On Uniformly S-Multiplication Modules and Rings

Abstract

In this article, we introduce and study the notions of uniformly S-multiplication modules and rings that are generalizations of multiplication modules and rings. Some examples are given to distinguish the new conceptions with the old classical ones.

1. Introduction

Throughout this article, R is always a commutative ring with an identity. For a subset U of an R-module M, we denote by $\langle U\rangle$ the submodule of M generated by U. A subset S of R is said to be multiplicative if $1\in S$ and $s_1{s}_2\in S$ for any $s_1\in S$, $s_2\in S$. Let N be a submodule of M, and denote by $\left(N{:}_{R}M\right)=\{r\in R\mid rM\subseteq N\}.$

The notion of multiplication rings was introduced by Krull [1] early in 1925. A ring R is called a multiplication ring if, for every pair of ideals $J\subseteq K$ of R, there exists an ideal I of R such that $J=IK$. Note that an integral domain is a multiplication ring if and only if it is a Dedekind domain (see [2]). Some characterizations of multiplication rings were given by Mott [3]. In 1974, Mehdi [4] first introduced the notion of multiplication modules. An R-module M is said to be a multiplication module if, for every pair of submodules $L\subseteq N$ of M, there exists an ideal I of R such that $L=IN$. Latter in 1988, Barnard [5] alternatively called an R-module M a multiplication if each submodule N of M is of the form $N=IM$ for some ideal I of R, or equivalently, $N=\left(N{:}_{R}M\right)M$. Some more studies on multiplication modules can be found in [5,6,7].

At the beginning of this century, Anderson et al. [8] introduced the notion of S-Noetherian rings, which are a generalization of classical Noetherian rings in terms of a multiplicative set S. Since then, some well-known notions of rings and modules have been investigated. In 2020, Anderson, Arabaci, Tekir, and Koç [9] introduced and studied the notion of S-multiplication modules. An R-module M is called an S-multiplication module if, for each submodule N of M, there exist $s\in S$ and an ideal I of R such that $sN\subseteq IM\subseteq N$. They generalized some known results on multiplication modules to S-multiplication modules and studied S-multiplication modules in terms of S-prime submodules. Recently, Chhiti and Moindze [10] studied the notion of S-multiplication rings. A ring R is called an S-multiplication ring if each ideal of R is of the S-multiplication type. They generalized some properties of multiplication rings to S-multiplication rings and then studied the transfer of S-multiplication rings to trivial ring extensions and amalgamated algebras.

In 2021, the second author of this paper first introduced and studied the uniformly S-torsion theory in [11]. Recently, the first author et al. [12] considered the notions of uniformly S-Noetherian rings and modules, which can be seen as “uniform” versions of S-Noetherian rings and modules. The motivation of this article is to introduce and study the notions of uniformly S-multiplication modules and rings, which are “uniform” versions of the S-multiplication modules and rings given in [9,10]. This paper is arranged as follows. In Section 2, we introduce and study the notion of uniformly S-multiplication modules. We transfer the uniformly S-multiplication modules to finite direct products, localizations, u-S-isomorphisms, and idealizations. In Section 3, we investigate uniformly S-multiplication rings. We also study uniformly S-multiplication rings under finite direct products, localizations, and idealizations. Furthermore, we connect and distinguish the notions of multiplication modules and rings, uniformly S-multiplication modules and rings, and S-multiplication modules and rings.

2. Uniformly $\mathit{S}$-Multiplication Modules

Recall from [5] that an R-module M is said to be a multiplication module if each submodule N of M is of the form $N=IM$ for some ideal I of R, or equivalently, $N=\left(N{:}_{R}M\right)M$. Let S be a multiplicative subset of R. Recently, Anderson et al. [9] introduced the concept of S-multiplication modules; an R-module M is called an S-multiplication module if, for each submodule N of M, there exist $s\in S$ and an ideal I of R such that $sN\subseteq IM\subseteq N$. Note that the “s” in this definition is not uniform, i.e., it is decided by the submodule N. To keep it in “uniformity”, we introduce the following notion.

Definition 1.

Let M be an R-module and let S be a multiplicative subset of R. Then, M is called a u-S-multiplication (uniformly S-multiplication) module (with respect to $s$) if there exists an element $s\in S$ such that, for each submodule N of M, there is an ideal I of R satisfying $sN\subseteq IM\subseteq N$.

From the definition, one can easily verify that an R-module M is a u-S-multiplication if and only if there exists $s\in S$ such that, for each submodule N of M, we have $sN\subseteq \left(N{:}_{R}M\right)M\subseteq N$.

