EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 1, 2022, 328-334
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On ϕ-β-Absorbing Submodules

Thawatchai Khumprapussorn

Department of Mathematics, School of Science, King Mongkut’s Institute of Technology
Ladkrabang, Bangkok 10520, THAILAND

Abstract. In this paper, we extend the concept of β-absorbing submodules to ϕ-β-absorbing
submodules over a commutative ring with nonzero identity which is a generalization of 2-absorbing
submodules. Let S(M) be the set of all submodules ofM and ϕ : S(M)→ S(M)∪{∅} be a function.
A proper submodule P of M is called a ϕ-β-absorbing submodule, if for each r, s ∈ R and m ∈M
with rsm ∈ P\ϕ(P ), then rs + rs ∈ (P : M) or r(m + m) ∈ P or s(m + m) ∈ P . Some of the
properties and characterizations of ϕ-β-absorbing submodules are investigated.

2020 Mathematics Subject Classifications: 13C05, 16D80, 16D99

Key Words and Phrases: 2-absorbing submodules, ϕ-2-absorbing submodules, β-absorbing
submodules, ϕ-β-absorbing submodules

1. Introduction

Throughout this paper, R will denote a commutative ring with identity and all modules
are unital left R-modules. We recall that a proper submodule P of a left R-module M
is called a prime submodule of M if for every r ∈ R and m ∈ M , rm ∈ P implies that
m ∈ P or r ∈ (P : M). Various generalizations of prime submodules have been studied.
For example, see [5], [1] and [6], a proper submodule P of a left R-module M is called a
2-absorbing (resp. weakly 2-absorbing, almost 2-absorbing) submodule if for each r, s ∈ R
and every m ∈M such that rsm ∈ P (resp. rsm ∈ P\{0}, rsm ∈ P\(P : M)P ), we have
rs ∈ (P : M) or rm ∈ P or sm ∈ P . According to [4], nZ is a 2-absorbing submodule of
Z if and only if n = 0 or n is a prime number or n = pq where p and q are prime numbers.

Let S(M) be the set of all submodules of M and ϕ : S(M) → S(M) ∪ {∅} be a
function. In this paper, we assume that ϕ(P ) ⊆ P . In [3], the authors introduced the
concept of ϕ-2-absorbing submodule which is a generalization of 2-absorbing submodules.
A proper submodule P of a left R-module M is a ϕ-2-absorbing submodule if whenever
a, b ∈ R,m ∈M with abm ∈ P and abm /∈ ϕ(P ), then am ∈ P or bm ∈ P or ab ∈ (P : M).
In addition, the notion of ϕ-2-absorbing submodule is also a generalization of both weakly
2-absorbing submodule and almost 2-absorbing submodule which depends on the definition
of ϕ.

DOI: https://doi.org/10.29020/nybg.ejpam.v15i1.4262

Email address: thawatchai.kh@kmitl.ac.th (T.Khumprapussorn)

http://www.ejpam.com 328 © 2022 EJPAM All rights reserved.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 15 (1) (2022), 328-334 329

Let (G,+) be a group and H is a subgroup of G. We denote the symbol β(H) by
{h + h | h ∈ H} and α(H) by {h | h + h ∈ H}. We see that β(H) ⊆ H ⊆ α(H). If I is
an ideal of R, then both of α(I) and β(I) are ideals of R. Moreover, if N is a submodule
of M , then both of α(N) and β(N) are submodules of M .

In [2], a proper submodule P of a left R-module M is called β-absorbing if for any
element r, s ∈ R and m ∈ M such that rsm ∈ P , we have rs + rs ∈ (P : M) or
r(m + m) ∈ P or s(m + m) ∈ P . The characterization of β-absorbing submodule of
Z-module Z was also given. On the Z-module Z, nZ is a β-absorbing submodule of Z if
and only if n = 0 or n = 32 or n is a prime number or n = pq where p and q are prime
numbers or n = 23p where p is prime number or n = 2pq where p and q are prime numbers.
The characterization of β-absorbing submodule of Z-module Z explains that β-absorbing
submodules need not to be 2-absorbing submodules. Also, in [2], a proper submodule P
of M is a weakly β-absorbing submodule of M if for each r, s ∈ R and every m ∈M such
that rsm ∈ P\{0}, we have rs+ rs ∈ (P : M) or r(m+m) ∈ P or s(m+m) ∈ P .

