EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 1, 2018, 35-50
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On Generalizations of φ-2-absorbing primary
submodules

Pairote Yiarayong1,∗, Manoj Siripitukdet1

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

Abstract. Let φ : S(M)→ S(M)∪ {∅} be a function where S(M) is the set of all submodules of
M . In this paper, we extend the concept of φ-2-absorbing primary submodules to the context of
φ-2-absorbing semi-primary submodules. A proper submodule N of M is called a φ-2-absorbing
semi-primary submodule, if for each m ∈ M and a1, a2 ∈ R with a1a2m ∈ N − φ(N), then
a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive integer n. Those are extended from

2-absorbing primary, weakly 2-absorbing primary, almost 2-absorbing primary, φn-2-absorbing
primary, ω-2-absorbing primary and φ-2-absorbing primary submodules, respectively. Some char-
acterizations of 2-absorbing semi-primary, φn-2-absorbing semi-primary and φ-2-absorbing semi-
primary submodules are obtained. Moreover, we investigate relationships between 2-absorbing
semi-primary, φn-2-absorbing semi-primary and φ-primary submodules of modules over commuta-
tive rings. Finally, we obtain necessary and sufficient conditions of a φ-2-absorbing semi-primary
in order to be a 2-absorbing semi-primary submodule.

2010 Mathematics Subject Classifications: 13C05, 13C13.

Key Words and Phrases: 2-absorbing semi-primary submodule, φα-2-absorbing semi-primary
submodule, φ-2-absorbing primary submodule, φ-primary ideal, φ-2-absorbing ideal.

1. Introduction

Throughout this paper, we assume that all rings are commutative with a nonzero
identity Suppose that R is a ring and M is an R-module The concept of ϕ-prime ideals,
a generalization of prime ideals was introduced and investigated in [2]. Let ϕ : J (R) →
J (R) ∪ {∅} be a function where J (R) is a set of ideals of R. A proper ideal I of R is
said to be ϕ-prime if whenever a1, a2 ∈ R and a1a2 ∈ I − ϕ(I), then a1 ∈ I or a2 ∈ I.
Since I − ϕ(I) = I − (I ∩ ϕ(I)), so without loss of generality, throughout this paper we
will consider ϕ(I) ⊆ I. Later, Darani [7] gave a generalization of primary ideals which
covers all the above mentioned definitions. He defined the a proper ideal I of R is said
to be ϕ-primary if for a1, a2 ∈ R with a1a2 ∈ I − ϕ(I), either a1 ∈ I or an2 ∈ I for some

∗Corresponding author.

Email addresses: pairote0027@hotmail.com (Pai. Yiarayong), manojs@nu.ac.th (M. Siripitukdet)

http://www.ejpam.com 35 c© 2018 EJPAM All rights reserved.



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 36

positive integer n. Thus a ϕ-prime ideal is just a ϕ-primary ideal. In [9], Ebrahimpour and
Nekooei called a proper ideal I of a commutative ring R to be ϕ-2-absorbing if whenever
a1, a2, a3 ∈ R and a1a2a3 ∈ I − ϕ(I), either a1a2 ∈ I or a2a3 ∈ I or a1a3 ∈ I. Badawi,
et al. [1] generalized the concept of 2-absorbing primary ideals to ϕ-2-absorbing primary
ideals. According to their definition, a proper ideal I of R is called a ϕ-2-absorbing primary
ideal if whenever a1a2a3 ∈ I − ϕ(I) for a1, a2, a3 ∈ R, then a1a2 ∈ I or a2a3 ∈

√
I or

a1a3 ∈
√
I. Clearly a ϕ-2-absorbing ideal of R is also a ϕ-2-absorbing primary ideal of R.

Other generalizations of prime ideals have recently been studied in [4, 3, 5, 6].
The notion of φ-prime submodule, which is a generalization of prime submodule, was

introduced by Zamani in [11]. Let φ : S(M)→ S(M) ∪ {∅} be a function where S(M) is
a set of all submodules of M . A proper submodule N of M is called φ-prime submodule
of M if whenever a ∈ R and am ∈ N − φ(N), then m ∈ N or a ∈ (N : M). Since
N − φ(N) = N − (N ∩ φ(I)), without loss of generality we may assume that φ(N) ⊆ N .
Recall that a proper submodule N of M is called a φ-primary submodule submodule
of M as in [11] if whenever am ∈ N − φ(N) for some a ∈ R,m ∈ M , then m ∈ N
or an ∈ (N : M) for some positive integer n. In 2017, Ebrahimpour and Mirzaee [8]
generalized the concept of semiprime submodules to φ-semiprime submodules. According
to their definition, a proper submodule N of M is called a φ-semiprime submodule if
whenever a2m ∈ N − φ(N) for a ∈ R,m ∈ M , then am ∈ N . In [10], the concept of
φ-prime and φ-primary submodules generalized to φ-2-absorbing primary submodule of
a module over a commutative ring. Let N be a proper submodule of M . N is said to
be a φ-2-absorbing primary submodule of M if whenever a1, a2 ∈ R and m ∈ M with
a1a2m ∈ N − φ(N), then a1a2 ∈

√
(N : M) or a1m ∈ N or a2m ∈ N . Moreover, recall

from [10] that a proper submodule N of M is said to be a φ-2-absorbing submodule of M
if whenever a1, a2 ∈ R and m ∈ M with a1a2m ∈ N − φ(N) implies a1a2 ∈ (N : M) or
a1m ∈ N or a2m ∈ N . Thus a φ-2-absorbing submodule is just a φ-2-absorbing primary
submodule.

In this paper, we extend the concept of φ-2-absorbing primary submodule to the con-
text of φ-2-absorbing semi-primary submodule. Let φ : S(M) → S(M) ∪ {∅} be a func-
tion where S(M) is the set of all submodules of M . A proper submodule N of M is
called a φ-2-absorbing semi-primary submodule if for each m ∈ M and a1, a2 ∈ R with
a1a2m ∈ N − φ(N), then a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive

integer n. Let N be a φ-2-absorbing semi-primary submodule of M .

