2-irreducible and strongly 2-irreducible submodules of a module

TL;DR

This paper introduces and explores the properties of 2-irreducible and strongly 2-irreducible submodules in modules over commutative rings, generalizing existing concepts of irreducibility.

Contribution

It defines new classes of submodules, 2-irreducible and strongly 2-irreducible, and investigates their properties as a generalization of irreducible submodules.

Findings

Defined 2-irreducible and strongly 2-irreducible submodules.

Analyzed properties and relationships of these submodules.

Extended the theory of submodules in module theory.

Abstract

Let R be a commutative ring with identity and M be an R-module. In this paper, we will introduce the concept of 2-irreducible (resp., strongly 2- irreducible) submodules of M as a generalization of irreducible (resp., strongly irreducible) submodules of M and investigated some properties of these classes of modules.

Equations

$$(Ry+Rz)\cap (Rx+Rz)\cap (Rx+Ry)\subseteq H_1\cap H_2\cap H_3\subseteq N.$$

$$An{n}_R(N)\subseteq An{n}_R(H_1)+An{n}_R(H_2)+An{n}_R(H_3).$$

$$(0{:}_M{I}_1)\cap (0{:}_M{I}_2)\cap (0{:}_M{I}_3)\subseteq (0{:}_{M}I).$$

$$J_{1}M\cap J_{2}M\cap J_{3}M\subseteq (N{:}_{R}M)M=N$$

$$(H_1\cap H_2\cap H_3{:}_{R}M)M=((H_1{:}_{R}M)\cap (H_2{:}_{R}M)\cap (H_3{:}_{R}M))M\subseteq (N{:}_{R}M)M.$$

$$K\cap IM\cap JM\subseteq rad(K)\cap rad(IM)\cap rad(JM)=rad(IJK)\subseteq rad(N)=N$$

$$K\cap IM\cap JM=IJK\subseteq N.$$

$$(H_1{:}_{R}M)\cap (H_2{:}_{R}M)\cap (H_3{:}_{R}M)=(H_1\cap H_2\cap H_3{:}_{R}M)\subseteq (N{:}_{R}M).$$

$$(H_1{:}_{R}M)M\cap (H_2{:}_{R}M)M\subseteq (N{:}_{R}M)M\subseteq N.$$

$$N=N+(H_1\cap H_2\cap H_3)=(N+H_1)\cap (N\cap H_2)\cap (N\cap H_3).$$

$$(H_1\times M_2)\cap (M_1\times 0)\cap (K_1\times M_2)\subseteq (H_1\cap K_1)\times 0\subseteq N_1\times N_2.$$

Full text

1. Introduction

Throughout this paper, $R$ will denote a commutative ring with identity and $\mathbb{Z}$ will denote the ring of integers.

An ideal $I$ of $R$ is said to be irreducible if $I=J_{1}\cap J_{2}$ for ideals $J_{1}$ and $J_{2}$ of $R$ implies that either $I=J_{1}$ or $I=J_{2}$. A proper ideal $I$ of $R$ is said to be strongly irreducible if for ideals $J_{1},J_{2}$ of $R$, $J_{1}\cap J_{2}\subseteq I$ implies that $J_{1}\subseteq I$ or $J_{2}\subseteq I$ [12]. An ideal $I$ of $R$ is said to be 2-irreducible if whenever $I=J_{1}\cap J_{2}\cap J_{3}$ for ideals $J_{1},J_{1}$ and $J_{3}$ of $R$, then either $I=J_{1}\cap J_{2}$ or $I=J_{1}\cap J_{3}$ or $I=J_{2}\cap J_{3}$. Clearly, any irreducible ideal is a 2-irreducible ideal [21].

A proper submodule $N$ of an $R$-module $M$ is said to be irreducible (resp., strongly irreducible) if for submodules $H_{1}$ and $H_{2}$ of $M$, $N=H_{1}\cap H_{2}$ (resp., $H_{1}\cap H_{2}\subseteq N$) implies that $N=H_{1}$ or $N=H_{2}$.( resp., $H_{1}\subseteq N$ or $H_{2}\subseteq N$).

The main purpose of this paper is to introduce the concept of 2-irreducible and strongly 2-irreducible submodules of an $R$-module $M$ as a generalization of irreducible and strongly irreducible submodules of $M$ and obtain some related results.

A submodule $N$ of an $R$-module $M$ is said to be a 2-irreducible submodule if whenever $N=H_{1}\cap H_{2}\cap H_{3}$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$, then either $N=H_{1}\cap H_{2}$ or $N=H_{2}\cap H_{3}$ or $N=H_{1}\cap H_{3}$ (Definition 2.1).

A proper submodule $N$ of an $R$-module $M$ is said to be a strongly 2-irreducible submodule if whenever $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$, then either $H_{1}\cap H_{2}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ (Definition 2.4).