In this research, we extend the notion of β-absorbing submodules to ϕ-β-absorbing
submodules. First, we introduce notions of ϕ-β-absorbing submodules. A proper submod-
ule P of a left R-module M is called a ϕ-β-absorbing submodule of M if for any element
r, s ∈ R and m ∈M such that rsm ∈ P\ϕ(P ), we have rs+rs ∈ (P : M) or r(m+m) ∈ P
or s(m + m) ∈ P . In case ϕ0(N) = {0} for all submodule N of M , we have weakly β-
absorbing submodules and ϕ0-β-absorbing submodules are equivalent. This case inspired
us to investigate some basic properties of ϕ-β-absorbing submodules in section 2, whereas
section 3 contains the characterizations of ϕ-β-absorbing submodules.

2. On ϕ-β-absorbing submodules

In this section, we define ϕ-β-absorbing submodules and obtain some related results.

Definition 1. A proper submodule P of an R-module M is said to be a ϕ-β-absorbing
submodule of M if whenever r, s ∈ R and m ∈M such that rsm ∈ P\ϕ(P ), then rs+rs ∈
(P : M) or r(m+m) ∈ P or s(m+m) ∈ P .

Every β-absorbing is a weakly β-absorbing submodule but the converse does not nec-
essarity hold. As mentioned above, β-absorbing submodules and weakly β-absorbing sub-
modules are special cases of ϕ-β-absorbing submodules.

Theorem 1. If P is a ϕ-β-absorbing submodule of M and (P : M)2β(P ) ⊈ ϕ(P ), then P
is a β-absorbing submodule of M .

Proof. Assume that P is a ϕ-β-absorbing submodule of M and (P : M)2β(P ) ⊈ ϕ(P ).
Let r, s ∈ R and m ∈ M be such that rsm ∈ P . If rsm /∈ ϕ(P ), then rs + rs ∈ (P : M)
or r(m+m) ∈ P or s(m+m) ∈ P . Next, assume that rsm ∈ ϕ(P ).

Case 1. rsP ⊈ ϕ(P ).
Then rsp0 /∈ ϕ(P ) for some p0 ∈ P . Hence rs(m + p0) ∈ P\ϕ(P ). Since P is a

ϕ-β-absorbing submodule of M , rs + rs ∈ (P : M) or r(m + p0 + m + p0) ∈ P or



T. Khumprapussorn / Eur. J. Pure Appl. Math, 15 (1) (2022), 328-334 330

s(m + p0 +m + p0) ∈ P . Since p0 ∈ P , we have rs + rs ∈ (P : M) or r(m +m) ∈ P or
s(m+m) ∈ P .

Case 2. rsP ⊆ ϕ(P ).
Subcase 2.1 s(P : M)m ⊈ ϕ(P ).
There exists an element a0 ∈ (P : M) such that sa0m /∈ ϕ(P ). Thus (r + a0)sm =

rsm+a0sm ∈ P\ϕ(P ). Since P is a ϕ-β-absorbing submodule of M , (r+a0)s+(r+a0)s ∈
(P : M) or (r + a0)(m+m) ∈ P or s(m+m) ∈ P . Then rs+ a0s+ rs+ a0s ∈ (P : M)
or r(m +m) + a0(m +m) ∈ P or s(m +m) ∈ P . Since a0 ∈ (P : M), a0M ⊆ P . This
implies that a0s + a0s ∈ (P : M) and a0(m + m) ∈ P . Therefore rs + rs ∈ (P : M) or
r(m+m) ∈ P or s(m+m) ∈ P .

Subcase 2.2 s(P : M)m ⊆ ϕ(P ).
Since (P : M)2β(P ) ⊈ ϕ(P ), we have that kt(n+n) /∈ ϕ(P ) for some k, t ∈ (P : M) and

n ∈ P . If rkm /∈ ϕ(P ), then r(k + s)m = rkm+ rsm /∈ ϕ(P ). Since P is a ϕ-β-absorbing
submodule of M , r(k + s) + r(k + s) ∈ (P : M) or r(m+m) ∈ P or (k + s)(m+m) ∈ P .
Since k ∈ (P : M), rs + rs ∈ (P : M) or r(m +m) ∈ P or s(m +m) ∈ P . Similarly, if
rtm /∈ ϕ(P ), then r(t+ s)m = rtm+ rsm /∈ ϕ(P ). Since P is a ϕ-β-absorbing submodule
of M , r(t + s) + r(t + s) ∈ (P : M) or r(m + m) ∈ P or (t + s)(m + m) ∈ P . Since
t ∈ (P : M), rs + rs ∈ (P : M) or r(m + m) ∈ P or s(m + m) ∈ P . From now on, we
assume that

rkm ∈ ϕ(P ) and rtm ∈ ϕ(P ). (1)