• If φ(N) = ∅ for every N ∈ S(M), then we say that φ = φ∅ and N is called a φ∅-2-
absorbing semi-primary submodule of M , and hence N is a 2-absorbing semi-primary
submodule of M .

• If φ(N) = {0} for every N ∈ S(M), then we say that φ = φ0 and N is called a
φ0-2-absorbing semi-primary submodule of M , and hence N is a weakly 2-absorbing
semi-primary submodule of M .

• If φ(N) = N for every N ∈ S(M), then we say that φ = φ1 and N is called a
φ1-2-absorbing semi-primary submodule of M . It is easy to see that every proper



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 37

submodule is φ1-2-absorbing semi-primary.

• If φ(N) = (N : M)N for every N ∈ S(M), then we say that φ = φ2 and N is
called a φ2-2-absorbing semi-primary submodule of M , and hence N is an almost
2-absorbing semi-primary submodule of M .

• If φ(N) = (N : M)n−1N for every N ∈ S(M), then we say that φ = φn≥2 and
N is called a φn-2-absorbing semi-primary submodule of M , and hence N is a n-2-
absorbing semi-primary submodule of M .

• If φ(N) =
∞⋂
i=1

(N : M)iN for every N ∈ S(M), then we say that φ = φω and N

is called a φω-2-absorbing semi-primary submodule of M , and hence N is a ω-2-
absorbing semi-primary submodule of M .

In section 2, we give some basic properties of φ-2-absorbing semi-primary submodules.
Among many results in this paper, it is shown that N is a φ-2-absorbing semi-primary
submodule of M if and only if for every a1, a2 ∈ R− (N : M) and a1a2 ∈ R−

√
(N : M),

(N : a1a2) ⊆ (φ(N) : a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n.
In Section 3, we study the stability of φα-2-absorbing semi-primary submodules. More-

over, we investigate relationships between 2-absorbing semi-primary, φ0-2-absorbing semi-
primary, φn-2-absorbing semi-primary and φ-primary submodules of modules over com-
mutative rings. Finally, we obtain necessary and sufficient conditions of a φ-2-absorbing
semi-primary in order to be a 2-absorbing semi-primary.

2. Properties of φ-2-Absorbing Semi-primary Submodules

The results of the following theorems seem to play an important role to study φ-
classical 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 1. Let M be an R-module and let φ : S(M) → S(M) ∪ {∅} be a function
where S(M) be a set of all submodules of M . A proper submodule N of M is called a
φ-2-absorbing semi-primary submodule, if for each m ∈ M and a1, a2 ∈ R with a1a2m ∈
N − φ(N), then a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive integer n.

Remark 1. It is easy to see that every φ-2-absorbing primary submodule is φ-2-absorbing
semi-primary.

The following example shows that the converse of Remark 1 is not true.

Example 1. Let R = Z and M = Z. Consider the submodule N = 12Z of M . Define
φ : S(M)→ S(M)∪{∅} by φ(N) = {0} for every N ∈ S(M). It is easy to see that N is a
φ-2-absorbing semi-primary submodule of M . Notice that 2 ·2 ·3 ∈ N−φ(N), but 2 ·3 6∈ N
and (2 · 2)n 6∈ (N : M) for all positive integer n. Therefore N is not a φ-2-absorbing
primary submodule of M .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 38

Theorem 1. Let φ : S(M)→ S(M) ∪ {∅} and ϕ : J (R)→ J (R) ∪ {∅} be two functions.

(i) If N is a φ-2-absorbing semi-primary submodule of M , then (N : m) is a ϕ-2-
absorbing primary ideal of R with m ∈M −N and (φ(N) : m)) ⊆ ϕ(N : m).

(ii) For every m ∈M−N if (N : m) is a ϕ-primary ideal of R, then N is a φ-2-absorbing
semi-primary submodule of M with ϕ(N : m) ⊆ (φ(N) : m)).

Proof. 1. Let a1, a2, a3 ∈ R such that a1a2a3 ∈ (N : m)−ϕ((N : m)). By assumption,
a1a3(a2m) ∈ N −φ(N). Then by Definition 1, a1a3 ∈

√
(N : M) ⊆

√
(N : m) or a1a2m ∈

N or an3a2m ∈ N for some positive integer n. Therefore a1a2 ∈ (N : m) or a2a3 ∈√
(N : m) or a1a3 ∈

√
(N : m). This completes the proof.

2. Let a1, a2 ∈ R such that a1a2m ∈ N − φ(N). Then a1a2 ∈ (N : m) and a1a2 6∈
(φ(N) : m). By assumption, a1a2 ∈ (N : m) − ϕ((N : m)). Again, by assumption,
a1 ∈ (N : m) or an2 ∈ (N : m) for some positive integer n. This completes the proof.

The following example shows that the converse of Theorem 1 is not true.

Example 2. 1. Let M = Z × Z × Z be an Z-module. Define ϕ : J (R) → J (R) ∪ {∅}
by ϕ(I) = {0} for every I ∈ J (R). Consider the submodule N = {0} × 12Z × Z of
M . Clearly, (N : (m1,m2,m3)) = {0} is a ϕ-2-absorbing primary ideal of J (R), where
(m1,m2,m3) ∈ M − N . Define φ : S(M) → S(M) ∪ {∅} by φ(N) = {(0, 0, 0)} for every
N ∈ S(M). Notice that 3 ·4(0, 1, 1) ∈ N−φ(N), but (3 ·4) 6∈

√
(N : M), 3(0, 1, 1) 6∈ N and

4n(0, 1, 1) 6∈ N for all positive integer n. Hence N is not a φ-2-absorbing semi-primary
submodule of M .

2. Let M = Z12 be an Z12-module. Define φ : S(M)→ S(M)∪{∅} by φ(N) = {[0]} for
every N ∈ S(M). Consider the submodule N = {[0]} of M . Clearly, N is a φ-2-absorbing
semi-primary submodule of M . Define ϕ : J (R) → J (R) ∪ {∅} by ϕ(I) = ∅ for every
I ∈ J (R). Notice that [4][3] ∈ {[0]} = (N : [1]) − ϕ((N : [1])), but [4] ∈ (N : [1]) and
[3]n ∈ (N : [1]) for all positive integer n.

