(m, n)-Closed Submodules of Modules over Commutative Rings

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(m, n)-Closed Submodules of Modules over Commutative Rings

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26 February 2024

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26 February 2024

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Abstract

Let R be a commutative ring and m, n be positive integers. We define a proper submodule N of an R-module M to be (m,n)-closed if for r∈R and b∈M, r^{m}b∈N implies rⁿ∈(N:_{R}M) or b∈N. This class of submodules lies properly between the classes of prime and primary submodules. Many characterizations, properties and supporting examples concerning this class of submodules are provided. The notion of (m,n)-modules is introduced and characterized. Furhermore, the (m,n)-closed avoidance theorem is proved. Finally, the (m,n)-closed submodules in amalgamated modules are studied.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

Throughout, we assume that all rings are commutative with identity and all modules are unital. Let R be a ring, M an R-module, N a proper submodule of M and

m, n, k

be positive integers. By M-

r a d (N)

and

(N :_{R} M)

, we denote the radical of N (the intersection of all prime submodules containing N) and the ideal of all elements r of R for which

r M \subseteq N

, respectively. Moreover, by

N i l (M)

we mean the set

{r \in R :

r^{k} M = 0_{M}

for some positive integer

k}

of all nilpotent elements in M. An R-module M is called multiplication if every submodule N of M has the form

I M

for some ideal I of R. Moreover, M is called faithful if it has a zero annihilator in R, that is

a n n_{R} (M) = (0 :_{R} M) = 0

The concepts of prime and primary submodules, which are important subjects of module theory have been widely studied by various authors. Recall that a proper submodule N of an R-module M is a prime (resp. primary) submodule if for

r \in R

and

b \in M

r b

\in N

implies

r \in (N :_{R} M)

(resp.

r \in \sqrt{(N :_{R} M)}

) or

b \in N

. The concepts of prime and primary submodules have been generalized in several ways (see, for example, [2,5,6,10,11,20,24].)

Following [11], N is called an n-absorbing submodule (resp. semi-k-absorbing submodule [23]) if whenever

r_{1} r_{2}

···

r_{n} b \in N

for

r_{1}, r_{2},

...,

r_{n}

\in R

and

b \in M

(resp.

r^{k} b \in N

for

r \in R

and

b \in M)

, then either

r_{1} r_{2}

···

r_{n}

\in (N :_{R} M)

or there are

n - 1

r_{i}

’s whose product with b is in

N .

(resp.

r^{k} \in (N :_{R} M)

r^{k - 1} b \in N

). In particular, as a subclass of 2-absorbing primary submodules defined in [18], the notion of 1-absorbing primary (resp. 1-absorbing prime) submodules are initiated in [24] (resp. in [4]). N is said to be a 1-absorbing primary (resp. 1-absorbing prime) submodule if whenever non-unit elements

r, s \in R

and

b \in M

with

r s b \in N

, then either

r s \in (N :_{R} M)

b \in M

r a d (N)

(resp.

b \in N

). The n-absorbing

(n \geq 1)

structures have received a significant amount of attention; see for instance [2,4,5,7,24].

As one of the generalizations of prime submodules, in [20], the concept of semiprime submodules is introduced. N is called a semiprime (resp. an

(m, n)

-semiprime [19]) submodule if for

r \in R

and

b \in M

r^{m} b \in N

implies

r b \in N

(resp.

r^{n} b \in N)

. In 2017, Anderson and Badawi introduced the class of

(m, n)

-closed ideals. According to [3], a proper ideal I of R is called an

(m, n)

-closed ideal if whenever

r^{m} \in I

for some

r \in R

, then

r^{n} \in I

. In a recent work [13], the class of

(m, n)

-prime ideals which is a subclass of the previous structure is defined and studied. A proper ideal I of R is called

(m, n)

-prime if for

r, s \in R

r^{m} s \in I

implies either

r^{n} \in I

s \in I .

Note that this subclass is proper in the class of

(m, n)

-closed ideals. For example, the ideal

I = 16 Z

(3, 2)

-closed in the ring of integers which is not

(3, 2)

-prime as

2^{3} \cdot 2 \in I

but

2^{2},

2 \notin I

. For more classes of ideals related to

(m, n)

-closed ideals, one can see [14] and [15].

The main objective of this work is to describe the structure of

(m, n)

-closed submodules as a submodule synonym of

(m, n)

-prime ideals. A proper submodule N of an R-module M is called an

(m, n)

-closed submodule if for

r \in R

and

b \in M

r^{m} b \in N

implies either

r^{n} \in (N :_{R} M)

b \in N

. In this case,

P = \{r \in R : r^{n} \in (N :_{R} M)\}

is a prime ideal of R and we sometimes refer to N as P-

(m, n)

-closed. We clarify that this class of submodules lies properly between the classes of prime and primary submodules. The paper is organized as follows. In Section 2, first, we determine the exact place of this class of submodules in the literature. Many supporting examples and counterexamples are presented. Various characterizations of this class of submodules are provided. (see Theorems 4, 10 and 18, Corollary 6) We obtain a characterization for faithful multiplication modules for which every proper submodule is P-

(m, n)

-closed. (see Corollary 17) Furthermore, we give a characterization for P-

(m, n)

-closed modules that is an R-module whose zero submodule is P-

(m, n)

-closed. (see Theorem 30). For any class of submodules, it is natural to investigate its behavior under homomorphisms, localizations, direct sums and extensions. (see Propositions 20-28, Theorems 26 and 27). Moreover, as generalization of ZPI-rings, we introduce and characterize

(m, n)

-modules M for which every proper submodule N is either an

(m, n)

-closed submodule or

N = I_{1} I_{2} \dots I_{k} K

where

I_{i}

’s are

P_{i}

(m, n)

-prime ideals of R and K is a Q-

(m, n)

-closed submodule, (see Theorem 30) We also justify (see Proposition 28) the relationship between the

(m, n)

-closed submodules in an R-module M and the

(m, n)

-prime ideals of the idealization ring

R (+) M

. At the end of this section, the

(m, n)

-closed avoidance theorem is verified (see Theorem 33).

Let

f : R_{1} \to R_{2}

be a ring homomorphism, J be an ideal of

R_{2}

M_{1}

be an

R_{1}

-module,

M_{2}

be an

R_{2}

-module (which is an

R_{1}

-module induced naturally by f) and

φ : M_{1} \to M_{2}

be an

R_{1}

-module homomorphism. We conclude section 3 by investigating some kinds of

(m, n)

-closed submodules in the the amalgamation

(R_{1} ⋈^{f} J)

-module

M_{1} ⋈^{φ} J M_{2}

M_{1}

and

M_{2}

along J with respect to

φ

(see Theorems 34 and 35).

2. $(m, n)$ -closed Submodules

In this section, we introduce the class of

(m, n)

-closed submodules and clarify its relationship with some other classes of submodules. We determine some of properties and characterizations for

(m, n)

-closed submodules. Moreover, we investigate the behavior of this structure under module homomorphisms, localizations, quotient modules, Cartesian products and idealizations.

Definition 1.

Let N be a proper submodule of an R-module M and

m, n

be positive integers. Then N is called an

(m, n)

-closed submodule if for

r \in R

and

b \in M

r^{m} b \in N

implies either

r^{n} \in (N :_{R} M)

b \in N

It is clear that the

(m, n)

-prime ideals of a ring R coincides with the

(m, n)

-closed submodules of the R-module R. If N is an

(m, n)

-closed submodule of an R-module M, then clearly N is a primary submodule of M and so

P = \sqrt{(N :_{R} M)}

is a prime ideal of R. In this case, we call N a P-

(m, n)

-closed submodule of M. Moreover, we note by [13] that

P = \{r \in R : r^{n} \in (N :_{R} M)\}

In the following remark, we clarify the relationship between

(m, n)

-closed submodules and some other kinds of submodules.

Remark 2.

Let N be a proper submodule of an R-module M and

n, m

be positive integers.

If N is a prime submodule of M, then clearly it is $(m, n)$ -closed for all $m, n \in N$ . Moreover, N is a prime submodule of M if and only if N is $(1, 1)$ -closed.
If N is a 1-absorbing prime submodule of M, then N is an $(m, n)$ -closed submodule for $n \geq 2$ . Indeed, let $r^{m} b \in N$ and $b \notin N$ for $r \in R$ and $b \in M$ . If r is a unit, then $b \in N$ , a contradiction. Hence, assume that r is nonunit. Then by assumption, $r^{m - 1} r \in (N :_{R} M)$ and so either $r \in (N :_{R} M)$ or $r^{m - 2} r \in (N :_{R} M)$ as $(N :_{R} M)$ is 1-absorbing prime in R. By continuing this process, we get $r^{2} \in (N :_{R} M)$ and so $r^{n} \in (N :_{R} M)$ , as needed. The converse is also true if $(N :_{R} M)$ a radical ideal of R.
In general, N can be a 2-absorbing submodule of M that is not an $(m, n)$ -closed submodule for all integers m and n. For example, the ideal $I = 〈4, 2 x, 2 y, x y, x z, x^{2}〉$ is a 2-absorbing submodule of the R-module $R = Z [x, y, z]$ which is not a primary submodule, see [5, Example 2.12]. Thus, I is not an $(m, n)$ -closed submodule of R for all integers m and n. For another example, the submodule $〈\bar{0}〉$ of the $Z$ -module $Z_{6}$ is a 2-absorbing submodule as it is an intersection of two distinct prime submodules, [10, Theorem 2.3]. However, for all integers m and n, $〈\bar{0}〉$ is not $(m, n)$ -closed as also it is not primary.
If N is $(m, n)$ -closed in M, then N is a semi n-absorbing submodule of M. Indeed, let $r \in R$ and $b \in M$ such that $r^{n} b \in N$ and $r^{n - 1} b \notin N$ (so that $b \notin N$ ). If $m \geq n$ , then $r^{m} b \in N$ implies $r^{n} \in (N :_{R} M)$ as N is $(m, n)$ -closed. If $m ≨ n$ , then $r^{m} r^{n - m} b \in N$ and $r^{n - m} b \notin N$ as $n - 1 \geq n - m$ . Thus, again $r^{n} \in (N :_{R} M)$ and N is a semi n-absorbing submodule of M.
If N is $(m, n)$ -closed in M, then it is $(m^{'}, n^{'})$ -closed for $n \leq n^{'}$ and $m^{'} \leq m$ .

