International Journal of Analysis and Applications

Volume 16, Number 3 (2018), 400-413

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-16-2018-400

ON WEAKLY 2-ABSORBING SEMI-PRIMARY SUBMODULES OF MODULES OVER

COMMUTATIVE RINGS

PAIROTE YIARAYONG∗ AND MANOJ SIRIPITUKDET

Department of Mathematics, Faculty of Science, Naresuan University, Phitsanuloke 65000, Thailand

∗Corresponding author: pairote0027@hotmail.com

Abstract. Let R be a commutative ring with identity and let M be a unitary R-module. We say that

a proper submodule N of M is a weakly 2-absorbing semi-primary submodule if a1,a2 ∈ R,m ∈ N with

0 6= a1a2m ∈ N, then a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some positive integer n. In

this paper, we study weakly 2-absorbing semi-primary submodules and we prove some basic properties of

these submodules. Also, we give a characterization of weakly 2-absorbing semi-primary submodules and we

investigate weakly 2-absorbing semi-primary submodules of some well-known modules.

1. Introduction

Throughout this paper, we assume that all rings are commutative with 1 6= 0. Let R be a commutative

ring and let M be an R-module. We will denote by (N : M) a residual of N by M, that is, the set of all

r ∈ R such that rM ⊆ N. Clearly,
√
I = {r ∈ R : rn ∈ I for some positive integer n} denotes the radical

ideal of R.

In 2003, Anderson and Smith [1] introduced the concept of a weakly prime ideal of a commutative ring.

They said that a proper ideal P of the commutative ring R is weakly prime if a, b ∈ R and 0 6= ab ∈ P ,

then a ∈ P or b ∈ P . A weakly primary ideals were first introduced and studied by Atani and Farzalipour

in [2]. Recall that a proper ideal P of R is called a weakly primary ideal of R as in [2] if for a, b ∈ R with

0 6= ab ∈ P , then a ∈ P or bn ∈ P for some positive integer n. Clearly, a weakly prime ideal of R is also a

2010 Mathematics Subject Classification. 13A15,13F05.

Key words and phrases. weakly 2-absorbing semi-primary submodule; 2-absorbing semi-primary submodule; absorbing semi-

primary triple-zero; weakly 2-absorbing primary ideal.

c©2018 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

400

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-400


Int. J. Anal. Appl. 16 (3) (2018) 401

weakly primary ideal of R. The concept of weakly 2-absorbing ideals, which is a generalization of 2-absorbing

ideals, was introduced by Badawi and Darani in [3]. Recall from [3] that a proper ideal I of R is said to be

a weakly 2-absorbing ideal of R if whenever a, b, c ∈ R with 0 6= abc ∈ I, then ab ∈ I or ac ∈ I or bc ∈ I.

In [4], Badawi et. al. defined a proper ideal I of a commutative ring R to be a weakly 2-absorbing primary

ideal if whenever a, b, c ∈ R and 0 6= abc ∈ I, then ab ∈ I or ac ∈
√
I or bc ∈

√
I.

The concept of weakly prime submodule was introduced and studied by Behboodi and Koohi [5]. We recall

that a proper submodule N of M is called a weakly prime submodule, if 0 6= rm ∈ N, where r ∈ R, m ∈ M,

then m ∈ N or r ∈ (N : M). The idea of decomposition of submodules into weakly primary submodules

were introduced by Atani and Farzalipour in [2]. A weakly primary submodule N of M to be a proper

submodule of M and if r ∈ R, m ∈ M and 0 6= rm ∈ N, then m ∈ N or rn ∈ (N : M) for some positive

integer n. Clearly, every primary submodule of a module is a weakly primary submodule. In [6], the concept

of weakly 2-absorbing submodule generalized to 2-absorbing submodule of a module over a commutative

ring. A proper submodule N of M is called a weakly 2-absorbing submodule, if whenever a, b ∈ R and

m ∈ M with 0 6= abm ∈ N, then ab ∈ (N : M) or am ∈ N or bm ∈ N. In 2016, Mostafanasab et al. [11]

introduced the concept of weakly 2-absorbing primary submodules of modules over commutative rings with

identities. Recall that a proper submodule N of M is called a weakly 2-absorbing primary submodule of M

as in [11] if whenever 0 6= abm ∈ N for some a, b ∈ R and m ∈ M, then ab ∈ (N : M) or am ∈ M −rad(N)

or bm ∈ M − rad(N). The concept of weakly classical prime submodule, which is a generalization of

classical prime submodule, was introduced by Mostafanasab et al. in [10]. Recall from [10] that a proper

submodule N of M is said to be a weakly classical prime submodule of M if whenever a, b ∈ R and m ∈ M

with 0 6= abm ∈ N, then am ∈ N or bm ∈ N. The concept of weakly classical primary submodule, a

generalization of primary submodules was introduced and investigated in [9]. He weakly classical primary

submodule N of M to be a proper submoduleof R and if a, b ∈ R and 0 6= abm ∈ N, then am ∈ N or

mbn ∈ N for some positive integer n.

Motivated and inspired by the above works, the purposes of this paper are to introduce generalization-

s of weakly 2-absorbing primary submodule to the context of weakly 2-absorbing semi-primary submod-

ule. A proper submodule N of M to be a weakly 2-absorbing semi-primary submodule of M if whenever

0 6= a1a2m ∈ N for a1, a2 ∈ R, m ∈ M, then a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some positive

integer n. Some characterizations of weakly 2-absorbing semi-primary submodules are obtained. More-

over, we investigate relationships between 2-absorbing semi-primary and weakly 2-absorbing semi-primary

submodules of modules over commutative rings.



Int. J. Anal. Appl. 16 (3) (2018) 402

2. Properties of weakly 2-Absorbing Semiprimary Submodules

The results of the following theorems seem to play an important role to study weakly 2-absorbing semi-

primary submodules of modules over commutative rings; these facts will be used frequently and normally

we shall make no reference to this definition.

