I. Akray and H. S. Hussein

#  $I$-primary submodules

Ismael Akray<EMAIL_ADDRESS>Department of Mathematics, Soran
University, Erbil city, Kurdistan region, Iraq  Halgurd S. Hussein
<EMAIL_ADDRESS>Department of Mathematics, Soran University, Erbil
city, Kurdistan region, Iraq

###### Abstract

In this paper, we give a generalization for weakly primary submodules called
$I$-primary submodule and we study some properties of it. We give some
characterizations of $I$-primary submodules. Also we establish the situation
of $I$-primary submodules in module localizations and decomposable modules.

###### :

primary 13A15; secondary 13C99; 13F05

###### keywords:

Primary ideal, primary submodule, weakly primary submodule

##  1 Introduction

Throughout this paper $R$ will be a commutative ring with nonzero identity and
$I$ a fixed ideal of $R$ and $M$ a unitary left $R$-module. The concept of
weakly prime ideals was introduced by Anderson and Smith (2003), where a
proper ideal $P$ is called weakly prime if, for $a,b\in R$ with $0\neq ab\in
P$, either $a\in P$ or $b\in P$, 6-6 . The radical of an ideal $I$ of $R$ is
defined to be the set of all $a\in R$ for which $a^{n}\in I$ for some positive
integer $n$. Primary ideals have an important role in commutative ring theory.
A proper ideal $Q$ of $R$ is said to be primary provided that for $a,b\in
R,ab\in Q$ implies that either $a\in Q$ or $b\in\sqrt{Q}$. We can generalize
the concept of primary ideals by restricting the set where $ab$ lies. A proper
ideal $Q$ of $R$ is weakly primary if for $a,b\in R$ with $0\neq ab\in Q$,
either $a\in Q$ or $b\in\sqrt{Q}$. Weakly primary ideals were first introduced
and studied by Ebrahimi Atani and Farzalipour in 2005, 5-5 .

An $R$-module $M$ is called a multiplication module if every submodule $N$ of
$M$ has the form $IM$ for some ideal $I$ of $R$, see 34 . Note that, since
$I\subseteq(N:_{R}M)$ then $N=IM\subseteq(N:_{R}M)M\subseteq N$. So that
$N=(N:_{R}M)M$. Let $N$ and $K$ be submodules of a multiplication $R$-module
$M$ with $N=I_{1}M$ and $K=I_{2}M$ for some ideals $I_{1}$ and $I_{2}$ of $R$.
The product of $N$ and $K$ denoted by $NK$ is defined by $NK=I_{1}I_{2}M$.
Then by (123, , Theorem 3.4), the product of $N$ and $K$ is independent of
presentations of $N$ and $K$. An $R$-module $M$ is called faithful if it has
zero annihilator. Let $N$ be a proper submodule of a nonzero $R$-module $M$.
Then the $M$-radical of $N$, denoted by $M-rad(N)$, is defined to be the
intersection of all prime submodules of $M$ containing $N$. If $M$ has no
prime submodule containing $N$, then we say $M-rad(N)=M$. It is shown in
(1234, , Theorem 2.12) that if $N$ is a proper submodule of a multiplication
$R$-module $M$, then $M-rad(N)=\sqrt{(N:_{R}M)}M$.

The class of prime submodules of modules was introduced and studied in 1992 as
a generalization of the class of prime ideals of rings. Then, many
generalizations of prime submodules were studied such as primary, classical
prime, weakly prime and classical primary submodules, see 34 ; 124 ; 134 ; 164
and 114 . The authors in 11-1  and 22-2  introduced the notions $I$-prime
ideals and $I$-prime submodules. A proper ideal of $R$ is called $I$-prime
ideal of $R$ if $rs\in P-IP$ for all $r,s\in R$ implies that either $r\in P$
or $s\in P$. A proper submodule $P$ of $M$ is called $I$-prime submodule of
$M$ if $rm\in P-IP$ for all $r\in R$ and $m\in M$ implies that either $m\in P$
or $r\in(P:M)$. In this article, we define and study $I$-primary submodules
which are generalizations of weakly primary (and weakly prime) submodules. We
generalize some basic properties of weakly prime and weakly primary to
$I$-primary submodules. We give some characterizations of $I$-primary
submodules.