We illustrate the place of the class of

(m, n)

-closed submodules for all positive integers n and m by the following diagram:

\begin{matrix} semiprime & ⟶ & semi n - absorbing & ⟵ & n - absorbing \\ ↑ & ↑ \\ prime & ⟶ & (m, n) - closed & ⟶ & primary \\ ↓ & ↗ & ↓ \\ 1 - absorbing prime & (m, n) - semiprime \end{matrix}

However, the implications in the above diagram are proper as we can see in the following example.

Example 3.

The submodule $N = \bar{0}$ of the $Z$ -module $M = Z_{4}$ is not prime in M. On the other hand, N is $(m, 2)$ -closed in M for all $m \in N$ . Indeed, let $r \in Z$ and $b \in Z_{4}$ such that $r^{m} b \in N$ and $b \notin N$ . Then clearly, $r^{m} \in 2 Z$ and so $r \in 2 Z$ . Thus, $r^{2} \in 4 Z = (N :_{R} M)$ and N is $(m, 2)$ -closed in M.
Semiprime submodules and $(m, n)$ -closed submodules are not comparable in general. The submodule $N = \bar{0}$ of the $Z$ -module $M = Z_{4}$ is also not semiprime in M as $2^{2} \cdot \bar{1} \in N$ but $2 \cdot \bar{1} \notin N$ . Moreover, any non primary radical ideal of any ring R is an example of a semiprime submodule of the R-module R that is not $(m, n)$ -closed for any $n, m \in N$ .
The submodule $N = \bar{8} Z_{16}$ of the $Z$ -module $M = Z_{16}$ is clearly a primary submodule. However, N is not $(m, 2)$ -closed in M for all $m \in N$ . For example, $2^{m} . \bar{4} \in N$ for all $m \in N$ but $\bar{4} \notin N$ and $2^{2} \notin 8 Z = (N :_{Z} M)$ .
The ideal $I = 6 Z$ is a semi-2-absorbing submodule of the $Z$ -module $Z$ since for $r, b \in Z$ , $r^{2} b \in 6 Z$ implies $r b \in 2 Z \cap 3 Z = 6 Z$ . On the other hand, for all $m, n \in N$ , I is not an $(m, n)$ -closed submodule of $Z$ since it is not primary.
The submodule $I = 8 Z$ is $(m, 3)$ -closed in the $Z$ -module $Z$ for all $m \in N$ , [13, Theorem 3]. However, $8 Z$ is not 2-absorbing (and so not 1-absorbing prime) in $Z$ since $2 \cdot 2 \cdot 2 \in I$ but $2 \cdot 2 \notin I$ .
One can easily verify that the submodule $\bar{6} Z_{12}$ is $(2, 1)$ -semiprime in the $Z$ -module $Z_{12}$ . However, $\bar{6} Z_{12}$ is not primary and so not $(2, 1)$ -closed in $Z_{12}$ .

Let N be a proper submodule of an R-module M and I be an ideal of R. The residual of N by I is the set

(N :_{M} I) = {m \in M : I m \subseteq N}

. It is clear that

(N :_{M} I)

is a submodule of M containing N. More generally, for any subset

A \subseteq R

(N :_{M} A)

is a submodule of M containing N. We next give one characterization of

(m, n)

-closed submodules of an R-module.

Theorem 4.

Let N be a proper submodule of an R-module M and

n, m

be positive integers. The following are equivalent.

N is an $(m, n)$ -closed submodule of $M .$
For all $r \in R$ with $r^{n} \notin (N :_{R} M)$ , $(N :_{M} r^{m}) = N$ .
Whenever $r \in R$ and K is a submodule of M with $r^{m} K \subseteq N$ , then $r^{n} \in (N :_{R} M)$ or $K \subseteq N$ .

Proof.

(1)⇒(2) Let

b \in (N :_{M} r^{m})

so that

r^{m} b \in N

. Since

r^{n} \notin (N :_{R} M)

, we have by assumption

b \in N

and so

(N :_{M} r^{m}) \subseteq N

. The other containment is clear.

(2)⇒(3) Let

r \in R

and K be a submodule of M with

r^{m} K \subseteq N .

r^{n} \notin (N :_{R} M)

, then by assumption we conclude

K \subseteq (N :_{M} r^{m}) = N

(3)⇒(1) Suppose that

r^{m} b \in N

for some

r \in R

and

b \in M

. The result follows by putting

K = R b

in (3). □

Corollary 5.

Let N be an

(m, n)

-closed submodule of an R-module

M .

Then

$(N :_{R} M)$ is a $(m, n)$ -prime ideal of $R .$
For any subset $A \subseteq R$ , either $N = A M$ or $(N :_{M} A)$ is an $(m, n)$ -closed submodule of M.

Proof.

(1) Clearly,

(N :_{R} M)

is proper in R. Let

a, b \in R

such that

a^{m} b \in (N :_{R} M)

. Then

a^{m} (b M) \subseteq N

and so by Theorem 4,

a^{n} \in (N :_{R} M)

b M \subseteq N

. Thus,

a^{n} \in (N :_{R} M)

b \in (N :_{R} M)

, as required.

(2) Suppose that

N \neq A M .

Then

(N :_{M} A)

is proper in M. Let

r \in R

and

b \in M

such that

r^{m} b \in (N :_{M} A) .

Then

r^{m} (A b) \subseteq N

and so by Theorem 4, we have

r^{n} \in (N :_{R} M) \subseteq ((N :_{M} A) :_{R} M)

A b \subseteq N

, that is

b \in (N :_{M} A)

. □

Note that the converse of (2) of Corollary 5 need not be true. For example,

N = \bar{8} Z_{16}

is not

(m, 2)

-closed in the

Z

-module

M = Z_{16}

for all

m \in N

(see Example 3(3)). But, for

I = 4 Z

(N :_{M} I) = \bar{2} Z_{16}

is clearly

(m, 2)

-closed in M.

Corollary 6.

Let N be a submodule of a multiplication R-module M and

n, m

be positive integers. Then

N is $(m, n)$ -closed in M if and only if $(N :_{R} M)$ is a $(m, n)$ -prime ideal of $R .$
If N is $(m, n)$ -closed, then M- $r a d (N)$ is a prime submodule of $M .$
If $P = \{r \in R : r^{n} \in (N :_{R} M)\}$ is a maximal ideal of R, then N is P- $(m, n)$ -closed in M.

Proof.

(1)

\Rightarrow)

Follows by (1) of Corollary 5.

\Leftarrow)

Let

r \in R

and K be a submodule of M with

r^{m} K \subseteq N

. Since M is multiplication, there is a presentation ideal I in R such that

K = I M

. Hence,

r^{m} I \subseteq (N :_{R} M)

and [13, Corollary 2(3)] yields either

r^{n} \in (N :_{R} M)

I \subseteq (N :_{R} M) .

Thus,

r^{n} \in (N :_{R} M)

K = I M \subseteq (N :_{R} M) M = N .

(2) Since by (1),

(N :_{R} M)

is a

(m, n)

-prime ideal of R, then

\sqrt{(N :_{R} M)}

is clearly a prime ideal of R. It is verified in [21, Theorem 2.12] that if M is a multiplication R-module, then M-

r a d (N) = \sqrt{(N :_{R} M)} M

. Thus, M-

r a d (N) = \sqrt{(N :_{R} M)} M

is a prime submodule of

M .

(3) If

P = \{r \in R : r^{n} \in (N :_{R} M)\}

is maximal in R, then

(N :_{R} M)

(m, n)

-prime in R by [13, Proposition 1]. Now, the result follows by (1). □

Unless M is multiplication, being

(N :_{R} M)

(m, n)

-prime ideal does not imply that N is an

(m, n)

-closed submodule. For example, the submodule

N = 0 \times \bar{4} Z_{8}

of the

Z

-module

M = Z \times Z_{8}

is clearly not prime (and so not

(1, 1)

-closed). On the other hand,

(N :_{R} M) = 0

is a prime (and so

(1, 1)

-prime) ideal of

Z

Following [22], a proper submodule N of an R-module M is called an n-submodule if

r b \in N

for

r \in R

and

b \in M

implies

r \in N i l (M)

b \in N

. Next, we determine the relationship between n-submodules and

(m, n)

-closed submodules.

Proposition 7.

Let N be a submodule of an R-module M and

n, m

be positive integers.

Let N be a P- $(m, n)$ -closed submodule of M. Then N is an n-submodule if and only if $P = N i l (M) .$
If N is an n-submodule of M and $P = N i l (M)$ , then N is a P- $(m, n)$ -closed submodule of M.

Proof.

(1) Suppose N is an n-submodule of M and let

r \in P

. Then

r^{n} \in (N :_{R} M)

and so

r^{n} b \in N

for some

b \in M ∖ N

as N is proper. Since N is n-submodule, we have

r^{n} \in N i l (M)

and so

r \in N i l (M)

. Thus,

P = N i l (M \dot{)}

as the reverse inclusion always holds. Conversely, suppose

P = N i l (M)

and let

r \in R

b \in M

such that

r b \in N

. Since

r^{m} b \in N

and N is P-

(m, n)

-closed in M, we conclude that

r \in P

b \in N

. Thus, N is an n-submodule as

P = N i l (M)

(2) Suppose that

r^{m} b \in N

where

r \in R

and

b \in M .

Then by assumption,

r^{m} \in N i l (M)

b \in N

and so clearly,

r \in P

b \in N

P = N i l (M)

. Thus, N is an

(m, n)

-closed submodule of M. □

As it is shown in Example 3(1), there are

(m, n)

-closed submodules which are not prime. In the following, we conclude a condition for an

(m, n)

-closed submodule to be prime.

Theorem 8.

Let N be a maximal

(m, n)

-closed submodule with respect to inclusion. Then N is a prime submodule of M.

Proof.

Let

r \in R

b \in M

such that

r b \in N

and

r \notin (N :_{R} M)

. Then

r M ⊈ N

and so

(N :_{M} r)

is proper in M. Moreover,

(N :_{M} r)

is an

(m, n)

-closed submodule of M by Corollary 5(2). Since

N \subseteq (N :_{M} r)

and by the maximality of N, we conclude that

N = (N :_{M} r) .

