$I$-prime ideals

TL;DR

This paper introduces the concept of $I$-prime ideals as a generalization of weakly prime ideals in commutative rings, providing characterizations, properties, and conditions for their relation to prime and weakly prime ideals.

Contribution

It defines $I$-prime ideals, explores their properties, and establishes conditions linking them to prime and weakly prime ideals in decomposite rings.

Findings

$I$-prime ideals generalize weakly prime ideals.

Characterizations of $I$-prime ideals are provided.

Conditions for $I$-prime ideals to be prime or weakly prime are established.

Abstract

In this paper, we introduce a new generalization of weakly prime ideals called $I$-prime. Suppose $R$ is a commutative ring with identity and $I$ a fixed ideal of $R$. A proper ideal $P$ of $R$ is $I$-prime if for $a, b \in R$ with $ab \in P-IP$ implies either $a \in P$ or $b \in P$. We give some characterizations of $I$-prime ideals and study some of its properties. Moreover, we give conditions under which $I$-prime ideals becomes prime or weakly prime and we construct the view of $I$-prime ideal in decomposite rings.

Full text

MSC(2010): Primary: 13A15;

Keywords: Prime ideal, weakly prime ideal, almost prime ideal, radical of the ideal.

1. Introduction

Throughout this article, $R$ will be a commutative ring with identity. Prime ideals play an essential role in ring theory. A prime ideal $P$ of $R$ is a proper ideal $P$ with the property that for $a,b\in R$, $ab\in P$ implies $a\in P$ or $b\in P$. There are several ways to generalize the notion of a prime ideal. We could either restrict or enlarge where $a$ and/or $b$ lie or restrict or enlarge where $ab$ lies. In this article we will interested in a generalization obtained by restricting where $ab$ lies. A proper ideal $P$ of $R$ is weakly prime if for $a,b\in R$ with $ab\in P-0$, either $a\in P$ or $b\in P$. Weakly prime ideals were studied in some detail by Anderson and Smith (2003) in [1]. Thus any prime ideal is weakly prime. Bhatwadekar and Sharma (2005) in [2] recently defined a proper ideal $I$ of an integral domain $R$ to be almost prime if for $a,b\in R$ with $ab\in I-I^{2}$, then either $a\in I$ or $b\in I$. This definition can obviously be made for any commutative ring $R$. Thus a weakly prime ideal is almost prime and any proper idempotent ideal is almost prime. Moreover, an ideal $I$ of $R$ is almost prime if and only if $I/I^{2}$ is a weakly prime ideal of $R/I^{2}$. Also almost prime ideals were generalized to $n$-almost prime as follows; a proper ideal $I$ is called $n$-almost prime ideal if for any $a,b\in R$ with $ab\in I-I^{n}$, then either $a\in I$ or $b\in I$. With weakly prime ideals and almost prime ideals in mind, we make the following definition. Let $R$ be a commutative ring and $I$ be a fixed ideal of $R$. Then a proper ideal $P$ of $R$ is called $I$-prime ideal if for $a,b\in R$, $ab\in P-IP$, implies $a\in P$ or $b\in P$. So every weakly prime and $n$-almost prime ideal is $I$-prime where $I$ taken to be zero or $P^{n-1}$ respectively. If $I=R$, then every ideal is $I$-prime, so we can assume $I$ to be proper ideal of $R$. For more details see [3].

Example 1.1**.**

Consider the ring $\textbf{Z}_{12}$ and take $P=I=<4>$. Then $P$ is $I$-prime which is not prime nor weakly prime.

2. Main Results

We begin with the following lemma.

Lemma 2.1**.**

Let $P$ be a proper ideal of a ring $R$. Then $P$ is an $I$-prime ideal if and only if $P/IP$ is weakly prime in $R/IP$.

Proof.