In Section 2 of this paper, for an $R$-module $M$, among other results, we prove that if $M$ is a Noetherian $R$-module and $N$ is a 2-irreducible submodule of $M$, then either $N$ is irreducible or $N$ is an intersection of exactly two irreducible submodules of $M$ (Theorem 2.20). In Theorem 2.7, we provide a characterization for strongly 2-irreducible submodules of $M$. Also, it is shown that if $M$ is a strong comultiplication $R$-module, then every non-zero proper submodule of $R$ is a strongly sum 2-irreducible $R$-module if and only if every non-zero proper submodule of $M$ is a strongly 2-irreducible submodule of $M$ (Theorem 2.9). Further, it is proved that if $N$ is a submodule of a finitely generated multiplication $R$-module $M$, then $N$ is a strongly 2-irreducible submodule of $M$ if and only if $(N:_{R}M)$ is a strongly 2-irreducible ideal of $R$ (Theorem 2.10). In Theorem 2.17 and 2.19, we provide some useful characterizations for strongly 2-irreducible submodules of some special classes of modules. Example 2.12 shows that the concepts of strongly irreducible submodules and strongly 2-irreducible submodules are different in general. Finally, let $R=R_{1}\times R_{2}\times\cdots\times R_{n}$ ($2\leq n<\infty$) be a decomposable ring and $M=M_{1}\times M_{2}\cdots\times M_{n}$ be an $R$-module, where for every $1\leq i\leq n$, $M_{i}$ is an $R_{i}$-module, respectively, it is proved that a proper submodule $N$ of $M$ is a strongly 2-irreducible submodule of $M$ if and only if either $N=\times^{n}_{i=1}N_{i}$ such that for some $k\in\{1,2,...,n\}$, $N_{k}$ is a strongly 2-irreducible submodule of $M_{k}$, and $N_{i}=M_{i}$ for every $i\in\{1,2,...,n\}\setminus\{k\}$ or $N=\times^{n}_{i=1}N_{i}$ such that for some $k,m\in\{1,2,...,n\}$, $N_{k}$ is a strongly irreducible submodule of $M_{k}$, $N_{m}$ is a strongly irreducible submodule of $M_{m}$, and $N_{i}=M_{i}$ for every $i\in\{1,2,...,n\}\setminus\{k,m\}$ (Theorem 2.26).

2. Main results

Definition 2.1**.**

We say that a submodule $N$ of an $R$-module $M$ is a 2-irreducible submodule if whenever $N=H_{1}\cap H_{2}\cap H_{3}$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$, then either $N=H_{1}\cap H_{2}$ or $N=H_{2}\cap H_{3}$ or $N=H_{1}\cap H_{3}$.

Theorem 2.2**.**

Let $M$ be a Noetherian $R$-module. If $N$ is a 2-irreducible submodule of $M$, then either $N$ is irreducible or $N$ is an intersection of exactly two irreducible submodules of $M$.

Proof.

Let $N$ be a 2-irreducible submodule of $M$. By [17, Exercise 9.31], $N$ can be written as a finite irredundant irreducible decomposition $N=N_{1}\cap N_{2}\cap...\cap N_{k}$. We show that either $k=1$ or $k=2$. If $k>3$, then since $N$ is 2-irreducible, $N=N_{i}\cap N_{j}$ for some $1\leq i,j\leq k$, say $i=1$ and $j=2$. Therefore $N_{1}\cap N_{2}\subseteq N_{3}$, which is a contradiction. ∎

Corollary 2.3**.**

Let $M$ be a Noetherian multiplication $R$-module. If $N$ is a 2-irreducible submodule of $M$, then $N$ a 2-absorbing primary submodule of $M$.

Proof.

Let $N$ be a 2-irreducible submodule of $M$. By the fact that every irreducible submodule of a Noetherian $R$-module is primary and regarding Theorem 2.20, we have either $N$ is a primary submodule or is a sum of two primary submodules. It is clear that every primary submodule is 2-absorbing primary, also the sum of two primary submodules is a 2-absorbing primary submodule, by [15, Theorem 2.20]. ∎

Definition 2.4**.**

We say that a proper submodule $N$ of an $R$-module $M$ is a strongly 2-irreducible submodule if whenever $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$, then either $H_{1}\cap H_{2}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$.

Example 2.5**.**

[21, Corollary 2] Consider the $\mathbb{Z}$-module $\mathbb{Z}$. Then $n\mathbb{Z}$ is a strongly 2-irreducible submodule of $\mathbb{Z}$ if $n=0$, $p^{t}$ or $p^{r}q^{s}$, where $p,q$ are prime integers and $t,r,s$ are natural numbers.

Proposition 2.6**.**

The strongly 2-irreducible submodules of a distributive $R$-module are precisely the 2-irreducible submodules.

Proof.

This is straightforward. ∎

Theorem 2.7**.**

Let $N$ be a proper submodule of an $R$-module $M$. Then the following conditions are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule;*

• (b)

For all elements $x,y,z$ of $M$, we have $(Rx+Ry)\cap(Rx+Rz)\cap(Ry+Rz)\subseteq N$ implies that either $(Rx+Ry)\cap(Rx+Rz)\subseteq N$ or $(Rx+Ry)\cap(Ry+Rz)\subseteq N$ or $(Rx+Rz)\cap(Ry+Rz)\subseteq N$.

Proof.

$(a)\Rightarrow(b)$ This ia clear.

$(b)\Rightarrow(a)$ Let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. If $H_{1}\cap H_{2}\not\subseteq N$, $H_{1}\cap H_{3}\not\subseteq N$, and $H_{2}\cap H_{3}\not\subseteq N$, then there exist elements $x,y,z$ of $M$ such that $x\in H_{2}\cap H_{3}$, $y\in H_{1}\cap H_{3}$, and $z\in H_{1}\cap H_{2}$ but $x\not\in N$, $y\not\in N$, and $z\not\in N$. Therefore,

[TABLE]

Hence by the part (a), either $(Ry+Rz)\cap(Rx+Rz)\subseteq N$ or $(Ry+Rz)\cap(Rx+Ry)\subseteq N$ or $(Rx+Rz)\cap(Rx+Ry)\subseteq N$. Thus either $z\in N$ or $y\in N$ or $x\in N$. This contradiction completes the proof. ∎

Recall that a waist submodule of an $R$-module $M$ is a submodule that is comparable to any other submodules of $M$.