If S is composed of units, then an R-module is a u-S-multiplication if and only if it is an S-multiplication; if $0\in S$, then every R-module is a u-S-multiplication. In general, we have the following implications.

$$\begin{array}{|c|}\hline \mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{e}\\ \hline\end{array}⟹\begin{array}{|c|}\hline u-S-\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{e}\\ \hline\end{array}⟹\begin{array}{|c|}\hline S-\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{e}\\ \hline\end{array}$$

Proposition 1.

Let $M_i$ be an $R_i$-module and let $S_i\subseteq R_i$ be a multiplicative subset $(i=1,2)$. Set $R=R_1\times R_2,S=S_1\times S_2$, and $M=M_1\times M_2$. Then, M is a u-S-multiplication module if and only if $M_1$ is a u-$S_1$-multiplication module and $M_2$ is a u-$S_2$-multiplication module.

Proof. 

For the “only if” part, suppose M is a u-S-multiplication module with respect to some $s=(s_1,s_2)\in S_1\times S_2$. Then, $(s_1,s_2)(N_1\times \left\{0\right\})\subseteq [(N_1\times \left\{0\right\}):M]M$ for any $R_1$-submodule $N_1$ of $M_1$. Therefore, $s_1{N}_1\subseteq (N_1:M)M$. It follows that $M_1$ is a u-S-multiplication module with respect to some $s_1\in S_1$. Similarly, $M_2$ is a u-S-multiplication module with respect to some $s_2\in S_2$.

For the “if” part, suppose $M_1$ is a u-S-multiplication module with respect to some $s_1\in S_1$ and $M_2$ is a u-S-multiplication module with respect to some $s_2\in S_2$. Set $s=(s_1,s_2)\in S$. Let N be an R-module. Then, $N=N(R_1\times R_2)\cong N_1\times N_2$, where $N_i=N{R}_i$$(i=1,2)$. Therefore, $s_i{N}_i\subseteq (N_i:M_i)M_i$ for each $i=1,2$. Consequently, $(s_1,s_2)(N_1\times N_2)\subseteq [(N_1\times N_2):(M_1\times M_2)](M_1\times M_2)$. It follows that $M=M_1\times M_2$ is a u-S-multiplication module with respect to s. □

Note that u-S-multiplication modules need not be a multiplication module. Indeed, let $R_1$ and $R_2$ be two commutative rings and let $M_1$ be a multiplication $R_1$-module; however, $M_2$ is not a multiplication $R_2$-module. Set $R=R_1\times R_2,S=\left\{1\right\}\times \left\{0\right\}$ and $M=M_1\times M_2$. Then. M is not a multiplication R-module, but it is a u-S-multiplication R-module by Proposition 1.

The following example shows that an S-multiplication module need not be a u-S-multiplication module.

Example 1

([9], Example 3). Consider the $\mathbb{Z}$-module $E\left(p\right)=\{\gamma :=\frac{r}{p^m}+\mathbb{Q}\in \mathbb{Q}/\mathbb{Z}\mid r\in \mathbb{Z},m\ge 0\}$, where p is a prime number. Take the multiplicative closed subset $S=\{p^n:n\in \mathbb{N}\cup \left\{0\right\}\}$ of $\mathbb{Z}$. Then, the $\mathbb{Z}$-module $E\left(p\right)$ is an S-multiplication module (see ([9], Example 3)).

We claim that $E\left(p\right)$ is not a u-S-multiplication. Indeed, assume that $E\left(p\right)$ is a u-S-multiplication with respect to $p^n\in S$ for some $n\ge 0$. All proper submodules of $E\left(p\right)$ are of the form $G_t=\{\gamma :=\frac{r}{p^t}+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}\mid \gamma \in \mathbb{Z}\}$ for every $t\in \mathbb{N}\cup \left\{0\right\}$. Assume that $t\ge n+1$. Then, $\left(G_t{:}_{\mathbb{Z}}E\left(p\right)\right)=0$. Therefore, $0\ne p^n{G}_t\ne \left(G_t{:}_{\mathbb{Z}}E\left(p\right)\right)E\left(p\right)=0_{E\left(p\right)}$. Hence, $E\left(p\right)$ is not a u-S-multiplication module.

Let S be a multiplicative subset of R. The saturation $S^{*}$ of S is defined as $S^{*}=\{s\in R\mid s_1=s{s}_2$ for some $s_1,s_2\in S\}$. A multiplicative subset S of R is called saturated if $S=S^{*}$. Note that $S^{*}$ is always a saturated multiplicative subset containing S.

Proposition 2.

Let M be an R-module. Then, the following statements hold.