Let N be a submodule of an R-module M and let φ : S(M) → S(M) ∪ {∅} be a
function. Define φN : S(M/N)→ S(M/N) ∪ {∅} by

φN (K/N) =

{
(φ(K) +N)/N ; φ(K) 6= ∅
∅ ;φ(K) = ∅,

for every submodule K of M with N ⊆ K[11]. In, 2010 Zamani in [11] gives relations
between φ-prime submodules of M and φN -prime submodules of M/N . This leads us to
give relations between φ-2-absorbing semi-primary submodules of M and φN -2-absorbing
semi-primary submodules of M/N .

Theorem 2. Let φ : S(M)→ S(M) ∪ {∅} be a function and let N,K be two submodules
of M with N ⊆ K. If K is a φ-2-absorbing semi-primary submodule of M , then K/N is
a φN -2-absorbing semi-primary submodule of M/N .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 39

Proof. Let a1, a2 ∈ R and m ∈ M such that a1a2(m + N) ∈ (K/N) − φN (K/N).
Then a1a2m ∈ K − φ(K). By Definition 1, a1a2 ∈

√
(K : M) or a1m ∈ K or an2m ∈ K

for some positive integer n. Clearly, a1a2 ∈
√

(K/N : M/N) or a1(m + N) ∈ K/N or
an2 (m+N) ∈ K/N for some positive integer n. This completes the proof.

Theorem 3. Let φ : S(M)→ S(M) ∪ {∅} be a function and let N,K be two submodules
of M . If N ⊆ φ(K) and K/N is a φN -2-absorbing semi-primary submodule of M/N , then
K is a φ-2-absorbing semi-primary submodule of M .

Proof. Let a1, a2 ∈ R and m ∈M such that a1a2m ∈ K −φ(K). Then a1a2(m+N) ∈
(K − φ(K))/N . By Definition 1, a1a2 ∈

√
(K/N : M/N) or a1(m + N) ∈ K/N or

an2 (m+N) ∈ K/N for some positive integer n. Clearly, a1a2 ∈
√

(K : M) or a1m ∈ K or
an2m ∈ K for some positive integer n.

Now, by Theorem 2 and Theorem 3, we have the following corollary.

Corollary 1. Let φ : S(M)→ S(M) ∪ {∅} be a function and let N,K be two submodules
of M with N ⊆ φ(K). Then K is a φ-2-absorbing semi-primary submodule of M if and
only if K/N is a φN -2-absorbing semi-primary submodule of M/N .

Proof. The proof follows from Theorem 2, 3.

Zamani in [11] gives relations between φ-prime submodules of M and φS-prime sub-
modules of S−1M . This leads us to give relations between φ-2-absorbing semi-primary
submodules of M and φS-2-absorbing semi-primary submodules of S−1M .

Theorem 4. Let S be a multiplicative closed subset of R and let φ : S(M)→ S(M)∪{∅}
be a function. If N is a φ-2-absorbing semi-primary submodule of M , then S−1N is a
φS-2-absorbing semi-primary submodule of S−1M .

Proof. Let a1, a2 ∈ R, s1, s2, s3 ∈ S and m ∈M such that a1
s1
a2
s2
m
s3
∈ S−1N−φS(S−1N).

Then there exists s ∈ S such that sa1a2m ∈ N . If sa1a2m ∈ φ(N), then a1
s1
a2
s2
m
s3

= sa1a2m
ss1s2s3

∈
S−1φ(N) = φS(S−1N), a contradiction. Now if sa1a2m 6∈ φ(N), then a1a2(sm) ∈ N −
φ(N). By Definition 1, a1a2 ∈

√
(N : M) or a1sm ∈ N or an2sm ∈ N for some positive

integer n. If a1sm ∈ N or an2sm ∈ N , then a1
s1
m
s3

= a1sm
s1ss3

∈ S−1N or (a2s2 )n ms3 =
an2 sm
sn2 s3s

∈
S−1N . Now if a1a2 ∈

√
(N : M), then there exists positive integer n1 such that (a1a2)

n1 ∈
(N : M). Clearly, (a1s1

a2
s2

)n ∈ S−1(N : M). This completes the proof.

Theorem 5. Let S be a multiplicative closed subset of R and let φ : S(M)→ S(M)∪{∅}
be a function. If S−1N is a φS-2-absorbing semi-primary submodule of S−1M such that
S ∩ Zd(N/φ(N)) = ∅ and S ∩ Zd(M/N) = ∅, then N is a φ-2-absorbing semi-primary
submodule of M .

Proof. Let a1, a2 ∈ R and m ∈M such that a1a2m ∈ N−φ(N). Then a1
1
a2
1
m
1 = abm

1 ∈
S−1N . If a1

1
a2
1
m
1 ∈ φS(S−1N) = S−1φ(N), then there exists s ∈ S such that sa1a2m ∈

φ(N) which is a contradiction. If a1
1
a2
1
m
1 6∈ φS(S−1N), then a1

1
a2
1
m
1 ∈ S

−1N − φS(S−1N).



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 40

By Definition 1, a1
1
a2
1 ∈

√
(S−1N : S−1M) or a1

1
m
1 ∈ S

−1N or (a21 )nm1 ∈ S
−1N for some

positive integer n. If a1
1
a2
1 ∈

√
(S−1N : S−1M), then (a11

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. By assumption, (a1a2)
nM ⊆ N so a1a2 ∈

√
(N : M).

In view of Theorem 4 and Theorem 5, we have the following result.

Corollary 2. Let S be a multiplicative closed subset of R and let φ : S(M)→ S(M)∪{∅}
be a function with S∩Zd(N/φ(N)) = ∅ and S∩Zd(M/N) = ∅. Then N is a φ-2-absorbing
semi-primary submodule of M if and only if S−1N is a φS-2-absorbing semi-primary
submodule of S−1M .