Proposition 2.8**.**

Let $N$ be a proper submodule of an $R$-module $M$. Then we have the following.

• (a)

If $N$ is a strongly 2-irreducible submodule of $M$, then it is also a 2-irreducible submodule of $M$.

• (b)

If $N$ is a strongly 2-irreducible submodule of $M$, then $N$ is a strongly 2-irreducible submodule of $T$ and $N/K$ is a strongly 2-irreducible submodule of $M/K$ for any $K\subseteq N\subseteq T$.

• (c)

If for all elements $x,y,z$ of $M$ we have $Rx\cap Ry\cap Rz\subseteq N$ implies that either $Rx\cap Ry\subseteq N$ or $Rx\cap Rz\subseteq N$ or $Ry\cap Rz\subseteq N$, then $N$ is a strongly 2-irreducible submodule of $M$.

• (d)

If $N$ is a waist submodule of $M$, then $N$ is strongly 2-irreducible submodule of $M$ if and only if $N$ is 2-irreducible module.

• (e)

If $N$ satisfies $(N+T)\cap(N+K)=N+(T\cap K)$, whenever $T\cap K\subseteq N$, then $N$ is strongly 2-irreducible submodule of $M$ if and only if $N$ is a 2-irreducible module.

Proof.

(a) Let $N$ be a strongly 2-irreducible submodule of $M$ and let $N=H_{1}\cap H_{2}\cap H_{3}$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then by assumption, either $H_{1}\cap H_{2}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$. Now the result follows from the fact that the reverse of inclusions are clear.

The parts (b), (d), and (e) are straightforward.

(c) Let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. If $H_{1}\cap H_{2}\not\subseteq N$, $H_{1}\cap H_{3}\not\subseteq N$, and $H_{2}\cap H_{3}\not\subseteq N$, then there exist elements $x,y,z$ of $M$ such that $x\in H_{2}\cap H_{3}$, $y\in H_{1}\cap H_{3}$, and $z\in H_{1}\cap H_{2}$ but $x\not\in N$, $y\not\in N$, and $z\not\in N$. Now the result follows by assumption. ∎

An $R$-module $M$ is said to be a comultiplication module if for every submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N=(0:_{M}I)$, equivalently, for each submodule $N$ of $M$, we have $N=(0:_{M}Ann_{R}(N))$ [2].

An $R$-module $M$ satisfies the double annihilator conditions (DAC for short) if for each ideal $I$ of $R$ we have $I=Ann_{R}(0:_{M}I)$ [9].

An $R$-module $M$ is said to be a strong comultiplication module if $M$ is a comultiplication $R$-module and satisfies the DAC conditions [4].

A submodule $N$ of an $R$-module $M$ is said to be a strongly sum 2-irreducible submodule if whenever $N\subseteq H_{1}+H_{2}+H_{3}$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$, then either $N\subseteq H_{1}+H_{2}$ or $N\subseteq H_{2}+H_{3}$ or $N\subseteq H_{1}+H_{3}$. Also, $M$ is said to be a strongly sum 2-irreducible module if and only if $M$ is a strongly sum 2-irreducible submodule of itself [10].

Theorem 2.9**.**

Let $M$ be a strong comultiplication $R$-module. Then every non-zero proper submodule of $R$ is a strongly sum 2-irreducible $R$-module if and only if every non-zero proper submodule of $M$ is a strongly 2-irreducible submodule of $M$.

Proof.

$"\Rightarrow"$ Let $N$ be a non-zero proper submodule of $M$ and let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then by using [11, 2.5],

[TABLE]

This implies that either $Ann_{R}(N)\subseteq Ann_{R}(H_{1})+Ann_{R}(H_{2})$ or $Ann_{R}(N)\subseteq Ann_{R}(H_{1})+Ann_{R}(H_{3})$ or $Ann_{R}(N)\subseteq Ann_{R}(H_{2})+Ann_{R}(H_{3})$ since by assumption, $Ann_{R}(N)$ is a strongly sum 2-irreducible $R$-module. Therefore, either $H_{1}\cap H_{2}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$ since $M$ is a comultiplication $R$-module.

$"\Leftarrow"$ Let $I$ be a non-zero proper submodule of $R$ and let $I\subseteq I_{1}+I_{2}+I_{3}$. Then

[TABLE]

Thus by assumption, either $(0:_{M}I_{1})\cap(0:_{M}I_{2})\subseteq(0:_{M}I)$ or $(0:_{M}I_{1})\cap(0:_{M}I_{3})\subseteq(0:_{M}I)$ or $(0:_{M}I_{2})\cap(0:_{M}I_{3})\subseteq(0:_{M}I)$. This implies that either $(0:_{M}I_{1}+I_{2})\subseteq(0:_{M}I)$ or $(0:_{M}I_{1}+I_{3})\subseteq(0:_{M}I)$ or $(0:_{M}I_{2}+I_{3})\subseteq(0:_{M}I)$. Thus either $I\subseteq I_{1}+I_{2}$ or $I\subseteq I_{1}+I_{3}$ or $I\subseteq I_{2}+I_{3}$ since $M$ is a strong comultiplication $R$-module. ∎

An $R$-module $M$ is said to be a multiplication module if for every submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N=IM$ [6].

Theorem 2.10**.**

Let $N$ be a submodule of a finitely generated multiplication $R$-module $M$. Then $N$ is a strongly 2-irreducible submodule of $M$ if and only if $(N:_{R}M)$ is a strongly 2-irreducible ideal of $R$.

Proof.