(1) 

If $S\subseteq T$ are multiplicative subsets of R and M is a u-S-multiplication module, then M is a u-T-multiplication module.

(2) 

M is a u-S-multiplication module if and only if M is a u-$S^{*}$-multiplication module, where $S^{*}$ is the saturation of S.

Proof. 

$\left(1\right)$: Obvious. $\left(2\right)$: Let M be a u-S-multiplication module. Since $S\subseteq S^{*}$, by $\left(i\right)$, M is a u-$S^{*}$-multiplication module. For the converse, assume that M is an $S^{*}$-multiplication module with some $s\in S^{*}$. Then, $sN\subseteq \left(N{:}_{R}M\right)M$ for any submodule N of M. Suppose $s_1=s{s}_2$ with some $s_1,s_2\in S$. Then, $s_{1}N=s{s}_{2}N\subseteq s_2\left(N{:}_{R}M\right)M\subseteq \left(N{:}_{R}M\right)M$. Therefore, M is a u-S-multiplication module with respect to $s_1\in S$. □

Let $\mathfrak{p}$ be a prime ideal of R. We say an R-module E is a u-$\mathfrak{p}$-multiplication shortly provided that E is a u-$(R\setminus \mathfrak{p})$-multiplication.

Theorem 1.

Let M be an R-module. Then, the following statements are equivalent.

(1) 

M is a multiplication module.

(2) 

M is a u-$\mathfrak{p}$-multiplication module for each $\mathfrak{p}\in \mathrm{Spec}\left(R\right)$.

(3) 

M is a u-$\mathfrak{m}$-multiplication module for each $\mathfrak{m}\in \mathrm{Max}\left(R\right)$.

(4) 

M is a u-$\mathfrak{m}$-multiplication module for each $\mathfrak{m}\in \mathrm{Max}\left(R\right)$ with $M_{\mathfrak{m}}\ne 0_{\mathfrak{m}}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right):$ Follows by their definitions.

$\left(2\right)\Rightarrow \left(3\right):$ This follows the assumption that every maximal ideal is a prime ideal.

$\left(3\right)\Rightarrow \left(4\right):$ This is trivial.

$\left(4\right)\Rightarrow \left(1\right):$ Suppose M is a u-$\mathfrak{m}$-multiplication module with respect to some $s_{\mathfrak{m}}\notin \mathfrak{m}$ for each $\mathfrak{m}\in \mathrm{Max}\left(R\right)$ with $M_{\mathfrak{m}}\ne 0_{\mathfrak{m}}$. Take a maximal ideal $\mathfrak{m}$ of R with $M_{\mathfrak{m}}\ne 0_{\mathfrak{m}}$. Since M is a u-$\mathfrak{m}$-multiplication module with respect to $s_{\mathfrak{m}}$, we have $s_{\mathfrak{m}}N\subseteq \left(N{:}_{R}M\right)M$ for every submodule N of M. Then, $N_{\mathfrak{m}}={\left(s_{\mathfrak{m}}N\right)}_{\mathfrak{m}}\subseteq {\left(\left(N{:}_{R}M\right)M\right)}_{\mathfrak{m}}\subseteq N_{\mathfrak{m}}$. If $M_{\mathfrak{m}}=0_{\mathfrak{m}}$, certainly $N_{\mathfrak{m}}={\left(\left(N{:}_{R}M\right)M\right)}_{\mathfrak{m}}$. Thus, we conclude that $N_{\mathfrak{m}}={\left(\left(N{:}_{R}M\right)M\right)}_{\mathfrak{m}}$ for each maximal ideal $\mathfrak{m}$ of R, and this yields $N=\left(N{:}_{R}M\right)M$. Therefore, M is a multiplication module. □

Recall from [11] that an R-sequence $M\stackrel{f}{\to }N\stackrel{g}{\to }L$ is called u-S-exact provided that there is an element $s\in S$ such that $s\mathrm{Ker}\left(g\right)\subseteq \mathrm{Im}\left(f\right)$ and $s\mathrm{Im}\left(f\right)\subseteq \mathrm{Ker}\left(g\right)$. An R-homomorphism $f:M\to N$ is a u-S-monomorphism(respectively, a u-S-epimorphism or an S-isomorphism) provided $0\to M\stackrel{f}{\to }N$(respectively, $M\stackrel{f}{\to }N\to 0$ or $0\to M\stackrel{f}{\to }N\to 0$) is u-S-exact. It is easy to verify that an R-homomorphism $f:M\to N$ is a u-S-monomorphism (respectively, u-S-epimorphism) if and only if $\mathrm{Ker}\left(f\right)$(respectively, $\mathrm{Coker}\left(f\right))$ is a u-S-torsion module.