Definition 2.1. A proper submodule N of an R-module M is called a weakly 2-absorbing semi-primary

(2-absorbing semi-primary) submodule, if for each m ∈ M and a1, a2 ∈ R, 0 6= a1a2m ∈ N(a1a2m ∈ N),

then a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some positive integer n.

Remark 2.1. It is easy to see that every weakly 2-absorbing primary submodule (2-absorbing semi-primary)

submodule is weakly 2-absorbing semi-primary submodule.

The following example shows that the converse of Definition 2.1 is not true.

Example 2.1. Let R = Z and M = Z. Consider the submodule N = 12Z of M. It is easy to see that N is

a 2-absorbing semi-primary submodule of M. Notice that 2 · 2 · 3 ∈ N, but 2 · 3 6∈ N and (2 · 2)n 6∈ (N : M)

for all positive integer n. Therefore N is not a 2-absorbing primary submodule of M.

Example 2.2. Let R = Z and M = Z30. Consider the submodule N = {[0]} of M. It is easy to see that N is

a weakly 2-absorbing semi-primary submodule of M. Notice that (2·3)[5] ∈{[0]}, but 2·3 6∈
√

(N : M), 2[5] 6∈

{[0]} and 3n[5] 6∈ {[0]} for all positive integer n. Therefore N is not a 2-absorbing semi-primary submodule

of M.

Theorem 2.1. Let N be a proper submodule of an R-module M. Then the following statements hold:

(1) If N is a weakly 2-absorbing semi-primary submodule of M, then (N : m) is a weakly 2-absorbing

primary ideal of R for every m ∈ M −N.

(2) For every m ∈ M − N if (N : m) is a weakly primary ideal of R, then N is a weakly 2-absorbing

semi-primary submodule of M.

Proof. 1. Let a1, a2, a3 ∈ R such that 0 6= a1a2a3 ∈ (N : m). Clearly, 0 6= a1a3(a2m) ∈ N. By Definition

2.1, a1a3 ∈
√

(N : M) ⊆
√

(N : m) or a1a2m ∈ N or an3 a2m ∈ N for some positive integer n. Therefore

a1a2 ∈ (N : m) or a2a3 ∈
√

(N : m) or a1a3 ∈
√

(N : m). Hence (N : m) is a weakly 2-absorbing primary

ideal of R.

2. Let a1, a2 ∈ R such that 0 6= a1a2m ∈ N. Then 0 6= a1a2 ∈ (N : m). By assumption, a1 ∈ (N : m)

or an2 ∈ (N : m) for some positive integer n. Therefore a1m ∈ N or an2 m ∈ N for some positive integer n.

Hence N is a weakly 2-absorbing semi-primary submodule of M. �

But the converse of the above theorem is not true. For every m ∈ M − N, if (N : m) is weakly 2-

absorbing primary ideal, then N may not be weakly 2-absorbing semi-primary. Let M = Z × Z × Z be an



Int. J. Anal. Appl. 16 (3) (2018) 403

Z-module. Consider the submodule N = {0}× 6Z × Z of M. Clearly, (N : (m1, m2, m3)) = {0} is a weakly

2-absorbing primary ideal of R, where (m1, m2, m3) ∈ M −N. Notice that (0, 0, 0) 6= (2 ·3)(0, 1, 1) ∈ N, but

2 · 3 6∈
√

(N : M), 2(0, 1, 1) 6∈ N and 3n(0, 1, 1) 6∈ N for all positive integer n. Therefore N is not a weakly

2-absorbing semi-primary submodule of M.

Theorem 2.2. If N is a weakly 2-absorbing semi-primary submodule of an R-module M, then (N : r) is a

weakly 2-absorbing semi-primary submodule of M containing N for every r ∈ R − (N : M).

Proof. Let a1, a2 ∈ R and m ∈ M such that 0 6= a1a2m ∈ (N : r). Then 0 6= a1a2(rm) = ra1a2m ∈ N.

By Definition 2.1, a1a2 ∈
√

(N : M) or a1rm ∈ N or an2 rm ∈ N for some positive integer n. Therefore

a1a2 ∈
√

(N : M) or a1m ∈ (N : r) or an2 ∈ (N : r) for some positive integer n. Hence (N : r) is a weakly

2-absorbing semi-primary submodule of M. �

Theorem 2.3. Let {0} be a 2-absorbing semi-primary submodule of an R-module M. Then N is a weakly

2-absorbing semi-primary submodule of M if and only if N is a 2-absorbing semi-primary submodule of M.

Proof. Suppose that N is a 2-absorbing semi-primary submodule of M. Clearly, N is a weakly 2-absorbing

semi-primary submodule of M.

Conversely, assume that N is a weakly 2-absorbing semi-primary submodule of M. Let a1, a2 ∈ R and

m ∈ M such that a1a2m ∈ N. If a1a2m 6∈ {0}, then 0 6= a1a2m ∈ N. By Definition 2.1, a1a2 ∈
√

(N : M)

or a1m ∈ N or an2 m ∈ N for some positive integer n. Now if a1a2m ∈ {0}, then a1a2 ∈
√

(N : M) or

a1m ∈ N or an2 m ∈ N for some positive integer n. Hence N is a 2-absorbing semi-primary submodule of

M. �

Theorem 2.4. Let M and Ḿ be two R-modules and f : M → Ḿ be an epimorphism of an R-module.

If N is a weakly 2-absorbing semi-primary submodule of M such that kerf ⊆ N, then f(N) is a weakly

2-absorbing semi-primary submodule of Ḿ.

Proof. Let a1, a2 ∈ R and ḿ ∈ Ḿ such that 0 6= a1a2ḿ ∈ f(N). Thus 0 6= a1a2ḿ = ḿ0 for some ḿ0 ∈ f(N).