$(\Rightarrow)$ Let $P$ be an $I$-prime in $R$. Let $a,b\in R$ with $0\neq(a+IP)(b+IP)=ab+IP\in P/IP$. Then $ab\in P-IP$ implies $a\in P$ or $b\in P$, hence $a+IP\in P/IP$ or $b+IP\in P/IP$. So $P/IP$ is weakly prime ideal in $R/IP$.

$(\Leftarrow)$ Suppose that $P/IP$ is weakly prime in $R/IP$ and take $r,s\in R$ such that $rs\in P-IP$. Then $0\neq rs+IP=(r+IP)(s+IP)\in P/IP$ so $r+IP\in P/IP$ or $s+IP\in P/IP$. Therefore $r\in P$ or $s\in P$. Thus $P$ is an $I$-prime ideal in $R$. ∎

Theorem 2.2**.**

*(1) Let $I\subseteq J$. If $P$ is $I$-prime ideal of a ring $R$, then it is $J$-prime.

(2) Let $R$ be commutative ring and $P$ an $I$-prime ideal that is not prime, then $P^{2}\subseteq IP$. Thus, an $I$-prime ideal $P$ with $P^{2}\nsubseteq IP$ is prime.*

Proof.

(1) The proof come from the fact that if $I\subseteq J$, then $P-JP\subseteq P-IP$. (2) Suppose that $P^{2}\nsubseteq IP$, we show that $P$ is prime. Let $ab\in P$ for $a,b\in R$. If $ab\notin IP$, then $P$ $I$-prime gives $a\in P$ or $b\in P$. So assume that $ab\in IP$. First, suppose that $aP\nsubseteq IP$; say $ax\notin IP$ for some $x\in P$. Then $a(x+b)\in P-IP$. So $a\in P$ or $x+b\in P$ and hence $a\in P$ or $b\in P$. So we can assume that $aP\subseteq IP$ in similar way we can assume that $bP\subseteq IP$. Since $P^{2}\nsubseteq IP$, there exist $y,z\in P$ with $yz\notin IP$. Then $(a+y)(b+z)\in P-IP$. So $P$ $I$-prime gives $a+y\in P$ or $b+z\in P$; Hence $a\in P$ or $b\in P$. Therefore $P$ is prime. ∎

In the following we give a counter example on the converse of part (1) of Theorem 2.2.

Example 2.3**.**

In the ring $\textbf{Z}_{12}[x]$, put $I=0$, $J=<4>$ and $P=<4x>$. Then $P-IP=<4x>-0$ and $P-JP=<4x>-<4><4x>=\phi$. Hence $P$ is $J$-prime but not $I$-prime.

Corollary 2.4**.**

Let $P$ be an $I$-prime ideal of a ring $R$ with $IP\subseteq P^{3}$. Then $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime.

Proof.

If $P$ is prime, then $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime. Assume that $P$ is not prime. By Theorem 2.2 $P^{2}\subseteq IP\subseteq P^{3}$. Thus $IP=P^{n}$ for each $n\geq 2$. So $\cap_{i=1}^{\infty}P^{i}=P\cap P^{2}=P^{2}$ and $(\cap_{i=1}^{\infty}P^{i})P=P^{2}P=P^{3}=IP$. Hence $P$ $I$-prime implies $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime. ∎

Remark 2.5*.*

Let $P$ be $I$-prime ideal. Then $P\subseteq\sqrt{IP}$ or $\sqrt{IP}\subseteq P$. If $P\subsetneqq\sqrt{IP}$, then $P$ is not prime since otherwise $IP\subseteq P$ implies $\sqrt{IP}\subseteq\sqrt{P}=P$. While if $\sqrt{IP}\subsetneqq P$, then $P$ is prime.

Corollary 2.6**.**

Let $P$ be $I$-prime ideal of a ring $R$ which is not prime. Then $\sqrt{P}=\sqrt{IP}$

Proof.

By Theorem 2.2, $P^{2}\subseteq IP$ and hence $\sqrt{P}=\sqrt{P^{2}}\subseteq\sqrt{IP}$. The other containment always holds.