Proposition 3.

Let M and $M^{\prime }$ be R-modules. Suppose M is u-S-isomorphic to $M^{\prime }$. Then, M is a u-S-multiplication module if and only if $M^{\prime }$ is a u-S-multiplication module.

Proof. 

Let $f:M\to M^{\prime }$ be a u-S-isomorphism. Then, there exists $s\in S$ such that $s\mathrm{Ker}\left(f\right)=s\mathrm{Coker}\left(f\right)=0$ and M is a u-S-multiplication module with respect to s. Let N be a submodule of $M^{\prime }$. Then, there is an ideal I of R such that $s{f}^{-1}\left(N\right)\subseteq IM\subseteq f^{-1}\left(N\right)$. Therefore, $f\left(s{f}^{-1}\left(N\right)\right)\subseteq f\left(IM\right)\subseteq f\left(f^{-1}\left(N\right)\right)$, i.e., $sN\subseteq I\mathrm{Im}\left(f\right)\subseteq N$. Since $s\mathrm{Coker}\left(f\right)=s{M}^{\prime }/\mathrm{Im}\left(f\right)=0$, we have $s{M}^{\prime }\subseteq \mathrm{Im}\left(f\right)$. Note that $s^{2}N\subseteq sI\mathrm{Im}\left(f\right)\subseteq sI{M}^{\prime }$. Consequently, $s^{2}N\subseteq \left(sI\right)M^{\prime }\subseteq N$. It follows that $M^{\prime }$ is a u-S-multiplication module with respect to $s^2$. The converse follows by ([13], Proposition 1.1). □

Proposition 4.

Let M and $M^{\prime }$ be R-modules. Suppose that S is a multiplicative subset of R and $f:M\twoheadrightarrow M^{\prime }$ is a u-S-epimorphism. If M is a u-S-multiplication module, then $M^{\prime }$ is a u-S-multiplication module. Conversely, suppose that $M^{\prime }$ is an S-multiplication module and $t\mathrm{Ker}\left(f\right)=0$ for some $t\in S$; then, M is a u-S-multiplication module.

Proof. 

By Proposition 3, we can assume that f is an epimorphism. Suppose M is a u-S-multiplication module with respect to some $s\in S$. Then, $sN\subseteq \left(N{:}_{R}M\right)M\subseteq N$ for any submodule N of M. Therefore, $f\left(sN\right)\subseteq f\left(\right(N:M\left)M\right)\subseteq f\left(N\right)$. Let $N^{\prime }$ be a submodule of $M^{\prime }$. Then, $N:=f^{-1}\left(N^{\prime }\right)$ is a submodule of M. It follows that $s{N}^{\prime }=sf\left(N\right)\subseteq (N:M)f\left(M\right)=(N:M)M^{\prime }\subseteq N^{\prime }.$ Thus, $s{N}^{\prime }\subseteq (N:M)M^{\prime }\subseteq N^{\prime }$ for any submodule $N^{\prime }$ of $M^{\prime }$. Hence, $M^{\prime }$ is a u-S-multiplication module with respect to s.

On the other hand, suppose that $M^{\prime }=f\left(M\right)$ is a u-S-multiplication module with respect to s. Then, for any submodule N of M, there is an ideal I of R with $sf\left(N\right)\subseteq If\left(M\right)\subseteq f\left(N\right)$. Hence, $sN+\mathrm{Ker}\left(f\right)\subseteq N+\mathrm{Ker}\left(f\right).$ Since $t\mathrm{Ker}\left(f\right)=0$, we have $\left(st\right)N\subseteq \left(tI\right)M\subseteq tN\subseteq N$. Consequently, M is a u-S-multiplication module with respect to $st$. □

Proposition 5.

Let R be a commutative ring and let S and T be multiplicative subsets of R. Set $\tilde{S}=\{\frac{s}{1}\in T^{-1}R|s\in S\}$, a multiplicative subset of $T^{-1}R$. Suppose M is a u-S-multiplication R-module. Then, $T^{-1}M$ is a u-$\tilde{S}$-multiplication $T^{-1}R$-module.

Proof. 

Suppose M is a u-S-multiplication R-module with respect to some $s\in S$. Then, for any submodule N of M, there is an ideal I of R such that $sN\subseteq IM\subseteq N$. Let L be an submodule of $T^{-1}M$. Then, $L=T^{-1}{N}^{\prime }$ for some submodule $N^{\prime }$ of M. It follows that $\frac{s}{1}L=T^{-1}\left(s{N}^{\prime }\right)\subseteq \left(T^{-1}I\right)\left(T^{-1}M\right)\subseteq T^{-1}{N}^{\prime }=L$. Therefore, $T^{-1}M$ is a u-$\tilde{S}$-multiplication $T^{-1}R$-module with respect to $\frac{s}{1}\in \tilde{S}$. □

A multiplicative subset S of R is said to satisfy the maximal multiple condition if there exists an $s\in S$ such that $t|s$ for each $t\in S$. Both finite multiplicative subsets and the multiplicative subsets that consist of units satisfy the maximal multiple condition.