Since f is an epimorphism, there exist m ∈ M and m0 ∈ N such that ḿ = f(m) and ḿ0 = f(m0). This

implies that 0 6= a1a2f(m) = f(m0). Therefore f(a1a2m−m0) = 0 and so a1a2m−m0 ∈ kerf ⊆ N. Also,

0 6= a1a2m ∈ N, because if a1a2m = 0, then m0 ∈ kerf. It follows that f(m0) = 0, a contradiction. Now,

since N is a weakly 2-absorbing semi-primary, we have a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some

positive integer n. Therefore a1a2 ∈
√

(f(N) : Ḿ) or a1ḿ ∈ f(N) or an2 ḿ ∈ f(N) for some positive integer

n. Hence f(N) is a 2-absorbing semi-primary submodule of Ḿ. �

Theorem 2.5. Let M be an R-module and N ⊆ K be two submodules of M. If K is a weakly 2-absorbing

semi-primary submodule of M, then K/N is a weakly 2-absorbing semi-primary submodule of M/N.



Int. J. Anal. Appl. 16 (3) (2018) 404

Proof. Let a1, a2 ∈ R and m ∈ M such that N 6= a1a2(m + N) ∈ (K/N). Then 0 6= a1a2m ∈ K.

By Definition 2.1, a1a2 ∈
√

(K : M) or a1m ∈ K or an2 m ∈ K for some positive integer n. Therefore

a1a2 ∈
√

(K/N : M/N) or a1(m + N) ∈ K/N or an2 (m + N) ∈ K/N for some positive integer n. Hence

K/N is a weakly 2-absorbing semi-primary submodule of M/N. �

Theorem 2.6. Let M be an R-module and N ⊆ K be two submodules of M. Suppose that N is a weakly 2-

absorbing semi-primary submodule of M. If K/N is a weakly 2-absorbing semi-primary submodule of M/N,

then K is a weakly 2-absorbing semi-primary submodule of M.

Proof. Let a1, a2 ∈ R and m ∈ M such that 0 6= a1a2m ∈ K. If a1a2m ∈ N, then 0 6= a1a2m ∈ N. By

Definition 2.1, a1a2 ∈
√

(N : M) ⊆
√

(K : M) or a1m ∈ N ⊆ K or an2 m ∈ N ⊆ K for some positive integer

n. If a1a2m 6∈ N, then N 6= a1a2(m + N) ∈ N. Again, by Definition 2.1, a1a2 ∈
√

(K/N : M/N) or

a1(m + N) ∈ K/N or an2 (m + N) ∈ K/N for some positive integer n. Thus a1a2 ∈
√

(K : M) or a1m ∈ K or

an2 m ∈ K for some positive integer n. Hence K is a weakly 2-absorbing semi-primary submodule of M. �

Corollary 2.1. Then N is a weakly 2-absorbing semi-primary submodule of an R-module M if and only if

N/{0} is a weakly 2-absorbing semi-primary submodule of an R-module M/{0}.

Proof. It is straightforward by Theorem 2.5 and Theorem 2.6. �

Theorem 2.7. Let N be a submodule of an R-module M and S be a multiplicative subset of R. If N is

a weakly 2-absorbing semi-primary submodule of M such that (N : M) ∩ S = ∅, then S−1N is a weakly

2-absorbing semi-primary submodule of S−1M.

Proof. Clearly, S−1N is a proper submodule of S−1M. Let a1, a2 ∈ R, s1, s2, s3 ∈ S and m ∈ M such that

0 6= a1
s1

a2
s2

m
s3
∈ S−1N. Then there exists s ∈ S such that sa1a2m ∈ N. If sa1a2m = 0, then a1s1

a2
s2

m
s3

=

sa1
ss1

a2
s2

m
s3

= 0
1
, a contradiction. If sa1a2m 6= 0, then 0 6= a1a2(sm) ∈ N. By Definition 2.1, a1a2 ∈

√
(N : M)

or a1sm ∈ N or an2 sm ∈ N for some positive integer n. Thus
a1
s1

a2
s2
∈
√

(S−1N : S−1M) or a1
s1

m
s3

= a1sm
s1s3s

∈

S−1N or ( a2
s2

)n m
s3

=
an2 sm
sn2 s3s

∈ S−1N for some positive integer n. Hence S−1N is a weakly 2-absorbing

semi-primary submodule of S−1M. �

Theorem 2.8. Let N be a submodule of an R-module M and S be a multiplicative subset of R. If S−1N is

a weakly 2-absorbing semi-primary submodule of S−1M such that S ∩ Zd(N) = ∅ and S ∩ Zd(M/N) = ∅,

then N is a weakly 2-absorbing semi-primary submodule of M.

Proof. Let a1, a2 ∈ R and m ∈ M such that 0 6= a1a2m ∈ N. Then a11
a2
1

m
1
∈ S−1N. If a1

1
a2
1

m
1

= 0
1
, then

there exists s ∈ S such that sa1a2m = 0 which is a contradiction. If a11
a2
1

m
1
6= 0

1
, then 0

1
6= a1

1
a2
1

m
1
∈ S−1N.

By Definition 2.1, a1
1

a2
1
∈
√

(S−1N : S−1M) or a1
1

m
1
∈ S−1N or ( a2

1
)n m

1
∈ S−1N for some positive integer



Int. J. Anal. Appl. 16 (3) (2018) 405

n. If a1
1

a2
1
∈

√
(S−1N : S−1M), then ( a1

1
a2
1

)n ∈ (S−1N : S−1M) for some positive integer n. Thus there

exists s ∈ S such that s(a1a2)nM ⊆ N for some positive integer n. Since S ∩ Zd(M/N) = ∅, we have

(a1a2)
nM ⊆ N so a1a2 ∈

√
(N : M). If a1

1
m
1
∈ S−1N, there exists s ∈ S such that sa1m ∈ N. Thus

s(a1m + N) = sa1m + N = N. But S ∩ Zd(M/N) = ∅, a1m ∈ N. If ( a21 )
n am

1
∈ N, there exists s ∈ S such

that such that san1 m ∈ N for some positive integer n. Thus s(an2 m + N) = san2 m + N = N for some positive

integer n. Since S ∩Zd(M/N) = ∅, we have an2 m ∈ N for some positive integer n. Therefore N is a weakly

2-absorbing semi-primary submodule of M. �

Theorem 2.9. Let N be a proper submodule of an R-module M. The following conditions are equivalent:

(1) N is a weakly 2-absorbing semi-primary submodule of M.