##  2 Main results

Let $I$ be an ideal of $R$ and $M$ an $R$-module. A proper submodule $P$ of
$M$ is called an $I$ -primary submodule of $M$ if $rm\in P-IP$ for all $r\in
R$ and $m\in M$ implies that either $m\in P$ or $r\in\sqrt{(P:M)}$. In view of
the above definition, in all what follows, $I$ is an ideal of $R$. Now we
begin with the following result.

######  Theorem 2.1.

Let $M$ be an $R$-module. Let $P$ be an $I$-primary submodule of $M$. If
$(P:M)P\nsubseteq IP$, then $P$ is a primary submodule of $M$.

###### Proof.

Suppose that $r\in R$ and $m\in M$ such that $rm\in P$. If $rm\notin IP$,
since $P$ is $I$-primary submodule of $M$, then $r^{n}\in(P:M)$ for some
positive integer $n$ or $m\in P$. Now suppose that $rm\in IP$. Let
$rP\subseteq IP$. Because if $rP\nsubseteq IP$, then there exists $p\in P$
such that $rp\notin IP$, then $r(m+p)\in P-IP$. Therefore $r^{n}\in(P:M)$ for
some positive integer $n$ or $m+p\in P$. Hence $r^{n}\in(P:M)$ for some
positive integer $n$ or $m\in p$. Now suppose that $(P:M)m\subseteq IP$.
Because if $(P:M)m\nsubseteq IP$, then there exists $t\in(P:M)$ such that
$tm\notin IP$ and so $(r+t)m\in P-IP$. Then we have $r+t\in\sqrt{(P:M)}$ or
$m\in P$ . Hence $r\in\sqrt{(P:M)}$ or $m\in P$. Since $(P:M)P\nsubseteq IP$,
so there exist $b\in(P:M)$ and $p\in P$ such that $bp\notin IP$. Then
$(r+b)(m+p)\in P-IP$. Therefore $r+b\in\sqrt{(P:M)}$ or $m+p\in P$. Hence
$r\in\sqrt{(P:M)}$ or $m\in P$. Thus $P$ is primary submodule in $M$. ∎

######  Corollary 2.2.

Let $P$ be a $0$-primary of $M$ such that $(P:M)P\neq 0$. Then $P$ is a
primary submodule of $M$.

######  Corollary 2.3.

Let $P$ be an $I$-primary submodule of $M$ such that $IP\subseteq(P:M)^{2}P$
and that $P$ is not primary submodule. Then $IP=(P:M)^{2}P$.

In what follows we give some charactrizations for $I$-primary submodules.

######  Theorem 2.4.

Let $P$ be a proper submodule of $M$, then the following are equivalent:
(1) $P$ is an $I$-primary submodule of $M$.
(2)For $r\in R-\sqrt{(P:M)},\ (P:r)=P\cup(IP:r)$.
(3) For $r\in R-\sqrt{(P:M)},\ (P:r)=P$ or $(P:r)=(IP:r)$.

###### Proof.

(1)$\Rightarrow$ (2) Suppose that $P$ is an $I$-primary submodule of $M$ such
that $r\notin\sqrt{(P:M)}$. Let $a\in(P:r)$. So $ra\in P$. If $ra\notin IP$,
then $a\in P$. Because $P$ is an $I$-primary submodule in $M$. If $ra\in IP$,
then $a\in(IP:r)$. So $(P:r)\subseteq P\cup(IP:r)$. Now since $IP\subseteq P$,
the other inclusion is hold.
(2)$\Rightarrow$ (3) It is clear because $(P:r)$ is a submodule of $M$.
(3)$\Rightarrow$ (1) Let $r\in R$ and $x\in M$ such that $rx\in P-IP$. If
$r\notin\sqrt{(P:M)}$, then by assumption, either $(P:r)=P$ or $(P:r)=(IP:r)$.
Since $rx\notin IP$ then $x\notin(IP:r)$, and since $rx\in P$, then
$x\in(P:r)$. So $(P:r)=P$. Therefore $x\in P$. Thus $P$ is an $I$-primary
submodule of $M$. ∎

The quotient and localization of primary submodules are again primary
submodules. But in case of $I$-primary submodules, we give a condition under
which the localization becomes true as we see in the following theorem.