Proposition 6.

Let M be an R-module and let S be a multiplicative subset of R satisfying the maximal multiple condition. Then, the following statements hold:

(1) 

M is a u-S-multiplication module.

(2) 

M is an S-multiplication module.

(3) 

$S^{-1}M$ is a multiplication $S^{-1}R$-module.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$: Trivial.

$\left(2\right)\Rightarrow \left(3\right)$: It follows by ([9], Corollary 2).

$\left(3\right)\Rightarrow \left(1\right)$: Assume that $S^{-1}M$ is a multiplication $S^{-1}R$-module. Take a submodule N of M. We have $S^{-1}N=\left(S^{-1}I\right)\left(S^{-1}M\right)=S^{-1}\left(IM\right)$ for any submodule N of M. Choose $s\in S$ such that $t|s$ for every $t\in S$. Note that for each $n\in N$, we have $\frac{n}{1}\in S^{-1}N=S^{-1}\left(IM\right)$, and so there exists $t\in S$ such that $tn\in IM$ and, hence, $sn\in IM$. Thus, $sN\subseteq IM$. Similarly, we have $sIM\subseteq N$. Therefore, we obtain $s^{2}N\subseteq \left(sI\right)M\subseteq N$. Hence, M is a u-S-multiplication module with respect to $s^2$. □

Recall from [12] the conception of u-S-Noetherian modules. Let ${\left\{M_j\right\}}_{j\in \mathrm{\Gamma }}$ be a family of R-modules and let $N_j$ be a submodule of $M_j$ generated by ${\left\{m_{i,j}\right\}}_{i\in {\mathrm{\Lambda }}_j}\subseteq M_j$ for each $j\in \mathrm{\Gamma }$. A family of R-modules ${\left\{M_j\right\}}_{j\in \mathrm{\Gamma }}$ is u-S-generated (with respective to s) by ${\left\{{\left\{m_{i,j}\right\}}_{i\in {\mathrm{\Lambda }}_j}\right\}}_{j\in \mathrm{\Gamma }}$ provided that there exists an element $s\in S$ such that $s{M}_j\subseteq N_j$ for each $j\in \mathrm{\Gamma }$, where $N_j=\langle {\left\{m_{i,j}\right\}}_{i\in {\mathrm{\Lambda }}_j}\rangle$. We say a family of R-modules ${\left\{M_j\right\}}_{j\in \mathrm{\Gamma }}$ is u-S-finite (with respective to s) if the set ${\left\{m_{i,j}\right\}}_{i\in {\mathrm{\Lambda }}_j}$ can be chosen as a finite set for each $j\in \mathrm{\Gamma }$.

Definition 2

([12]). Let R be a ring and let S be a multiplicative subset of R. An R-module M is called a u-S-Noetherian R-module provided the set of all submodules of M is u-S-finite. A ring R is called a u-S-Noetherian if R itself is a u-S-Noetherian R-module.

Let R be a ring, let S be a multiplicative subset of R, and let M be an R-module. Denote by $M^{•}$ an ascending chain $M_1\subseteq M_2\subseteq \cdots$ of submodules of M. An ascending chain $M^{•}$ is called stationary with respective to s if there exists $k\ge 1$ such that $s{M}_n\subseteq M_k$ for any $n\ge k$. Following ([12], Theorem 2.7), M is u-S-Noetherian if and only if there exists an element $s\in S$ such that any ascending chain of submodules of M is stationary with respective to s.

Proposition 7.

Let R be a u-S-Noetherian ring and let M be a u-S-multiplication R-module. Then, M is a u-S-Noetherian R-module.

Proof. 

