Some generalizations of strongly prime ideals

TL;DR

This paper introduces new classes of ideals called strongly 2-absorbing primary and strongly 2-absorbing ideals, extending the concept of strongly prime ideals, and explores their fundamental properties.

Contribution

It presents the definitions and basic properties of these new classes of ideals, broadening the understanding of ideal generalizations in algebra.

Findings

Defined strongly 2-absorbing primary ideals and strongly 2-absorbing ideals.

Established fundamental properties of these classes.

Extended the concept of strongly prime ideals.

Abstract

In this paper, we introduce the concepts of strongly 2-absorbing primary ideals (resp., submodules) and strongly 2-absorbing ideals (resp., submodules) as generalizations of strongly prime ideals. Furthermore, we investigate some basic properties of these classes of ideals.

Equations

$$\begin{gathered} E(S)=\{x\in K: \\ {x}^n\notin Sforeachn\ge 1\}. \end{gathered}$$

$$a+b=(x)(y)(x^{-1}{y}^{-1}(a+b))\in I$$

$$(\sum _{i\in I}(N_i{:}_{R}M)M{:}_{R}M)=\sum _{i\in I}(N_i{:}_{R}M)$$

$$(K_m{:}_{R}M)\subseteq (K_n{:}_{R}M)\subseteq (K_t{:}_{R}M).$$

Full text

1. Introduction

Throughout this paper, $R$ will denote an integral domain with quotient field $K$. Further, $\mathbb{Z}$, $\mathbb{Q}$, and $\mathbb{N}$ will denote respectively the ring of integers, the field of rational numbers, and the set of natural numbers.

A proper ideal $I$ of $R$ is said to be strongly prime if, whenever $xy\in I$ for elements $x,y\in K$, then $x\in I$ or $y\in I$ [8]. A proper ideal $I$ of $R$ is said to be strongly primary if, whenever $xy\in I$ for elements $x,y\in K$, then $x\in I$ or $y^{n}\in I$ for some $n\geq 1$ [4].

The concept of $2$-absorbing ideals was introduced in [3]. 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 $bc\in I$. In [5], Badawi, et al. introduced the concept of 2-absorbing primary ideal which is a generalization of primary ideal. A proper ideal $I$ of $R$ is called a 2-absorbing primary ideal of $R$ if whenever $a,b,c\in R$ and $abc\in I$, then $ab\in I$ or $ac\in\sqrt{I}$ or $bc\in\sqrt{I}$.

The purpose of this paper is to introduce the concepts of strongly 2-absorbing primary ideals (resp., submodules) and strongly 2-absorbing ideals (resp., submodules) as generalizations of strongly prime ideals. Furthermore, we investigate basic properties of these classes of ideals.

Let $R$ be an integral domain with quotient field $K$. An ideal $I$ of $R$ is said to be a strongly 2-absorbing primary ideal if, whenever $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $(yz)^{n}\in I$ or $(xz)^{m}\in I$ for some $n,m\in\mathbb{N}$ (Definition 2.1). A 2-absorbing ideal $I$ of $R$ is said to be a strongly 2-absorbing ideal if, whenever $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $yz\in I$ or $xz\in I$ (Definition 3.1). Moreover, a submodule $N$ of an $R$-module $M$ is said to be strongly 2-absorbing primary (resp., strongly 2-absorbing) if $(N:_{R}M)$ is a strongly 2-absorbing primary (resp., strongly 2-absorbing) ideal of $R$ (Definition 2.11 and 3.23).

Let $R$ be an integral domain with quotient field $K$. In Section 2 of this paper, among other results, we prove that if $I$ is a strongly primary ideal of $R$, then $I$ is a strongly 2-absorbing primary ideal of $R$ (Proposition 2.2). Example 2.3, shows that the converse of Proposition 2.2 is not true in general. In Theorem 2.5, we provide a useful characterization for strongly 2-absorbing primary ideals of $R$, where $R$ is a rooty domain. In Theorem 2.7, we show that for a strongly 2-absorbing primary ideal $I$ of $R$:

• (a)

If $J$ and $H$ are radical ideals of $R$, then $JH\subseteq I$ or $I^{2}\subseteq J\cup H$;

• (b)

If $J$ and $I$ are prime ideals of $R$, then $J$ and $I$ are comparable.

Furthermore, it is shown that if $P$ and $Q$ are non-zero strongly primary ideals of $R$, then $P\cap Q$ is a strongly 2-absorbing primary ideal of $R$ (Theorem 2.9).

In Section 3 of this paper, among other results, we prove that if $I$ is a strongly prime ideal of $R$, then $I$ is a strongly 2-absorbing ideal of $R$ (Proposition 3.2). But the converse of Proposition 3.2 is not true in general (Example 3.5, Proposition 3.6, and Example 3.7). In Theorem 3.3, we provide a useful characterization for a strongly 2-absorbing ideal of $R$. Also, we see that if $P$ and $Q$ are non-zero strongly prime ideals of $R$, then $P\cap Q$ is a strongly 2-absorbing ideal of $R$ (Theorem 3.16). Finally, it is proved that if $M$ is a Noetherian $R$-module, then $M$ contains a finite number of minimal strongly 2-absorbing submodules (Theorem 3.30).