######  Theorem 2.5.

Let $M$ be an $R$-module. Let $P$ be an $I$-primary submodule of $M$. Then,
(i) If $N\subseteq P$ is a submodule of $M$, then $\frac{P}{N}$ is an
$I$-primary submodule of $\frac{M}{N}$.
(ii) Suppose that $S$ is a multiplicatively closed subset of $R$ such that
$S^{-1}P\neq S^{-1}M$ and $S^{-1}(IP)\subseteq(S^{-1}I)(S^{-1}P)$. Then
$S^{-1}P$ is an $(S^{-1}I)$-primary submodule of $S^{-1}M$.

###### Proof.

(i) Let $r\in R$ and $m\in M$ such that
$r(m+N)=rm+N\in\frac{P}{N}-I\frac{P}{N}$. Then $rm+N\in\frac{P-IP}{N}$. So
$rm\in P-IP$. As $P$ is $I$-primary submodule, then either $r^{n}\in(P:M)$ for
some positive integer $n$ or $m\in P$. Therefore
$r^{n}\in(\frac{P}{N}:\frac{M}{N})$ for some positive integer $n$ or
$m+N\in\frac{P}{N}$. Hence $\frac{P}{N}$ is $I$-primary submodule of
$\frac{M}{N}$.

(ii) For all $\frac{r}{s}\in S^{-1}R$ and $\frac{m}{t}\in S^{-1}M$, let
$\frac{r}{s}.\frac{m}{t}=\frac{rm}{st}\in S^{-1}P-(S^{-1}I)(S^{-1}P)\subseteq
S^{-1}P-S^{-1}(IP)=S^{-1}(P-IP)$. Then $\frac{rm}{st}=\frac{x}{u}$ for $x\in
P-IP$ and $u\in S$. So for some $q\in S,qurm=qstx\in P-IP$. As $P$ is
$I$-primary submodule , $(qru)^{n}\in(P:M)$ for some positive integer $n$ or
$m\in P$. So $\frac{(qur)^{n}}{(qus)^{n}}=(\frac{r}{s})^{n}\in
S^{-1}(P:M)=(S^{-1}P:_{S^{-1}R}S^{-1}M)$ or $\frac{m}{t}\in S^{-1}P$. Hence
$S^{-1}P$ is $(S^{-1}I)$-primary submodule of $S^{-1}M$. ∎

######  Proposition 2.6.

Let $M$ be an $R$-module and $P$ be a submodule of $M$.
(i) If $I_{1}\subseteq I_{2}$. Then $P$ is $I_{1}$-primary implies $P$ is
$I_{2}$-primary.
(ii) If $P$ is $I$-prime then $P$ is $I$-primary.

###### Proof.

(1) Suppose that $P$ is $I_{1}$-primary.Let $r\in R,m\in M$ with $rm\in
P-I_{2}P$. Since $I_{1}\subseteq I_{2},\ P-I_{2}P\subseteq P-I_{1}P$. Then
$rm\in P-I_{1}P$. But $P$ is $I_{1}$-primary. So $r\in\sqrt{(P:M)}$ or $m\in
P$. Thus $P$ is $I_{2}$-primary.
(2) The proof is trivial. ∎

An ideal $I$ is called radical if $I=\sqrt{I}$. In the following we give a
condition under which $I$-prime and $I$-primary be equivalent.

######  Proposition 2.7.

Let $P$ be a proper submodule of an $R$-module $M$ such that $(P:M)$ is
radical ideal. Then $P$ is $I$-primary if and only if $P$ is $I$-prime.

###### Proof.

The proof comes from the definitions. ∎

######  Theorem 2.8.

Let $M_{1}$ and $M_{2}$ be modules over $R_{1}$ and $R_{2}$ respectively with
$M=M_{1}\times M_{2}$. Suppose that $P_{1}$ is an $I_{1}$-primary submodule of
$M_{1}$ such that $IP_{1}\times M_{2}\subseteq I(P_{1}\times M_{2})$. Then
$P_{1}\times M_{2}$ is an $I$-primary submodule of $M_{1}\times M_{2}$.