$"\Rightarrow"$ Let $N$ be a strongly 2-irreducible submodule of $M$ and let $J_{1}\cap J_{2}\cap J_{3}\subseteq(N:_{R}M)$ for some ideals $J_{1},J_{2}$, and $J_{3}$ of $R$. Then

[TABLE]

by [8, Corollary 1.7]. Thus by assumption, either $J_{1}M\cap J_{2}M\subseteq N$ or $J_{1}M\cap J_{3}M\subseteq N$ or $J_{2}M\cap J_{3}M\subseteq N$. Hence, either $(J_{1}\cap J_{2})M\subseteq(N:_{R}M)M$ or $(J_{1}\cap J_{3})M\subseteq(N:_{R}M)M$ or $(J_{2}\cap J_{3})M\subseteq(N:_{R}M)M$. Therefore, either $J_{1}\cap J_{2}\subseteq(N:_{R}M)$ or $J_{1}\cap J_{3}\subseteq(N:_{R}M)$ or $J_{2}\cap J_{3}\subseteq(N:_{R}M)$ by [18, Corollary of Theorem 9].

$"\Leftarrow"$ Let $(N:_{R}M)$ is a strongly 2-irreducible ideal of $R$ and let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for some submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then we have

[TABLE]

Thus $(H_{1}:_{R}M)\cap(H_{2}:_{R}M)\cap(H_{3}:_{R}M)\subseteq(N:_{R}M)$ by [18, Corollary of Theorem 9]. Hence, either $(H_{1}:_{R}M)\cap(H_{2}:_{R}M)\subseteq(N:_{R}M)$ or $(H_{1}:_{R}M)\cap(H_{3}:_{R}M))\subseteq(N:_{R}M)$ or $(H_{2}:_{R}M)\cap(H_{3}:_{R}M)\subseteq(N:_{R}M)$ since $(N:_{R}M)$ is a strongly 2-irreducible ideal of $R$. Therefore, either $H_{1}\cap H_{2}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$ by [8, Corollary 1.7]. ∎

Example 2.11**.**

Consider the $\mathbb{Z}$-module $\mathbb{Z}_{p^{t}q^{n}r^{m}}$, where $p,q,r$ are prime integers and $t,n,m$ are natural numbers.

• (a)

By using Theorem 2.10 and Example 2.5, one can see that $\bar{p^{t}}\mathbb{Z}_{p^{t}q^{n}r^{m}}$ and $\bar{q^{n}r^{m}}\mathbb{Z}_{p^{t}q^{n}r^{m}}$ are strongly 2-irreducible submodules of $\mathbb{Z}_{p^{t}q^{n}r^{m}}$.

• (b)

$\bar{pqr}\mathbb{Z}_{p^{3}qr}=\bar{pq}Z_{p^{3}qr}\cap\bar{pr}\mathbb{Z}_{p^{3}qr}\cap\bar{qr}\mathbb{Z}_{p^{3}qr}$ implies that $\bar{pqr}\mathbb{Z}_{p^{3}qr}$ is not a 2-irreducible submodule of $\mathbb{Z}_{p^{3}qr}$.

The following example shows that the concepts of strongly irreducible submodules and strongly 2-irreducible submodules are different in general.

Example 2.12**.**

Consider the $\mathbb{Z}$-module $\mathbb{Z}_{6}$. Then $0=\bar{3}\mathbb{Z}_{6}\cap\bar{2}\mathbb{Z}_{6}$ implies that the [math] submodule of $\mathbb{Z}_{6}$ is not strongly irreducible. But $(0:_{\mathbb{Z}}\mathbb{Z}_{6})=6\mathbb{Z}$ is a strongly 2-irreducible ideal of $\mathbb{Z}$ by Example 2.5. Since the $\mathbb{Z}$-module $\mathbb{Z}_{6}$ is a finitely generated multiplication $\mathbb{Z}$-module, [math] is a strongly 2-irreducible submodule of $\mathbb{Z}_{6}$ by Theorem 2.10.

Lemma 2.13**.**

Let $M$ be an $R$-module. If $N_{1}$ and $N_{2}$ are strongly irreducible submodules of $M$, then $N_{1}\cap N_{2}$ is a strongly 2-irreducible submodule of $M$.

Proof.

This is straightforward. ∎

A proper submodule $P$ of an $R$-module $M$ is said to be prime if for any $r\in R$ and $m\in M$ with $rm\in P$, we have $m\in P$ or $r\in(P:_{R}M)$ [7].

Proposition 2.14**.**

Let $M$ be a multiplication $R$-module and let $N_{1}$, $N_{2}$, and $N_{3}$ be prime submodules of $M$ such that $N_{1}+N_{2}=N_{1}+N_{3}=N_{2}+N_{3}=M$. Then $N_{1}\cap N_{2}\cap N_{3}$ is not a strongly 2-irreducible submodule of $M$.

Proof.

Assume on the contrary that $N_{1}\cap N_{2}\cap N_{3}$ is a strongly 2-irreducible submodule of $M$. Then $N_{1}\cap N_{2}\cap N_{3}\subseteq N_{1}\cap N_{2}\cap N_{3}$ implies that either $N_{1}\cap N_{2}\subseteq N_{1}\cap N_{2}\cap N_{3}$ or $N_{1}\cap N_{3}\subseteq N_{1}\cap N_{2}\cap N_{3}$ or $N_{2}\cap N_{3}\subseteq N_{1}\cap N_{2}\cap N_{3}$. We can assume without loss of generality that $N_{1}\cap N_{2}\subseteq N_{1}\cap N_{2}\cap N_{3}$. Then $N_{1}\cap N_{2}\subseteq N_{3}$. It follows that $(N_{1}:_{R}M)N_{2}\subseteq N_{3}$. As $N_{3}$ is a prime submodule of $M$, we have $N_{2}\subseteq N_{3}$ or $(N_{2}:_{R}M)\subseteq(N_{3}:_{R}M)$. Thus $N_{2}\subseteq N_{3}$ or $N_{1}\subseteq N_{3}$ since $M$ is a multiplication $R$-module. Therefore, $N_{3}=M$, which is a contradiction. ∎

Corollary 2.15**.**

Let $M$ be a multiplication $R$-module such that every proper submodule of $M$ is strongly 2-irreducible. Then $M$ has at most two maximal submodules.

