} }/\mathfrak q=\dim\hat{N_{\mathfrak{p}}}$. Whence, in view of the Lichtenbaum-Hartshorne Theorem, $H^d_{IR_\mathfrak{p}}(N_{\mathfrak{p}})\neq 0,$ and so $\mathfrak{p}\in {\operatorname{Supp}}(H^d_I(N)).$

\[3.6\] Let $(R, \mathfrak{m})$ be a local (Noetherian) ring, $N$ a finitely generated $R$-module of dimension $d$ and $I$ an ideal of $R.$ Then $${\operatorname{Supp}}(H^{d-1}_I(N))\,\subseteq\,\bar{Q}^*(I, N) \cup\,
\{\mathfrak{m}\}.$$ Therefore ${\operatorname{Ass}}_R H^{d-1}_I(N)$ is a finite set.

Let $\mathfrak{p}\in {\operatorname{Supp}}(H^{d-1}_I(N))$ such that $\mathfrak{p}\neq \mathfrak{m}.$ Then $H^{d-1}_{IR_{\mathfrak{p}}}(N_{\mathfrak{p}})\neq 0$, and so $\dim N_{\mathfrak{p}} = d-1.$ Now, by the proof of Corollary \[3.5\] we have $\mathfrak{p}\in \bar{Q}^*(I, N)$, as required.

[**Acknowledgments**]{}

The authors are deeply grateful to the referee for providing numerous suggestions to improve the readability of the paper and for drawing the authors’ attention to Lemma 2.3 and Theorem 2.5. Also, we would like to thank Dr. Monireh Sedghi for a careful reading of the original manuscript and helpful suggestions.

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[^1]: 2010 [*Mathematics Subject Classification*]{}: 13E05, 13B2.\
    The first author thanks the Martin-Lüther-Universität Halle-Wittenberg for the hospitality and facilities offered during the preparation of this paper.\
    $^*$Corresponding author: e-mail: [*naghipour@ipm.ir*]{} (Reza Naghipour)