###### Proof.

Let $(r_{1},r_{2})\in R$, and $(x_{1},x_{2})\in M$ with
$(r_{1},r_{2})(x_{1},x_{2})=(r_{1}x_{1},r_{2}x_{2})\in P_{1}\times
M_{2}-I(P_{1}\times M_{2})$, and $P_{1}\times M_{2}-I(P_{1}\times
M_{2})\subseteq P_{1}\times M_{2}-IP_{1}\times M_{2}=(P_{1}-IP_{1})\times
M_{2}$. We have $r_{1}x_{1}\in P_{1}-IP_{1}$ but $P_{1}$ is $I$-primary
submodule of $M_{1}$. Then $r_{1}^{n}\in(P_{1}:_{R_{1}}M_{1})$ for some
positive integer $n$ or $x_{1}\in P_{1}$. So
$(r_{1},r_{2})^{n}=(r_{1}^{n},r_{2}^{n})\in(P_{1}:_{R_{1}}M_{1})\times
R_{2}=(P_{1}\times M_{2}:_{R_{1}\times R_{2}}M_{1}\times M_{2})$ for some
positive integer $n$ or $(x_{1},x_{2})\in P_{1}\times M_{2}$. Hence
$P_{1}\times M_{2}$ is an $I$-primary submodule of $M_{1}\times M_{2}$. ∎

We illustrate the formation of the $I$-primary submodules in decomposition
modules. In other manner the formation of $I$-primary submodules in
decomposable module have one of the five types stated in the following
theorem.

######  Theorem 2.9.

Let $R=R_{1}\times R_{2}$. Let $M_{1}$ and $M_{2}$ be $R_{1}$ and
$R_{2}$-modules respectively with $M=M_{1}\times M_{2}$. Let $I_{1}$ and
$I_{2}$ be ideals of $R_{1}$ and $R_{2}$ respectively with $I=I_{1}\times
I_{2}$. Then all the following types are $I$-primary submodule of $M_{1}\times
M_{2}$.

  1. 1.

$P_{1}\times P_{2}$ where $P_{i}$ is a proper submodule of $M_{i}$ with
$I_{i}P_{i}=P_{i}$ for $i=1,2$.

  2. 2.

$P_{1}\times M_{2}$ where $P_{1}$ is a primary submodule of $M_{1}$.

  3. 3.

$P_{1}\times M_{2}$ where $P_{1}$ is an $I_{1}$-primary submodule of $M_{1}$
and $I_{2}M_{2}=M_{2}$.

  4. 4.

$M_{1}\times P_{2}$ where $P_{2}$ is a primary submodule of $M_{2}$.

  5. 5.

$M_{1}\times P_{2}$ where $P_{2}$ is an $I_{2}$-primary submodule of $M_{2}$
and $I_{1}M_{1}=M_{1}$.

###### Proof.

  1. 1.

Since $I_{1}P_{1}=P_{1}$ and $I_{2}P_{2}=P_{2}$. Then $I_{1}P_{1}\times
I_{2}P_{2}=I{{}_{1}}\times I_{2}(P_{1}\times P_{2})=I(P_{1}\times
P_{2})=P_{1}\times P_{2}$. So $P_{1}\times P_{2}=I(P_{1}\times P_{2})=\phi$.
Thus there is nothing to prove.

  2. 2.

Let $P_{1}$ be a primary submodule of $M_{1}$. Then $P_{1}\times M_{2}$ is a
primary submodule of $M_{1}\times M_{2}$. Thus $P_{1}\times M_{2}$ is
$I$-primary submodule of $M$.

  3. 3.

Suppose that $P_{1}$ is an $I_{1}$-primary submodule of $M_{1}$ and
$I_{2}M_{2}=M_{2}$. Let $(r_{1},r_{2})\in R$ and $(m_{1},m_{2})\in M$ such
that $(r_{1},r_{2})(m_{1},m_{2})=(r_{1}m_{1},r_{2}m_{2})\in P_{1}\times
M_{2}-I(P_{1}\times M_{2})=P_{1}\times M_{2}-(I_{1}\times I_{2})(P_{1}\times
M_{2})=P_{1}\times M_{2}-(I_{1}P_{1}\times I_{2}M_{2})=P_{1}\times
M_{2}-(I_{1}P_{1}\times M_{2})=(P_{1}-I_{1}P_{1})\times M_{2}$. Then
$r_{1}m_{1}\in P_{1}-I_{1}P_{1}$ and $P_{1}$ is $I_{1}$-primary submodule of
$M_{1}$, so $r_{1}^{n}\in(P_{1}:_{R_{1}}M_{1})$ for some positive integer
$n\in N$ or $m_{1}\in P_{1}$. Therefore
$(r_{1},r_{2})^{n}=(r_{1}^{n},r_{2}^{n})\in(P_{1}:_{R_{1}}M_{1})\times
R_{2}=(P_{1}\times M_{2}:_{R_{1}\times R_{2}}M_{1}\times M_{2})$ for some
positive integer $n\in N$ or $(m_{1},m_{2})\in P_{1}\times M_{2}$. So
$P_{1}\times M_{2}$ is an $I$-primary submodule of $M_{1}\times M_{2}$.