We may assume R is a u-S-Noetherian ring and M is a u-S-multiplication R-module with respect to $s\in S$. Let $M_1\subseteq M_2\subseteq \cdots$ be an ascending chain of submodules of M. Set $A_i=(M_i:M)$. Then, $A_1\subseteq A_2\subseteq \cdots$ is an ascending chain of ideals of R. Then there exists n such that $s{A}_k\subseteq A_n\subseteq A_k$ for any $k\ge n$. Since M is a u-S-multiplication, $s{M}_i\subseteq (M_i:M)M=A_{i}M$ for all i. Hence, $s^2{M}_k\subseteq s{A}_{k}M\subseteq A_{n}M\subseteq M_n$. It follows that M is a u-S-Noetherian R-module with respect to $s^2$. □

Let M be an R-module. The idealization construction $R(+)M=R\oplus M$ of M is a commutative ring with componentwise additions and multiplications $(a,m)(b,m^{\prime })=(ab,a{m}^{\prime }+bm)$ for each $a,b\in R;m,m^{\prime }\in M$ (see [14]). If S is a multiplicative subset of R and N is a submodule of M, then $S(+)N$ is a multiplicative subset of $R(+)M$. Now, we transfer the uniformly S-multiplication properties to idealization constructions.

Theorem 2.

Let M be an R-module, let N be a submodule of M, and let S be a multiplicative subset of R. Then, the following statements are equivalent.

(1) 

N is a u-S-multiplication R-module.

(2) 

$0(+)N$ is a u-$S(+)0$-multiplication ideal of $R(+)M$.

(3) 

$0(+)N$ is a u-$S(+)M$-multiplication ideal of $R(+)M$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right):$ Suppose N is a u-S-multiplication R-module with respect to some $s\in S$. Let J be an ideal of $R(+)M$ contained in $0(+)N$. Then, $J=0(+)N^{\prime }$ for some submodule $N^{\prime }$ of N. Since N is a u-S-multiplication R-module with respect to s, there exists an ideal I of R such that $s{N}^{\prime }\subseteq IN\subseteq N^{\prime }$. Hence,

$$(s,0)J=(s,0)0(+)N=0(+)s{N}^{\prime }\subseteq 0(+)IN=I(+)M·0(+)N\subseteq 0(+)N^{\prime }=J.$$

It follows that $0(+)N$ is a u-$S(+)0$-multiplication ideal of $R(+)M$.

$\left(2\right)\Rightarrow \left(3\right):$ Since $S(+)0\subseteq S(+)M$, (3) follows by Proposition 2.

$\left(3\right)\Rightarrow \left(1\right):$ Suppose that $0(+)N$ is a u-$S(+)M$-multiplication ideal of $R(+)M$ with respective to some $(s,m)\in S(+)M$. Let $N^{\prime }$ be a submodule of N. Then, $0(+)N^{\prime }$ is an ideal of $R(+)M$ with $0(+)N^{\prime }\subseteq 0(+)N$. Since $0(+)N$ is a u-$S(+)M$-multiplication ideal of $R(+)M$ with respect to $(s,m)$, then there exists $J^{\prime }$ of $R(+)M$ such that $(s,m)0(+)N^{\prime }\subseteq J^{\prime }·0(+)N\subseteq 0(+)N^{\prime }.$ Set $J=J^{\prime }+0(+)M.$ Then, $J=I(+)M$ for some ideal I of R. Note that

$$J^{\prime }·0(+)N=J^{\prime }·0(+)N+0(+)M·0(+)N=(J^{\prime }+0(+)M)·0(+)N=J·0(+)N.$$

So $(s,m)0(+)N^{\prime }\subseteq J·0(+)N\subseteq 0(+)N^{\prime }.$ This implies that $s{N}^{\prime }\subseteq IN\subseteq N^{\prime }$. So N is a u-S-multiplication R-module with respect to s. □

3. Uniformly $\mathit{S}$-Multiplication Rings

Let R be a ring and let S be a multiplicative subset of R. Recall from [10] that an ideal I of R is an S-multiplication ideal if I is an S-multiplication R-module, and a ring R is an S-multiplication ring if each ideal of R is an S-multiplication. Equivalently, for each pair of ideals $J\subseteq K$ of R, there exist $s\in S$ and an ideal I of R satisfying $sJ\subseteq IK\subseteq J$. Now, we introduce the notion of uniformly S-multiplication rings.

Definition 3.

Let R be a ring and let S be a multiplicative subset of R. Then, R is called a u-S-multiplication (uniformly S-multiplication) ring (with respect to $s)$ if there exists $s\in S$ such that each ideal of R is a u-S-multiplication with respect to s, equivalently, if there exists $s\in S$ such that, for each pair of ideals $J\subseteq K$ of R, there exists an ideal I of R satisfying $sJ\subseteq IK\subseteq J$.

If S is composed of units, then a ring R is a u-S-multiplication if and only if it is an S-multiplication; if $0\in S$, then every ring R is a u-S-multiplication. In general, we have the following implications.

$$\begin{array}{|c|}\hline \mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\\ \hline\end{array}⟹\begin{array}{|c|}\hline u-S-\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\\ \hline\end{array}⟹\begin{array}{|c|}\hline S-\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} \mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\\ \hline\end{array}$$

Proposition 8.