Proof. The proof follows from Theorem 4, 5.

In the following result, we give an equivalent definition of φ-2-absorbing semi-primary
submodules.

Theorem 6. Let φ : S(M) → S(M) ∪ {∅} be a function. The following conditions are
equivalent:

(i) N is a φ-2-absorbing semi-primary submodule of M .

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

(N : M), then (N : a1a2) ⊆ (φ(N) :
a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n.

Proof. (i ⇒ ii) Let m ∈ (N : a1a2). Then a1a2m ∈ N . If a1a2m ∈ φ(N), then
m ∈ (φ(N) : a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive integer n. If a1a2m 6∈ φ(N),
then a1a2m ∈ N − φ(N). By Definition 1, a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for

some positive integer n. By assumption, m ∈ (N : a1) or m ∈ (N : an2 ) for some positive
integer n. Therefore (N : a1a2) = (φ(N) : a1a2) ∪ (N : a1) ∪ (N : an2 ) for some positive
integer n.

(ii⇒ i) It is obvious.

Corollary 3. Let φ : S(M) → S(M) ∪ {∅} be a function. The following conditions are
equivalent:

(i) N is a φ-2-absorbing semi-primary submodule of M .

(ii) 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) ⊆ (φ(N) : aI) ∪ (N : a) ∪ (N : In) for some positive

integer n.

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

(N : M), then
(N : IJ) ⊆ (φ(N) : IJ) ∪ (N : I) ∪ (N : Jn) for some positive integer n.

Proof. The proof is similar to Theorem 6.

The following theorem offers a characterization of φ-2-absorbing semi-primary sub-
modules.



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 41

Theorem 7. Let φ : S(M) → S(M) ∪ {∅} be a function. The following conditions are
equivalent:

(i) N is a φ-2-absorbing semi-primary submodule of M .

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

√
((N : M) : a) ∪

√
(N : m).

Proof. (i ⇒ ii) Let a ∈ R − (N : M) and m ∈ M such that am 6∈ N . Assume that
r ∈ (N : am). If ram 6∈ φ(N), then ram ∈ N − φ(N). By Definition 1, ar ∈

√
(N : M)

or am ∈ N or rnm ∈ N for some positive integer n. By assumption, r ∈ (
√

(N : M) :
a) ∪

√
(N : m) ⊆ (φ(N) : am) ∪ (

√
((N : M) : a) ∪

√
(N : m). Now if ram ∈ φ(N), then

r ∈ (φ(N) : am) ⊆ (φ(N) : am) ∪ (
√

((N : M) : a) ∪
√

(N : m).
(ii⇒ i) It is obvious.

Corollary 4. Let φ : S(M) → S(M) ∪ {∅} be a function. The following conditions are
equivalent:

(i) N is a φ-2-absorbing semi-primary submodule of M .

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

√
(N : M) : I) ∪

√
(N : m).

Proof. The proof is similar to Theorem 7.

3. Properties of φα-2-Absorbing Semi-primary Submodules

We start with the following theorem that gives a relation between φα-2-absorbing semi-
primary and φ-2-absorbing semi-primary submodule. Our starting points are the following
definitions:

Definition 2. Let M be an R-module and let S(M) be the set of all submodules of M .
Define the following functions φα : S(M) → S(M) ∪ {∅} and the corresponding φα-2-
absorbing semi-primary submodules:

• If φ(N) = ∅ for every N ∈ S(M), then we say that φ = φ∅ and N is called a φ∅-2-
absorbing semi-primary submodule of M , and hence N is a 2-absorbing semi-primary
submodule of M .

• If φ(N) = {0} for every N ∈ S(M), then we say that φ = φ0 and N is called a
φ0-2-absorbing semi-primary submodule of M , and hence N is a weakly 2-absorbing
semi-primary submodule of M .

• If φ(N) = N for every N ∈ S(M), then we say that φ = φ1 and N is called a
φ1-2-absorbing semi-primary submodule of M .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 42

• If φ(N) = (N : M)N for every N ∈ S(M), then we say that φ = φ2 and N is
called a φ2-2-absorbing semi-primary submodule of M , and hence N is an almost
2-absorbing semi-primary submodule of M .

• If φ(N) = (N : M)n−1N for every N ∈ S(M), then we say that φ = φn≥2 and
N is called a φn-2-absorbing semi-primary submodule of M , and hence N is a n-2-
absorbing semi-primary submodule of M .

• If φ(N) =
∞⋂
i=1

(N : M)iN for every N ∈ S(M), then we say that φ = φω and N

is called a φω-2-absorbing semi-primary submodule of M , and hence N is a ω-2-
absorbing semi-primary submodule of M .

Remark 2. Let M be an R-module and let S(M) be a set of all submodules of M . For
two functions φα, φβ : S(M) → S(M) ∪ {∅}. We define φα ≤ φβ, if φα(N) ⊆ φβ(N) for
all N ∈ S(M)[11]. Observe that φ∅ ≤ φ0 ≤ φω ≤ . . . ≤ φn+1 ≤ φn ≤ . . . ≤ φ2 ≤ φ1.

Notice that for an R-module M, the zero submodule {0} is always a φ0-2-absorbing
semi-primary submodule. In the following example, we give a module in which a φ0-2-
absorbing semi-primary submodule is not φ-2-absorbing semi-primary .

Example 3. Let R = Z and M = Z30 × Z30. Consider the submodule N = {[0]} × Z30

of M . Define φ : S(M) → S(M) ∪ {∅} by φ(N) = {[0]} × {[0]} for every N ∈ S(M).
It is easy to see that N is a φ0-2-absorbing semi-primary submodule of M . Notice that
(2 · 3)([5], [1]) ∈ N − φ(N), but 2 · 3 6∈

√
(N : M), 2([5], [1]) 6∈ {[0]}×Z30 and 3n([5], [1]) 6∈

{[0]} × Z30 for all positive integer n. Therefore N is not a φ-2-absorbing semi-primary
submodule of M .