2. Strongly 2-absorbing primary ideals and submodules

Definition 2.1**.**

Let $R$ be an integral domain with quotient field $K$. We say that an ideal $I$ of $R$ is a strongly 2-absorbing primary ideal if, whenever $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $(yz)^{n}\in I$ or $(xz)^{m}\in I$ for some $n,m\in\mathbb{N}$.

Proposition 2.2**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly primary ideal of $R$. Then $I$ is a strongly 2-absorbing primary ideal of $R$.

Proof.

Let $xyz\in I$ for some $x,y,z\in K$. Then by assumption, either $xy\in I$ or $z^{n}\in I$ for some $n\geq 1$. If $xy\in I$, then we are done. If $z^{n}\in I$, then $(zx)^{n}(zy)^{n}=(zxy)^{n}z^{n}\in I$. Thus again by assumption, either $(zx)^{n}\in I$ or $(yz)^{ns}\in I$ for some $s\geq 1$ as desired. ∎

The following example shows that the converse of Proposition 2.2 is not true in general.

Example 2.3**.**

Let $K$ be a field of characteristic 2, and put $I=(X^{2})K[[X^{2},X^{3}]]$, where $K[[X^{2},X^{3}]]$ is the ring of formal power series over the indeterminates $X^{2}$ and $X^{3}$. By considering the elements $X^{3}$ and $1/X$ in the quotient field $K((X))$, it is clear that $I$ is not strongly primary. Now, let $fgh\in I$, where $f,g,h\in K((X))$. Then there exist units $u,v,w$ of the $DVR$ $K[[X]]$ and integers $\alpha$, $\beta$, $\gamma$ for which $f=uX^{\alpha}$, $g=vX^{\beta}$, and $h=wX^{\gamma}$. Then $fgh\in I$ implies that $\alpha+\beta+\gamma\geq 2$; hence, $(\beta+\gamma)+(\alpha+\gamma)+(\alpha+\beta)\geq 4$. Now, if one of $\beta+\gamma$ or $\alpha+\gamma$ is at least one, then correspondingly either $(gh)^{2}\in I$ or $(fh)^{2}\in I$. On the other hand, if both $\beta+\gamma$ and $\alpha+\gamma$ are at most 0, then $\alpha+\beta\geq 4$. However, this would mean that $fg\in I$. Therefore, $I$ must be a strongly 2-absorbing primary ideal of $K[[X^{2},X^{3}]]$.

Notation 2.4**.**

For a subset $S$ of $R$, we define $E(S)$ by

[TABLE]

Let $R$ be an integral domain with quotient field $K$. An ideal $I$ of $R$ is called strongly radical if whenever $x\in K$ satisfies $x^{n}\in I$ for some $n\geq 1$, then $x\in I$ [1].

Following [9], an integral domain $R$ is called rooty if each radical ideal of $R$ is strongly radical (equivalently, each prime ideal of $R$ is strongly radical. Thus valuation domains are rooty domains [2]).

Theorem 2.5**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be an ideal of $R$. Consider the following statements:

• (a)

$I$* is a 2-absorbing primary ideal of $R$ and for each $x,y\in K$ with $xy\not\in I$ we have $x^{-1}I\cap E(I)=\emptyset$ or $y^{-1}I\cap E(I)=\emptyset$.*

• (b)

$I$* is a strongly 2-absorbing primary ideal of $R$.*

Then $(a)\Rightarrow(b)$. Moreover, if $K\setminus E(I)$ is closed under addition (in particular, if $R$ is rooty), then $(b)\Rightarrow(a)$.

Proof.

$(a)\Rightarrow(b)$ Let $xyz\in I$ for some $x,y,z\in K$ and $xy\not\in I$. Then by part (a), either $x^{-1}I\cap E(I)=\emptyset$ or $y^{-1}I\cap E(I)=\emptyset$. If $x^{-1}I\cap E(I)=\emptyset$, then $yz=yzxx^{-1}=(yzx)x^{-1}\in x^{-1}I$ implies that $(yz)^{n}\in I$ for some $n\geq 1$. Similarly, if $y^{-1}I\cap E(I)=\emptyset$, then we have $(xz)^{m}\in I$ for some $m\geq 1$, as needed.

$(b)\Rightarrow(a)$ Assume on the contrary that $x,y\in K$ with $xy\not\in I$ and $x^{-1}I\cap E(I)\not=\emptyset$ and $y^{-1}I\cap E(I)\not=\emptyset$. Then there exist $a,b\in I$ such that $x^{-1}a\in E(I)$ and $y^{-1}b\in E(I)$. Now as $I$ is a strongly 2-absorbing primary ideal of $R$, we have $(x)(y)(x^{-1}y^{-1}a)=a\in I$ implies that $(y^{-1}a)^{n}\in I$ for some $n\geq 1$. In a similar way we have $(x^{-1}b)^{m}\in I$ for some $m\geq 1$. On the other hand,

[TABLE]

implies that either $xy\in I$ or $(x^{-1}(a+b))^{s}\in I$ or $(y^{-1}(a+b))^{t}\in I$. Therefore, as $K\setminus E(I)$ is closed under addition, either $xy\in I$ or $x^{-1}a\not\in E(I)$ or $y^{-1}b\not\in E(I)$, which is a contradiction. ∎

Theorem 2.6**.**

Let $R$ be an integral domain with quotient field $K$ and $I$ be an ideal of $R$. Consider the following:

• (a)

If $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $yz\in\sqrt{I}$ or $xz\in\sqrt{I}$.