Proof.

This follows from Proposition 2.14 ∎

Let $N$ be a submodule of an $R$-module $M$. The intersection of all prime submodules of $M$ containing $N$ is said to be the (prime) radical of $N$ and denote by $rad_{M}N$ (or simply by $rad(N)$). In case $N$ does not contained in any prime submodule, the radical of $N$ is defined to be $M$. Also, $N\not=M$ is said to be a radical submodule of $M$ if $rad(N)=N$ [14]

Lemma 2.16**.**

Let $I$ be an ideal of $R$ and $N$ be a submodule of an $R$-module $M$. Then $rad(IN)=rad(N)\cap rad(IM)$.

Proof.

By [13, Corollary of Theorem 6], we have $rad(N\cap IM))=rad(N)\cap rad(IM)$. Since $IN\subseteq IM\cap N$, $rad(IN)\subseteq rad(IM\cap N)$. Thus $rad(IN)\subseteq rad(N)\cap rad(IM)$. Now let $P$ be a prime submodule of $M$ such that $IN\subseteq P$. As $P$ is prime, $N\subseteq P$ or $I\subseteq(P:_{R}M)$. Hence $N\cap IM\subseteq P$. This in tourn implies that $rad(N)\cap rad(IM)\subseteq rad(IN)$, as desired. ∎

A proper ideal $I$ of $R$ is said to be a 2-absorbing ideal of $R$ if whenever $a,b,c\in R$ and $abc\in I$, then $ab\in I$ or $ac\in I$ or $bc\in I$ [5].

A proper submodule $N$ of an $R$-module $M$ is said to be a 2-absorbing primary submodule of $M$ if whenever $a,b\in R$, $m\in M$, and $abm\in N$, then $am\in rad(N)$ or $bm\in rad(N)$ or $ab\in(N:_{R}M)$ [15].

A proper submodule $N$ of an $R$-module $M$ is called a *2-absorbing submodule *of $M$ if whenever $abm\in N$ for some $a,b\in R$ and $m\in M$, then $am\in N$ or $bm\in N$ or $ab\in(N:_{R}M)$ [19] and [16].

Theorem 2.17**.**

Let $M$ be a finitely generated multiplication $R$-module and $N$ be a radical submodule of $M$. Then the following conditions are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule of $M$;*

• (b)

$N$* is a 2-absorbing submodule of $M$;*

• (c)

$N$* is a 2-absorbing primary submodule of $M$;*

• (d)

$N$* is either a prime submodule of $M$ or is an intersection of exactly two prime submodules of $M$.*

Proof.

$(a)\Rightarrow(b)$ Let $I$, $J$ be ideals of $R$ and $K$ be a submodule of $M$ such that $IJK\subseteq N$. Then by using Lemma 2.16,

[TABLE]

Hence by part (a), either $K\cap IM\subseteq N$ or $K\cap JM\subseteq N$ or $IM\cap JM\subseteq N$. Thus either $IK\subseteq N$ or $JK\subseteq N$ or $IJM\subseteq N$ as needed.

$(b)\Rightarrow(c)$ This is clear.

$(c)\Rightarrow(b)$ This is clear by using [15, Theorem 2.6].

$(b)\Rightarrow(d)$ Since $N$ is a 2-absorbing submodule of $M$, $(N:_{R}M)$ is a 2-absorbing ideal of $R$ by [20, Proposition 1]. Hence $\sqrt{(N:_{R}M)}=P$ is a prime ideal of $R$ or $\sqrt{(N:_{R}M)}=P\cap Q$, where $P$ and $Q$ are distinct prime ideals of $R$ that are minimal over $(N:_{R}M)$ by [5, Theorem 2.4]. We have $\sqrt{(N:_{R}M)}M=rad(N)$ by [14, Theorem 4]. If $\sqrt{(N:_{R}M)}=P$, then $PM=rad(N)$. Since $M$ is a multiplication $R$-module, $PM$ is a prime submodule of $M$ by [8, Corollary 2.11]. Now let $\sqrt{Ann_{R}(N)}=P\cap Q$, where $P$ and $Q$ are distinct prime ideals of $R$. Then $(P\cap Q)M=rad(N)$. By [8, Corollary 1.7], $(P\cap Q)M=PM\cap QM$. Thus by [8, Corollary 2.11], $rad(N)$ is an intersection of two prime submodules of $M$. Now, we prove the claim by assumption that $N$ is a radical submodule of $M$.

$(d)\Rightarrow(a)$ This follows from Lemma 2.13. ∎

The following example shows that parts $(a)$ and $(b)$ of Theorem 2.17 are not equivalent in general.