We are finding additional condition to show that a 2-absorbing semi-primary submod-
ule is a φ-2-absorbing semi-primary submodule of an R-module.

Theorem 8. Let φ : S(M) → S(M) ∪ {∅} be a function and let φ(N) is a 2-absorbing
semi-primary submodule of M . Then N is a φ-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 φ-2-absorbing semi-primary submodule of M . Conversely, assume that N is a
φ-2-absorbing semi-primary submodule of M . Let a1, a2 ∈ R and m ∈ M such that
a1a2m ∈ N . If a1a2m 6∈ φ(N), then a1a2m ∈ N−φ(N). By Definition 1, a1a2 ∈

√
(N : M)

or a1m ∈ N or an2m ∈ N for some positive integer n. Now if a1a2m ∈ φ(N), then
a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive integer n.

Further, we give another characterization of φα-2-absorbing semi-primary submodule
of M .

Theorem 9. Let φ2 : S(M) → S(M) ∪ {∅} be a function and let (0 : rk) ⊆ rkM 6= M ,
where r ∈ R. Then rkM is a φ2-2-absorbing semi-primary submodule of M if and only if
it is a 2-absorbing semi-primary submodule of M .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 43

Proof. Suppose that rkM is a 2-absorbing semi-primary submodule of M . Clearly,
rkM is a φ2-2-absorbing semi-primary submodule of M . Conversely, assume that rkM
is a φ2-2-absorbing semi-primary submodule of M . Let a1, a2 ∈ R and m ∈ M such
that a1a2m ∈ rkM . If a1a2m 6∈ φ2(rkM), then a1a2m ∈ rkM − φ2(rkM). By Definition
1, a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive integer n. Assume

that a1a2m ∈ φ2(rkM). Since a1a2m, r
ka2m ∈ rkM , we have (a1 + rk)a2m ∈ rkM . If

(a1 + rk)a2m 6∈ φ2(rkM), then (a1 + rk)a2m ∈ rkM − φ2(rkM). Then by Definition 1,
a1a2 ∈

√
(rkM : M) or a1m ∈ rkM or an2m ∈ rkM for some positive integer n. Now if

(a1 + rk)a2m ∈ φ2(rkM), then rka2m ∈ φ2(rkM). Then there exists m0 ∈ (rkM : M)M
such that rka2m = rkm0. By assumption, a2m−m0 ∈ rkM . Hence a2m ∈ rkM .

Theorem 10. Let φ : S(M)→ S(M)∪{∅} be a function and let N,K be two submodules
of M with N ⊆ K. If φ(K) ⊆ N and K is a φ-2-absorbing semi-primary submodule of
M , then K/N is a φ0-2-absorbing semi-primary submodule of M/N .

Proof. Let a1, a2 ∈ R and m ∈ M such that a1a2m + N ∈ K/N − (φN )0(K/N).
Since φ(K) ⊆ N , we have a1a2m 6∈ φ(K). Clearly, a1a2m ∈ K − φ(K). By Definition
1, a1a2 ∈

√
(K : M) or a1m ∈ K or an2m ∈ K for some positive integer n. Thus a1a2 ∈√

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

Theorem 11. Let φ : S(M) → S(M) ∪ {∅} be a function, N be a φ-2-absorbing semi-
primary submodule of M and let K be a submodule of M with N ⊆ K. If φ(N) ⊆ φ(K)
and K/N is a φ0-2-absorbing semi-primary submodule of M/N , then K is a φ-2-absorbing
semi-primary submodule of M .

Proof. Let a1, a2 ∈ R and m ∈ M such that a1a2m ∈ K − φ(K). By assumption,
a1a2m 6∈ φ(N). If a1a2m ∈ N , then a1a2m ∈ N − φ(N). By Definition 1, a1a2 ∈√

(N : M) ⊆
√

(K : M) or a1m ∈ N ⊆ K or an2m ∈ N ⊆ K for some positive integer n.
If a1a2m 6∈ N , then a1a2(m+N) 6∈ φ0(K/N). Therefore a1a2(m+N) ∈ K/N −φ0(K/N).
By Definition 1, a1a2 ∈

√
(K/N : M/N) or a1(m+N) ∈ K/N or an2 (m+N) ∈ K/N for

some positive integer n. This completes the proof.

As an immediate consequence of Theorem 10 and Theorem 11 we have the next corol-
lary.

Corollary 5. Let φ : S(M)→ S(M) ∪ {∅} be a function and let N,K be two submodules
of M with N ⊆ K. Then N is a φ-2-absorbing semi-primary submodule of M if and only
if N/φ(N) is a φ0-2-absorbing semi-primary submodule of M/φ(N).

Proof. It is straightforward by Theorem 10 and Theorem 11.

Theorem 12. Let φα : S(M)→ S(M) ∪ {∅} be a function. Then the following hold.

(i) If N is a φβ-2-absorbing semi-primary submodule of M such that φβ ≤ φγ, then N
is a φγ-2-absorbing semi-primary submodule of M .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 44

(ii) If N is a φ∅-2-absorbing semi-primary submodule of M , then N is a φ0-2-absorbing
semi-primary submodule of M .

(iii) If N is a φ0-2-absorbing semi-primary submodule of M , then N is an ω-2-absorbing
semi-primary submodule of M .

(iv) If N is a φω-2-absorbing semi-primary submodule of M , then N is a φn-2-absorbing
semi-primary submodule of M .

Proof. i. Let a1, a2 ∈ R and m ∈ M such that a1a2m ∈ N − φγ(N). By assumption,
φβ(N) ⊆ φγ(N). Then a1a2m ∈ N − φγ(N) ⊆ N − φβ(N). Then by Definition 1,
a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N for some positive integer n.

ii, iii, iv. It is obvious.