Let $S\subseteq T$ be two multiplicative subsets of R and $S^{*}$ the saturation of S. Then the following statements hold.

(1) 

If R is a u-S-multiplication ring, then R is a u-T-multiplication ring.

(2) 

R is a u-S-multiplication ring if and only if R is a u-$S^{*}$-multiplication ring.

Proof. 

(1) It immediately follows from the definition of u-S-multiplication rings.

(2) Suppose R is an $S^{*}$-multiplication ring with some $s\in S^{*}$. Then for any pair of ideals $J\subseteq K$, there exists ideal I of R such that $sJ\subseteq IK\subseteq J$. Suppose $s_1=s{s}_2$ with some $s_1,s_2\in S$. Then $s_{1}J\subseteq IK\subseteq J$. So R is a u-S-multiplication ring with respect to $s_1\in S$. □

Corollary 1.

Every multiplication ring is a u-S-multiplication ring.

Proof. 

Remark that a multiplication ring is exactly a u-$\left\{1\right\}$-multiplication ring. Therefore, the result follows by Proposition 8(1). □

The proof of following result is similar to that of Proposition 1, and so we omit it.

Proposition 9.

Let $R=R_1\times R_2$ and $S=S_1\times S_2$. Then, R is a u-S-multiplication ring if and only if $R_1$ is a u-$S_1$-multiplication ring and $R_2$ is a u-$S_2$-multiplication ring.

The following example shows that u-S-multiplication rings are not necessary multiplication rings.

Example 2.

Let $R_1$ be a multiplication ring and let $R_2$ be a non-multiplication ring. Set $R=R_1\times R_2$ and $S=\left\{1\right\}\times \left\{0\right\}$. Then, R is not a multiplication ring but a u-S-multiplication ring by Proposition 9.

Trivially, every u-S-multiplication ring is an S-multiplication. Moreover, we have the following result.

Proposition 10.

Let S be a multiplicative subset of R that satisfies the maximal multiple condition. Then, R is a S-multiplication ring if and only if R is a u-S-multiplication ring.

Proof. 

If R is a u-S-multiplication ring, R is trivially an S-multiplication. On the other hand, suppose R is an S-multiplication ring. Then, each ideal I of R is an S-multiplication. Therefore, for each pair of ideals $J\subseteq K$ of R, there exist $t\in S$ and an ideal I of R such that $tJ\subseteq IK\subseteq J$. Since S satisfies the maximal multiple condition, there exists $s\in S$ such that $t|s$. Thus, $sJ\subseteq tJ\subseteq IK\subseteq J$. It follows that R is a u-S-multiplication ring with respect to s. □

Let R be a ring and let S be a multiplicative subset of R. For any $s\in S$, there is a multiplicative subset $S_s=\{1,s,s^2,\cdots \}$ of S. We denote by $M_s$ the localization of M at $S_s$ for an R-module M.

Proposition 11.

Suppose R is a u-S-multiplication ring. Then, there is an $s\in S$ such that $R_s$ is a multiplication ring.

Proof. 

Suppose R is a u-S-multiplication ring with respect to some $s\in S$. Let $J\subseteq K$ be a pair of ideals of $R_s$. Then, there are two ideals $J^{\prime }\subseteq K^{\prime }$ of R such that $J=J_s^{\prime }$ and $K=K_s^{\prime }$. There exists an ideal $I^{\prime }$ of R satisfying $s{J}^{\prime }\subseteq I^{\prime }{K}^{\prime }\subseteq J^{\prime }$. By localizing at s, we have $J\subseteq IK\subseteq J$, where $I=I_s^{\prime }$. It follows that $R_s$ is a multiplication ring. □

It follows from Proposition 9.13 in [2] that an integral domain is a multiplication ring if and only if it is a Dedekind domain. The following example shows that rings with each ideal u-S-multiplication are not necessary u-S-multiplication rings, and thus S-multiplication rings are u-S-multiplication rings in general.

Example 3.

Let D be an integral domain such that $D_s$ is not a Dedekind domain for any $0\ne s\in D$ (e.g., $D=k[x_1,x_2,\cdots ]$, the polynomial ring with infinite variables over a field k). Set $S=D-\left\{0\right\}$. Then D is not a u-S-multiplication ring by Proposition 11. However, every ideal of D is a u-S-multiplication, and thus, D is an S-multiplication ring. Indeed, let K be an ideal of R and let J be a sub-ideal of K. Suppose $K=0$. Then, $J=0$, and thus, $sJ\subseteq IK\subseteq J$ always holds. Otherwise, let $0\ne s\in K$ and $I=J$. Then, we also have $sJ\subseteq IK\subseteq J$. It follows that K is a u-S-multiplication ideal of R.