(2) For every a1, a2 ∈ R − (N : M) if a1a2 ∈ R −
√

(N : M), then (N : a1a2) ⊆ (0 : a1a2) ∪ (N :

a1) ∪ (N : an2 ) for some positive integer n.

(3) For every a1, a2 ∈ R − (N : M) if R is a u-ring and a1a2 ∈ R −
√

(N : M), then (N : a1a2) ⊆ (0 :

a1a2) or (N : a1a2) ⊆ (N : a1) or (N : a1a2) ⊆ (N : an2 ) for some positive integer n.

Proof. (1 ⇒ 2) Let m ∈ (N : a1a2). Then a1a2m ∈ N. If a1a2m = 0, then m ∈ (0 : a1a2) ⊆ (0 :

a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n. If a1a2m 6= 0, then 0 6= a1a2m ∈ N. By Definition

2.1, a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some positive integer n. But a1a2 ∈ R −
√

(N : M),

m ∈ (N : a1) or m ∈ (N : an2 ) for some positive integer n. Therefore m ∈ (N : a1) ∪ (N : an2 ) for some

positive integer n. Hence (N : a1a2) = (0 : a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n.

(2 ⇔ 3) It is obvious.

(2 ⇒ 1) Let a1, a2 ∈ R such that 0 6= a1a2m ∈ N. Then m ∈ (N : a1a2) and m 6∈ (N : 0). By assumption,

m ∈ (0 : a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n. Clearly, a1m ∈ N or an2 m ∈ N for some

positive integer n. Hence N is a weakly 2-absorbing semi-primary submodule of M. �

Corollary 2.2. Let N be a proper submodule of an R-module M. The following conditions are equivalent:

(1) N is a weakly 2-absorbing semi-primary submodule of M.

(2) For every a ∈ R − (N : M) and every ideal I of R such that I 6⊆ (N : M), if aI 6⊆
√

(N : M), then

(N : aI) ⊆ (0 : aI) ∪ (N : a) ∪ (N : In) for some positive integer n.

(3) For every a ∈ R − (N : M) and every ideal I of R such that I 6⊆ (N : M), if R is a u-ring and

aI 6⊆
√

(N : M), then (N : aI) ⊆ (0 : aI) or (N : aI) ⊆ (N : a) or (N : aI) ⊆ (N : In) for some

positive integer n.

(4) For every ideals I, J of R such that I, J 6⊆ (N : M), if IJ 6⊆
√

(N : M), then (N : IJ) ⊆ (0 :

IJ) ∪ (N : I) ∪ (N : Jn) for some positive integer n.

(5) For every ideals I, J of R such that I, J 6⊆ (N : M), if R is a u-ring and IJ 6⊆
√

(N : M), then

(N : IJ) ⊆ (0 : IJ) or (N : IJ) ⊆ (N : I) or (N : IJ) ⊆ (N : Jn) for some positive integer n.



Int. J. Anal. Appl. 16 (3) (2018) 406

Proof. It is clear from Theorem 2.9. �

Theorem 2.10. Let N be a proper submodule of an R-module M. The following conditions are equivalent:

(1) N is a weakly 2-absorbing semi-primary submodule of M.

(2) For every a ∈ R − (N : M) and m ∈ M, if am 6∈ N, then (N : am) ⊆ (0 : am) ∪ (
√

((N : M) :

a) ∪
√

(N : m).

Proof. (1 ⇒ 2) Let a ∈ R − (N : M) and m ∈ M such that am 6∈ N. Assume that r ∈ (N : am). Then

ram ∈ N. If ram 6= 0, then 0 6= ram ∈ N. By Definition 2.1, ar ∈
√

(N : M) or am ∈ N or rnm ∈ N

for some positive integer n. Since am 6∈ N, we have r ∈ (
√

(N : M) : a) or r ∈
√

(N : m). This implies

that r ∈ (
√

(N : M) : a) ∪
√

(N : m) ⊆ (0 : am) ∪ (
√

((N : M) : a) ∪
√

(N : m). Thus (N : am) ⊆ (0 :

am)∪(
√

((N : M) : a)∪
√

(N : m). If ram = 0, then r ∈ (0 : am) ⊆ (0 : am)∪(
√

((N : M) : a)∪
√

(N : m).

Therefore (N : am) ⊆ (0 : am) ∪ (
√

((N : M) : a) ∪
√

(N : m).

(2 ⇒ 1) It is clear. �

Corollary 2.3. Let N be a proper submodule of an R-module M. The following conditions are equivalent:

(1) N is a weakly 2-absorbing semi-primary submodule of M.

(2) For every ideal I of R such that I ⊆ R − (N : M) and m ∈ M, if Im 6⊆ N, then (N : Im) ⊆ (0 :

Im) ∪ (
√

(N : M) : I) ∪
√

(N : m).

Proof. It is clear from Theorem 2.10. �

Definition 2.2. Let N be a proper submodule of M. If N is a 2-absorbing semi-primary submodule and

a1a2m = 0, a1a2 6∈
√

(N : M), a1m 6∈ N and an2 m 6∈ N for all positive integer n, then (a1, a2, m) is called a

absorbing semi-primary triple-zero of N where a1, a2 ∈ R, m ∈ M.

Theorem 2.11. Let N be a weakly 2-absorbing semi-primary submodule of an R-module M. Suppose that

K is a submodule of M and a1, a2 ∈ R such that N ⊆ K and a1a2K ⊆ N. If (a1, a2, m) is not a absorbing

semi-primary triple-zero of N for every m ∈ K, then a1a2 ∈
√

(K : M) or a1K ⊆ N or an2 K ⊆ N for some

positive integer n.