From the above definitions we obtain immediately the following implication chart for
the considered types of submodules:

2-absorbing semi-primary ⇒ weakly 2-absorbing semi-primary ⇒ ω-2-absorbing
semi-primary ⇒ φn≥2-2-absorbing semi-primary ⇒ almost 2-absorbing semi-primary

Theorem 13. Let φ, φ3 : S(M) → S(M) ∪ {∅} be two functions and let N be a φ-2-
absorbing semi-primary submodule. If φ3 6≤ φ, then N is a 2-absorbing semi-primary
submodule of M .

Proof. Let a1, a2 ∈ R and m ∈ M such that a1a2m ∈ N . If a1a2m 6∈ φ(N), then
a1a2m ∈ N − φ(N). By Definition 1, a1a2 ∈

√
(N : M) or a1m ∈ N or an2m ∈ N

for some positive integer n. Next, let a1a2m ∈ φ(N). In this case, we may assume
that a1a2N ⊆ φ(N), because if a1a2N 6⊆ φ(N) then there exists m0 ∈ N such that
a1a2m0 6∈ φ(N). Clearly, a1m ∈ N or a1a2 ∈

√
(N : M) or an2m ∈ N for some positive

integer n. Second we may assume that (N : M)2m ⊆ φ(N). If this is not the case, there
exist r1, r2 ∈ (N : M) such that (a1 + r1)(a2 + r2)m 6∈ φ(N). By assumption, a1m ∈ N or
a1a2 ∈

√
(N : M) or an2 ∈

√
(N : m) for some positive integer n. Again, by assumption,

(N : M)2N 6⊆ φ(N). There exist r1, r2 ∈ (N : M) and m0 ∈ N such that r1r2m0 6∈ φ(N).
Thus by Definition 1, a1m ∈ N or a1a2 ∈

√
(N : M) or an2m ∈ N for some positive integer

n.

Corollary 6. Let φn : S(M)→ S(M) ∪ {∅} be a function and let N be a φ0-2-absorbing
semi-primary submodule of M . If φ3 6= φ0, then N is a 2-absorbing semi-primary sub-
module of M .

Proof. Similar to the proof of Theorem 13.

Theorem 14. Let φ, φ4 : S(M)→ S(M) ∪ {∅} be two functions. If N is a φ-2-absorbing
semi-primary submodule such that φ ≤ φ4, then N is a ω-2-absorbing semi-primary sub-
module of M .



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 45

Proof. If N is a 2-absorbing semi-primary submodule of M , then there is nothing
to prove. Assume that N is not a 2-absorbing semi-primary submodule of M . Then by
Theorem 13, (N : M)2N = φ3(N) ⊆ φ(N) ⊆ (N : M)3N . This implies that φ(N) = (N :
M)2N = (N : M)3N . Thus φ(N) = (N : M)iN for all i ≥ 3.

Theorem 15. Let M be a multiplication R-module and φ : S(M) → S(M) ∪ {∅} be a
function. Then the following properties hold.

(i) If N is a φ-2-absorbing semi-primary submodule of M with N3 6⊆ φ(N), then N is
a 2-absorbing semi-primary submodule of M .

(ii) If N is a φn-2-absorbing semi-primary submodule of M with N3 6= Nn for all n ≥ 3,
then N is a 2-absorbing semi-primary submodule of M .

Proof. 1. Suppose that N is a φ-2-absorbing semi-primary submodule of M that is
not 2-absorbing semi-primary. Clearly, N = (N : M)M . Then by Theorem 13, N3 = (N :
M)3M = (N : M)2((N : M)M) = (N : M)2N = φ3(N) ⊆ φ(N).

2. Suppose that N is a φn-2-absorbing semi-primary submodule of M that is not 2-
absorbing semi-primary. Clearly, Nn ⊆ N3, for all n ≥ 3. Then by parts 1, N3 ⊆ φn(N) =
(N : M)n−1N = (N : M)nM = Nn. Hence N3 = Nn. This completes the proof.

Theorem 16. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with φ = ψ1 × ψ2. Then the
following statements are equivalent:

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

(ii) (a) N1 is a ψ1-2-absorbing semi-primary submodule of M1.

(b) For each a1, a2 ∈ R and m ∈ M1 such that a1a2m ∈ ψ1(N1) if a1a2 6∈√
(N1 : M1) and a1m 6∈ N1, a

n
2m 6∈ N1 for all positive integer n, then a1a2 ∈

(ψ2(M2) : M2).

Proof. (i⇒ ii). (a). It is obvious.
(b). Let a1a2m ∈ ψ1(N1), a1m 6∈ N1 and an2m 6∈ N1, where a1, a2 ∈ R and m ∈ M1.

Suppose that a1a2 6∈ (ψ2(M2) : M2). There exists m2 ∈ M2 such that a1a2m2 /∈ ψ2(M2).
Thus a1a2(m,m2) ∈ N1 ×M2 − φ(N1 ×M2). By part (1), i.e., a1a2 ∈

√
(N1 : M1) or

a1m ∈ N1 or an2m ∈ N1 which is a contradiction.
(ii ⇒ i). Let a1, a2 ∈ R and (m1,m2) ∈ M1 ×M2 such that a1a2(m1,m2) ∈ N1 ×

M2 − φ(N1 × M2). If a1a2m1 6∈ ψ1(N1), then a1a2m1 ∈ N1 − ψ1(N1). By part (a),
a1a2 ∈

√
(N1 ×M2 : M1 ×M2) or a1(m1,m2) = (a1m1, a1m2) ∈ N1×M2 or an2 (m1,m2) =

(an2m1, a
n
2m2) ∈ N1 ×M2, and thus we are done. If a1a2m1 ∈ ψ1(N1), then a1a2m2 6∈

ψ2(M2). Therefore a1a2 6∈ (ψ2(M2) : M2). By part (b), a1a2 ∈
√

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

Corollary 7. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with φ = ψ1 × ψ2. Then the
following conditions are equivalent:



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 46

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

(ii) (a) N2 is a ψ2-2-absorbing semi-primary submodule of M2.

(b) For each a1, a2 ∈ R and m ∈ M2 such that a1a2m ∈ ψ2(N2), if a1a2 6∈√
(N2 : M2), a1m 6∈ N2 and an2m 6∈ N2 for all positive integer n, then a1a2 ∈

(ψ1(M1) : M1).