• (b)

If $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $(yz)^{n}\in I$ or $(xz)^{m}\in I$ for some $n,m\geq 1$ (i.e., $I$ is a strongly 2-absorbing primary ideal of $R$).

Then $(a)\Rightarrow(b)$. Moreover, if $R$ is a rooty domain, then $(b)\Rightarrow(a)$.

Proof.

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

$(b)\Rightarrow(a)$ Let $xyz\in I$ for elements $x,y,z\in K$. If $xy\not\in I$, then we have either $(yz)^{n}\in I$ or $(xz)^{m}\in I$ for some $n,m\geq 1$ by part (b). Since $R$ is a rooty domain, $yz\in\sqrt{I}$ or $xz\in\sqrt{I}$, as needed. ∎

Theorem 2.7**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly 2-absorbing primary ideal of $R$. Then we have the following:

• (a)

If $J$ and $H$ are radical ideals of $R$, then $JH\subseteq I$ or $I^{2}\subseteq J\cup H$.

• (b)

If $J$ and $I$ are prime ideals of $R$, then $J$ and $I$ are comparable.

Proof.

(a) Suppose that $J$ and $H$ are radical ideals of $R$ such that $JH\not\subseteq I$. Then there exist $a\in J$ and $b\in H$ such that $ab\in JH\setminus I$. Let $x,y\in I$. Then $(xy/ab)(a/x)(b/1)\in I$ implies that either $(a/x)(b/1)\in I$ or $((xy/ab)(a/x))^{n}\in I$ or $((xy/ab)(b/1))^{m}\in I$ for some $n,m\geq 1$. Thus either $x(ab/x)\in xR\subseteq I$ or $(b(y/b))^{n}\in b^{n}I\subseteq b^{n}R\subseteq H$ or $(a(xy/a))^{m}\in a^{m}I\subseteq a^{m}R\subseteq J$. Hence, either $ab\in I$ or $y^{n}\in H$ or $(xy)^{m}\in J$. Since $ab\not\in I$, we have either $y\in\sqrt{H}=H$ or $xy\in\sqrt{J}=J$. Therefore, $xy\in J\cup H$. This implies that $I^{2}\subseteq J\cup H$, as desired.

(b) The result follows from the fact that $J^{2}\subseteq I$ or $I^{2}\subseteq J$ by part (a). ∎

Corollary 2.8**.**

Let $R$ be an integral domain with quotient field $K$ and $Q$ be a maximal ideal of $R$. If $Q$ is a strongly 2-absorbing primary ideal of $R$, then $R$ is a local ring with maximal ideal $Q$.

Proof.

It follows from Theorem 2.7. ∎

Theorem 2.9**.**

Let $R$ be an integral domain with quotient field $K$ and let $P$ and $Q$ be nonzero strongly primary ideals of $R$. Then $P\cap Q$ is a strongly 2-absorbing primary ideal of $R$.

Proof.

Suppose $(xy)z\in P\cap Q$ and $x,y,z\in K$. Then $(xy)z\in P$ and $(xy)z\in Q$. Since $P$ is strongly primary, so either $xy\in P$ or $z^{n}\in P$ for some $n\geq 1$. If $xy\in P$, then either $x\in P$ or $y^{m}\in P$ for some $m\geq 1$. Similarly, $x\in Q$ or $y^{t}\in Q$ or $z^{s}\in Q$ for some $s,t\geq 1$. First assume that $x\in P$ and $x\in Q$. Then $(xy)y^{-1}=x\in P$ implies that $xy\in P$ or $(y^{-1})^{h}\in P$ for some $h\geq 1$. Similarly, $xy\in Q$ or $(y^{-1})^{g}\in Q$ for some $g\geq 1$. If $(y^{-1})^{g}\in Q\subseteq R$ or $(y^{-1})^{h}\in P\subseteq R$, then $(xz)^{h}=(xyz)^{h}(y^{-1})^{h}\in P\cap Q$ or $(xz)^{g}=(xyz)^{g}(y^{-1})^{g}\in P\cap Q$ by definition of an ideal. Otherwise, $xy\in P\cap Q$ as requested. If the statements above lead to different elements in $P$ and $Q$, we still have that the intersection is strongly 2-absorbing primary. For example, if $z^{n}\in P$ and $y^{t}\in Q$, then clearly $(xy)^{nt}\in P$ and $(xy)^{nt}\in Q$ by definition of an ideal, thus $(xy)^{nt}\in P\cap Q$. ∎

Proposition 2.10**.**

Let $R$ be an integral domain with quotient field $K$ and $S$ be a multiplicatively closed subset of $R$. If $I$ is a strongly 2-absorbing primary ideal of $R$ such that $S\cap I=\emptyset$, then $S^{-1}I$ is a strongly 2-absorbing primary ideal of $S^{-1}R$.