Thus,

b \in N

and N is a prime submodule of

M .

□

Corollary 9.

Let M be an R-module. If M has an

(m, n)

-closed submodule, then it has a prime submodule.

Proof.

Let N be an

(m, n)

-closed submodule and set

Ω

= {K : K

is an

(m, n)

-closed submodule containing

N} .

Then

Ω

is non-empty as

N \in Ω

. Let

K_{1} \subseteq K_{2} \subseteq \dots

be a chain in

Ω

. We show that

\cup K_{i}

(m, n)

-closed in M. Suppose that

r^{m} b \in \cup K_{i}

for some

r \in R

b \in M

and

b \notin \cup K_{i} .

Then

r^{m} b \in K_{i}

and

b \notin K_{i}

for some i which imply that

r^{n} \in (K_{i} :_{R} M) \subseteq (\cup K_{i} :_{R} M) .

Hence,

\cup K_{i}

(m, n)

-closed and an upper bound of the chain. Now, Zorn’s Lemma yields that

Ω

has a maximal element, say

L \in Ω

. Thus, L is a prime submodule of M by Theorem 8. □

Recall that the product of two submodules

N = I M

and

K = J M

of a multiplication module M is defined as

N K = I J M

. In particular, for

b_{1}, b_{2} \in M

b_{1} b_{2}

, we mean the product of the submodules

R b_{1}

and

R b_{2}

. Let M be a multiplication R-module where R is a principal ideal domain. In this case, we conclude further characterizations for

(m, n)

-closed submodules.

Theorem 10.

Let R be a principal ideal domain, N be a proper submodule of a multiplication R-module M and

n, m

be positive integers. The following statements are equivalent.

N is an $(m, n)$ -closed submodule of $M .$
If I is an ideal of R and $b \in M ∖ N$ with $I^{m} b \subseteq N$ , then $I^{n} \subseteq (N :_{R} M)$ .
For an ideal I of R and a submodule K of M, $I^{m} K \subseteq N$ implies either $I^{n} \subseteq (N :_{R} M)$ or $K \subseteq N$ .
For an ideal I of R, $I^{n} ⊈ (N :_{R} M)$ implies $(N :_{M} I^{m}) = N$ .
For $b_{1}, b_{2} \in M$ such that $b_{1}^{m} b_{2} \subseteq N$ , we have either $b_{1}^{n} \subseteq N$ or $b_{2} \in N$ .

Proof.

(1)⇒(2) Suppose that

I^{m} b \subseteq N

for an ideal I of R and

b \in

M ∖ N

. Put

I = 〈r〉

for some

r \in R .

Then

r^{m} b \in N

which implies

r^{n} \in (N :_{R} M)

and so

I^{n} \subseteq (N :_{R} M) .

(2)⇒(3) Suppose that

I^{m} K \subseteq N

but

K ⊈ N

. Then

I^{n} b \subseteq N

for some

b \in K ∖ N .

Thus, by (2), we conclude that

I^{n} \subseteq (N :_{R} M)

(3)⇒(4) Straightforward.

(4)⇒(5) Let

b_{1}, b_{2} \in M

such that

b_{1}^{m} b_{2} \subseteq N

and

b_{1}^{n} ⊈ N

. If I and J are the presentation ideals of

R b_{1}

and

R b_{2}

, respectively, then

I^{n} ⊈ (N :_{R} M)

and so by (4), we conclude that

b_{2} \in R b_{2} = J M \subseteq (N :_{M} I^{m}) = N

(5)⇒(1) Let

r^{m} b \in N

for some

r \in R

and

b \in M ∖ N

. Let

b^{'} = 〈r〉 M

and I be the presentation ideal of

R b

. Then

b^{' m} b \subseteq N

with

b \notin N

and so by assumption,

{〈r〉}^{n} M = b^{' n} \subseteq N

. Therefore,

r^{n} \in (N :_{R} M)

and the result follows. □

Corollary 11.

Let I be an ideal of a ring R and N be an

(m, 2)

-closed submodule of an R-module

M .

I^{m} \subseteq (N :_{R} M)

, then,

2 I^{2} \subseteq (N :_{R} M)

. Additionally, if 2 is unit in R, then

I^{2} \subseteq (N :_{R} M)

Proof.

Suppose that

I^{m} \subseteq (N :_{R} M)

. Then for all

r_{1}, r_{2} \in I

and

b \in M

, we have

r_{1}^{m} b,

r_{2}^{m} b,

{(r_{1} + r_{2})}^{m} b \in N .

Since N is

(m, 2)

-closed, we conclude that either

r_{1}^{2} \in (N :_{R} M)

(resp.

r_{2}^{2} \in (N :_{R} M)

{(r_{1} + r_{2})}^{2} \in (N :_{R} M)

) or

b \in N

. Thus,

r_{1}^{2} b,

r_{2}^{2} b,

{(r_{1} + r_{2})}^{2} b \in N

and so

2 r_{1} r_{2} b = ({(r_{1} + r_{2})}^{2} - r_{1}^{2} - r_{2}^{2}) b \in N

. Therefore,

2 I^{2} \subseteq (N :_{R} M)

. In particular, if 2 is unit, then

I^{2} \subseteq (N :_{R} M)

. □

Corollary 12.

Let I be an ideal of a ring R,

N_{1}

N_{2}

be submodules of an R-module M and

n, m

be positive integers.

If for $i = 1, 2$ , $N_{i}$ is a $P_{i}$ - $(m, n)$ closed submodule and $I ⊈ P_{1} \cup P_{2}$ , then $I N_{1} = I N_{2}$ implies $N_{1} = N_{2} .$
If $I N_{1}$ is a P- $(m, n)$ -closed submodule of $M$ with $I ⊈ P$ , then $I N_{1} = N_{1} .$

Proof.

(1) Suppose that

I ⊈ P_{1} \cup P_{2}

and

I N_{1} = I N_{2} .

Then

I ⊈ P_{1}

and so there is an element

r \in I

such that

r^{n} \notin (N_{1} :_{R} M)

. Now, since

r^{m} N_{2} \subseteq I N_{1} \subseteq N_{1}

and

N_{1}

(m, n)

-closed, we have

N_{2} \subseteq N_{1}

by Theorem 4. Similarly,

I ⊈ P_{2}

implies

N_{1} \subseteq N_{2}

and the required equality holds.

(2) Let

I N_{1}

be an

(m, n)

-closed submodule of

M

with

I ⊈ P .

Then there exists

r \in I

such that

r^{n} \notin (I N_{1} :_{R} M)

. Since

r^{m} N_{1} \subseteq I N_{1}

and

r^{n} \notin (I N_{1} :_{R} M)

, we have

N_{1} \subseteq I N_{1}

and so

I N_{1} = N_{1}

, as needed. □

We recall that a module M is torsion (resp. torsion-free) if

T (M) = M

(resp.

T (M) = {0}

) where

T (M) = {m \in M :

there exists

0 \neq r \in R

such that

r m = 0}

. A submodule N of an R-module M is called a pure submodule if for any

r \in R

(r M) \cap N = r N

Proposition 13.

Let N be a submodule of a torsion free R-module M such that

P = \{r \in R : r^{n} \in (N :_{R} M)\} = 0

. Then for all

m \in N

, N is a pure submodule of M if and only if N is

(m, n)

-closed.

Proof.

Suppose that N is a pure submodule of M and note that N is proper in M since otherwise

R = P = 0

. Let

r^{m} b \in N

for some

r \in R

b \in M .

Then

r^{m} b \in (r^{m} M) \cap N = r^{m} N

and so

r^{m} b = r^{m} t

for some

t \in N

which yields,

r^{m} (b - t) = 0

. If

r = 0

, then

r^{n} \in (N :_{R} M)

. If

r \neq 0

, then M is torsion free and

r^{m} (b - t) = 0

imply

r^{m - 1} (b - t) = 0

. Continue in this process to get

b = t \in N

Conversely, suppose that N is an

(m, n)

-closed submodule of M and note that

r N \subseteq r M \cap N

for any

r \in R

. If

r = 0

, then the containment

r M \cap N \subseteq r N

is clear. So, assume that r is nonzero. Let

r b \in r M \cap N

for some

b \in M

and assume on contrary that

b \notin N

. Then clearly

r^{m} b \in N

and since N is

(m, n)

-closed, we conclude

r \in P = 0

, a contradiction. Thus

b \in N

and

r M \cap N \subseteq r N,

as required. □

Lemma 14.

[1] For an ideal I of a ring R and a submodule N of a finitely generated faithful multiplication R-module M, the following hold.

1.

(I N :_{R} M) = I (N :_{R} M)

2.

If I is finitely generated faithful multiplication, then

(a): $(I N :_{M} I) = N$ .
(b): Whenever $N \subseteq I M$ , then $(J N :_{M} I) = J (N :_{M} I)$ for any ideal J of R.

Proposition 15.

Let M be a finitely generated faithful multiplication R-module, N a submodule of M and I a finitely generated faithful multiplication ideal of R. Then for

n, m \in N

, we have

N is $(m, n)$ -closed in $I M$ if and only if $(N :_{M} I)$ is $(m, n)$ -closed in M.
If $I N$ is $(m, n)$ -closed in M, then either I is $(m, n)$ -closed in R or N is $(m, n)$ -closed in M.

Proof.