  4. 4.

The proof is similar to part (2).

  5. 5.

The proof is similar to part (3).

∎

######  Proposition 2.10.

Let $M$ be an $R$-module and let $P$ be a proper submodule of $M$. Then $P$ is
$I$-primary in $M$ if and only if $\frac{P}{IP}$ is $0$-primary in
$\frac{M}{IP}$.

###### Proof.

Suppose that $P$ is $I$-primary in $M$. Let $r\in R,m\in M$ such that $0\neq
rm+IP=r(m+IP)\in\frac{P}{IP}$ in $\frac{M}{IP}$. Then $rm\in P-IP$. But by
assuming $P$ is $I$-primary, so $m\in P$ or $r\in\sqrt{(P:M)}$. Then
$r\in\sqrt{P:M}=\sqrt{(\frac{P}{IP}:\frac{M}{IP})}$ or $m+IP\in\frac{P}{IP}$.
Hence $\frac{P}{IP}$ is $0$-primary submodule in $\frac{M}{IP}$. Conversely
suppose that $\frac{P}{IP}$ is $0$-primary submodule in $\frac{M}{IP}$. Let
$r\in R,m\in M$ such that $rm\in P-IP$. So $0\neq
r(m+IP)=rm+IP\in\frac{P}{IP}$. But $\frac{P}{IP}$ is $\\{0\\}$-primary in
$\frac{M}{IP}$. Thus $m+IP\in\frac{P}{IP}$ or
$r\in\sqrt{(\frac{P}{IP}:\frac{M}{IP})}$ and so $m\in P$ or
$r\in\sqrt{(P:M)}$. Hence $P$ is $0$-primary. ∎

Now we give two other charactrizations of $I$-primary submodules.

######  Theorem 2.11.

Let $M$ be an $R$-module and $P$ be a proper submodule of $M$. Then $P$ is
$I$-primary in $M$ if and only if for any ideal $J$ of $R$ and submodule $N$
of $M$ such that $JN\subseteq P-IP$, we have $J\subseteq\sqrt{(P:M)}$ or
$N\subseteq P$

###### Proof.

Suppose that $P$ is $I$-primary submodule of $M$, and $JN\subseteq P-IP$ for
some ideal $J$ of $R$ and submodule $N$ of $M$. If $J\nsubseteq\sqrt{(P:M)}$
and $N\nsubseteq P$, so there exists $r\in J-\sqrt{(P:M)}$ and $x\in N-P$ such
that $rx\in P-IP$. By assuming $P$ is $I$-primary in $M$, either $x\in P$ or
$r\in\sqrt{(P:M)}$ which is a contradiction. Hence $J\subseteq\sqrt{(P:M)}$ or
$N\subseteq P$. Conversely suppose that $rm\in P-IP$ for $r\in R$ and $m\in
M$. Then $(r)(m)=(rm)\subseteq P-IP$. So by assumption, either
$(r)\subseteq\sqrt{(P:M)}$ or $(m)\subseteq P$. Therefore $r\in\sqrt{(P:M)}$
or $m\in P$. Thus $P$ is an $I$-primary submodule of $M$. ∎

The following theorem gives the relation between $I$-primary ideals and
$I$-primary submodules.

######  Theorem 2.12.

Let $M$ be a finitely generated faithful multiplication $R$-module and $P$ a
proper submodule of $M$ with $(IP:M)=I(P:M)$. Then $P$ is an $I$-primary
submodule of $M$ if and only if $(P:M)$ is an $I$-primary ideal of $R$.