∎

Now we give a way to construct $I$-prime ideals $P$ when $\cap_{i=1}^{\infty}P^{i}\subseteq IP\subseteq P^{3}$.

Remark 2.7*.*

Assume that $P$ is $I$-prime, but not prime. Then by Theorem 2.2, if $IP\subseteq P^{2}$, then $P^{2}=IP$. In particular, if $P$ is weakly prime ([math]-prime) but not prime, then $P^{2}=0$. Suppose that $IP\subseteq P^{3}$. Then $P^{2}\subseteq IP\subseteq P^{3}$; So $P^{2}=P^{3}$ and thus $P^{2}$ is idempotent.

Theorem 2.8**.**

*(1) Let $R,S$ be commutative rings and $P$ [math]-prime ideal of $R$. Then $P\times S$ is $I$-prime ideal of $R\times S$ for each ideal $I$ of $R\times S$ with $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)\subseteq P\times S$.

(2) Let $P$ be finitely generated proper ideal of commutative ring $R$. Assume $P$ is $I$-prime where $IP\subseteq P^{3}$. Then either $P$ is [math]-prime or $P^{2}\neq 0$ is idempotent and $R$ decomposes as $T\times S$ where $S=P^{2}$ and $P=J\times S$ where $J$ is [math]-prime. Thus $P$ is $I$-prime for $\cap_{i=1}^{\infty}P^{i}\subseteq IP\subseteq P$.*

Proof.

(1) Let $R$ and $S$ be commutative rings and $P$ be weakly prime ([math]-prime) ideal of $R$. Then $P\times S$ need not be a [math]-prime ideal of $R\times S$; In fact, $P\times S$ is [math]-prime if and only if $P\times S$ (or equivalently $P$) is prime [see Anderson 2003]. However, $P\times S$ is $I$-prime for each $I$ with $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)$. If $P$ is prime , then $P\times S$ is prime ideal and thus is $I$-prime for all $I$. Assume that $P$ is not prime. Then $P^{2}=0$ and $(P\times S)^{2}=0\times S$. Hence $\cap_{i=1}^{\infty}(P\times S)^{i}=\cap_{i=1}^{\infty}P^{i}\times S=0\times S$. Thus $P\times S-\cap_{i=1}^{\infty}(P\times S)^{i}=P\times S-0\times S=(P-0)\times S$. Since $P$ is weakly prime, $P\times S$ is $\cap_{i=1}^{\infty}(P\times S)^{i}$-prime and as $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)$, $P\times S$ is $I$-prime.

(2) If $P$ is prime, then $P$ is [math]-prime. So we can assume that $P$ is not prime. Then $P^{2}\subseteq IP$ and hence $P^{2}\subseteq IP\subseteq P^{3}$. So $P^{2}=P^{3}$. Hence $P^{2}$ is idempotent. Since $P^{2}$ is finitely generated, $P^{2}=<e>$ for some idempotent $e\in R$. Suppose $P^{2}=0$. Then $IP\subseteq P^{3}=0$. So $IP=0$ and hence $P$ is [math]-prime. So assume $P^{2}\neq 0$. Put $S=P^{2}=Re$ and $T=R(1-e)$, so $R$ decomposesas $T\times S$ where $S=P^{2}$. Let $J=P(1-e)$, so $P=J\times S$ where $J^{2}=(P(1-e))^{2}=P^{2}(1-e)^{2}=<e>(1-e)=0$. We show that $J$ is [math]-prime. Let $ab\in J^{2}-0$, so $(a,1)(b,1)=(ab,1)\in J\times S-(J\times S)^{2}=J\times S-0\times S\subseteq P-IP$ since $IP\subseteq P^{3}$ implies $IP\subseteq P^{3}=(J\times S)^{3}=0\times S$. Hence $(a,1)\in P$ or $(b,1)\in P$ so $a\in J$ or $b\in J$. Therefore $J$ is weakly prime. ∎

Corollary 2.9**.**

Let $R$ be an indecomposable commutative ring and $P$ a finitely generated $I$-prime ideal of $R$, where $IP\subseteq P^{3}$. Then $P$ is weakly prime.