Proof. Similar to the proof of Theorem 16.

Theorem 17. Let ψi : S(Mi)→ S(Mi) ∪ {∅} is a function with φ = ψ1 × . . .× ψk. Then
the following conditions are equivalent:

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

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

(b) For each a1, a2 ∈ R and m ∈ Mi such that a1a2m ∈ ψi(Ni), if a1a2 6∈√
(Ni : Mi), a1m 6∈ Ni and an2m 6∈ Ni for all positive integer n, then there

exists j ∈ {1, 2, . . . , k} such that a1a2 ∈ (ψj(Mj) : Mj).

Proof. Similar to the proof of Theorem 16.

Next, let Ri be a commutative ring with identity and let Mi be an Ri-module. Then
M1×M2 is an R1×R2-module and each submodule of M1×M2 is of the form N1×N2 for
some submodules N1 of M1 and N2 of M2. Next we show that, if N1 is a (ψ1)0-2-absorbing
semi-primary submodule of M1, then N1×M2 is a φ-2-absorbing semi-primary submodule
if {0}×M2 ⊆ ψ1×ψ2(N1×M2). First, we would like to show that, N1 is a ψ1-2-absorbing
semi-primary submodule of M1 if N1 ×M2 is a φ-2-absorbing semi-primary submodule of
M1 ×M2.

Theorem 18. Let ψi : S(Mi)→ S(Mi)∪{∅} be a function with φ = ψ1×ψ2. If N1×M2

is a φ-2-absorbing semi-primary submodule of M1 × M2, then N1 is a ψ1-2-absorbing
semi-primary submodule of M1.

Proof. Let a1, a2 ∈ R1 and m ∈ M1 such that a1a2m ∈ N1 − ψ1(N1). Then
(a1, 0)(a2, 0)(m, 0) ∈ N1 ×M2 − ψ1(N1) × ψ2(M2). By Definition 1, a1a2 ∈

√
(N1 : M1)

or a1m ∈ N1 or an2m ∈ N1.

Lemma 1. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function. If N1 is a (ψ1)0-2-absorbing
semi-primary submodule of M1 such that {0}×M2 ⊆ ψ1×ψ2(N1×M2), then N1×M2 is
a ψ1 × ψ2-2-absorbing semi-primary submodule of M1 ×M2.

Proof. Let (a1, b1), (a2, b2) ∈ R1 ×R2 and (m1,m2) ∈M1 ×M2 such that

(a1, b1)(a2, b2)(m1,m2) ∈ N1 ×M2 − φ(N1 ×M2).



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 47

By assumption, (a1a2m1, b1b2m2) 6∈ {0} ×M2. Clearly, a1a2m1 ∈ N1 − (ψ1)0(N1). By
Definition 1, a1a2 ∈

√
(N1 : M1) or a1m1 ∈ N1 or an2m1 ∈ N1 for some positive integer n.

Therefore N1 ×M2 is a φ-2-absorbing semi-primary submodule of M1 ×M2.

Corollary 8. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with φ = ψ1 × ψ2. If N2 is a
(ψ2)0-2-absorbing semi-primary submodule of M2 such that M1×{0} ⊆ ψ1×ψ2(M1×N2),
then M1 ×N2 is a φ-2-absorbing semi-primary submodule of M1 ×M2.

Proof. Similar to the proof of Lemma 1.

Theorem 19. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with φ = ψ1 × . . . × ψk
and M1 × . . . × Mi−1 × {0} × Mi+1 × . . . × Mk ⊆ φ(M1 × . . . × Mi−1 × Ni × Mi+1 ×
. . . ×Mk). Then Nj is a (ψj)0-2-absorbing semi-primary submodule of Mj if and only if
M1×M2× . . .×Mi−1×Ni×Mi+1× . . .×Mk is a φ-2-absorbing semi-primary submodule
of M1 ×M2 × . . .×Mk.

Proof. Similar to the proof of Theorem 18 and Lemma 1.

Theorem 20. Let (ψi)n : S(Mi) → S(Mi) ∪ {∅} be a function with φn = (ψ1)n × (ψ2)n.
If N1 is a (ψ1)0-2-absorbing semi-primary submodule of M1 such that (ψ2)3(M2) = M2,
then N1 ×M2 is a φ3-2-absorbing semi-primary submodule of M1 ×M2.

Proof. If N1 is a 2-absorbing semi-primary submodule of M1, then N1 ×M2 is a 2-
absorbing semi-primary submodule of M1 ×M2. Clearly, N1 ×M2 is a φ3-2-absorbing
semi-primary submodule of M1 ×M2. Assume that N1 is not 2-absorbing semi-primary.
By Corollary 6, (ψ1)3 ≤ (ψ1)0 so (N1 : M1)

2N1 = {0}. Therefore (ψ1)3×(ψ2)3(N1×M2) =
(ψ1)3(N1) × (ψ2)3(M2) = {0} ×M2. Now, by Lemma 1, N1 ×M2 is a (ψ1)3 × (ψ2)3-2-
absorbing semi-primary submodule of M1 ×M2.

Corollary 9. Let (ψi)n : S(Mi) → S(Mi) ∪ {∅} be a function with φn = (ψ1)n × (ψ2)n.
If N2 is a (ψ2)0-2-absorbing semi-primary submodule of M2 such that (ψ1)3(M1) = M1,
then M1 ×N2 is a φ3-2-absorbing semi-primary submodule of M1 ×M2.

Proof. Similar to the proof of Theorem 20.

Theorem 21. Let (ψi)n : S(Mi)→ S(Mi)∪{∅} be a function with φn = (ψ1)n×. . .×(ψk)n.
If Nj is a (ψj)0-2-absorbing semi-primary submodule of Mj such that (ψj)3(Mj) = Mj,
then M1 ×M2 × . . . ×Mi−1 × Ni ×Mi+1 × . . . ×Mk is a φ3-2-absorbing semi-primary
submodule of M1 × . . .×Mk.