###### Proof.

Assume that $P$ is $I$-primary submodule in $M$. Let $r,s\in R$, such that
$rs\in(P:M)-I(P:M)$. Then $rsM\subseteq P$. If $rsM\subseteq IP$. Then
$rs\in(IP:M)=I(P:M)$ which is contradiction. Assume $rsM\nsubseteq IP$. Then
$rsM\subseteq P-IP$. So $r\in\sqrt{(P:M)}$ or $sM\subseteq P$. Therefore
$r\in\sqrt{(P:M)}$ or $s\in(P:M)$. Hence $(P:M)$ is $I$-primary ideal.
Conversely assume that $(P:M)$ is $I$-primary ideal. Let $r\in R,m\in M$ such
that $rm\in P-IP$. $r(Rm:M)=(rRm:M)\subseteq(P:M)$ and $r(Rm:M)\nsubseteq
I(P:M)$ otherwise $rRm=r(Rm:M)M\subseteq I(P:M)M=IP$. That is
$r(Rm:M)\subseteq(P:M)-I(P:M)$. Therefore $r\in\sqrt{(P:M)}$ or
$(Rm:M)\subseteq(P:M)$. Hence $r\in\sqrt{(P:M)}$ or $Rm\subseteq P$. Thus
$r\in\sqrt{(P:M)}$ or $m\in P$. This means that $P$ is $I$-primary submodule.
∎

######  Theorem 2.13.

Let $M$ be a finitely generated faithful multiplication $R$-module and $P$ be
a proper submodule of $M$ such that $I(P:M)=(IP:M)$. Then $P$ is $I$-primary
in $M$ if and only if whenever $N$ and $K$ are submodules of $M$ such that
$NK\subseteq P-IP$, we have either $N\subseteq P$ or $K\subseteq\sqrt{P}$.

###### Proof.

Suppose that $P$ is an $I$-primary submodule in $M$. So $(N:P)$ is an
$I$-primary ideal in $R$. As $M$ is a multiplication $R$-module, $N=(N:M)M$,
$K=(K:M)M$ and so $N.K=(N:M)(K:M)M$. Suppose that $NK\subseteq P-IP$, but
$N\nsubseteq P$ and $K\nsubseteq\sqrt{P}$. Thus $(N:M)\nsubseteq(P:M)$ and
$(K:M)\nsubseteq\sqrt{(P:M)}$. As $(P:M)$ is $I$-primary in $R$. So either
$(N:M)(K:M)\nsubseteq(P:M)$ or $(N:M)(K:M)\subseteq I(P:M)$. In the first
case, we have $N.K=(N:M)(K:M)M\subseteq(P:M)M=P$ which is contradiction. Now
if $(N:M)(K:M)\subseteq I(P:M)$, then $N.K=(N:M)(K:M)M\subseteq I(P:M)M=IP$,
which is again a contradiction. Hence either $N\subseteq P$ or $K\subseteq P$.
To prove the converse part, by Theorem 2.12, it is enough to prove that
$(P:M)$ is $I$-primary ideal in $R$. Let $a,b\in R$ such that
$ab\in(P:M)-I(P:M)$ with $a\notin(P:M)$ and $b\notin(P:M)$. Take $N=aM,K=bM$.
Then $N.K=abM\subseteq(P:M)M=P$. If $NK=abM\subseteq IP$, then
$ab\in(IP:M)=I(P:M)$, which is contradiction. Hence $NK\subseteq P-IP$. By
hypothesis and since $M$ is a faithful multiplication module, we have either
$N=aM\subseteq P$ or $K=bM\subseteq\sqrt{P}=\sqrt{(P:M)M}$ and so $a\in(P:M)$
or $b\in\sqrt{(P:M)}$. Therefore $(P:M)$ is an $I$-primary ideal of $R$. ∎

######  Proposition 2.14.

Let $P$ be a submodule of an $R$-module $M$ and $N$ be any $R$-module. Then
$P$ is an $I$-primary submodule of $M$ if and only if $P\oplus N$ is an
$I$-primary submodule of $M\oplus N$.