Corollary 2.10**.**

Let $R$ be a Noetherian integral domain. A proper ideal $P$ of $R$ is prime if and only if $P$ is $P^{2}$-prime.

Theorem 2.11**.**

*Let $a$ be a non-unit element in $R$.

(1) Let $(0:a)\subseteq(a)$. Then $(a)$ is $I$-prime for $I(a)\subseteq(a)^{2}$ if and only if $(a)$ is prime.

Let $(R,m)$ be quasi-local ring. Then

(2) $(a)$ is $I$-prime for $I(a)\subseteq(a)^{3}$ if and only if $(a)$ is [math]-prime.

(3) $(a)$ is $m$-prime if and only if $a$ is irreducible.*

Proof.

(1) Suppose that (a) is $I$-prime and $bc\in(a)$. If $bc\notin I(a)$, then $b\in(a)$ or $c\in(a)$. So suppose that $bc\in I(a)$. Now $(b+a)c\in(a)$. If $(b+a)c\notin I(a)$, then $b+a\in(a)$ or $c\in(a)$ and hence $b\in(a)$ or $c\in(a)$. So assume that $(b+a)c\in I(a)$. Then $ac\in I(a)$ and hence $ac\in(a)^{2}$. So $ac=za^{2}$ and hence $c-za\in(0:a)$. Thus $c\in(0:a)+(a)=(a)$. The converse part is trivial since every prime ideal is $I$-prime.

(2) If $(a)$ is [math]-prime, then $(a)$ is $I$-prime for each $I$ with $I(a)\subseteq(a)^{3}$. Conversely, let $(a)$ be $I$-prime for $I(a)\subseteq(a)^{3}$. Since a quasi local ring has no nontrivial idempotents, $(a)$ is [math]-prime by Theorem 2.8 part(2).

(3) If $a$ is irreducible means that $a=xy$ implies that $x\in(a)$ or $y\in(a)$ and $(a)$ is $m$-prime means that $xy\in(a)-m(a)$ which implies that $x\in(a)$ or $y\in(a)$. But $xy\in(a)-m(a)$ if and only if $xy=za$ for some unit $z\in R$ if and only if $a=z^{-1}xy$ for some unit $z^{-1}\in R$. Thus $(a)$ is $m$-prime if and only if $a=xy$ implies $x\in(a)$ or $y\in(a)$. ∎

We now give some characterizations of $I$-prime ideals.

Theorem 2.12**.**

*Let $P$ be a proper ideal of $R$. Then the following are equivalent:

(1) $P$ is $I$-prime.

(2) For $x\in R-P$, $(P:x)=P\cup(IP:x)$

(3) For $x\in R-P$, $(P:x)=P$ or $(P:x)=(IP:x)$

(4) For ideals $J$ and $K$ of $R$, $JK\subseteq P$ and $JK\nsubseteq IP$ imply $J\subseteq P$ or $K\subseteq P$.*

Proof.

(1) $\Rightarrow$ (2) Suppose $r\in R-P$. Let $s\in(P:r)$, so $rs\in P$. If $rs\in P-IP$, then $s\in P$. If $rs\in IP$, then $s\in(IP:r)$, So $(P:r)\subseteq P\cup(IP:r)$. The other containment always holds.

(2) $\Rightarrow$ (3) Note that if an ideal is a union of two ideals, then it is equal to one of them.

(3) $\Rightarrow$ (4) Let $J$ and $K$ be two ideals of $R$ with $JK\subseteq P$. Assume that $J\nsubseteq P$ and $K\nsubseteq P$. We claim that $JK\subseteq IP$. Suppose $r\in J$. First, Let $r\notin P$. Then $rK\subseteq P$ gives $K\subseteq(P:r)$. Now $K\nsubseteq P$, so $(P:r)=(IP:r)$. Thus $rK\subseteq IP$. Next, let $r\in J\cap P$. Choose $s\in J-P$. Then $r+s\in J-P$. So by the first case $sK\subseteq IP$ and so $(r+s)K\subseteq IP$. Let $t\in K$. Then $rt=(r+s)t-st\in IP$. So $rK\subseteq IP$. Hence $JK\subseteq IP$.