Proof.

Assume that $a,b,c\in K$ such that $abc\in S^{-1}I$. Then there exists $s\in S$ such that $(sa)(b)c=sabc\in I$. Since $I$ is a strongly 2-absorbing primary ideal of $R$, this implies that either $(sa)c\in I$ or $((b)c)^{n}=(bc)^{n}\in I$ or $((sa)(b))^{m}=(sab)^{m}\in I$ for some $n,m\geq 1$. Thus $ac=(sa)c/s\in s^{-1}I$ or $(bc)^{n}=((b)c/1)^{n}\in s^{-1}I$ or $(ab)^{m}=((sa)(b)/s)^{m}\in s^{-1}I$ as needed. ∎

Definition 2.11**.**

Let $R$ be an integral domain with quotient field $K$ and $M$ be an $R$-module. We say that a submodule $N$ of $M$ is a strongly 2-absorbing primary if, $(N:_{R}M)$ is a strongly 2-absorbing primary ideal of $R$.

Proposition 2.12**.**

Let $R$ be an integral domain with quotient field $K$, $M$ be an $R$-module, and $N_{1}$, $N_{2}$ be two submodules of $M$ with $(N_{1}:_{R}M)$ and $(N_{2}:_{R}M)$ strongly primary ideals of $R$. Then $N_{1}\cap N_{2}$ is a strongly 2-absorbing primary submodule of $M$.

Proof.

Since $(N_{1}\cap N_{2}:_{R}M)=(N_{1}:_{R}M)\cap(N_{2}:_{R}M)$, the result follows from Proposition 2.9. ∎

Proposition 2.13**.**

Let $R$ be an integral domain with quotient field $K$, $N$ be submodule of a finitely generated $R$-module $M$, and let $S$ be a multiplicatively closed subset of $R$. If $N$ is a strongly 2-absorbing primary submodule and $(N:_{R}M)\cap S=\emptyset$, then $S^{-1}N$ is a strongly 2-absorbing primary $S^{-1}R$-submodule of $S^{-1}M$.

Proof.

As $M$ is finitely generated, $(S^{-1}N:_{S^{-1}R}S^{-1}M)=S^{-1}(N:_{R}M)$ by [10, Lemma 9.12]. Now the result follows from Proposition 2.10. ∎

Proposition 2.14**.**

Let $R$ be an integral domain with quotient field $K$ and $M$ be an $R$-module. Let $N$ be a strongly 2-absorbing primary submodule of $M$. Then we have the following.

• (a)

If $r\in K$ such that $r^{-1}\in R$, then $(N:_{M}r)$ is a strongly 2-absorbing primary submodule of $M$.

• (b)

If $f:M\rightarrow\acute{M}$ is a monomorphism of $R$-modules, then $N$ is a strongly 2-absorbing primary submodule of $M$ if and only if $f(N)$ is a strongly 2-absorbing primary submodule of $f(M)$.

Proof.

(a) Let $xyz\in((N:_{M}r):_{R}M)$ for some $x,y,z\in K$. Then $rxyz\in(N:_{R}M)$. Thus as $N$ is a strongly 2-absorbing primary submodule, either $rxy\in(N:_{R}M)$ or $(rxz)^{n}\in(N:_{R}M)$ or $(yz)^{m}\in(N:_{R}M)$ for some $n,m\geq 1$. Hence either $xy=r^{-1}rxy\in r^{-1}(N:_{R}M)\subseteq(N:_{R}M)$ or $(xz)^{n}=(r^{-1}rxz)^{n}\in r^{-1}(N:_{R}M)\subseteq(N:_{R}M)$ or $(yz)^{m}\in(N:_{R}M)$, as needed.

(b) This follows from the fact that $(N:_{R}M)=(f(N):_{R}f(M))$. ∎

3. Strongly 2-absorbing ideals and submodules

Definition 3.1**.**

Let $R$ be an integral domain with quotient field $K$. We say that a 2-absorbing ideal $I$ of $R$ is a strongly 2-absorbing ideal if, whenever $xyz\in I$ for elements $x,y,z\in K$, we have either $xy\in I$ or $yz\in I$ or $xz\in I$.

Proposition 3.2**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly prime ideal of $R$. Then $I$ is a strongly 2-absorbing ideal of $R$.

Proof.

Let $xyz\in I$ for some $x,y,z\in K$. Then by assumption, either $xy\in I$ or $z\in I$. If $xy\in I$, then we are done. If $z\in I$, then $zxyz\in I$. Thus again by assumption, either $zx\in I$ or $yz\in I$ as desired. ∎

The following theorem is a characterization for a strongly 2-absorbing ideal of $R$.

Theorem 3.3**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a 2-absorbing ideal of $R$. Then the following statements are equivalent:

• (a)

$I$* is a strongly 2-absorbing ideal of $R$;*

• (b)

For each $x,y\in K$ with $xy\not\in R$ we have either $x^{-1}I\subseteq I$ or $y^{-1}I\subseteq I$.