(1) Suppose N is an

(m, n)

-closed submodule of

I M

. If

(N :_{M} I) = M

, then by Lemma 14,

N = (I N :_{M} I) = I (N :_{M} I) = I M

, a contradiction. Thus,

(N :_{M} I)

is proper in M. Let

r^{m} b \in

(N :_{M} I)

for

r \in R

and

b \in M

such that

r^{n} \notin ((N :_{M} I) :_{R} M)

. If

r^{n} \in (N :_{R} I M)

, then

r^{n} M = r^{n} (I M :_{M} I) \subseteq (r^{n} I M :_{M} I) \subseteq (N :_{M} I)

, a contradiction. Thus,

r^{m} I b \subseteq N

and

r^{n} \notin (N :_{R} I M)

. By Theorem 4, we have

I b \subseteq N

and so

b \in

(N :_{M} I)

. Conversely, suppose

(N :_{M} I)

is an

(m, n)

-closed submodule of M. Then N is proper in

I M

since otherwise Lemma 14 implies

(N :_{M} I) = (I M : I) = M

, a contradiction. Let

r \in R

and

a b \in I M

(

a \in I

) such that

r^{m} a b \in N

and

r^{n} \notin (N :_{R} I M)

. Then

r^{m} (〈a b〉 :_{M} I) \subseteq (〈r^{m} a b〉 :_{M} I) \subseteq (N :_{M} I)

. Moreover, since clearly

r^{n} \notin ((N :_{M} I) :_{R} M)

, then by Theorem 4,

(〈a b〉 :_{M} I) \subseteq (N :_{M} I)

. Thus, by Lemma 14,

a b \in (〈a b〉 I :_{M} I) = I (〈a b〉 :_{M} I) \subseteq I (N :_{M} I) = (I N :_{M} I) = N

. Therefore, N is

(m, n)

-closed in

I M

(2) Suppose

I N

(m, n)

-closed in M. If

N = M

, then

I = I (N :_{R} M) = (I N :_{R} M)

is an

(m, n)

-closed ideal of R by Corollary 6. Suppose N is proper in M. By Lemma 14,

N = (I N :_{M} I)

and so

(N :_{R} M) = ((I N :_{M} I) :_{R} M) = (I (N :_{R} M) :_{M} I)

. Now, let

r \in R,

b \in M

such that

r^{m} b \in N

and

r^{n} \notin (N :_{R} M)

. Then

r^{m} I b \subseteq I N

and clearly

r^{n} \notin (I N :_{R} M)

. By assumption,

I b \subseteq I N

and so again by Lemma 14,

b \in (I N :_{M} I) = N

. Therefore, N is an

(m, n)

-closed submodule of M. □

As we can see in [13, Remark 2], if I is an

(m, n)

-closed ideal of R and N is an

(m, n)

-closed submodule of M, then

I N

need not be

(m, n)

-closed in M.

Proposition 16.

Let M be a faithful multiplication R-module, I be an ideal of R and

n, m

be positive integers. Then I is a P-

(m, n)

-prime ideal of R if and only if

I M

is a P-

(m, n)

-closed submodule of M.

Proof.

\Rightarrow)

I M = M

, then

I = (I M :_{R} M) = R

, a contradiction. Let

r \in R,

b \in M

with

r^{m} b \in I M

and

b \notin I M

. Since I is

(m, n)

-prime, it is primary which implies that

I M

is a primary submodule of

M .

Thus, clearly we have

r \in \sqrt{(I M :_{R} M)} = \sqrt{I} = P

. By [13, Lemma 1], we have

P = \{r \in R : r^{n} \in I\}

, and so

r^{n} \in I = (I M :_{R} M)

. Therefore,

I M

is a P-

(m, n)

-closed submodule of M.

\Leftarrow)

I = (I M :_{R} M)

, then clearly I is proper. Now, let

r, s \in R

such that

r^{m} s \in I

and

r^{n} \notin I = (I M :_{R} M)

. By Theorem 4,

r^{m} s M \subseteq I M

implies

s M \subseteq I M

and so

s \in (I M :_{R} M) = I

, as needed. □

Corollary 17.

Let M be a faithful multiplication R-module and

n, m

be positive integers. The following are equivalent.

Every proper submodule of M is P- $(m, n)$ -closed.
Every proper ideal of R is P- $(m, n)$ -prime.
R has no non-trivial idempotents, $dim (R) = 0$ and $r^{n} = 0$ for all $r \in N i l (R)$ .

Proof.

(1) \Leftrightarrow (2)

: Follows directly by Proposition 16.

(2) \Leftrightarrow (3)

: [13, Theorem 2]. □

Now, we are ready to present a general characterization for P-

(m, n)

-closed submodules in multiplication modules.

Theorem 18.

Let M be a multiplication R-module, N be a submodule of M and

n, m

be positive integers. The following statements are equivalent.

N is a P- $(m, n)$ -closed submodule of $M .$
$(N :_{R} M)$ is a P- $(m, n)$ -prime ideal of $R .$
$N = I M$ for some P- $(m, n)$ -prime ideal I of R including $a n n_{R} (M) .$

Proof.

(1)⇒(2) Corollary 6.

(2)⇒(3) We choose

I = (N :_{R} M)

(3)⇒(1) Suppose that

N = I M

for some P-

(m, n)

-prime ideal I of R including

a n n_{R} (M) .

It is well-known that M is faithful as an

R / a n n_{R} (M)

-module. Hence,

(I / a n n_{R} (M)) M = I M = N

is a P-

(m, n)

-closed submodule of the

R / a n n_{R} (M)

-module M by Proposition 16. We show that N is a P-

(m, n)

-closed submodule of the R-module

M .

Suppose

r^{m} b \in N

for some

r \in R

and

b \in M .

Then

{(r + a n n_{R} (M))}^{m} b \in N

which implies that

{(r + a n n_{R} (M))}^{n} \in (N :_{R / a n n_{R} (M)} M)

b \in N .

In the case

{(r + a n n_{R} (M))}^{n} \in (N :_{R / a n n_{R} (M)} M),

we have

r^{n} M = (r^{n} + a n n_{R} (M)) M \subseteq N

, that is,

r^{n} \in (N :_{R} M)

. Since also,

\sqrt{(N :_{R} M)} = \sqrt{(I M :_{R} M)} = \sqrt{(I / a n n_{R} (M)) M :_{R} M)} = \sqrt{I} = P

, then N is a P-

(m, n)

-closed submodule of

M .

□

Recall that for an R-module M,

Z_{R} (M)

denotes the set of all zero divisors on M. Following [16], a submodule N of an R-module M is called an r-submodule if whenever

r b \in N

for

r \in R

and

b \in M

with

a n n_{M} (r) = 0

, then

b \in N

. We call an R-module M a P-

(m, n)

-closed module if the submodule

{0}

is P-

(m, n)

-closed. Next, we give a characterization for

(m, n)

-closed modules.

Theorem 19.

Let M be an R-module and

n, m \in N

. If

P = \{r \in R : r^{n} \in (0 :_{R} M)\}

, then the following are equivalent.

M is a P- $(m, n)$ -closed module.
$P = Z_{R} (M) .$
Every r-submodule of M is $(m, n)$ -closed.

Proof.

(1)⇒(2) Let

r \in P

so that

r^{n} M = 0 .

Let k be the least positive integer such that

r^{k} M = 0

. Then

r (r^{k - 1} M) = 0

and

r^{k - 1} M \neq 0

. Thus,

r \in Z_{R} (M)

and

P \subseteq Z_{R} (M) .

Conversely, let

r \in Z_{R} (M) .

Then

r b = 0

for some

0 \neq b \in M .

Since

r^{m} b = 0

and

b \notin {0}

, we have

r^{n} \in (0 :_{R} M)

, and so

r \in P

. Thus,

Z_{R} (M) \subseteq P

and the equality holds.

(2)⇒(3) Suppose

P = Z_{R} (M)

and N is an r-submodule of M. Let

r \in R

and

b \in M

such that

r^{m} b \in N

and

r^{n} \notin (N :_{R} M)

. Since

r \notin P = Z_{R} (M)

, then clearly

r^{m} \notin Z_{R} (M)

and so

a n n_{M} (r^{m}) = 0 .

It follows that that

b \in N

as N is an r-submodule. Therefore, N is an

(m, n)

-closed submodule of

M .

(3)⇒(1) Note that the submodule

{0}

is always an r-submodule. Hence, the claim follows from (3). □

Let I be an ideal of a ring R and N be a submodule of an R-module M. By

Z_{I} (R)

and

Z_{N} (M)

, we denote the sets

{r \in R :

r a \in I

for some

a \in R ∖ I}

and

{r \in R : r m \in N

for some

m \in M ∖ N}

Proposition 20.

Let N be a proper submodule of an R-module M and S be a multiplicatively closed subset of R such that

(N :_{R} M) \cap S = \emptyset .

Then for

n, m \in N

, we have

If N is an $(m, n)$ -closed submodule of M, then $S^{- 1} N$ is an $(m, n)$ -closed submodule of $S^{- 1} M .$
If $S^{- 1} N$ is an $(m, n)$ -closed submodule of $S^{- 1} M$ and $Z_{(N :_{R} M)} (R) \cap S = Z_{N} (M) = ϕ,$ then N is an $(m, n)$ -closed submodule of M.

Proof.

(1) Let

{(\frac{r}{s_{1}})}^{m} (\frac{b}{s_{2}}) \in S^{- 1} N

for

\frac{r}{s_{1}} \in S^{- 1} R

and

\frac{b}{s_{2}} \in S^{- 1} M

. Then

{(u r)}^{m} b \in N

for some

u \in S

. Since N is

(m, n)

-closed, then either

{(u r)}^{n} \in (N :_{R} M)

b \in N

. Thus,

{(\frac{r}{s_{1}})}^{n} = \frac{u^{n} r^{n}}{u^{n} s_{1}^{n}} \in S^{- 1} (N :_{R} M) = (S^{- 1} N :_{S^{- 1} R} S^{- 1} M)

(\frac{b}{s_{2}}) \in S^{- 1} N

(2) Suppose that

r^{m} b \in N

for some

r \in R

and

b \in M

. Then

{(\frac{r}{1})}^{m} (\frac{b}{1}) \in S^{- 1} N

which implies either

{(\frac{r}{1})}^{n} \in (S^{- 1} N :_{S^{- 1} R} S^{- 1} M) = S^{- 1} (N :_{R} M)

(\frac{b}{1}) \in S^{- 1} N .

Thus,

u r^{n} \in (N :_{R} M)

v b \in N

for some

u, v \in S .

Since

Z_{(N :_{R} M)} (R) \cap S = Z_{N} (M) = \emptyset,

we have

r^{n} \in (N :_{R} M)

b \in N

and we are done. □

Proposition 21.

Let M be a P-

(m, n)

-closed R-module, S be a multiplicatively closed subset of R and

φ : M \to S^{- 1} M

be a natural homomorphism defined by

m \mapsto \frac{m}{1}

for all

m \in M

. Then φ is a monomorphism or

S^{- 1} M = 0

Proof.