Proof. Assume that a1a2 6∈
√

(K : M), a1K 6⊆ N and an2 K 6⊆ N for all positive integer n. Then there are

k1, k2 ∈ K such that a1k1 6∈ N and an2 k2 6∈ N for all positive integer n. If a1a2k1 6= 0, then 0 6= a1a2k1 ∈ N.

By Definition 2.1, an12 k1 ∈ N for some positive integer n1. So let a1a2k1 = 0. By Definition 2.2, a
n2
2 k1 ∈ N

for some positive integer n2. Now if a1a2k2 6= 0, then 0 6= a1a2k2 ∈ N. Again, by Definition 2.1, ak2 ∈ N.

Next let a1a2k2 = 0. Now by Definition 2.2, a1k2 ∈ N. Let n0 = max{n1, n2}. Then an02 k1, a1k2 ∈ N.

Since a1a2K ⊆ N, we have a1a2(k1 + k2) ∈ N. If a1a2(k1 + k2) 6= 0, then 0 6= a1a2(k1 + k2) ∈ N. Thus by

Definition 2.1, a1(k1 + k2) ∈ N or an32 (k1 + k2) ∈ N for some positive integer n3. This implies that a1k1 ∈ N



Int. J. Anal. Appl. 16 (3) (2018) 407

or an42 k2 ∈ N where n4 = max{n0, n3} and we get a contradiction. Assume that a1a2(k1 + k2) = 0.

New since (a1, a2, k1 + k2) is not a absorbing semi-primary triple-zero of N, we have a1(k1 + k2) ∈ N or

an52 (k1 + k2) ∈ N for some positive integer n5. Clearly, a1k1 ∈ N or a
n6
2 k2 ∈ N, where n6 = max{n0, n5},

which again is a contradiction. Hence a1a2 ∈
√

(K : M) or a1K ⊆ N or an2 K ⊆ N for some positive integer

n. �

Theorem 2.12. Let N be a weakly 2-absorbing semi-primary submodule of an R-module M. Suppose that

(a1, a2, m) is a absorbing semi-primary triple-zero of N for some a1, a2 ∈ R and m ∈ M. Then

(1) a1a2N = {0};

(2) a1(N : M)m = {0};

(3) (N : M)a2m = {0};

(4) (N : M)2m = {0};

(5) a1(N : M)N = {0};

(6) (N : M)a2N = {0}.

Proof. 1. Suppose that a1a2N 6= {0}. Then there exists m0 ∈ N such that a1a2m0 6∈ {0}. Thus a1a2m +

a1a2m0 6= 0 so 0 6= a1a2(m + m0) ∈ N. By Definition 2.1, a1a2 ∈
√

(N : M) or a1(m + m0) ∈ N or

an2 (m + m0) ∈ N for some positive integer n. Therefore a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some

positive integer n. This is a contradiction. Hence a1a2N = {0}.

2. Suppose that a1(N : M)m 6= {0}. Then there exists r ∈ (N : M) such that a1rm 6= 0. Since rm ∈ N,

we have 0 6= a1(a2 + r)m ∈ N. By Definition 2.1, a1(a2 + r) ∈
√

(N : M) or a1m ∈ N or (a2 + r)nm ∈ N for

some positive integer n. Thus a1a2 ∈
√

(N : M) or a1m ∈ N or an2 ∈ N for some positive integer n. This is

a contradiction. Hence a1(N : M)m = {0}.

3. The proof is similar to part 2.

4. Assume that (N : M)2m 6= {0}. Then there exist r, s ∈ (N : M) such that rsm 6= 0. Then

by parts 1 and 2, (a1 + r)(a2 + s)m 6= 0. Clearly, 0 6= (a1 + r)(a2 + s)m ∈ N. By Definition 2.1,

(a1 + r)(a2 + s) ∈
√

(N : M) or (a1 + r)m ∈ N or (a2 + s)nm ∈ N for some positive integer n. Therefore

a1a2 ∈
√

(N : M) or a1m ∈ N or an2 ∈ N for some positive integer n. This is a contradiction. Hence

(N : M)2m = {0}.

5. Suppose that a1(N : M)N 6= {0}. Then there exist r ∈ (N : M) and m0 ∈ N such that a1rm0 6= 0.

Therefore by parts 1 and 2 we conclude that a1(a2 + r)(m + m0) 6= 0. Clearly, 0 6= a1(a2 + r)(m + m0) ∈ N.

By Definition 2.1, a1(a2 + r) ∈
√

(N : M) or a1(m + m0) ∈ N or (a2 + r)n(m + m0) ∈ N for some positive

integer n. Therefore a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N for some positive integer n. This is a

contradiction. Hence a1(N : M)N = {0}.

6. The proof is similar to part 5. �



Int. J. Anal. Appl. 16 (3) (2018) 408

Theorem 2.13. Let M be an R-module. If N is a weakly 2-absorbing semi-primary submodule of M that

is not 2-absorbing semi-primary, then (N : M)2N = {0}.

Proof. Suppose that N is a weakly 2-absorbing semi-primary submodule of M that is not 2-absorbing

semi-primary submodule. Then there exists a absorbing semi-primary triple-zero (a1, a2, m) of N for some

a1, a2 ∈ R and m ∈ M. Assume that (N : M)2N 6= {0}. Then there exist r, s ∈ (N : M) and m0 ∈ N

such that rsm0 6= 0. Since (a1 + r)(a2 + s)(m + m0) 6= 0, we have 0 6= (a1 + r)(a2 + s)(m + m0) ∈ N. By

Definition 2.1, (a1 + r)(a2 + s) ∈
√

(N : M) or (a1 + r)(m + n) ∈ N or (a2 + s)n(m + n) ∈ N for some

positive integer n. Therefore a1a2 ∈
√

(N : M) or a1m ∈ N or an2 m ∈ N. This is a contradiction. Hence

(N : M)2N = {0}. �

Corollary 2.4. Let M be a multiplication R-module. If N is a weakly 2-absorbing semi-primary submodule

of M that is not 2-absorbing semi-primary submodule, then N3 = {0}.