(4) $\Rightarrow$ (1) Let $rs\in P-IP$. Then $(r)(s)\subseteq P$ but $(r)(s)\nsubseteq IP$. So $(r)\subseteq P$ or $(s)\subseteq P$; i-e. $r\in P$ or $s\in P$. ∎

Corollary 2.13**.**

Suppose $P$ is $I$-prime ideal that is not prime. Then $P\sqrt{IP}\subseteq IP$.

Proof.

Let $r\in\sqrt{IP}$. If $r\in P$, then $rP\subseteq P^{2}\subseteq IP$ by Theorem 2.2. So assume that $r\notin P$ by Theorem 2.12, $(P:r)=P$ or $(P:r)=(IP:r)$. As $P\subseteq(P:r)$, the last gives $rP\subseteq IP$. So assume that $(P:r)=P$. Let $r^{n}\in IP$, but $r^{n-1}\notin IP$. Then $r^{n}\in P$, so $r^{n-1}\in(P:r)=P$. Thus $r^{n-1}\in P-IP$, so $r\in P$ a contradiction. ∎

It is known that if $S$ is a multiplicatively closed subset of a commutative ring $R$ and $P$ as a prime ideal of $R$ with $P\cap S=\phi$, then $S^{-1}P$ is a prime ideal of $S^{-1}R$ and $S^{-1}P\cap R=P$. The first result extended to weakly prime ideals in [2, Proposition 13] and to almost prime ideals in [5, Lemma 2.13]. Fix an ideal $I$ of $R$ we prove that if $P$ is $I$-prime with $P\cap S=\phi$, then $S^{-1}P$ is $S^{-1}I$-prime. Note that for an ideal $J$ of $R$ with $J\subseteq P$, $I(P/J)=(IP+J)/J$. If $P$ is prime (respectively, weakly prime, $n$-almost prime), then so is $P/J$. We generalize this result to $I$-prime ideals in the following proposition.

Proposition 2.14**.**

*Let $R$ be a ring and $I$ be an ideal of $R$. Let $P$ be $I$-prime ideal of $R$. Then the following are true.

(1) If $J$ is an ideal of $R$ with $J\subseteq P$, then $P/J$ is $I$-prime ideal of $R/J$.

(2) Assume that $S$ is multiplicatively closed subset of $R$ with $P\cap S=\phi$. Then $S^{-1}P$ is a $S^{-1}I$-prime ideal of $S^{-1}R$. If $S^{-1}P\neq S^{-1}(IP)$, then $S^{-1}P\cap R=P$.*

Proof.

(1) Let $x,y\in R$ with $\bar{x}\bar{y}\in P/J-I(P/J)=P/J-(IP+J)/J$. Thus $xy\in P-(IP+J)$. Hence $xy\in P-IP$, so $x\in P$ or $y\in P$. Therefore $\bar{x}\in P/J$ or $\bar{y}\in P/J$; so $P/J$ is $I$-prime.

(2) Let $\frac{a}{s}\frac{b}{t}\in S^{-1}P-S^{-1}I.S^{-1}P\subseteq S^{-1}P-S^{-1}(IP)=S^{-1}(P-IP)$. So $abk\in P-IP$ for some $k\in S$. Thus $P$ $I$-prime gives $a\in P$ or $bk\in P$. So $a/s\in S^{-1}P$ or $b/t\in S^{-1}P$. Hence $S^{-1}P$ is $S^{-1}I$-prime. Let $a\in S^{-1}P\cap R$, so there exists $u\in S$ with $au\in P$. If $au\notin IP$, then $au\in P-IP$, so $a\in P$. If $au\in IP$, then $a\in S^{-1}(IP)\cap R$. So $S^{-1}P\cap R\subseteq P\cup(S^{-1}(IP)\cap R)$. Hence $S^{-1}P\cap R=P$ or $S^{-1}P\cap R=S^{-1}(IP)\cap R$. But the second case gives $S^{-1}P=S^{-1}(IP)$. ∎