Proof.

$(a)\Rightarrow(b)$ Assume on the contrary that $x,y\in K$ with $xy\not\in R$ and neither $x^{-1}I\not\subseteq I$ nor $y^{-1}I\not\subseteq I$. Then there exist $a,b\in I$ such that $x^{-1}a\not\in I$ and $y^{-1}b\not\in I$. Now as $I$ is a strongly 2-absorbing ideal of $R$, we have $(x)(y)(x^{-1}y^{-1}a)=a\in I$ implies that $y^{-1}a\in I$. In the similar way we have $x^{-1}b\in I$. On the other hand,

[TABLE]

implies that either $xy\in I$ or $x^{-1}(a+b)\in I$ or $y^{-1}(a+b)\in I$. Therefore, either $xy\in R$ or $x^{-1}a\in I$ or $y^{-1}b\in I$, a contradiction.

$(b)\Rightarrow(a)$ Let $xyz\in I$ for some $x,y,z\in K$. If $xy\in R$, $xz\in R$, and $yz\in R$, then we are done since $I$ is a 2-absorbing ideal of $R$. So suppose without loss of generality that $xy\not\in R$. Then by part (b), either $x^{-1}I\subseteq I$ or $y^{-1}I\subseteq I$. If $x^{-1}I\subseteq I$, then $yz=yzxx^{-1}=(yzx)x^{-1}\in x^{-1}I\subseteq I$. Similarly, if $y^{-1}I\subseteq I$, then we have $xz\in I$, as desired. ∎

Corollary 3.4**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly 2-absorbing ideal of $R$. Then for each $x,y\in K$ with $xy\not\in R$ we have either $I\subseteq Rx$ or $I\subseteq Ry$.

Proof.

Let $x,y\in K$ with $xy\not\in R$. Then by Theorem 3.3 $(a)\Rightarrow(b)$, we have either $x^{-1}I\subseteq I$ or $y^{-1}I\subseteq I$. Thus either $I\subseteq Ix\subseteq Rx$ or $I\subseteq Ix\subseteq Ry$. ∎

Example 3.5, Proposition 3.6, and Example 3.7 show that the converse of Proposition 2.2 is not true in general.

Example 3.5**.**

If $Q$ is the maximal ideal of a non-trivial $DVR$, $V$, then $Q^{2}$ is a strongly 2-absorbing ideal of $V$ that is not a strongly prime ideal, since $Q^{2}$ is not even a prime ideal of $V$.

Proposition 3.6**.**

Let $R$ be an integral domain with a prime ideal $P$ such that there exists a discrete valuation overring $(V,Q)$ of $R$ centered at $P$ (that is, $Q\cap R=P$), where $Q=xV$. Suppose that $ux^{k}\in P$ for all units $u$ of $V$ and natural numbers $k\geq 2$, but there is no unit $u$ of $V$ for which $ux\in P$. Then $P$ is a strongly 2-absorbing ideal of $R$ that is not a strongly prime ideal.

Proof.

The fact that $P$ is not a strongly prime ideal of $R$ is immediate from the fact that $x^{2}\in P$, but $x\not\in P$, by assumption. Now, since $P$ is a prime ideal of $R$, it is necessarily a 2-absorbing ideal of $R$. Let $y$ and $z$ be elements of the quotient field of $R$ for which $yz\not\in R$. By Theorem 3.3, it suffices to show that either $y^{-1}P\subseteq P$ or $z^{-1}P\subseteq P$. Observe that there exist units $u$ and $v$ of $V$ and integers $\alpha$ and $\beta$ for which $y=ux^{\alpha}$ and $z=vx^{\beta}$. Since $yz\not\in R$, it must be the case that $\alpha+\beta\leq 1$. However, this means that either $\alpha\leq 0$ or $\beta\leq 0$. As such, either $-\alpha+\gamma\geq 2$ or $-\beta+\gamma\geq 2$ for all integers $\gamma\geq 2$, from which it follows that either $y^{-1}P\subseteq P$ or $z^{-1}P\subseteq P$ as needed. ∎

Example 3.7**.**

If $K$ is a field, then the ideal $(X^{2},X^{3})$ in $K[[X^{2},X^{3}]]$ the ring of formal power series in the indeterminates $X^{2}$ and $X^{3}$ over $K$ is an example of a strongly 2-absorbing prime ideal that is not strongly prime.

Proposition 3.8**.**

Let $R$ be an integral domain with quotient field $K$, $I$ be a strongly 2-absorbing ideal of $R$, and $Q$ be a prime ideal of $R$ which is properly contained in $I$. Then $I/Q$ is a strongly 2-absorbing ideal of $R/Q$.

Proof.