Proof. Suppose that N is a weakly 2-absorbing semi-primary submodule of M that is not 2-absorbing semi-

primary submodule. By assumption, N = (N : M)M. Then by Theorem 2.13, N3 = (N : M)3M = (N :

M)2((N : M)M) = (N : M)2N = {0}. �

Lemma 2.1. Suppose that N is a weakly 2-absorbing semi-primary submodule of an R-module M and

(0 : m2) is a 2-absorbing primary ideal of a ring R where m2 ∈ M − N. For all m1 ∈ M, if rs ∈ (N :

m1) −
√

(N : m2), then (N : rsm2) ⊆ (N : rm2) ∪
√

(N : snm2) for some positive integer n.

Proof. Suppose that rs ∈ (N : m1) − (N : m2) where m1 ∈ M and m2 ∈ M − N. Let a ∈ (N : rsm2).

Then (ars)m2 = a(rsm2) ∈ N so ars ∈ (N : m2). If arsm2 6= 0, then 0 6= ars ∈ (N : m2). By

assumption, ar ∈ (N : m2) or as ∈
√

(N : m2) or rs ∈
√

(N : m2). By the assumption, ar ∈ (N : m2) or

as ∈
√

(N : m2). Thus a ∈ (N : rm2) or a ∈
√

(N : snm2) for some positive integer n. This implies that

(N : rsm2) ⊆ (N : rm2)∪
√

(N : snm2) for some positive integer n. Now if arsm2 = 0, then ars ∈ (0 : m2).

Thus ar ∈ (0 : m2) or as ∈
√

(N : m2) or rs ∈
√

(N : m2). Therefore (N : rsm2) ⊆ (N : rm2)∪
√

(N : snm2)

for some positive integer n. �

Proposition 2.1. Let N be an irreducible submodule of an R-module M. For all r ∈ R if (N : r) = (N : r2),

then N is a weakly 2-absorbing semi-primary submodule of M.

Proof. Let a1, a2 ∈ R and m ∈ M such that 0 6= a1a2m ∈ N. Suppose that a1a2 6∈
√

(N : M), a1m 6∈ N and

an2 m 6∈ N for all positive integer n. Clearly, N ⊆ (N + a1a2M)∩(N + Ra1m)∩(N + Ran2 m) for all positive

integer n. Let m0 ∈ (N + a1a2M) ∩ (N + Ra1m) ∩ (N + Ran2 m). This implies that m0 ∈ N + a1a2M, m0 ∈

N +Ra1m and m0 ∈ N +Ran2 m. Then there exist r1, r2 ∈ R, m1 ∈ M and n1, n2 ∈ N such that n1+a1a2m1 =

m0 = n2 + r1a1m = m0 = n3 + b
n
2 m. Since a1n1 + a

2
1a2m1 = a1m0 = a1n2 + r1a

2
1m = a1m0 = a1n3 + a1b

n
2 m,



Int. J. Anal. Appl. 16 (3) (2018) 409

we have a21r1m ∈ N. It follows that r1m ∈ (N : a21). By the assumption, r1m ∈ (N : a1), so that r1a1m ∈ N.

Thus N = (N +a1a2M)∩(N +Ra1m)∩(N +Ran2 m). Now since N is an irreducible, we have N +a1a2M ⊆ N

or a1m ∈ N + Ra1m ⊆ N or an2 m ∈ N + Ran2 m ⊆ N, a contradiction. Hence N is a weakly 2-absorbing

semi-primary submodule of M. �

Theorem 2.14. Let Mi be an Ri-module and Ni be a proper submodule of Mi, for i = 1, 2. If N1 × M2 is

a weakly 2-absorbing semi-primary submodule of M1 × M2, then N1 is a weakly 2-absorbing semi-primary

submodule of M1.

Proof. Suppose that N1×M2 is a weakly 2-absorbing semi-primary submodule of M1×M2. Let a1, a2 ∈ R1

and m ∈ M1 such that 0 6= a1a2m ∈ N1. Then (0, 0) 6= (a1, 0)(a2, 0)(m, 0) = (a1a2m, 0) ∈ N1 × M2. By

Definition 2.1, (a1a2, 0) = (a1, 0)(a2, 0) ∈
√

(N1 ×M2 : M1 ×M2) or (a1m, 0) = (a1, 0)(m, 0) ∈ N1 × M2

or (an2 m, 0) = (a2, 0)
n(m, 0) ∈ N1 × M2 for some positive integer n. This implies that a1a2 ∈

√
(N1 : M1)

or a1m ∈ N1 or an2 m ∈ N1 for some positive integer n. Hence N1 is a weakly 2-absorbing semi-primary

submodule of M1. �

Corollary 2.5. Let Mi be an Ri-module and Ni be a proper submodule of Mi, for i = 1, 2. If M1 × N2 is

a weakly 2-absorbing semi-primary submodule of M1 × M2, then N2 is a weakly 2-absorbing semi-primary

submodule of M2.

Proof. It is clear from Theorem 2.14. �

Corollary 2.6. Let Mi be an Ri-module and Ni be a proper submodule of Mi, for i = 1, 2, . . . , k. If M1 ×

M2×. . .×Mj−1×Nj×Mj+1×. . .×Mk is a weakly 2-absorbing semi-primary submodule of M1×M2×. . .×Mk,

then Nj is a weakly 2-absorbing semi-primary submodule of Mj.

Proof. It is clear from Theorem 2.14 and Corollary 2.5. �

Theorem 2.15. Let Mi be an R-module and let Ni be a proper submodule of Mi, for i = 1, 2. Then the

following conditions are equivalent:

(1) N1 ×M2 is a weakly 2-absorbing semi-primary submodule of M1 ×M2.

(2) (a) N1 is a weakly 2-absorbing semi-primary submodule of M1.

(b) For each a1, a2 ∈ R and m ∈ M1 such that a1a2m = 0, if a1a2 6∈
√

(N1 : M1) and a1m 6∈

N1, a
n
2 m 6∈ N1 for all positive integer n, then a1a2 ∈ (0 : M2).