###### Proof.

Assume that $P$ is an $I$-primary submodule in $M$. Let $r\in R,(m,n)\in
M\oplus N$ such that $r(m,n)\in(P\oplus N)-I(P\oplus N)$. Then $rm\in P-IP$.
But $P$ is $I$-primary in $M$, so $m\in P$ or $r\in\sqrt{(P:M)}$, that is
$(m,n)\in P\oplus N$ or $r\in\sqrt{(P\oplus N:M\oplus N)}$. Therefore $P\oplus
M$ is $I$-primary in $M\oplus N$. Conversely suppose that $P\oplus N$ is
$I$-primary in $M\oplus N$. Let $r\in R,m\in M$ such that $rm\in P-IP$. Then
$r(m,0)\in(P\oplus N)-I(P\oplus N)$. But by assuming $P\oplus N$ is
$I$-primary, so $(m,0)\in P\oplus N$ or $r\in\sqrt{(P\oplus N:M\oplus N)}$.
Therefore $m\in P$ or $r\in\sqrt{(P:M)}$. Hence $P$ is $I$-primary in $M$. ∎

Let $J$ be an ideal. We show that under a certain condition the
$I$-primaryness of submodules $P$ and $(P:J)$ are equivalent. First we give
the following lemma.

######  Lemma 2.15.

Let $P$ be a submodule of a faithful multiplication $R$-module $M$ and $J$ be
finitely generated faithful multiplication ideal of $R$, then:
(1) $P=(JP:J)$;
(2) If $P\subseteq JM$, then $(KP:J)=K(P:J)$ for any ideal $K$ of $R$.

######  Proposition 2.16.

Let $P$ be a submodule of a faithful multiplication $R$-module $M$ and $J$ be
a finitely generated faithful multiplication ideal of $R$. Then $P$ is an
$I$-primary submodule of $JM$ if and only if $(P:J)$ is $I$-primary in $M$.

###### Proof.

Suppose $P$ is $I$-primary in $JM$. Take $r\in R,m\in M$ such that
$rm\in(P:J)-I(P:J)$. Then $rJm\subseteq P-IP$. If $rJm\nsubseteq IP$, then by
Lemma 2.15 , $rm\in(IP:J)=I(P:J)$ which is a contradiction. As $P$ is an
$I$-primary in $JM$, then $Jm\subseteq P$ or $r\in\sqrt{(P:JM)}$. So
$m\in(P:J)$ or $r\in\sqrt{((P:J):M)}$, because $(P:JM)=((P:J):M)$. Hence
$(P:J)$ is an $I$-primary submodule in $M$. Conversely, assume that $(P:J)$ is
$I$-primary in $M$. Let $K$ be an ideal of $R$, and $N$ be a submodule of $JM$
such that $KN\subseteq P-IP$. Then $K(N:J)\subseteq(KN:J)\subseteq(P:J)$.
Moreover, if $K(N:J)\subseteq I(P:J)$, then by Lemma 2.15
$KN=K(JN:J)=K(N:J)J\subseteq I(P:J)J=IP$ which is a contradiction. Hence
$K(N:J)\subseteq(P:J)-I(P:J)$. As $(P:J)$ is $I$-primary in $M$, either
$K\subseteq\sqrt{((P:J):M)}=\sqrt{(P:JM)}$ or $(N:J)\subseteq(P:J)$, which
implies that $N=(N:J)J\subseteq(P:J)P=P$. Hence $P$ is an $I$-primary
submodule in $JM$. ∎

Let $M$ and $F$ be $R$-modules and $r\in R$. Then it is clear that for any
submodule $P$ of $M$, $F\otimes(P:r)\subseteq(F\otimes P:r)$. In the following
lemma we give a condition under which the equality holds.

######  Lemma 2.17.

Let $r\in R$ and $P$ a submodule of $M$. Then for every flat $R$-module $F$,
we have $F\otimes(P:r)=(F\otimes P:r)$.

###### Proof.

Consider the exact sequence $0\longrightarrow(P:r)\longrightarrow
M\stackrel{{\scriptstyle f_{r}}}{{\longrightarrow}}\frac{M}{P}$ where
$f_{r}(m)=rm+P$. As $F$ is flat , the exactness of the sequence
$0\longrightarrow P\longrightarrow M\longrightarrow\frac{M}{P}\longrightarrow
0$ implies the exactness of the sequence $0\longrightarrow F\otimes
P\longrightarrow F\otimes M\longrightarrow F\otimes\frac{M}{P}\longrightarrow
0$ which gives $F\otimes\frac{M}{P}\cong\frac{F\otimes M}{F\otimes P}$. So the
exactness of the sequence $0\longrightarrow(P:r)\longrightarrow
M\longrightarrow\frac{M}{P}$ imply the exactness of the sequence
$0\longrightarrow F\otimes(P:r)\longrightarrow F\otimes
M\stackrel{{\scriptstyle
1\otimes\acute{f_{r}}}}{{\longrightarrow}}\frac{F\otimes M}{F\otimes P}$ where
$(1\otimes\acute{f_{r}})(n\otimes m)=r.(n\otimes m)+F\otimes P$ for $n\in F$.
Therefore $F\otimes(P:r)=ker(1\otimes\acute{f_{r}})=(F\otimes P:_{F\otimes
M}r)$. ∎