Remark 1.

Note that the converse of Proposition 11 is not true in general. Indeed, let D be a valuation domain with valuation group $\mathbb{Z}\times \mathbb{Z}$. It follows by ([15], Chapter II, Exercise 3.4) that the maximal ideal $\mathfrak{m}$ of R is principally generated, say generated as $s\ne 0$. Let $S=D-\left\{0\right\}$. Then, D is not a u-S-multiplication ring by Example 3. However $D_s$ is a discrete valuation domain, and hence, it is a multiplication ring.

Let $\mathfrak{p}$ be a prime ideal of R. We say a ring R is a u-$\mathfrak{p}$-multiplication provided that R is a u-$(R\setminus \mathfrak{p})$-multiplication.

Theorem 3.

Let R be a ring. Then, the following statements are equivalent:

(1) 

R is a multiplication ring.

(2) 

R is a u-$\mathfrak{p}$-multiplication ring for each $\mathfrak{p}\in \mathrm{Spec}\left(R\right)$.

(3) 

R is a u-$\mathfrak{m}$-multiplication ring for each $\mathfrak{m}\in \mathrm{Max}\left(R\right)$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)\Rightarrow \left(3\right):$ Trivial.

$\left(3\right)\Rightarrow \left(1\right):$ Suppose R is a u-$\mathfrak{m}$-multiplication ring with respect to some $s_{\mathfrak{m}}\notin \mathfrak{m}$ for each $\mathfrak{m}\in \mathrm{Max}\left(R\right)$. Let $J\subseteq K$ be a pair of ideals of R. Then, there exists an ideal $I^{\mathfrak{m}}$ of R such that $s_{\mathfrak{m}}J\subseteq I^{\mathfrak{m}}K\subseteq J$. Since $\{s^{\mathfrak{m}}\mid \mathfrak{m}\in \mathrm{Max}\left(R\right)\}$ generates R, there exist finite elements $s^{{\mathfrak{m}}_1},...,s^{{\mathfrak{m}}_n}$ such that $J=\langle s^{{\mathfrak{m}}_1},...,s^{{\mathfrak{m}}_n}\rangle J\subseteq \left({\displaystyle \sum _{i=1}^n}{I}^{\mathfrak{m}}\right)K\subseteq J$. Setting $I={\displaystyle \sum _{i=1}^n}{I}^{\mathfrak{m}}$, we have $IK=J$. Consequently, R is a multiplication ring. □

Proposition 12.

Let R be a ring, let M be an R-module, and let S be a multiplicative subset of R. Suppose $R(+)M$ is a u-$S(+)M$-multiplication ring with respect to some $(s,m)\in S(+)M$. Then, R is a u-S-multiplication ring with respect to s, and each submodule of M is a u-S-multiplication R-module with respect to s.

Proof. 

Let $M^{\prime }$ be a submodule of M and let N be a submodule of $M^{\prime }$. Then, $0(+)N$ is a sub-ideal of $0(+)M^{\prime }$. Hence, there exists an ideal $I^{\prime }$ of $R(+)M$ such that $(s,m)0(+)N\subseteq I^{\prime }0(+)M^{\prime }\subseteq 0(+)N$. Set $I=\{r\in R\mid \mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e} \mathrm{e}\mathrm{x}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{s}(r,m)\in I^{\prime }\}.$ Then, $sN\subseteq I{M}^{\prime }\subseteq N$, and hence, $M^{\prime }$ is a u-S-multiplication R-module with respect to s.

Let $J\subseteq K$ be a pair of ideals of R. Then, $J(+)M\subseteq K(+)M$ is a pair of ideals of $R(+)M$. Hence, there exists an ideal $L^{\prime }$ of $R(+)M$ such that $(s,m)J(+)M\subseteq L^{\prime }K(+)M\subseteq J(+)M$. Set $L=\{r\in R\mid \mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e} \mathrm{e}\mathrm{x}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{s}(r,m)\in L^{\prime }\}.$ Then, $sJ\subseteq LK\subseteq J$. Hence, R is a u-S-multiplication ring with respect to s. □

Remark 2.

We do not know whether the converse of Proposition 12 is true. That is, suppose R is a u-S-multiplication ring with respect to s and each submodule of M is a u-S-multiplication R-module with respect to s. Do we have $R(+)M$ is a u-$S(+)M$-multiplication ring with respect to some $(s,m)\in S(+)M$?