Correction to: On S-multiplication modules

Abstract

In this corrigendum, we give an example showing that the implication $(2)\Rightarrow (1)$ of Proposition 4 is not true in general. Also, we provide the correct version of Proposition 4.

Keywords:

• Multiplication module
• prime submodule
• S-multiplication module
• S-prime submodule

2010 MATHEMATICS SUBJECT CLASSIFICATION:

• 16P40
• 13A15

This article refers to:

On S-multiplication modules

The following example shows that the implication $(2)\Rightarrow (1)$ of [1, Proposition 4] is not true in general.

Example 1.

Consider the $\mathbb{Z}$-module $E(p)=\left\{\gamma :\gamma =\frac{m}{p^n}+\mathbb{Z},m\in \mathbb{Z},n\ge 0\right\}$ and the multiplicatively closed subset $S=\{p^n:n\ge 0\}$ of $\mathbb{Z},$ where p is a prime number$.$ Then E(p) is an S-multiplication module (See, [1, Example 3]). Also, we know that all submodules of E(p) is of the form $G_t=\left\{\gamma :\gamma =\frac{m}{p^t}+\mathbb{Z},m\in \mathbb{Z}\right\}$ for some $t\ge 0.$ Note that $(G_t:E(p))=(0)$ is an S-prime ideal of $\mathbb{Z}.$ However, $G_t$ is not an S-prime submodule of $E(p).$ To see this, choose $s\in S.$ Then $s=p^k$ for some $k\ge 0.$ Since $p^{k+t+1}(\frac{1}{p^{k+t+1}}+\mathbb{Z})\in G_t,s{p}^{k+t+1}\notin (G_t:E(p))$ and $s(\frac{1}{p^{k+t+1}}+\mathbb{Z})=\frac{1}{p^{t+1}}+\mathbb{Z}\notin G_t,$ it follows that G_t is not an S-prime submodule of $E(p).$

Now, we give correct version of Proposition 4 as follows:

Proposition 1.

Let $M$ be an S-multiplication R-module, where $S\subseteq R$ is a multiplicatively closed set, and $N$ a submodule of $M.$ Suppose that $(N:tM)\subseteq (N:uM)$ implies that $(N:t)\subseteq (N:u)$ for each $t,u\in S.$Then the following statements are equivalent.

1. $N$ is an S-prime submodule of M.

2. $(N:M)$ is an S-prime ideal of R.

Proof.

$(i)\Rightarrow (ii):$ Follows from [2, Proposition 2.9 (i)].

$(ii)\Rightarrow (i):$ Suppose that $(N:M)$ is an S-prime ideal of $R.$ Then there exists an $s\in S$ so that $xy\in (N:M)$ implies $sx\in (N:M)$ or $sy\in (N:M)$ for all $x,y\in R.$ First, we will show that $(N:tM)\subseteq (N:sM)$ for each $t\in S.$ Let $b\in (N:tM).$ Then we have $bt\in (N:M),$ which implies that $sb\in (N:M)$ or $st\in (N:M).$ The latter case is impossible since $(N:M)\cap S=\varnothing .$ Thus, we have $sb\in (N:M),$ that is, $b\in (N:sM).$ Then, we conclude $(N:tM)\subseteq (N:sM)$ for each $t\in S.$ Suppose that $am\in N$ for some $a\in R$ and $m\in M.$ Now, we will show that $sa\in (N:M)$ or $sm\in N.$ Assume that $sa\notin (N:M).$ For every $x\in (Rm:M),$ we have $ax\in (N:M);$ so $sx\in (N:M),$ which implies that $s(Rm:M)\subseteq (N:M).$ Since $M$ is an S-multiplication module, choose $t\in S$ such that $\mathit{tRm}\subseteq (Rm:M)M.$ This gives that $\mathit{stRm}\subseteq s(Rm:M)M\subseteq N,$ so we have $Rm\subseteq (N:st).$ Since $(N:\mathit{stM})\subseteq (N:sM),$ by assumption, we conclude that $Rm\subseteq (N:st)\subseteq (N:s).$ Thus, we have $sm\in N,$ which completes the proof. □

Remark 1.

In the previous proposition, the condition “$(N:tM)\subseteq (N:uM)$ implies that $(N:t)\subseteq (N:u)$ for each $t,u\in S$” is necessary. Consider the $\mathbb{Z}$-module E(p). Let S be the multiplicatively closed set as in Example 1. Take the submodule $N=G_m$ for some $m\ge 1$ as in Example 1$.$ Then $(N:E(p))=(0)$ is an S-prime ideal of $\mathbb{Z}.$ On the other hand, put $t=p^{k+1},u=p^k\in S$ for some $k\ge 1.$ Then note that $(N:tE(p))=(0)\subseteq (N:uE(p)).$ Since $\frac{1}{p^{k+m+1}}+\mathbb{Z}\in (N:t)-(N:u),$ it follows that the aforementioned condition fails in S-multiplication $\mathbb{Z}$-module $E(p).$ Also, by Example 1, $N$ is not an S-prime submodule of $E(p).$

Acknowledgment

The authors would like to thank Farkhonde Farzalipour for his careful reading of the paper and for drawing to their attention the fact that Proposition 4 seems to be not true.