Suppose that

φ

is not a monomorphism. Choose

0 \neq b \in K e r (φ)

. Since

φ (b) = \frac{b}{1} = 0

, there exists some

s \in S

such that

s b = 0

. Since

s^{m} b = 0

and 0 is

(m, n)

-closed, we have

s^{n} \in (0 :_{R} M) .

Thus,

s^{n} m = 0

for all

m \in M

and so

S^{- 1} M = 0 .

□

Proposition 22.

Let

M_{1}

and

M_{2}

be R-modules,

f : M_{1} \to M_{2}

be an R-module homomorphism and

n, m

be positive integers.

If $N_{2}$ is an $(m, n)$ -closed submodule of $M_{2}$ , then $f^{- 1} (N_{2})$ is an $(m, n)$ -closed submodule of $M_{1}$ .
If f is onto and $N_{1}$ is an $(m, n)$ -closed submodule of $M_{1}$ containing $K e r (f)$ , then $f (N_{1})$ is an $(m, n)$ -closed submodule of $M_{2} .$

Proof.

(1) Let

r^{m} b \in f^{- 1} (N_{2})

for some

r \in R

and

b \in M_{1} .

Then

r^{m} f (b) \in N_{2}

implies either

r^{n} \in (N_{2} :_{R} M_{2})

f (b) \in N_{2} .

Suppose

r^{n} \in (N_{2} :_{R} M_{2})

so that

r^{n} f (M_{1}) \subseteq N_{2}

. Then

r^{n} M_{1} \subseteq f^{- 1} (N_{2})

and so

r^{n} \in (f^{- 1} (N_{2}) :_{R} M_{1})

. If

f (b) \in N_{2}

, then

b \in f^{- 1} (N_{2})

, as required.

(2) Suppose that

r^{m} b_{2} \in f (N_{1})

for some

r \in R

and

b_{2} \in M_{2} .

Let

b_{2} : = f (b_{1})

for some

b_{1} \in M_{1} .

Then

r^{m} b_{1} \in N_{1}

K e r (f) \subseteq N_{1}

and so either

r^{n} \in (N_{1} :_{R} M_{1})

b_{1} \in N_{1} .

Therefore,

r^{n} M_{2} = r^{n} f (M_{1}) = f (r^{n} M_{1}) \subseteq f (N_{1})

b_{2} \in f (N_{1})

and

f (N_{1})

is an

(m, n)

-closed submodule of

M_{2} .

□

As a direct consequence of Proposition 22, we have the following.

Corollary 23.

Let R be a ring,

M_{1}

and

M_{2}

be R-modules and

n, m

be positive integers.

If $M_{1} \subseteq M_{2}$ and N is an $(m, n)$ -closed submodule of $M_{2},$ then $N \cap M_{1}$ is an $(m, n)$ -closed submodule of $M_{1} .$
Let $K \subseteq N$ be submodules of $M_{1}$ . Then $N / K$ is an $(m, n)$ -closed submodule of $M_{1} / K$ if and only if N is an $(m, n)$ -closed submodule of $M_{1}$ .

Proposition 24.

Let

{\{m_{i}, n_{i}\}}_{i = 1}^{k}

be positive integers and M be an R-module.

If ${\{N_{i}\}}_{i = 1}^{k}$ are P- $(m_{i}, n_{i})$ -closed submodules of an R-module M, then $⋂_{i = 1}^{k} N_{i}$ is a P- $(m, n)$ -closed submodule of M for all $m \leq min \{m_{1}, m_{2}, \dots, m_{k}\}$ and $n \geq max \{n_{1}, n_{2}, \dots, n_{k}\}$ .
If ${P_{i}}_{i \in I}$ is a family of prime submodules of M and $⋂_{i \in I} P_{i}$ is an $(m, n)$ -closed submodule of M (for $n, m \in N$ ), then $⋂_{i \in I} P_{i}$ is a prime submodule of M.

Proof.

(1) Suppose

N_{i}

is P-

(m_{i}, n_{i})

-closed in M for all

i \in \{1, 2, \dots, k\}

. Let

m \leq min \{m_{1}, m_{2}, \dots, m_{k}\}

such that

r^{m} b \in

⋂_{i = 1}^{k} N_{i}

and

b \notin

⋂_{i = 1}^{k} N_{i}

for

r \in R

and

b \in M

. Then

b \notin N_{j}

for some

j \in \{1, 2, \dots, k\}

. Since

r^{m_{j}} b \in N_{j}

, then by assumption

r^{n_{j}} \in (N_{j} :_{R} M)

and so

r \in \sqrt{(N_{j} :_{R} M)} = P

n \geq n_{j}

. By [13, Lemma 1], we have for all

i \in \{1, 2, \dots, k\}

P = \{r \in R : r^{n_{i}} \in (N_{i} :_{R} M)\}

. Thus,

r^{n} \in ⋂_{i = 1}^{k} (N_{i} :_{R} M) = (⋂_{i = 1}^{k} N_{i} :_{R} M)

n \geq max \{n_{1}, n_{2}, \dots, n_{k}\}

. Since also

\sqrt{⋂_{i = 1}^{k} (N_{i} :_{R} M)} = ⋂_{i = 1}^{k} \sqrt{(N_{i} :_{R} M)} = P

, then

⋂_{i = 1}^{k} N_{i}

is a P-

(m, n)

-closed submodule of M.

(2) Suppose that

r b \in ⋂_{i \in I} P_{i}

and

r \notin (⋂_{i \in I} P_{i} :_{R} M) = ⋂_{i \in I} (P_{i} :_{R} M) .

Then

r^{m} b \in ⋂_{i \in I} P_{i}

and

r \notin (P_{j} : M)

for some

j \in I

. Since

(P_{j} : M)

is a prime ideal of R, then

r^{n} \notin (P_{j} : M)

and so

r^{n} \notin (⋂_{i \in I} P_{i} :_{R} M) .

Since

⋂_{i \in I} P_{i}

(m, n)

-closed, we have

b \in

⋂_{i \in I} P_{i}

and we are done. □

Corollary 25.

Let N be a submodule of an R-module M and

n, m

be positive integers. Then M-rad

(N)

is an

(m, n)

-closed submodule of M if and only if M-rad

(N)

is prime.

In general, if

N_{1}

and

N_{2}

are two

(m, n)

-closed submodules of an R-module M with

\sqrt{(N_{1} :_{R} M)} \neq \sqrt{(N_{2} :_{R} M)}

, then

N_{1} \cap N_{2}

need not be

(m, n)

-closed, see [13], Remark 2.

Next, we discuss the

(m, n)

-closed submodules in a direct sum of modules.

Theorem 26.

Let

M_{1}

and

M_{2}

be R-modules,

N_{1}

and

N_{2}

be submodules of

M_{1}

and

M_{2}

, respectively and

n, m

be positive integers.

If $N_{1} \oplus N_{2}$ is an $(m, n)$ -closed submodule of $M_{1} \oplus M_{2}$ , then $N_{i}$ is an $(m, n)$ -closed submodule of $M_{i}$ for all $i = 1, 2$ such that $N_{i} \neq M_{i}$ .
If $N_{1}$ and $N_{2}$ are P- $(m, n)$ -closed submodules of $M_{1}$ and $M_{2}$ , respectively, then $N = N_{1} \oplus N_{2}$ is a P- $(m, n)$ -closed submodule of $M = M_{1} \oplus M_{2}$ .
Suppose $\sqrt{(N_{1} :_{R} M_{1})} \neq \sqrt{(N_{2} :_{R} M_{2})}$ . Then $N = N_{1} \oplus N_{2}$ is an $(m, n)$ -closed submodule of $M = M_{1} \oplus M_{2}$ if and only if one of the following statements holds:

$(a)$ $N = N_{1} \oplus M_{2}$ where $N_{1}$ is an $(m, n)$ -closed submodule of $M_{1}$ .

$(b)$ $N = M_{1} \oplus N_{2}$ where $N_{2}$ is an $(m, n)$ -closed submodule of $M_{2}$ .

Proof.

(1) Suppose

N_{1} \oplus N_{2}

is an

(m, n)

-closed submodule of

M_{1} \oplus M_{2}

and suppose, say,

N_{1} \neq M_{1}

. Let

r^{m} b \in N_{1}

such that

b \notin N_{1}

. Then

r^{m} (b, 0) \in N_{1} \oplus N_{2}

with

(b, 0) \notin N_{1} \oplus N_{2}

. Therefore,

r^{n} \in (N_{1} \oplus N_{2} :_{R} M_{1} \oplus M_{2})

and then clearly,

r^{n} \in (N_{1} :_{R} M_{1})

, as needed.

(2) Suppose that

N_{1}

and

N_{2}

are P-

(m, n)

-closed submodules of

M_{1}

and

M_{2}

, respectively. First, we show that

N_{1} \oplus M_{2}

is a P-

(m, n)

-closed submodule of M. Let

r \in R

and

(b_{1}, b_{2}) \in M

with

r^{m} (b_{1}, b_{2}) \in N_{1} \oplus M_{2}

. Then

r^{m} b_{1} \in N_{1}

which implies

r^{n} \in (N_{1} :_{R} M_{1})

b_{1} \in N_{1} .

Thus,

r^{n} \in (N_{1} \oplus M_{2} :_{R} M)

(b_{1}, b_{2}) \in N_{1} \oplus M_{2}

and

N_{1} \oplus M_{2}

is an

(m, n)

-closed submodule of M. By using the same manner, it can be verified that

M_{1} \oplus N_{2}

is a P-

(m, n)

-closed submodule of

M .

Since

N = N_{1} \oplus N_{2} = (N_{1} \oplus M_{2}) \cap (M_{1} \oplus N_{2})

, we are done by Proposition 24.

(3) Suppose

N = N_{1} \oplus N_{2}

is an

(m, n)

-closed submodule of

M .

Suppose that

N_{1}

is proper in

M_{1}

so that

(N_{1} :_{R} M_{1}) \neq R

. Choose

r \in \sqrt{(N_{2} :_{R} M_{2})} ∖ \sqrt{(N_{1} :_{R} M_{1})}

and let

b \in M_{2} .