Proof. (1 ⇒ 2). (a). This follows from Theorem 2.14.

(b). Let a1a2m = 0, a1m 6∈ N1 and an2 m 6∈ N1 for all positive integer n, where a1, a2 ∈ R and m ∈ M1.

Suppose that a1a2 6∈ (0 : M2). There exists m2 ∈ M2 such that a1a2m2 6= 0. Thus (0, 0) 6= a1a2(m, m2) =



Int. J. Anal. Appl. 16 (3) (2018) 410

(a1a2m, a1a2m2) ∈ N1 × M2. By part 1, i.e., a1a2 ∈
√

(N1 ×M2 : M1 ×M2) or a1(m, m2) ∈ N1 × M2 or

an2 (m, m2) ∈ N1 × M2 for some positive integer n. Thus a1a2 ∈
√

(N1 : M1) or a1m ∈ N1 or an2 m ∈ N1

which is a contradiction. Hence a1a2 ∈ (0 : M2).

(2 ⇒ 1). Let a1, a2 ∈ R and (m1, m2) ∈ M1 ×M2 such that (0, 0) 6= (a1a2m1, a1a2m2) = a1a2(m1, m2) ∈

N1 × M2. If a1a2m1 6= 0, then 0 6= a1a2m1 ∈ N1. By part (a), a1a2 ∈
√

(N1 : M1) or a1m1 ∈ N1 or

an2 m1 ∈ N1 for some positive integer n. So a1a2 ∈
√

(N1 ×M2 : M1 ×M2) or a1(m1, m2) = (a1m1, a1m2) ∈

N1×M2 or an2 (m1, m2) = (an2 m1, an2 m2) ∈ N1×M2, and thus we are done. If a1a2m1 = 0, then a1a2m2 6= 0.

Therefore a1a2 6∈ (0 : M2). By part (b), a1a2 ∈
√

(N1 : M1) or a1m1 ∈ N1 or an2 m1 ∈ N1 for some positive

integer n. Thus a1a2 ∈
√

(N1 ×M2 : M1 ×M2) or a1(m1, m2) ∈ N1 ×M2 or an2 (m1, m2) ∈ N1 ×M2. Hence

N1 ×M2 is a weakly 2-absorbing semi-primary submodule of M1 ×M2. �

Corollary 2.7. Let Mi be an R-module and let Ni be a proper submodule of Mi, for i = 1, 2. Then the

following conditions are equivalent:

(1) M1 ×N2 is a weakly 2-absorbing semi-primary submodule of M1 ×M2.

(2) (a) N2 is a weakly 2-absorbing semi-primary submodule of M2.

(b) For each a1, a2 ∈ R and m ∈ M2 such that a1a2m = 0, if a1a2 6∈
√

(N2 : M2), a1m 6∈ N2 and

an2 m 6∈ N2 for all positive integer n, then a1a2 ∈ (0 : M1).

Proof. This follows from Theorem 2.15. �

Corollary 2.8. Let Mi be an R-module and let Ni be a proper submodule of Mi, for i = 1, 2, . . . , k. Then

the following conditions are equivalent:

(1) M1 × M2 × . . . × Mi−1 × Ni × Mi+1 × Mk is a weakly 2-absorbing semi-primary submodule of

M1 ×M2 × . . .×Mk.

(2) (a) Ni is a weakly 2-absorbing semi-primary submodule of Mi.

(b) For each a1, a2 ∈ R and m ∈ M2 such that a1a2m = 0, if a1a2 6∈
√

(N2 : M2), a1m 6∈ N2 and

an2 m 6∈ N2 for all positive integer n, then there exists j ∈{1, 2, . . . , k} such that a1a2 ∈ (0 : Mj).

Proof. This follows from Theorem 2.15. �

Theorem 2.16. Let Ni be a proper submodule of an Ri-module Mi, for i = 1, 2. Then the following

conditions are equivalent:

(1) N1 is a 2-absorbing semi-primary submodule of M1.

(2) N1 ×M2 is a 2-absorbing semi-primary submodule of M1 ×M2.

(3) N1 ×M2 is a weakly 2-absorbing semi-primary submodule of M1 ×M2, where M2 6= {0}.

Proof. (1 ⇒ 2). This is clear, by Theorem 2.15.



Int. J. Anal. Appl. 16 (3) (2018) 411

(2 ⇒ 3). The proof is clear.

(3 ⇒ 1). Suppose that N1 × M2 is a weakly 2-absorbing semi-primary submodule of M1 × M2, where

M2 6= {0}. Let a1, a2 ∈ R1 and m ∈ M1 such that a1a2m ∈ N1. By assumption, there exists m2 ∈ M2 such

that m2 6= 0. Since (a1, 1)(a2, 1)(m, m2) = (a1a2m, m2) 6= (0, 0) , we have (0, 0) 6= (a1, 1)(a2, 1)(m, m2) ∈

N1 × M2. By Definition 2.1, (a1, 1)(a2, 1) ∈
√

(N1 ×M2 : M1 ×M2) or (a1, 1)(m, m2) ∈ N1 × M2 or

(a2, 1)
n(m, m2) ∈ N1 × M2 for some positive integer n. Therefore a1a2 ∈

√
(N1 : M1) or a1m ∈ N1 or

an2 m ∈ N1 for some positive integer n and hence N1 is a 2-absorbing semi-primary submodule of M1. �

Corollary 2.9. Let Ni be a proper submodule of an Ri-module Mi, for i = 1, 2. Then the following conditions

are equivalent:

(1) N2 is a 2-absorbing semi-primary submodule of M1.

(2) M1 ×N2 is a 2-absorbing semi-primary submodule of M1 ×M2.

(3) M1 ×N2 is a weakly 2-absorbing semi-primary submodule of M1 ×M2, where M1 6= {0}.

Proof. This follows from Theorem 2.16. �

Corollary 2.10. Let Ni be a proper submodule of an Ri-module Mi, for i = 1, 2, . . . , k. Then the following

conditions are equivalent:

(1) Ni is a 2-absorbing semi-primary submodule of M1.