Clearly, $I/Q$ is a 2-absorbing ideal of $R/Q$. Now let $\phi:R\rightarrow R/Q$ denote the canonical homomorphism. Suppose that $x_{1}=\phi(y_{1})/\phi(z_{1})$ and $x_{2}=\phi(y_{2})/\phi(z_{2})$ are elements of the quotient field of $R/Q$ such that $x_{1}x_{2}\not\in R/Q$. Then $(y_{1}/z_{1})(y_{2}/z_{2})\not\in R$. Hence if $a\in I$, we have $(z_{1}/y_{1})a\in I$ or $(z_{2}/y_{2})a\in I$ by using Theorem 3.3. We can assume without loss of generality that $(z_{1}/y_{1})a\in I$. It follows that $(\phi(z_{1})/\phi(y_{1}))\phi(a)\in I/Q$. Thus $x^{-1}(I/Q)\subseteq I/Q$, as needed. ∎

Remark 3.9**.**

Clearly, every strongly 2-absorbing ideal of $R$ is a 2-absorbing ideal of $R$. But the converse is not true in general. Because for example, if we consider the integral domain $\mathbb{Z}$, then $K=\mathbb{Q}$ and $(8/15)(3/2)(5/2)=2\in 2\mathbb{Z}$ implies that $2\mathbb{Z}$ is not a strongly 2-absorbing ideal of $\mathbb{Z}$. But $2\mathbb{Z}$ is a 2-absorbing ideal of $\mathbb{Z}$.

Definition 3.10**.**

We say that an integral domain $R$ is a 2-absorbing pseudo-valuation domain if every 2-absorbing ideal of $R$ is a strongly 2-absorbing ideal of $R$.

Proposition 3.11**.**

Every valuation domain is a 2-absorbing pseudo-valuation domain.

Proof.

Let $V$ be a valuation domain, and let $I$ be a 2-absorbing ideal of $V$. Suppose $xyz\in I$, where $x,y,z\in K$, the quotient field of $V$. If $x$, $y$, and $z$ are in $V$, we are done. Suppose without loss of generality that $x\not\in V$. Since $V$ is a valuation domain, we have $x^{-1}\in V$. Hence $yz=(x^{-1})(xyz)\in I$, as needed. ∎

Definition 3.12**.**

Let $R$ be an integral domain with quotient field $K$. We say that a non-zero prime ideal $P$ of $R$ is a strongly semiprime if whenever $x^{2}\in P$ for element $x\in K$, we have $x\in P$.

Remark 3.13**.**

Let $R$ be an integral domain with quotient field $K$. Clearly every non-zero strongly prime ideal of $R$ is a strongly semiprime ideal of $R$. But as we see in the following example the converse is not true in general.

Example 3.14**.**

Consider an integral domain $\mathbb{Z}$. Then $K=\mathbb{Q}$ and $(4/3)(3/2)=2\in 2\mathbb{Z}$ implies that $2\mathbb{Z}$ is not a strongly prime ideal of $\mathbb{Z}$. But $2\mathbb{Z}$ is a strongly semiprime ideal of $\mathbb{Z}$.

Proposition 3.15**.**

Let $R$ be an integral domain with quotient field $K$.

• (a)

If $P$ is a strongly semiprime and strongly 2-absorbing ideal of $R$, then $P$ is a strongly prime ideal of $R$.

• (b)

If $P_{1}$ and $P_{2}$ are strongly semiprime ideals of $R$, then $P_{1}\cap P_{2}$ is a strongly semiprime ideal of $R$.

Proof.

(a) Let $P$ be a strongly semiprime and 2-absorbing ideal of $R$ and let $x\in K\setminus R$. Then as $P$ is strongly semiprime $x^{2}\not\in P$. Since $P$ is strongly 2-absorbing, this implies that $x^{-1}P\subseteq P$ by Theorem 3.3. Now the result follows from [8, Proposition 1.2].

(b) This is clear. ∎

Theorem 3.16**.**

Let $R$ be an integral domain with quotient field $K$ and let $P$ and $Q$ be non-zero strongly prime ideals of $R$. Then $P\cap Q$ is a strongly 2-absorbing ideal of $R$.

Proof.

The proof is similar to that of Theorem 2.9. ∎

Proposition 3.17**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly 2-absorbing ideal of $R$. Then we have the following:

• (a)

$\sqrt{I}$ is a strongly 2-absorbing ideal of $R$ and $x^{2}\in I$ for every $x\in\sqrt{I}$.

• (b)

If $S$ is a multiplicatively closed subset of $R$ such that $S\cap I=\emptyset$, then $S^{-1}I$ is a strongly 2-absorbing ideal of $S^{-1}R$.

Proof.

(a) Since $I$ is a strongly 2-absorbing ideal of $R$, observe that $x^{2}\in I$ for every $x\in\sqrt{I}$. Let $x,y,z\in K$ such that $xyz\in\sqrt{I}$. Then $(xyz)^{2}=x^{2}y^{2}z^{2}\in I$. Since $I$ is a 2-absorbing ideal of $R$, we may assume without loss of generality that $x^{2}y^{2}\in I$. Now since $(xy)^{2}=x^{2}y^{2}\in I$, we have $xy\in\sqrt{I}$ as desired.

(b) The proof is similar to that of Proposition 2.10. ∎

Theorem 3.18**.**

Let $R$ be an integral domain with quotient field $K$ and let $I$ be a strongly 2-absorbing ideal of $R$. Then we have the following.