Proof. Similar to the proof of Theorem 20 and Corollary 9.

Theorem 22. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with φ = ψ1 × ψ2. Then the
following conditions are equivalent:



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 48

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

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

(iii) N1 ×M2 is a φ-2-absorbing semi-primary submodule of M1 ×M2, where ψ2(M2) 6=
M2.

Proof. (1 ⇒ 2). Let (a1, b1), (a2, b2) ∈ R1 × R2 and (m1,m2) ∈ M1 ×M2 such that
(a1, b1)(a2, b2)(m1,m2) ∈ N1 × M2. Clearly, a1a2m1 ∈ N1. By Definition 2, a1a2 ∈√

(N1 : M1) or a1m1 ∈ N1 or an2m1 ∈ N1 for some positive integer n. This completes the
proof.

(2⇒ 3). It is easy to see that every 2-absorbing primary submodule is φ-2-absorbing
semi-primary.

(3 ⇒ 1). Let a1, a2 ∈ R1 and m ∈ M1 such that a1a2m ∈ N1. By assumption, there
exists m2 ∈ M2 such that m2 6∈ ψ2(M2). Since φ(N1 ×M2) ⊆ M1 × ψ2(M2), we have
(a1, 1)(a2, 1)(m,m2) ∈ N1×M2−ψ1×ψ2(N1×M2). By Definition 1, a1a2 ∈

√
(N1 : M1)

or a1m ∈ N1 or an2m ∈ N1.

Corollary 10. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function φ = ψ1 × ψ2. Then the
following conditions are equivalent:

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

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

(iii) M1 ×N2 is a φ-2-absorbing semi-primary submodule of M1 ×M2, where ψ1(M1) 6=
M1.

Proof. Similar to the proof of Theorem 22.

Theorem 23. Let ψi : S(Mi)→ S(Mi)∪ {∅} be a function with φ = ψ1 × . . .×ψk. Then
the following conditions are equivalent:

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

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

(iii) M1 ×M2 × . . . ×Mi−1 × Ni ×Mi+1 × . . . ×Mk is a φ-2-absorbing semi-primary
submodule of M1 × . . .×M2 with ψj(Mj) 6= Mj.

Proof. Similar to the proof of Theorem 22 and Corollary 10.

Theorem 24. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with ψ2(M2) = M2 and
φ = ψ1 × ψ2. Then N1 ×M2 is a φ-2-absorbing semi-primary submodule of M1 ×M2 if
and only if N1 is a ψ1-2-absorbing semi-primary submodule of M1.



Pai. Yiarayong, M. Siripitukdet / Eur. J. Pure Appl. Math, 11 (1) (2018), 35-50 49

Proof. The proof is clear.

Corollary 11. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with ψ1(M1) = M1 and
φ = ψ1 × ψ2. Then M1 × N2 is a φ-2-absorbing semi-primary submodule of M1 ×M2 if
and only if N2 is a ψ2-2-absorbing semi-primary submodule of M2.

Proof. Similar to the proof of Theorem 24.

Theorem 25. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with ψj(Mj) = Mj and
φ = ψ1 × . . . × ψk. Then M1 × M2 × . . . × Mi−1 × Ni × Mi+1 × . . . × Mk is a φ-2-
absorbing semi-primary submodule of M1 × . . .×Mk if and only if Ni is a ψi-2-absorbing
semi-primary submodule of Mi.

Proof. Similar to the proof of Theorem 24 and Corollary 11.

Theorem 26. Let Ni be a proper submodule of Mi and let ψi : S(Mi)→ S(Mi) ∪ {∅} be
a function with φ = ψ1 × ψ2. If N1 × N2 is a φ-2-absorbing semi-primary submodule of
M1 ×M2, then

(i) N1 is a ψ1-2-absorbing semi-primary submodule of M1,

(ii) N2 is a ψ2-2-absorbing semi-primary submodule of M2.

Proof. The proof is clear.

The next theorem gives conditions for a φ-2-absorbing semi-primary to be 2-absorbing
semi-primary.

Theorem 27. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with ψi(Mi) 6= Mi, φ =
ψ1 × ψ2 × ψ3. If N is a φ-2-absorbing semi-primary submodule of M1 ×M2 ×M3, then
N = φ(N) or N is a 2-absorbing semi-primary submodule of M1 ×M2 ×M3.

Proof. Suppose that N is a φ-2-absorbing semi-primary submodule of M1 × M2 ×
M3 that is not 2-absorbing semi-primary. Now suppose that N1 × N2 × N3 = N 6=
ψ1 × ψ2 × ψ3(N). Thus Ni 6= ψi(Ni) for some i = 1, 2, 3. We may assume that
N1 6= ψ1(N1). There exists m1 ∈ N1 such that m1 /∈ ψ1(N1). 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) 6∈ ψ1 × ψ2 × ψ3(N1 × N2 × N3), we have
(1, 0, 1)(1, 1, 0)(m1,m2,m3) ∈ N −φ(N). By Definition 1, m2 ∈ N2 or m3 ∈ N3, a contra-
diction. 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 13, {0}×M2×{0} = (0, 1, 0)2N ⊆ (N : N1×
M2×N3)

2N = (ψ1)3× (ψ2)3× (ψ3)3(N) ⊆ ψ1×ψ2×ψ3(N) = ψ1(N1)×ψ2(M2)×ψ3(N3),
which is a contradiction. This completes the proof.

The above theorem shows the relationship between 2-absorbing semi-primary and φ-2-
absorbing semi-primary submodules in R1 ×R2 ×R3-modules. From the above theorem,
we have the following corollary.



REFERENCES 50

Corollary 12. Let ψi : S(Mi) → S(Mi) ∪ {∅} be a function with ψi(Mi) 6= Mi, φ =
ψ1 × ψ2 × ψ3 and N 6= φ(N). Then N is a φ-2-absorbing semi-primary submodule of
M1×M2×M3 if and only if N is a 2-absorbing semi-primary submodule of M1×M2×M3.

Proof. This follows from Theorem 27.

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