If ktm /∈ ϕ(P ), then (k+ r)(t+ s)m /∈ ϕ(P ). Since P is a ϕ-β-absorbing submodule of M ,
(k + r)(t + s) + (k + r)(t + s) ∈ (P : M) or (k + r)(m +m) ∈ P or (t + s)(m +m) ∈ P .
Since k, t ∈ (P : M), rs + rs ∈ (P : M) or r(m + m) ∈ P or s(m + m) ∈ P . Now, we
assume that

ktm ∈ ϕ(P ). (2)

Subsubcase 2.2.1 kr(n+ n) /∈ ϕ(P ) or st(n+ n) /∈ ϕ(P ).
Suppose that kr(n+ n) /∈ ϕ(P ). Then r(s+ k)(n+ n+m) /∈ ϕ(P ). Since P is a ϕ-β-

absorbing submodule of M , r(s+k)+ r(s+k) ∈ (P : M) or r(n+n+m+n+n+m) ∈ P
or (s+k)(n+n+m+n+n+m) ∈ P . This implies that rs+rs ∈ (P : M) or r(m+m) ∈ P
or s(m+m) ∈ P .

Next, suppose that st(n+ n) /∈ ϕ(P ). Then s(r + t)(n+ n+m) /∈ ϕ(P ). Since P is a
ϕ-β-absorbing submodule of M , s(r+t)+s(r+t) ∈ (P : M) or s(n+n+m+n+n+m) ∈ P
or (r+t)(n+n+m+n+n+m) ∈ P . This implies that rs+rs ∈ (P : M) or r(m+m) ∈ P
or s(m+m) ∈ P .

Subsubcase 2.2.2 kr(n+ n) ∈ ϕ(P ) and st(n+ n) ∈ ϕ(P ).
Then (s + k)(r + t)(n + n + m) /∈ ϕ(P ). Since P is a ϕ-β-absorbing submodule of

M , (s + k)(r + t) + (s + k)(r + t) ∈ (P : M) or (s + k)(n + n +m + n + n +m) ∈ P or
(r + t)(n+ n+m+ n+ n+m) ∈ P . Since k, t ∈ (P : M), we have rs+ rs ∈ (P : M) or
r(m+m) ∈ P or s(m+m) ∈ P .

Therefore P is a β-absorbing submodule of M .



T. Khumprapussorn / Eur. J. Pure Appl. Math, 15 (1) (2022), 328-334 331

Corollary 1. [1] If P is a weakly β-absorbing submodule of M and (P : M)2β(P ) ̸= {0},
then P is a β-absorbing submodule of M .

Next, we use the function ϕi : S(M)→ S(M) ∪ {∅} by the following meaning, for any
submodule N of an R-module M and natural number n with n ≥ 2,

ϕ∅(N) = ∅
ϕ0(N) = {0}
ϕ1(N) = (N : M)β(N)

ϕn(N) = (N : M)nβ(N)

ϕω(N) =

∞⋂
i=1

(N : M)iβ(N)

Let ϕ : S(M) → S(M) ∪ {∅} and φ : S(M) → S(M) ∪ {∅} be functions. We write
ϕ ≤ φ if ϕ(N) ⊆ φ(N) for all N ∈ S(M). Then ϕ∅ ≤ ϕ0 ≤ ϕω ≤ · · · ≤ ϕn+1 ≤ ϕn ≤ · · · ≤
ϕ2 ≤ ϕ1.

Proposition 1. Let ϕ : S(M) → S(M) ∪ {∅} and φ : S(M) → S(M) ∪ {∅} be functions
such that ϕ ≤ φ. If P is a ϕ-β-absorbing submodule of M , then P is a φ-β-absorbing
submodule of M .

Proof. This proof is straightforward.

Let M1 be a R1-module and M2 be a R2-module. Then M = M1×M2 is an R1×R2-
module by (r1, r2)(m1,m2) = (r1m1, r2m2). Next, let ϕ : S(M) → S(M) ∪ {∅} be a
function and P be a submodule of M1. Then

(P ×M2)\ϕ(P ×M2) ⊆ (P ×M2)\({0} ×M2) = (P\{0})×M2.

We have the following results.