Example 2.18**.**

Consider the submodule $G_{t}=\langle 1/p^{t}+\mathbb{Z}\rangle$ of the $\mathbb{Z}$-module $\mathbb{Z}_{p^{\infty}}$. Then $G_{t}$ is a strongly 2-irreducible submodule of $\mathbb{Z}_{p^{\infty}}$. But $G_{t}$ is not a 2-absorbing submodule of $\mathbb{Z}_{p^{\infty}}$. It should be note that the $\mathbb{Z}$-module $\mathbb{Z}_{p^{\infty}}$ is not a finitely genrated multiplication $\mathbb{Z}$-module.

A submodule $N$ of an $R$-module $M$ is said to be pure if $IN=IM\cap N$ for every ideal $I$ of $R$ [1]. Also, an $R$-module $M$ is said to be fully pure if every submodule of $M$ is pure [3].

Theorem 2.19**.**

Let $M$ be a fully pure multiplication $R$-module and $N$ be a submodule of $M$. Then the following conditions are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule of $M$;*

• (b)

$N$* is a 2-absorbing submodule of $M$;*

• (c)

$N$* is a 2-irreducible submodule of $M$.*

Proof.

$(a)\Rightarrow(b)$ Let $I$, $J$ be ideals of $R$ and $K$ be a submodule of $M$ such that $IJK\subseteq N$. Then since $M$ is fully pure,

[TABLE]

Hence by part (a), either $K\cap IM\subseteq N$ or $K\cap JM\subseteq N$ or $IM\cap JM\subseteq N$. Thus either $IK\subseteq N$ or $JK\subseteq N$ or $IJM\subseteq N$.

$(b)\Rightarrow(a)$ Let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then

[TABLE]

Thus either $(H_{1}:_{R}M)(H_{2}:_{R}M)\subseteq(N:_{R}M)$ or $(H_{1}:_{R}M)(H_{3}:_{R}M)\subseteq(N:_{R}M)$ or $(H_{2}:_{R}M)(H_{3}:_{R}M)\subseteq(N:_{R}M)$ since $(N:_{R}M)$ is a 2-absorbing ideal of $R$ by [20, Proposition 1]. We can assume without loss of generality that $(H_{1}:_{R}M)(H_{2}:_{R}M)\subseteq(N:_{R}M)$. Thus as $M$ is fully pure, we have

[TABLE]

Therefore, $H_{1}\cap H_{2}\subseteq N$ since $M$ is a multiplication $R$-module.

$(a)\Leftrightarrow(c)$ By [3, proof of Theorem 2.19], $M$ is a distributive $R$-module. Now the result follows from Proposition 2.6. ∎

Lemma 2.20**.**

Let $M$ be an $R$-module, $S$ a multiplicatively closed subset of $R$, and $N$ be a finitely generated submodule of $M$. If $S^{-1}N\subseteq S^{-1}K$ for a submodule $K$ of $M$, then there exists $s\in S$ such that $sN\subseteq K$.

Proof.

This is straightforward. ∎

Proposition 2.21**.**

Let $M$ be an $R$-module, $S$ be a multiplicatively closed subset of $R$ and $N$ be a finitely generated prime strongly 2-irreducible submodule of $M$ such that $(N:_{R}M)\cap S=\emptyset$. Then $S^{-1}N$ is a strongly 2-irreducible submodule of $S^{-1}M$ if $S^{-1}N\not=S^{-1}M$.

Proof.

Let $S^{-1}H_{1}\cap S^{-1}H_{2}\cap S^{-1}H_{3}\subseteq S^{-1}N$ for submodules $S^{-1}H_{1}$, $S^{-1}H_{2}$ and $S^{-1}H_{3}$ of $S^{-1}M$. Then $S^{-1}(H_{1}\cap H_{2}\cap H_{3})\subseteq S^{-1}N$. By Lemma 2.20, there exists $s\in S$ such that $s(H_{1}\cap H_{2}\cap H_{3})\subseteq N$. This implies that $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ since $N$ is prime and $(N:_{R}M)\cap S=\emptyset$. Now as $N$ is a strongly 2-irreducible submodule of $M$, we have either $H_{1}\cap H_{2}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$. Therefore, either $S^{-1}H_{1}\cap S^{-1}H_{2}\subseteq S^{-1}N$ or $S^{-1}H_{1}\cap S^{-1}H_{3}\subseteq S^{-1}N$ or $S^{-1}H_{2}\cap S^{-1}H_{3}\subseteq S^{-1}N$, as needed. ∎

Proposition 2.22**.**

Let $M$ be an $R$-module and $\{K_{i}\}_{i\in I}$ be a chain of strongly 2-irreducible submodules of $M$. Then $\cap_{i\in I}K_{i}$ is a strongly 2-irreducible submodule of $M$.

Proof.

Let $H_{1}\cap H_{2}\cap H_{3}\subseteq\cap_{i\in I}K_{i}$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Assume that $H_{1}+H_{2}\not\subseteq\cap_{i\in I}K_{i}$, $H_{1}+H_{3}\not\subseteq\cap_{i\in I}K_{i}$, and $H_{2}+H_{3}\not\subseteq\cap_{i\in I}K_{i}$. Then there are $m,n,t\in I$, where $H_{1}\cap H_{2}\not\subseteq K_{m}$, $H_{1}\cap H_{3}\not\subseteq K_{n}$, and $H_{2}\cap H_{3}\not\subseteq K_{t}$. Since $\{K_{i}\}_{i\in I}$ is a chain we can assume that $K_{m}\subseteq K_{n}\subseteq K_{t}$. But as $H_{1}\cap H_{2}\cap H_{3}\subseteq K_{m}$ and $K_{m}$ is a strongly sum 2-irreducible submodule of $M$, we have either $H_{1}\cap H_{2}\subseteq K_{m}$ or $H_{1}\cap H_{3}\subseteq K_{m}$ or $H_{2}\cap H_{3}\subseteq K_{m}$. In any case, we get a contradiction. ∎

Theorem 2.23**.**

Let $f:M\rightarrow\acute{M}$ be a epimorphism of R-modules. Then we have the following.