Then

r^{k} \in (N_{2} :_{R} M_{2})

for some positive integer k and hence

{(r^{k})}^{m} (0, b) \in N

. It follows that

r^{k n} \in (N :_{R} M)

(0, b) \in N

. In the first case, we get

r^{k n} \in (N_{1} :_{R} M_{1})

which is a contradiction. Thus,

b \in N_{2}

and so

N_{2} = M_{2} .

Similarly, if

N_{2}

is proper in

M_{2}

, then we must have

N_{1} = M_{1}

. Without loss of generality, we may assume that

N_{2} = M_{2}

and show that

N_{1}

is an

(m, n)

-closed submodule of

M_{1}

. Let

r^{m} b \in N_{1}

such that

b \notin N_{1}

. Then

r^{m} (b, 0) \in N

with

(b, 0) \notin N

. This yields that

r^{n} \in (N :_{R} M)

and then clearly,

r^{n} \in (N_{1} :_{R} M_{1})

, as needed. The converse part follows by the proof of (2). □

The direct sum of two

(m, n)

-closed submodules need not to be an

(m, n)

-closed submodule in general. Let p and q be distinct prime integers. Then the submodules

N_{1} = p^{2} Z

and

N_{2} = q^{2} Z

are

(m, 2)

-closed in the

Z

-module

M = Z

for all

m \in N

, [13, Theorem 3]. However,

p^{2} Z \oplus q^{2} Z

is not an

(m, 2)

-closed submodule of

Z \oplus Z

q^{m} (p^{2}, 1) \in p^{2} Z \oplus q^{2} Z

for all

m \geq 2

, but neither

q^{2} \in (p^{2} Z \oplus q^{2} Z :_{Z} Z \oplus Z) = p^{2} q^{2} Z

nor

(p^{2}, 1) \in p^{2} Z \oplus q^{2} Z .

Note that

\sqrt{(N_{1} :_{Z} M)} \neq \sqrt{(N_{2} :_{Z} M)}

We next generalize Theorem 26 to a finite direct sum of submodules.

Theorem 27.

Let

M_{1}, M_{2}, \dots, M_{k}

be R-modules,

N_{1}, N_{2}, \dots, N_{k}

be submodules of

M_{1}, M_{2}, \dots, M_{k}

, respectively and

n, m

be positive integers.

If $N_{1} \oplus N_{2} \oplus \dots \oplus N_{k}$ is an $(m, n)$ -closed submodule of $M_{1} \oplus M_{2} \oplus \dots \oplus M_{k}$ , then $N_{i}$ is an $(m, n)$ -closed submodule of $M_{i}$ for all i such that $N_{i} \neq M_{i}$ .
If $N_{1}, . . ., N_{k}$ are P- $(m, n)$ -closed submodules of $M_{1}, M_{2}, . . ., M_{k}$ , respectively, then $N = N_{1} \oplus N_{2} \oplus \dots \oplus N_{k}$ is a P- $(m, n)$ -closed submodule of $M = M_{1} \oplus M_{2} \oplus \dots \oplus M_{k}$ .
Suppose $⋂_{i \neq j} \sqrt{(N_{i} :_{R} M_{i})} ⊈ \sqrt{(N_{j} :_{R} M_{j})}$ for some $j \in {1, 2, . . ., k}$ . Then $N = N_{1} \oplus N_{2} \oplus \dots \oplus N_{k}$ is an $(m, n)$ -closed submodule of $M = M_{1} \oplus M_{2} \oplus \dots \oplus M_{k}$ if and only if $N = M_{1} \oplus \dots \oplus N_{j} \oplus \dots \oplus M_{k}$ where $N_{j}$ is an $(m, n)$ -closed submodule of $M_{j}$ ( $j = 1, 2, \dots, k$ ).

Proof.

The proofs of (1) and (2) are similar to those of (1) and (2) in Theorem 26.

(3) With no loss of generality, we may assume that

j = 1

. We use mathematical induction on k. Clearly, the result is true for

k = 2

by (1) of Theorem 26. Suppose the result is true for k and let

N = N_{1} \oplus N_{2} \oplus \dots \oplus N_{k} \oplus N_{k + 1} = N^{'} \oplus N_{k + 1}

where

⋂_{i = 2}^{k + 1} \sqrt{(N_{i} :_{R} M_{i})} ⊈ \sqrt{(N_{1} :_{R} M_{1})}

. Suppose N is an

(m, n)

-closed submodule of

M = M_{1} \oplus M_{2} \oplus \dots \oplus M_{k} \oplus M_{k + 1} = M^{'} \oplus M_{k + 1}

. Now, clearly

\sqrt{(N^{'} :_{R} M^{'})} \neq \sqrt{(N_{k + 1} :_{R} M_{k + 1})}

. If

N^{'} = M^{'}

, then

(N_{1} :_{R} M_{1}) = R

which is impossible. Thus, again by Theorem 26, we have

N_{k + 1} = M_{k + 1}

and

N^{'}

(m, n)

-closed in

M^{'}

. By induction hypothesis, we get

N^{'} = N_{1} \oplus M_{2} \oplus \dots \oplus M_{k}

(and so

N = N_{1} \oplus M_{2} \oplus \dots \oplus M_{k} \oplus M_{k + 1}

) where

N_{1}

(m, n)

-closed in

M_{1}

. Conversely, suppose

N = N_{1} \oplus M_{2} \oplus \dots \oplus M_{k} \oplus M_{k + 1}

where

N_{1}

(m, n)

-closed in

M_{1}

. Then by induction hypothesis,

L = N_{1} \oplus M_{2} \oplus \dots \oplus M_{k}

(m, n)

-closed in

M_{1} \oplus M_{2} \oplus \dots \oplus M_{k}

. Thus,

N = L \oplus M_{k + 1}

(m, n)

-closed in

M_{1} \oplus M_{2} \oplus \dots \oplus M_{k} \oplus M_{k + 1}

by Theorem 26(2). □

Note that the sum of two

(m, n)

-closed submodules is not an

(m, n)

-closed submodule in general. Consider the

Z

-module

Z \oplus Z

and the submodules

N = p Z \oplus Z

and

N = Z \oplus q Z

where p and q are prime integers. By Theorem 26, both N and K are

(m, n)

-closed submodules of

Z \oplus Z

. However,

N + K = Z \oplus Z

is not

(m, n)

-closed.

Recall that the idealization ring of an R-module M is the set

R (+) M = R \oplus M = \{(r, m) : r \in R, m \in M\}

with coordinate-wise addition and multiplication defined as

(r_{1}, m_{1}) (r_{2}, m_{2}) = (r_{1} r_{2}, r_{1} m_{2} + r_{2} m_{1})

. If I is an ideal of R and N a submodule of M, then

I (+) N

is an ideal of

R (+) M

if and only if

I M \subseteq N

. Next, we justify the relationship between the

(m, n)

-closed submodules of M and the

(m, n)

-prime ideals of the idealization ring

R (+) M

Proposition 28.

Let I be an ideal of a ring R, N be a proper submodule of an R-module M and

n, m

be positive integers.

If $I (+) N$ is an $(m, n)$ -closed ideal of $R (+) M$ , then N is an $(m, n)$ -closed submodule of M and I is a $(m, n)$ -prime ideal of R.
If $I = (N :_{R} M)$ is a $(m, n)$ -prime ideal of R and N is an $(m, n)$ -closed submodule of M, then $I (+) N$ is an $(m, n + 1)$ -closed ideal of $R (+) M$ .

Proof.

(1) Suppose

I (+) N

is an

(m, n)

-closed ideal of

R (+) M

. Then I is a

(m, n)

-prime ideal of R by [13, Proposition 6]. Now, let

r \in R

and

b \in M

such that

r^{m} b \in N

and

r^{n} \notin (N :_{R} M)

. Then

{(r, 0)}^{m} (0, b) = (0, r^{m} b) \in I (+) N

and

{(r, 0)}^{n} \notin I (+) N

since otherwise

r^{n} M \subseteq I M \subseteq N

, a contradiction. Therefore,

(0, b) \in I (+) N

and so

b \in N

, as required.

(2) Suppose

I = (N :_{R} M)

. Let

(r, x), (s, y) \in R (+) M

such that

{(r, x)}^{m} (s, y) = (r^{m} s, r^{m} y + m r^{m - 1} s x) \in I (+) N

. Then

r^{m} s \in I

and

r^{m} y + m r^{m - 1} s x \in N

. We have two cases:

Case I:

r^{n} \in I

. In this case, we have

{(r, x)}^{n + 1} = (r^{n + 1}, (n + 1) r^{n} x) \in I (+) N

I M \subseteq N

Case II:

r^{n} \notin I = (N :_{R} M)

. Since I is

(m, n)

-prime in R, then

s \in I

and so

m r^{m - 1} s x \in I M \subseteq N

. Therefore,

r^{m} y \in N

. Since N is

(m, n)

-closed in M and

r^{n} \notin (N :_{R} M)

, then

y \in N

and so

(s, y) \in I (+) N

. Therefore,

I (+) N

is an

(m, n + 1)

-closed ideal of

R (+) M

. □

We note that if N is an

(m, n)

-closed submodule of M and I is a

(m, n)

-prime ideal of R, then

I (+) N

need not be

(m, n)

-closed in

R (+) M

. For example, while

2 Z

is a

(2, 1)

-prime ideal and

\bar{2} Z_{4}

is a

(2, 1)

-closed submodule of the

Z

-module

Z_{4}

, the ideal

2 Z (+) \bar{2} Z_{4}

is not

(2, 1)

-prime in

Z (+) Z_{4}

. Indeed,

{(2, \bar{1})}^{2} = (4, \bar{0}) \in 2 Z (+) \bar{2} Z_{4}

but

(2, \bar{1}) \notin 2 Z (+) \bar{2} Z_{4}

Recall that a ring (resp. a domain) R is called a ZPI-ring (resp. a Dedekind domain) if every proper ideal of R is a product of prime ideals. As a generalization of this structure, we define

(m, n)

-rings and

(m, n)

-modules.

Definition 29.

Let R be a ring, M be an R-module and

n, m

be positive integers.