Proposition 2. Let R = R1×R2 and M = M1×M2 and let ϕ : S(M)→ S(M)∪ {∅} be
a function. If P is a ϕ0-β-absorbing submodule of M1 with {0} ×M2 ⊆ ϕ(P ×M2), then
P ×M2 is a ϕ-β-absorbing submodule of M .

Proof. Assume that P is a ϕ0-β-absorbing submodule of M1 with {0} × M2 ⊆
ϕ(P × M2). Let (r1, r2), (s1, s2) ∈ R1 × R2 and (m1,m2) ∈ M1 × M2 be such that
(r1, r2)(s1, s2)(m1,m2) ∈ (P × M2)\ϕ(P × M2). Then r1s1m1 ∈ P\{0}. Since P is
a ϕ0-β-absorbing submodule of M1, r1s1 + r1s1 ∈ (P : M) or r1(m1 + m1) ∈ P or
s1(m1+m1) ∈ P . This implies that (r1, r2)(s1, s2)+ (r1, r2)(s1, s2) ∈ (P ×M2 : M1×M2)
or (r1, r2)[(m1,m2) + (m1,m2)] ∈ P × M2 or (s1, s2)[(m1,m2) + (m1,m2)] ∈ P × M2.
Therefore P ×M2 is a ϕ-β-absorbing submodule of M .



T. Khumprapussorn / Eur. J. Pure Appl. Math, 15 (1) (2022), 328-334 332

Proposition 3. Let Mi be an Ri-module and ϕMi : S(Mi) → S(Mi) ∪ {∅} be a function
where i = 1, 2. For this result, we define ϕ : S(M1 × M2) → S(M1 × M2) ∪ {∅} by
ϕ = ϕM1 × ϕM2. If P1 × P2 is a ϕ-β-absorbing submodule of M1 × M2, then Pi is a
ϕMi-β-absorbing submodule of Mi.

Proof. Assume that P1 × P2 is a ϕ-β-absorbing submodule of M1 × M2. To show
that P1 is a ϕM1-β-absorbing submodule of M1, let r, s ∈ R1 and m ∈ M1 be such that
rsm ∈ P1\ϕM1(P1). Since ϕ(P1 × P2) = ϕM1(P1) × ϕM2(P2), we have (r, 1)(s, 1)(m, 0) =
(rsm, 0) ∈ (P1×P2)\ϕ(P1×P2). Since P1×P2 is a ϕ-β-absorbing submodule of M1×M2,
(r, 1)(s, 1) + (r, 1)(s, 1) ∈ (P1 ×M2 : M1 ×M2) or (r, 1)[(m, 0) + (m, 0)] ∈ P1 ×M2 or
(s, 1)[(m, 0) + (m, 0)] ∈ P1 ×M2. This implies that rs+ rs ∈ (P1 : M1) or r(m+m) ∈ P1

or s(m + m) ∈ P1. Hence P1 is a ϕM1-β-absorbing submodule of M1. Similarly, we can
show that P2 is a ϕM2-β-absorbing submodule of M2.

3. Characterizations of ϕ-β-absorbing submodules

The purpose of this section is to investigate characterizations of ϕ-β-absorbing sub-
modules. For a submodule N of a left R-module M and r ∈ R, we define the symbol Nr

by {m ∈M | rm ∈ N}.

Theorem 2. Let P be a submodule of M . Then the following statements are equivalent :

(i) P is a ϕ-β-absorbing submodule of M .

(ii) For all r, s ∈ R, if rs+ rs /∈ (P : M), then Prs ⊆ α(Pr) ∪ α(Ps) ∪ ϕ(P )rs.

Proof. (i) → (ii) Assume that P is a ϕ-β-absorbing submodule of M . Let r, s ∈ R
be such that rs + rs /∈ (P : M) and m ∈ Prs. Then rsm ∈ P . If rsm ∈ ϕ(P ), then
m ∈ ϕ(P )rs. Assume that rsm ∈ P\ϕ(P ). Since P is a ϕ-β-absorbing submodule of M
and rs+ rs /∈ (P : M), r(m+m) ∈ P or s(m+m) ∈ P . Thus m ∈ α(Pr) or m ∈ α(Ps).
This shows that Prs ⊆ α(Pr) ∪ α(Ps) ∪ ϕ(P )rs.

(ii)→ (i) Assume that (ii) holds. Let r, s ∈ R and m ∈M be such that rsm ∈ P\ϕ(P )
and rs + rs /∈ (P : M). Then m ∈ Prs. By our assumptions, m ∈ α(Pr) ∪ α(Ps). Hence
r(m+m) ∈ P or s(m+m) ∈ P . Therefore P is a ϕ-β-absorbing submodule of M .