• (a)

If $N$ is a strongly 2-irreducible submodule of $M$ such that $ker(f)\subseteq N$, then $f(N)$ is a strongly 2-irreducible submodule of $f(M)$.

• (b)

If $\acute{N}$ is a strongly 2-irreducible submodule of $\acute{M}$, then $f^{-1}(\acute{N})$ is a strongly 2-irreducible submodule of $M$.

Proof.

(a) Let $N$ be a strongly 2-irreducible submodule of $M$. If $f(N)=\acute{M}$, then we have $N+Ker(f)=f^{-1}(f(N))=f^{-1}(\acute{M})=f^{-1}(f(M))=M$. Now as $ker(f)\subseteq N$, we get that $N=M$, which is a contradiction. Therefore, $f(N)\not=\acute{M}$. Suppose that $\acute{H_{1}}\cap\acute{H_{2}}\cap\acute{H_{3}}\subseteq f(N)$ for submodules $\acute{H_{1}}$, $\acute{H_{2}}$ and $\acute{H_{3}}$ of $f(M)$. Then $f^{-1}(\acute{H_{1}})\cap f^{-1}(\acute{H_{2}})\cap f^{-1}(\acute{H_{3}})\subseteq f^{-1}(f(N))=N$ since $ker(f)\subseteq N$. Thus by assumption, either $f^{-1}(\acute{H_{1}})\cap f^{-1}(\acute{H_{2}})\subseteq N$ or $f^{-1}(\acute{H_{1}})\cap f^{-1}(\acute{H_{3}})\subseteq N$ or $f^{-1}(\acute{H_{2}})\cap f^{-1}(\acute{H_{3}})\subseteq N$. Now as $f$ is epimorphism, we have either $\acute{H_{1}}\cap\acute{H_{2}}\subseteq f(N)$ or $\acute{H_{1}}\cap\acute{H_{3}}\subseteq f(N)$ or $\acute{H_{2}}\cap\acute{H_{3}}\subseteq f(N)$, as needed.

(b) Let $\acute{N}$ be a strongly 2-irreducible submodule of $\acute{M}$. Since $\acute{N}\not=\acute{M}$ and $f$ is a epimorphism, we have $f^{-1}(\acute{N})\not=M$. Now let $H_{1}\cap H_{2}\cap H_{3}\subseteq f^{-1}(\acute{N})$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then $f(H_{1})\cap f(H_{2})\cap f(H_{3})\subseteq f(f^{-1}(\acute{N}))=\acute{N}$. Thus by assumption, either $f(H_{1})\cap f(H_{2})\subseteq\acute{N}$ or $f(H_{1})\cap f(H_{3})\subseteq\acute{N}$ or $f(H_{2})\cap f(H_{3})\subseteq\acute{N}$. Now we have either $H_{1}\cap H_{2}\subseteq f^{-1}(\acute{N})$ or $H_{1}\cap H_{3}\subseteq f^{-1}(\acute{N})$ or $H_{2}\cap H_{3}\subseteq f^{-1}(\acute{N})$, as required. ∎

Theorem 2.24**.**

Let $M$ be a finitely generated multiplication distributive $R$-module and let $N$ be a non-zero proper submodule of $M$. Then the following statements are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule of $M$;*

• (b)

$(N:_{R}M)$* is a strongly 2-irreducible ideal of $R$;*

• (c)

$(N:_{R}M)$* is a 2-irreducible ideal of $R$.*

Proof.

$(a)\Rightarrow(b)$ This follows from Theorem 2.10.

$(b)\Rightarrow(c)$ This follows from [21, Proposition 1].

$(c)\Rightarrow(a)$ Let $H_{1}\cap H_{2}\cap H_{3}\subseteq N$ for submodules $H_{1}$, $H_{2}$ and $H_{3}$ of $M$. Then as $M$ is a distributive $R$-module, we have

[TABLE]

This implies that $(N:_{R}M)=(N+H_{1}:_{R}M)\cap(N+H_{2}:_{R}M)\cap(N+H_{3}:_{R}M)$. Thus by assumption, either$(N:_{R}M)=(N+H_{1}:_{R}M)\cap(N+H_{2}:_{R}M)$ or $(N:_{R}M)=(N+H_{1}:_{R}M)\cap(N+H_{3}:_{R}M)$ or $(N:_{R}M)=(N+H_{2}:_{R}M)\cap(N+H_{3}:_{R}M)$. Therefore, by [8, Corollary 1.7], either $N=N+(H_{1}\cap H_{2})$ or $N=N+(H_{1}\cap H_{3})$ or $N=N+(H_{2}\cap H_{3})$, since $M$ is a finitely generated multiplication $R$-module. Thus either, $H_{1}\cap H_{2}\subseteq N$ or $H_{1}\cap H_{3}\subseteq N$ or $H_{2}\cap H_{3}\subseteq N$ as needed. ∎

Let $R_{i}$ be a commutative ring with identity and $M_{i}$ be an $R_{i}$-module, for $i=1,2$. Let $R=R_{1}\times R_{2}$. Then $M=M_{1}\times M_{2}$ is an $R$-module and each submodule of $M$ is in the form of $N=N_{1}\times N_{2}$ for some submodules $N_{1}$ of $M_{1}$ and $N_{2}$ of $M_{2}$.