We call R an $(m, n)$ -ring if every proper ideal I of R can be written as a finite product of $P_{i}$ - $(m, n)$ -prime ideals $I_{i}$ of $R,$ that is, $I = I_{1} I_{2} \dots I_{k}$ for some positive integer k.
We call M an $(m, n)$ -module if every proper submodule N of M is either an $(m, n)$ -closed submodule or $N = I_{1} I_{2} \dots I_{k} K$ where $I_{i}$ ’s are $P_{i}$ - $(m, n)$ -prime ideals of R and K is a Q- $(m, n)$ -closed submodule.

Theorem 30.

Let

n, m

be positive integers. Then every faithful multiplication module over an

(m, n)

-ring R is an

(m, n)

-module. In particular, a faithful multiplication module over a ZPI-ring is an

(m, n)

-module.

Proof.

Let N be a proper submodule of M. Since M is multiplication,

N = I M

for some ideal I of R. Now,

I = I_{1} I_{2} \dots I_{k}

where

I_{i}

is a

P_{i}

(m, n)

-prime ideal of R for

i = 1, 2, \dots, k

. Hence,

N =

I_{1} I_{2} \dots I_{k} M

where

I_{i} M \neq M

for some

i \in {1, . . ., k}

as N is proper. These proper submodules

I_{i} M

are

(m, n)

-closed by Proposition 16. Thus, N is either an

(m, n)

-closed submodule or it has an

(m, n)

-factorization. The particular part is clear as a ZPI-ring is an

(m, n)

-ring. □

In the following, we show that the converse of Theorem 30 also holds if M is finitely generated.

Theorem 31.

Let M be a finitely generated faithful multiplication R-module and

n, m

be positive integers. If M is an

(m, n)

-module, then R is an

(m, n)

-ring.

Proof.

Let I be a proper ideal of R. Then

I M

is a proper submodule of M since otherwise,

I = (I M :_{R} M) = R

by Lemma 14. By assumption,

I M = I_{1} I_{2} \dots I_{k} K

where

I_{i}

’s are

P_{i}

(m, n)

-prime ideals of R and K is a Q-

(m, n)

-closed submodule. Since M is multiplication,

K =

(K :_{R} M) M

and

(K :_{R} M)

is a

(m, n)

-prime ideal by Corollary 5. Thus,

I M =

I_{1} I_{2} \dots I_{k} (K :_{R} M) M

and

I = (I M : M) = N_{1} N_{2} \dots N_{k} (K :_{R} M)

again by Lemma 14. Therefore, R is an

(m, n)

-ring. □

We end up this section by the

(m, n)

-closed avoidance theorem. We assume in the rest of this section that M is a finitely generated multiplication R-module and

N,

N_{1}, . . ., N_{k}

are submodules of M. We recall that a covering

N \subseteq N_{1} \cup N_{2} \cup \dots \cup N_{n}

(resp.

N = N_{1} \cup N_{2} \cup \dots \cup N_{n})

is said to be an efficient covering (resp. efficient union) if no

N_{k}

is superfluous (resp. excluded), [17]. A covering (union) of a submodule by two submodules is never efficient.

Theorem 32.

Let

N \subseteq N_{1} \cup N_{2} \cup \dots \cup N_{k}

be an efficient covering of submodules

N_{1}, N_{2}, . . ., N_{k}

of an R-module M where

k ≩ 2 .

Suppose that

\sqrt{(N_{i} :_{R} M)} ⊈ \sqrt{(N_{j} : b)}

for all

b \in M ∖ M

r a d (N_{i})

whenever

i \neq j

. Then no

N_{i}

(

1 \leq i \leq k

) is an

(m, n)

-closed submodule of M for all positive integers

n, m

such that

n \leq m .

Proof.

Suppose on contrary that

N_{j}

is an

(m, n)

-closed submodule of M for some

1 \leq j \leq k

. It can be easily observed that

N \subseteq ⋃_{i = 1}^{k} (N_{i} \cap N)

is an efficient covering as

N \subseteq ⋃_{i = 1}^{k} N_{i}

is efficient. From [17, Lemma 2.1], we conclude the following inclusion

(⋂_{i \neq j} N_{i}) \cap N = (⋂_{i = 1}^{k} N_{i}) \cap N \subseteq N_{j} \cap N

Choose

r_{i} \in \sqrt{(N_{i} : M)} ∖ \sqrt{(N_{j} : b)}

where

i \neq j

and

b \in M ∖ M

r a d (N_{j})

. Then, there exists the least positive integer

m_{i}

such that

r_{i}^{m_{i}} \in (N_{i} :_{R} M)

for each

i \neq j

. Set

r : = r_{1} r_{2} \dots r_{j - 1}

s : = r_{j + 1} r_{j + 2} \dots r_{k}

and

m = max {m_{1}, m_{2}, . . ., m_{j - 1}, m_{j + 1}, . . ., m_{n}}

. Then

r^{m} s^{m} b \in (⋂_{i \neq j} N_{i}) \cap N .

We show that

r^{m} s^{m} b \notin (N_{j} \cap N)

. For this purpose, assume on the opposite that

r^{m} s^{m} b \in N_{j} \cap N .

Then

r^{m} s^{m} \in (N_{j} :_{R} b) \subseteq \sqrt{(N_{j} :_{R} b)} .

Since

\sqrt{(N_{j} : b)}

is a prime ideal and

r_{1}^{m} r_{2}^{m} \dots r_{j - 1}^{m} r_{j + 1}^{m} r_{j + 2}^{m} \dots r_{k}^{m} \in \sqrt{(N_{j} : b)},

we conclude

r_{i} \in \sqrt{(N_{j} :_{R} b)}

for some

i \neq j

, a contradiction. Consequently,

r^{m} s^{m} b \notin (N_{j} \cap N),

and so

r^{m} s^{m} b \in ((⋂_{i \neq j} N_{i}) \cap N) ∖ (N_{j} \cap N)

which contradicts the inclusion

(⋂_{i \neq j} N_{i}) \cap N \subseteq N_{j} \cap N

. Therefore, no

N_{i}

is an

(m, n)

-closed submodule. □

Theorem 33.

(

(m, n)

-closed avoidance Theorem) Let

n, m

be positive integers and let N,

N_{1}, N_{2}, . . ., N_{k}

(k \geq 2)

be submodules of an R-module M such that at most two of

N_{1}, N_{2}, . . ., N_{k}

are not

(m, n)

-closed and

\sqrt{(N_{i} :_{R} M)} ⊈ \sqrt{(N_{j} : b)}

for all

b \in M ∖ M

r a d (N_{i})

whenever

i \neq j

. If

N \subseteq N_{1} \cup N_{2} \cup \dots \cup N_{k}

, then

N \subseteq N_{j}

for some

1 \leq j \leq k .

Proof.

Since a covering of an ideal by two ideals is never efficient, assume that

k \geq 3 .

As any covering can be reduced to an efficient covering by omitting any unnecessary terms, we may suppose that

N \subseteq N_{1} \cup N_{2} \cup \dots \cup N_{k}

is an efficient covering of ideals of R. From Theorem 32, no

N_{j}

is an

(m, n)

-closed submodule of M. However, our assumption states that at most two of

N_{i}

’s are not

(m, n)

-closed. Consequently,

N \subseteq N_{j}

for some

1 \leq j \leq k .

□

3. $(m, n)$ -closed Submodules of Amalgamation Modules

Let R be a ring, J an ideal of R and M an R-module. The amalgamated duplication of R along J is defined as

R ⋈ J = \{(r, r + j) : r \in R, j \in J\}

which is a subring of

R \times R

, see [9]. The duplication of the R-module M along the ideal J denoted by

M ⋈ J

is defined recently in [8] as

M ⋈ J = \{(b, b^{'}) \in M \times M : b - b^{'} \in J M\}

which is an

(R ⋈ J)

-module with scalar multiplication defined by

(r, r + j) (b, b^{'}) = (r b, (r + j) b^{'})

for

r \in R

j \in J

and

(b, b^{'}) \in M ⋈ J

. Many properties and results concerning this kind of modules can be found in [8].

Let N be a submodule of an R-module M and J be an ideal of R. Then clearly

N ⋈ J = \{(a, b) \in N \times M : a - b \in J M\}

and

\bar{N} = \{(b, a) \in M \times N : b - a \in J M\}

are submodules of

M ⋈ J

In general, let

f : R_{1} \to R_{2}

be a ring homomorphism, J be an ideal of

R_{2}

M_{1}

be an

R_{1}

-module,

M_{2}

be an

R_{2}

-module (which is an

R_{1}

-module induced naturally by f) and

φ : M_{1} \to M_{2}

be an

R_{1}

-module homomorphism. The subring

R_{1} ⋈^{f} J = \{(r, f (r) + j) : r \in R_{1}, j \in J\}

R_{1} \times R_{2}

is called the amalgamation of

R_{1}

and

R_{2}

along J with respect to f. In [12], the amalgamation of

M_{1}

and

M_{2}

along J with respect to

φ

is defined as

M_{1} ⋈^{φ} J M_{2} = \{(b_{1}, φ (b_{1}) + b_{2}) : b_{1} \in M_{1} and b_{2} \in J M_{2}\}

which is an

(R_{1} ⋈^{f} J)

-module with the scalar product defined as

(r, f (r) + j) (b_{1}, φ (b_{1}) + b_{2}) = (r b_{1}, φ (r b_{1}) + f (r) b_{2} + j φ (b_{1}) + j b_{2})

For submodules

N_{1}

and

N_{2}

M_{1}

and

M_{2}

, respectively, clearly the sets

N_{1} ⋈^{φ} J M_{2} = \{(b_{1}, φ (b_{1}) + b_{2}) \in M_{1} ⋈^{φ} J M_{2} : b_{1} \in N_{1}\}

and

{\bar{N_{2}}}^{φ} = \{(b_{1}, φ (b_{1}) + b_{2}) \in M_{1} ⋈^{φ} J M_{2} : φ (b_{1}) + b_{2} \in N_{2}\}

are submodules of

M_{1} ⋈^{φ} J M_{2}

In the following two theorems, we justify conditions under which

N_{1} ⋈^{φ} J M_{2}

and

{\bar{N_{2}}}^{φ}

are

(m, n)

-closed in

M_{1} ⋈^{φ} J M_{2}

Theorem 34.