(2) M1×M2×. . .×Mi−1×Ni×Mi+1×Mk is a 2-absorbing semi-primary submodule of M1×M2×. . .×Mk.

(3) M1 × M2 × . . . × Mi−1 × Ni × Mi+1 × Mk is a weakly 2-absorbing semi-primary submodule of

M1 ×M2 × . . .×Mk, where Mj 6= {0}.

Proof. This follows from Theorem 2.16 and Corollary 2.9. �

Theorem 2.17. Let Ni be a proper submodule of an Ri-module Mi, for i = 1, 2. If N1 × N2 is a weakly

2-absorbing semi-primary submodule of M1 ×M2, then

(1) N1 is a weakly 2-absorbing semi-primary submodule of M1.

(2) N2 is a weakly 2-absorbing semi-primary submodule of M2.

Proof. (1). Suppose that N1×N2 is a weakly 2-absorbing semi-primary submodule of M1×M2. Let a1, a2 ∈

R1 and m ∈ M1 such that 0 6= a1a2m ∈ N1. Clearly, (0, 0) 6= (a1, 1)(a2, 1)(m, m2) = (a1a2m, m2) ∈ N1×N2.

By Definition 2.1, (a1a2, 1) = (a1, 1)(a2, 1) ∈
√

(N1 ×N2 : M1 ×M2) or (a1m, m2) = (a1, 1)(m, m2) ∈

N1×N2 or (an2 m, m2) = (a2, 1)n(m, m2) ∈ N1×N2 for some positive integer n. Therefore a1a2 ∈
√

(N1 : M1)

or a1m ∈ N1 or an2 m ∈ N1 for some positive integer n. Hence N1 is a weakly 2-absorbing semi-primary

submodule of M1.

(2). This follows from part 1. �



Int. J. Anal. Appl. 16 (3) (2018) 412

Example 2.3. Let M = Z×Z be an Z-module. Consider the submodule N = 5Z×12Z of M. It is easy to see

that 5Z and 12Z are weakly 2-absorbing semi-primary submodule of M. Notice that (0, 0) 6= 2 · 3(5, 2) ∈ N,

but 2 · 3 6∈
√

(M : N), 2(5, 2) 6∈ N, and (2 · 3)n 6∈ (N : M) for all positive integer n. Therefore N is not a

weakly 2-absorbing semi-primary submodule of M. This example shows that the converse of Theorem 2.17

is not true.

Theorem 2.18. Let Ni be a submodule of an Ri-module Mi, for i = 1, 2, 3. If N is a weakly 2-absorbing

semi-primary submodule of M1×M2×M3, then N = {(0, 0, 0)} or N is a 2-absorbing semi-primary submodule

of M1 ×M2 ×M3.

Proof. Suppose that N is a weakly 2-absorbing semi-primary submodule of M1 × M2 × M3 that is not

2-absorbing semi-primary. We will show that N = {(0, 0, 0)}. Now suppose that N1 × N2 × N3 = N 6=

{0}×{0}×{0}. Thus Ni 6= {0}, for some i = 1, 2, 3. We claim that N1 6= {0}. There exists m1 ∈ N1 such that

m1 6= 0. To show that N2 = M2 or N3 = M3. Assume that N2 6= M2 and N3 6= M3. Thus there exist m2 ∈

M2 and m3 ∈ M3 such that m2 6∈ N2 and m3 6∈ N3. Since (1, 0, 1)(1, 1, 0)(m1, m2, m3) = (m1, 0, 0) 6= (0, 0, 0),

we have (0, 0, 0) 6= (1, 0, 1)(1, 1, 0)(m1, m2, m3) ∈ N1 ×N2 ×N3. By Definition 2.1, we get (1, 0, 1)(1, 1, 0) ∈√
(N1 ×N2 ×N3 : M1 ×M2 ×M3) or (1, 0, 1)(m1, m2, m3) ∈ N or (1, 1, 0)n(m1, m2, m3) ∈ N, for some

positive integer n. So m2 ∈ N2 or m3 ∈ N3, a contradiction. Therefore N = N1 × M2 × N3 or N =

N1×N2×M3. If N = N1×M2×N3, then (0, 1, 0) ∈ (N : M1×M2×M3). By Theorem 2.13, {0}×M2×{0} =

(0, 1, 0)2N ⊆ (N : N1 ×M2 ×N3)2N = {(0, 0, 0)}, which is a contradiction. Hence N = {(0, 0, 0)}. �

Theorem 2.19. Let Ni be a submodule of an Ri-module Mi, for i = 1, 2, 3. If N 6= {(0, 0, 0)} and N

is a 2-absorbing semi-primary submodule of M1 × M2 × M3, then N is a weakly 2-absorbing semi-primary

submodule of M1 ×M2 ×M3.

Proof. Similar to the proof of Theorem 2.18 �

The above theorem shows the relationship between 2-absorbing semi-primary and weakly 2-absorbing

semi-primary submodules in R1×R2×R3-modules. From the above theorem, we have the following corollary.

Corollary 2.11. Let Ni be a submodule of an Ri-module Mi, for i = 1, 2, 3 with N 6= {(0, 0, 0)}. Then N

is a weakly 2-absorbing semi-primary submodule of M1 ×M2 ×M3 if and only if N is a weakly 2-absorbing

semi-primary submodule of M1 ×M2 ×M3.

Proof. This follows from Theorem 2.18. �

Corollary 2.12. Let Ni be a submodule of an Ri-module Mi, for i = 1, 2, . . . , k ≥ 3 with N 6= {(0, 0, . . . , 0)}.

Then N is a weakly 2-absorbing semi-primary submodule of M1×M2× . . .×Mk if and only if N is a weakly

2-absorbing semi-primary submodule of M1 ×M2 × . . .×Mk.



Int. J. Anal. Appl. 16 (3) (2018) 413

Proof. This follows from Theorem 2.19. �

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	1. Introduction
	2. Properties of weakly 2-Absorbing Semiprimary Submodules
	References