Theorem 2.25**.**

Let $R=R_{1}\times R_{2}$ be a decomposable ring and $M=M_{1}\times M_{2}$ be an $R$-module, where $M_{1}$ is an $R_{1}$-module and $M_{2}$ is an $R_{2}$-module. Suppose that $N=N_{1}\times N_{2}$ is a proper submodule of $M$. Then the following conditions are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule of $M$;*

• (b)

Either $N_{1}=M_{1}$ and $N_{2}$ strongly 2-irreducible submodule of $M_{2}$ or $N_{2}=M_{2}$ and $N_{1}$ is a strongly 2-irreducible submodule of $M_{1}$ or $N_{1}$, $N_{2}$ are strongly irreducible submodules of $M_{1}$, $M_{2}$, respectively.

Proof.

$(a)\Rightarrow(b)$. Let $N=N_{1}\times N_{2}$ be a strongly 2-irreducible submodule of $M$ such that $N_{2}=M_{2}$. From our hypothesis, $N$ is proper, so $N_{1}\not=M_{1}$. Set $\acute{M}=M/(0\times M_{2})$. One can see that $\acute{N}=N/(0\times M_{2})$ is a strongly 2-irreducible submodule of $\acute{M}$. Also, observe that $\acute{M}\cong M_{1}$ and $\acute{N}\cong N_{1}$. Thus $N_{1}$ is a strongly 2-irreducible submodule of $M_{1}$. By a similar argument as in the previous case, $N_{2}$ is a strongly 2-irreducible submodule of $M_{2}$, where, $N_{1}=M_{1}$. Now suppose that $N_{1}\not=M_{1}$ and $N_{2}\not=M_{2}$. We show that $N_{1}$ is a irreducible submodule of $M_{1}$. Suppose that $H_{1}\cap K_{1}\subseteq N_{1}$ for some submodules $H_{1}$ and $K_{1}$ of $M_{1}$. Then

[TABLE]

Thus by assumption, either $(H_{1}\times M_{2})\cap(M_{1}\times 0)\subseteq N_{1}\times N_{2}$ or $(H_{1}\times M_{2})\cap(K_{1}\times M_{2})\subseteq N_{1}\times N_{2}$ or $(M_{1}\times 0)\cap(K_{1}\times M_{2})\subseteq N_{1}\times N_{2}$. Therefore, $H_{1}\subseteq N_{1}$ or $K_{1}\subseteq N_{1}$ since $N_{2}\not=M_{2}$. Thus $N_{1}$ is a strongly irreducible submodule of $M_{1}$. Similarly, we can show that $N_{2}$ is strongly irreducible submodule of $M_{2}$.

$(b)\Rightarrow(a)$. Suppose that $N=N_{1}\times M_{2}$, where $N_{1}$ is a strongly 2-irreducible submodule of $M_{1}$. Then it is clear that $N$ is a strongly 2-irreducible submodule of $M$. Now, assume that $N=N_{1}\times N_{2}$, where $N_{1}$ and $N_{2}$ are strongly irreducible submodules of $M_{1}$ and $M_{2}$, respectively. Hence $(N_{1}\times M_{2})\cap(M_{1}\times N_{2})=N_{1}\times N_{2}=N$ is a strongly 2-irreducible submodule of $M$, by Lemma 2.13. ∎

Theorem 2.26**.**

Let $R=R_{1}\times R_{2}\times\cdots\times R_{n}$ ($2\leq n<\infty$) be a decomposable ring and $M=M_{1}\times M_{2}\cdots\times M_{n}$ be an $R$-module, where for every $1\leq i\leq n$, $M_{i}$ is an $R_{i}$-module, respectively. Then for a proper submodule $N$ of $M$ the following conditions are equivalent:

• (a)

$N$* is a strongly 2-irreducible submodule of $M$;*

• (b)

Either $N=\times^{n}_{i=1}N_{i}$ such that for some $k\in\{1,2,...,n\}$, $N_{k}$ is a strongly 2-irreducible submodule of $M_{k}$, and $N_{i}=M_{i}$ for every $i\in\{1,2,...,n\}\setminus\{k\}$ or $N=\times^{n}_{i=1}N_{i}$ such that for some $k,m\in\{1,2,...,n\}$, $N_{k}$ is a strongly irreducible submodule of $M_{k}$, $N_{m}$ is a strongly irreducible submodule of $M_{m}$, and $N_{i}=M_{i}$ for every $i\in\{1,2,...,n\}\setminus\{k,m\}$.

Proof.

We use induction on $n$. For $n=2$ the result holds by Theorem 2.25. Now let $3\leq n<\infty$ and suppose that the result is valid when $K=M_{1}\times\cdots\times M_{n-1}$. We show that the result holds when $M=K\times M_{n}$. By Theorem 2.25, $N$ is a strongly 2-irreducible submodule of $M$ if and only if either $N=L\times M_{n}$ for some strongly 2-irreducible submodule $L$ of $K$ or $N=K\times L_{n}$ for some strongly 2-irreducible submodule $L_{n}$ of $M_{n}$ or $N=L\times L_{n}$ for some strongly irreducible submodule $L$ of $K$ and some strongly irreducible submodule $L_{n}$ of $M_{n}$. Note that a proper submodule $L$ of $K$ is a strongly irreducible submodule of $K$ if and only if $L=\times^{n-1}_{i=1}N_{i}$ such that for some $k\in\{1,2,...,n-1\}$, $N_{k}$ is a strongly irreducible submodule of $M_{k}$, and $N_{i}=M_{i}$ for every $i\in\{1,2,...,n-1\}\setminus\{k\}$. Consequently the claim is now verified. ∎