Consider the

(R_{1} ⋈^{f} J)

-module

M_{1} ⋈^{φ} J M_{2}

defined as above. Let

N_{1}

be a submodule of

M_{1}

and

n, m

be positive integers. Then

N_{1} ⋈^{φ} J M_{2}

is an

(m, n)

-closed submodule of

M_{1} ⋈^{φ} J M_{2}

if and only if

N_{1}

is an

(m, n)

-closed submodule of

M_{1}

Proof.

First, we note that

N_{1}

is a proper submodule of

M_{1}

if and only if

N_{1} ⋈^{φ} J M_{2}

is a proper submodule of

M_{1} ⋈^{φ} J M_{2}

Suppose

N_{1} ⋈^{φ} J M_{2}

(m, n)

-closed in

M_{1} ⋈^{φ} J M_{2}

and let

r_{1}^{m} b_{1} \in N_{1}

for

r_{1} \in R_{1}

and

b_{1} \in M_{1}

. Then

(r_{1}, f (r_{1})) \in R_{1} ⋈^{f} J

and

(b_{1}, φ (b_{1})) \in M_{1} ⋈^{φ} J M_{2}

with

{(r_{1}, f (r_{1}))}^{m} (b_{1}, φ (b_{1})) = (r_{1}^{m} b_{1}, φ (r_{1}^{m} b_{1})) \in N_{1} ⋈^{φ} J M_{2}

. Thus, either

{(r_{1}, f (r_{1}))}^{n} \in (N_{1} ⋈^{φ} J M_{2} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

(b_{1}, φ (b_{1})) \in N_{1} ⋈^{φ} J M_{2}

. In the first case, for all

b \in M_{1}

{(r_{1}, f (r_{1}))}^{n} (b, φ (b)) \in N_{1} ⋈^{φ} J M_{2}

and so

r_{1}^{n} M_{1} \subseteq N_{1}

. In the second case,

b_{1} \in N_{1}

and so

N_{1}

is an

(m, n)

-closed submodule of

M_{1}

. Conversely, let

(r_{1}, f (r_{1}) + j) \in R_{1} ⋈^{f} J

and

(b_{1}, φ (b_{1}) + b_{2}) \in M_{1} ⋈^{φ} J M_{2}

such that

{(r_{1}, f (r_{1}) + j)}^{m} (b_{1}, φ (b_{1}) + b_{2}) \in N_{1} ⋈^{φ} J M_{2}

. Then

r_{1}^{m} b_{1} \in N_{1}

and hence either

r_{1}^{n} \in (N_{1} :_{R_{1}} M_{1})

b_{1} \in N_{1}

. If

r_{1}^{n} \in (N_{1} :_{R_{1}} M_{1})

, then clearly

{(r_{1}, f (r_{1}) + j)}^{n} \in (N_{1} ⋈^{φ} J M_{2} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

and if

b_{1} \in N_{1}

, then

(b_{1}, φ (b_{1}) + b_{2}) \in N_{1} ⋈^{φ} J M_{2}

. Therefore,

N_{1} ⋈^{φ} J M_{2}

is an

(m, n)

-closed submodule of

M_{1} ⋈^{φ} J M_{2}

. □

Theorem 35.

Consider the

(R_{1} ⋈^{f} J)

-module

M_{1} ⋈^{φ} J M_{2}

defined as in Theorem 34 where f and φ are epimorphisms. Let

N_{2}

be a submodule of

M_{2}

and

n, m

be positive integers. Then

$N_{2}$ is an $(m, n)$ -closed submodule of $M_{2}$ if and only if ${\bar{N_{2}}}^{φ}$ is an $(m, n)$ -closed submodule of $M_{1} ⋈^{φ} J M_{2}$ .
If ${\bar{N_{2}}}^{φ}$ is an $(m, n)$ -closed submodule of $M_{1} ⋈^{φ} J M_{2}$ and $J ⊈ (N_{2} :_{R_{2}} M_{2})$ , then $(N_{2} :_{M_{2}} J)$ is an $(m, n)$ -closed submodule of $M_{2}$ .

Proof.

(1) Suppose

N_{2}

is an

(m, n)

-closed submodule of

M_{2}

. Suppose

{\bar{N_{2}}}^{φ} = M_{1} ⋈^{φ} J M_{2}

and let

b_{2} = φ (b_{1}) \in M_{2}

for some

b_{1} \in M_{1}

. Then

(b_{1}, b_{2}) \in M_{1} ⋈^{φ} J M_{2} = {\bar{N_{2}}}^{φ}

and so

b_{2} \in N_{2}

. Thus,

N_{2} = M_{2}

which is a contradiction. Therefore,

{\bar{N_{2}}}^{φ}

is proper in

M_{1} ⋈^{φ} J M_{2}

. Let

(r_{1}, f (r_{1}) + j) \in R_{1} ⋈^{f} J

and

(b_{1}, φ (b_{1}) + b_{2}) \in M_{1} ⋈ J M_{2}

such that

{(r_{1}, f (r_{1}) + j)}^{m} (b_{1}, φ (b_{1}) + b_{2}) \in {\bar{N_{2}}}^{φ}

. Then

{(f (r_{1}) + j)}^{m} (φ (b_{1}) + b_{2}) \in N_{2}

and so

{(f (r_{1}) + j)}^{n} \in (N_{2} :_{R_{2}} M_{2})

(φ (b_{1}) + b_{2}) \in N_{2}

. If

{(f (r_{1}) + j)}^{n} \in (N_{2} :_{R_{2}} M_{2})

, then for all

(b_{1}, φ (b_{1}) + b_{2}) \in M_{1} ⋈^{φ} J M_{2}

, clearly

{(r_{1}, f (r_{1}) + j)}^{n} (b_{1}, φ (b_{1}) + b_{2}) \in {\bar{N_{2}}}^{φ}

and so

{(r_{1}, f (r_{1}) + j)}^{n} \in ({\bar{N_{2}}}^{φ} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

. If

(φ (b_{1}) + b_{2}) \in N_{2}

, then

(b_{1}, φ (b_{1}) + b_{2}) \in {\bar{N_{2}}}^{φ}

and the result follows.

Conversely, suppose

{\bar{N_{2}}}^{φ}

is an

(m, n)

-closed submodule of

M_{1} ⋈^{φ} J M_{2}

. Clearly,

N_{2}

is proper in

M_{2}

. Let

r_{2} = f (r_{1}) \in R_{2}

and

b_{2} = φ (b_{1}) \in M_{2}

such that

r_{2}^{m} b_{2} \in N_{2}

. Then

(r_{1}, r_{2}) \in R_{1} ⋈^{f} J

and

(b_{1}, b_{2}) \in M_{1} ⋈^{φ} J M_{2}

with

{(r_{1}, r_{2})}^{m} (b_{1}, b_{2}) \in {\bar{N_{2}}}^{φ}

. Thus,

{(r_{1}, r_{2})}^{n} \in ({\bar{N_{2}}}^{φ} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

(b_{1}, b_{2}) \in {\bar{N_{2}}}^{φ}

. If

{(r_{1}, r_{2})}^{n} (M_{1} ⋈^{φ} J M_{2}) \subseteq {\bar{N_{2}}}^{φ}

, then for all

b = φ (b^{'}) \in M_{2}

, we have

{(r_{1}, r_{2})}^{n} (b^{'}, b) \in {\bar{N_{2}}}^{φ}

and so

r_{2}^{n} \in (N_{2} :_{R_{2}} M_{2})

. If

(b_{1}, b_{2}) \in {\bar{N_{2}}}^{φ}

, then

b_{2} \in N_{2}

and we are done.

(2) Since

J ⊈ (N_{2} :_{R_{2}} M_{2})

(N_{2} :_{M_{2}} J)

is proper in

M_{2}

. Suppose

{\bar{N_{2}}}^{φ}

is an

(m, n)

-closed submodule of

M_{1} ⋈^{φ} J M_{2}

. Let

r_{2} = f (r_{1}) \in R_{2}

b_{2} \in M_{2}

such that

r_{2}^{m} b_{2} \in (N_{2} :_{M_{2}} J)

. Then

r_{2}^{m} J b_{2} \subseteq N_{2}

and so we have

{(r_{1}, r_{2})}^{m} (0, J b_{2}) \subseteq {\bar{N_{2}}}^{φ}

. By assumption,

{(r_{1}, r_{2})}^{n} \in ({\bar{N_{2}}}^{φ} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

(0, J b_{2}) \subseteq {\bar{N_{2}}}^{φ}

. If

{(r_{1}, r_{2})}^{n} \in ({\bar{N_{2}}}^{φ} :_{R_{1} ⋈^{f} J} M_{1} ⋈^{φ} J M_{2})

, then simple computations give

r_{2}^{n} \in (N_{2} :_{R_{2}} J M_{2}) = ((N_{2} :_{M_{2}} J) :_{R_{2}} M_{2})

. If

(0, J b_{2}) \subseteq {\bar{N_{2}}}^{φ}

, then

J b_{2} \subseteq N_{2}

and so

b_{2} \in (N_{2} :_{M_{2}} J)

, as required. □

In particular, we have the following corollaries of the previous two theorems.

Corollary 36.

Let N be a submodule of an R-module M, J be an ideal of R and

n, m

be positive integers. Then

N ⋈ J

is an

(m, n)

-closed submodule of

M ⋈ J

if and only if N is an

(m, n)

-closed submodule of M.

Corollary 37.

Let N be a submodule of an R-module M, J be an ideal of R and

n, m

be positive integers. Then

N is an $(m, n)$ -closed submodule of M if and only if $\bar{N}$ is an $(m, n)$ -closed submodule of $M ⋈ J$ .
If $\bar{N}$ is an $(m, n)$ -closed submodule of $M ⋈ J$ and $J ⊈ (N :_{R} M)$ , then $(N :_{M} J)$ is an $(m, n)$ -closed submodule of M.

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(m, n)-Closed Submodules of Modules over Commutative Rings

Abstract

1. Introduction

2. ( m , n ) -closed Submodules

3. ( m , n ) -closed Submodules of Amalgamation Modules

References

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2. $(m, n)$ -closed Submodules

3. $(m, n)$ -closed Submodules of Amalgamation Modules