Let $R_{1}$ and $R_{2}$ be two rings. It is known that the prime ideals of $R_{1}\times R_{2}$ have the form $P\times R_{2}$ or $R_{1}\times Q$, where $P$ is a prime ideal of $R_{1}$ and $Q$ is a prime ideal of $R_{2}$. We next, generalize this result to $I$-primes.

Theorem 2.15**.**

*For $i=1,2$ let $R_{i}$ be ring and $I_{i}$ ideal of $R_{i}$. Let $I=I_{1}\times I_{2}$. Then the $I$-prime ideals of $R_{1}\times R_{2}$ have exactly one of the following three types:

(1) $P_{1}\times P_{2}$, where $P_{i}$ is a proper ideal of $R_{i}$ with $I_{i}P_{i}=P_{i}$.

(2) $P_{1}\times R_{2}$ where $P_{1}$ is an $I_{1}$-prime of $R_{1}$ and $I_{2}R_{2}=R_{2}$.

(3) $R_{1}\times P_{2}$, where $P_{2}$ is an $I_{2}$-prime of $R_{2}$ and $I_{1}R_{1}=R_{1}$.*

Proof.

We first prove that an ideal of $R_{1}\times R_{2}$ having one of these three types is $I$-prime. The first type is clear since $P_{1}\times P_{2}-I(P_{1}\times P_{2})=P_{1}\times P_{2}-(I_{1}P_{1}\times I_{2}P_{2})=\phi$. Suppose that $P_{1}$ is $I_{1}$-prime and $I_{2}R_{2}=R_{2}$. Let $(a,b)(x,y)\in P_{1}\times R_{2}-I_{1}P_{1}\times I_{2}R_{2}=P_{1}\times R_{2}-I_{1}P_{1}\times R_{2}=(P_{1}-I_{1}P_{1})\times R_{2}$. Then $ax\in P_{1}-I_{1}P_{1}$ implies that $a\in P_{1}$ or $x\in P_{1}$, so $(a,b)\in P_{1}\times R_{2}$ or $(x,y)\in P_{1}\times R_{2}$. Hence $P_{1}\times R_{2}$ is $I$-prime. Similarly we can prove the last case.

Next, let $P_{1}\times P_{2}$ be $I$-prime and $ab\in P_{1}-I_{1}P_{1}$. Then $(a,0)(b,0)=(ab,0)\in P_{1}\times P_{2}-I(P_{1}\times P_{2})$, so $(a,0)\in P_{1}\times P_{2}$ or $(b,0)\in P_{1}\times P_{2}$, i-e, $a\in P_{1}$ or $b\in P_{1}$. Hence $P_{1}$ is $I_{1}$-prime. Likewise, $P_{2}$ is $I_{2}$-prime.

Assume that $P_{1}\times P_{2}\neq I_{1}P_{1}\times I_{2}P_{2}$. Say $P_{1}\neq I_{1}P_{1}$. Let $x\in P_{1}-I_{1}P_{1}$ and $y\in P_{2}$. Then $(x,1)(1,y)=(x,y)\in P_{1}\times P_{2}$. So $(x,1)\in P_{1}\times P_{2}$ or $(1,y)\in P_{1}\times P_{2}$. Thus $P_{2}=R_{2}$ or $P_{1}=R_{1}$. Assume that $P_{2}=R_{2}$. So $P_{1}\times R_{2}$ is $I$-prime, where $P_{1}$ is $I_{1}$-prime. ∎

Acknowledgments

The author is deeply grateful to referee for his careful reading of the manuscript. Whose comments made the paper more readable.