• (a)

If $J$ and $H$ are ideals of $R$, then $JH\subseteq I$ or $I^{2}\subseteq J\cup H$.

• (b)

If $J$ and $I$ are prime ideals of $R$, then $J$ and $I$ are comparable.

Proof.

The proof is similar to that of Theorem 2.7. ∎

Corollary 3.19**.**

Let $R$ be an integral domain with quotient field $K$ and $Q$ be a maximal ideal of $R$. If $Q$ is a strongly 2-absorbing ideal of $R$, then $R$ is a local ring with maximal ideal $Q$.

Proof.

This follows from Theorem 3.18 (b). ∎

Recall that if $K$ is the field of fractions of $R$, then an intermediate ring in the extension $R\subseteq K$ is called an overring of $R$.

Proposition 3.20**.**

Let $R$ be an integral domain with quotient field $K$, $I$ be a strongly 2-absorbing ideal of $R$, and let $T$ be an overring of $R$. Then $IT$ is a strongly 2-absorbing ideal of $T$.

Proof.

Let $x,y\in K$ and $xy\not\in T$. Then $xy\not\in R$. Thus by Theorem 3.3, either $x^{-1}I\subseteq I$ or $y^{-1}I\subseteq I$. Therefore, either $x^{-1}IT\subseteq IT$ or $y^{-1}IT\subseteq IT$. Hence $IT$ is a strongly 2-absorbing ideal of $T$, again by Theorem 3.3. ∎

Proposition 3.21**.**

Let $R$ be an integral domain with quotient field $K$ and let $\{I_{\lambda}\}_{\lambda\in\Lambda}$ be a chain of strongly 2-absorbing ideals of $R$. Then $\sum_{\lambda\in\Lambda}I_{\lambda}$ is a strongly 2-absorbing ideal of $R$.

Proof.

Suppose that $x,y\in K$ with $xy\not\in R$ and we have $x^{-1}\sum_{\lambda\in\Lambda}I_{\lambda}\not\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$ and $y^{-1}\sum_{\lambda\in\Lambda}I_{\lambda}\not\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$. Then there exist $\alpha,\beta\in\Lambda$ such that $x^{-1}I_{\alpha}\not\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$ and $y^{-1}I_{\beta}\not\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$. Hence, $x^{-1}I_{\alpha}\not\subseteq I_{\alpha}$ and $y^{-1}I_{\beta}\not\subseteq I_{\beta}$. Thus $y^{-1}I_{\alpha}\subseteq I_{\alpha}$ and $x^{-1}I_{\beta}\subseteq I_{\beta}$. By assumption, $I_{\alpha}\subseteq I_{\beta}$ or $I_{\beta}\subseteq I_{\alpha}$. This implies that $x^{-1}I_{\alpha}\subseteq x^{-1}I_{\beta}\subseteq I_{\beta}\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$ or $y^{-1}I_{\beta}\subseteq y^{-1}I_{\alpha}\subseteq I_{\alpha}\subseteq\sum_{\lambda\in\Lambda}I_{\lambda}$. This is a contradiction. Thus by Theorem 3.3, $\sum_{\lambda\in\Lambda}I_{\lambda}$ is a strongly 2-absorbing ideal of $R$. ∎

Recall that a chained ring is any ring whose set of ideals is totally ordered by inclusion.

Corollary 3.22**.**

If $R$ is a chained ring and contains a strongly 2-absorbing ideal, then $R$ contains a unique largest strongly 2-absorbing ideal.

Proof.

This is proved easily by using Zorn’s Lemma and Proposition 3.21. ∎

Definition 3.23**.**

Let $R$ be an integral domain with quotient field $K$ and $M$ be an $R$-module. We say that a submodule $N$ of $M$ is a strongly 2-absorbing if $(N:_{R}M)$ is a strongly 2-absorbing ideal of $R$.

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].

Proposition 3.24**.**

Let $R$ be an integral domain which is a chained ring with quotient field $K$ and $M$ be a faithful finitely generated multiplication $R$-module. If $\{N_{i}\}_{i\in I}$ is a family of strongly 2-absorbing submodules of $M$, then $\sum_{i\in I}N_{i}$ is a strongly 2-absorbing submodule of $M$.

Proof.

This follows from Proposition 3.21 and the fact that

[TABLE]

by [7, Theorem 3.1]. ∎

Proposition 3.25**.**

Let $R$ be an integral domain with quotient field $K$ and $M$ be an $R$-module. Then we have the following:

• (a)

If $N$ is a strongly 2-absorbing submodule of $M$ and $r\in K$ such that $r^{-1}\in R$, then $(N:_{M}r)$ is a strongly 2-absorbing submodule of $M$.

• (b)

If $f:M\rightarrow\acute{M}$ is a monomorphism of $R$-modules, then $N$ is a strongly 2-absorbing submodule of $M$ if and only if $f(N)$ is a strongly 2-absorbing submodule of $f(M)$.

• (c)

If $N_{1}$, $N_{2}$ are two submodules of $M$ with $(N_{1}:_{R}M)$ and $(N_{2}:_{R}M)$ strongly prime ideals of $R$, then $N_{1}\cap N_{2}$ is a strongly 2-absorbing submodule of $M$.