######  Theorem 2.18.

Let $P$ be an $I$-primary submodule of an $R$-module $M$ and $F$ be a flat
$R$-module with $F\otimes P\neq F\otimes M$. Then $F\otimes P$ is an
$I$-primary submodule of $F\otimes M$.

###### Proof.

Suppose that $P$ is an $I$-primary and $r\in R-\sqrt{(P:M)}$. Then by Theorem
2.4 $(P:r)=P$ or $(P:r)=(IP:r)$. So by Lemma 2.17, $(F\otimes
P:r)=F\otimes(P:r)=F\otimes P$ or $(F\otimes
P:r)=F\otimes(P:r)=F\otimes(IP:r)=(F\otimes IP:r)=(I(F\otimes P):r)$ and
consequently $F\otimes P$ is an $I$-primary submodule of $F\otimes M$. ∎

######  Proposition 2.19.

Let $F$ be a faithfully flat $R$-module. Then a submodule $P$ of an $R$-module
$M$ is $I$-primary if and only if $F\otimes P$ is an $I$-primary submodule of
$F\otimes M$.

###### Proof.

Suppose that $P$ is an $I$-primary submodule of an $R$-module $M$ and $F$ a
faithfully flat $R$-module. If $F\otimes P=F\otimes M$, then the exactness of
the sequence $0\longrightarrow F\otimes P\longrightarrow F\otimes
M\longrightarrow 0$ imply the exactness of $0\longrightarrow P\longrightarrow
M\longrightarrow 0$ and hence $P=M$ which is a contradiction. So $F\otimes
P\neq F\otimes M$ and by Theorem 2.18 $F\otimes P$ is an $I$-primary submodule
of $F\otimes M$. Conversely, let $F\otimes P$ be an $I$-primary submodule of
$F\otimes M$. Hence we obtain $F\otimes P\neq F\otimes M$ and so $P\neq M$. By
Lemma 2.17, for every $r\in R-\sqrt{(P:M)}$ we have $r\notin(F\otimes
P:F\otimes M)$ and $F\otimes(P:r)=(F\otimes P:r)=F\otimes P$ or
$F\otimes(P:r)=(F\otimes P:r)=(I(F\otimes P):r)=(F\otimes
IP:r)=F\otimes(IP:r)$. Suppose $F\otimes(P:r)=F\otimes P$. Then
$0\longrightarrow F\otimes(P:r)\longrightarrow F\otimes P\longrightarrow 0$ is
an exact sequence and as $F$ is faithfully flat,
$0\longrightarrow(P:r)\longrightarrow P\longrightarrow 0$ is an exact sequence
and consequently $(P:r)=P$. The other case can be proved similarly. Thus by
Theorem 2.4, $P$ is an $I$-primary submodule of $M$. ∎

It is well-known that $I\otimes F\cong IF$ for any ideal $I$ of $R$ and flat
$R$-module $F$. Thus by Theorem 2.18 and Proposition 2.19 we conclude the
following.

######  Corollary 2.20.

Let $F$ be a flat $R$-module and $P$ be an $I$-primary ideal of $R$ with
$PF\neq F$. Then $PF$ is an $I$-primary submodule of $F$. In the case $F$ is
faithfully flat, the converse is also true.

We know that every polynomial ring $R[x]$ is a flat $R$-module and that
$R[x]\otimes M\cong M[x]$. Hence as an immediate consequence of the Theorem
2.18 we give the following corollary.

######  Corollary 2.21.

Let $M$ be an $R$-module and $x$ an indeterminate. If $P$ is an $I$-primary
submodule of $M$, then $P[x]$ is an $I$-primary submodule of $M[x]$.

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