Proof.

(a) The proof is similar to that of Proposition 2.14 (a).

(b) The proof is similar to that of Proposition 2.14 (b).

(c) Since $(N_{1}\cap N_{2}:_{R}M)=(N_{1}:_{R}M)\cap(N_{2}:_{R}M)$, the result follows from Proposition 3.16. ∎

Proposition 3.26**.**

Let $R$ be an integral domain with quotient field $K$, $N$ be a submodule of a finitely generated $R$-module $M$, and let $S$ be a multiplicatively closed subset of $R$. If $N$ is a strongly 2-absorbing submodule and $(N:_{R}M)\cap S=\emptyset$, then $S^{-1}N$ is a strongly 2-absorbing $S^{-1}R$-submodule of $S^{-1}M$.

Proof.

As $M$ is finitely generated, $(S^{-1}N:_{S^{-1}R}S^{-1}M)=S^{-1}(N:_{R}M)$ by [10, Lemma 9.12]. .Now the result follows from Proposition 3.17. ∎

Proposition 3.27**.**

Let $R$ be an integral domain with quotient field $K$, $M$ be an $R$-module, and let $\{K_{i}\}_{i\in I}$ be a chain of strongly 2-absorbing submodules of $M$. Then $\cap_{i\in I}K_{i}$ is a strongly 2-absorbing submodule of $M$.

Proof.

Let $a,b,c\in K$ and $abc\in(\cap_{i\in I}K_{i}:_{R}M)=\cap_{i\in I}(K_{i}:_{R}M)$. Assume to the contrary that $ab\not\in\cap_{i\in I}(K_{i}:_{R}M)$, $bc\not\in\cap_{i\in I}(K_{i}:_{R}M)$, and $ac\not\in\cap_{i\in I}(K_{i}:_{R}M)$. Then there are $m,n,t\in I$ where $ab\not\in(K_{n}:_{R}M)$, $bc\not\in(K_{m}:_{R}M)$, and $ac\not\in(K_{t}:_{R}M)$. Since $\{K_{i}\}_{i\in I}$ is a chain, we can assume without loss of generality that $K_{m}\subseteq K_{n}\subseteq K_{t}$. Then

[TABLE]

As $abc\in(K_{m}:_{R}M)$, we have $ab\in(K_{m}:_{R}M)$ or $ac\in(K_{m}:_{R}M)$ or $bc\in(K_{m}:_{R}M)$. In any case, we have a contradiction. ∎

Definition 3.28**.**

Let $R$ be an integral domain with quotient field $K$. We say that a strongly 2-absorbing submodule $N$ of an $R$-module $M$ is a minimal strongly 2-absorbing submodule of a submodule $H$ of $M$, if $H\subseteq N$ and there does not exist a strongly 2-absorbing submodule $T$ of $M$ such that $H\subset T\subset N$.

It should be noted that a minimal strongly 2-absorbing submodule of $M$ means that a minimal strongly 2-absorbing submodule of the submodule [math] of $M$.

Lemma 3.29**.**

Let $R$ be an integral domain with quotient field $K$ and let $M$ be an $R$-module. Then every strongly 2-absorbing submodule of $M$ contains a minimal strongly 2-absorbing submodule of $M$.

Proof.

This is proved easily by using Zorn’s Lemma and Proposition 3.27. ∎

Theorem 3.30**.**

Let $R$ be an integral domain with quotient field $K$ and let $M$ be a Noetherian $R$-module. Then $M$ contains a finite number of minimal strongly 2-absorbing submodules.

Proof.

Suppose that the result is false. Let $\Sigma$ denote the collection of all proper submodules $N$ of $M$ such that the module $M/N$ has an infinite number of minimal strongly 2-absorbing submodules. Since $0\in\Sigma$, we have $\Sigma\not=\emptyset$. Therefore $\Sigma$ has a maximal member $T$, since $M$ is a Noetherian $R$-module. Clearly, $T$ is not a strongly 2-absorbing submodule. Therefore, there exist $a,b,c\in K$ such that $abc(M/T)=0$ but $ab(M/T)\not=0$, $ac(M/T)\not=0$, and $bc(M/T)\not=0$. The maximality of $T$ implies that $M/(T+abM)$, $M/(T+acM)$, and $M/(T+bcM)$ have only finitely many minimal strongly 2-absorbing submodules. Suppose $P/T$ is a minimal strongly 2-absorbing submodule of $M/T$. So $abcM\subseteq T\subseteq P$, which implies that $abM\subseteq P$ or $acM\subseteq P$ or $bcM\subseteq P$. Thus $P/(T+abM)$ is a minimal strongly 2-absorbing submodule of $M/(T+abM)$ or $P/(T+bcM)$ is a minimal strongly 2-absorbing submodule of $M/(T+bcM)$ or $P/(T+acM)$ is a minimal strongly 2-absorbing submodule of $M/(T+acM)$. Thus, there are only a finite number of possibilities for the submodule $M/T$. This is a contradiction. ∎

Acknowledgments. The author would like to thank Professor Andrew Hetzel for his helpful suggestions and useful comments.