For a submodule N of a left R-module M and m ∈M , we define the symbol (N : m)
by {r ∈ R | rm ∈ N}.

Theorem 3. Let P be a submodule of M . Then the following statements are equivalent :

(i) P is a ϕ-β-absorbing submodule of M .

(ii) For all s ∈ R and m ∈M , if sm /∈ α(P ), then

(P : sm) ⊆ α((P : sM)) ∪ α((P : m)) ∪ (ϕ(P ) : sm).



T. Khumprapussorn / Eur. J. Pure Appl. Math, 15 (1) (2022), 328-334 333

Proof. (i) → (ii) Assume that P is a ϕ-β-absorbing submodule of M . Let s ∈ R and
m ∈ M be such that sm /∈ α(P ). Let r ∈ (P : sm). Then rsm ∈ P . If rsm ∈ ϕ(P ), then
r ∈ (ϕ(P ) : sm). Assume that rsm ∈ P\ϕ(P ). Since P is a ϕ-β-absorbing submodule
of M and sm /∈ α(P ), rs + rs ∈ (P : M) or r(m + m) ∈ P . Thus r ∈ α((P : sM)) or
r ∈ α((P : m)). This proves that (P : sm) ⊆ α((P : sM)) ∪ α((P : m)) ∪ (ϕ(P ) : sm).

(ii) → (i) Assume that (ii) holds. Let r, s ∈ R and m ∈ M be such that rsm ∈
P\ϕ(P ) and sm /∈ α(P ). Then r ∈ (P : sm) and r /∈ (ϕ(P ) : sm). This implies that
r ∈ α((P : sM)) ∪ α((P : m)). Hence rs + rs ∈ (P : M) or r(m +m) ∈ P . This shows
that P is a ϕ-β-absorbing submodule of M .

By the definition of ϕ1, a proper submodule P of M is said to be ϕ1-β-absorbing
provided for each r, s ∈ R and m ∈M , if rsm ∈ P\(P : M)β(P ), then rs+ rs ∈ (P : M)
or r(m +m) ∈ P or s(m +m) ∈ P . For an element m of an R-module M , we recall the
symbol that ({0} : m) = {r ∈ R | rm = {0}} and denote ({0} : m) by (0 : m) for short.

Finally, we show some assumptions which β-absorbing submodules and ϕ1-β-absorbing
submodules are equivalent.

Theorem 4. Let m be a nonzero element of an R-module M such that (0 : m) = {0}
and Rm ̸= M . Then Rm is a β-absorbing submodule of M if and only if Rm is a ϕ1-β-
absorbing submodule of M .

Proof. (→) This is obvious.
(←) Assume that Rm is not a β-absorbing submodule of M . Then there are r, s ∈ R

and x ∈M such that rsx ∈ Rm and rs+rs /∈ (Rm : M) and r(x+x) /∈ Rm and s(x+x) /∈
Rm. If rsx /∈ (Rm : M)β(Rm), then we are done. Assume that rsx ∈ (Rm : M)β(Rm).
Since r(x + x) /∈ Rm, r(x + x + m + m) /∈ Rm. Since rsx ∈ Rm and rsm ∈ Rm,
rs(x + m) ∈ Rm. If rs(x + m) /∈ (Rm : M)β(Rm), then we are done. Suppose that
rs(x + m) ∈ (Rm : M)β(Rm). Since rsx ∈ (Rm : M)β(Rm), rsm ∈ (Rm : M)β(Rm).
Note that β(Rm) = β(R)m. Hence rsm ∈ (Rm : M)β(R)m. Thus rsm = tm for
some t ∈ (Rm : M)β(R). So rs = t ∈ (Rm : M)β(R) ⊆ (Rm : M). Consequently,
rs+ rs ∈ (Rm : M) which is a contradiction.

For each r ∈ R, we would like to remind that {0}r = {m ∈M | rm = 0}.

Theorem 5. Let M be an R-module and r ∈ R be such that rM ̸= M and {0}r ⊆ β(rM).
Then rM is a β-absorbing submodule of M if and only if rM is a ϕ1-β-absorbing submodule
of M .

Proof. (→) This is obvious.
(←) Assume that rM is a ϕ1-β-absorbing submodule of M . Let a, b ∈ R and m ∈ M

such that abm ∈ rM . There are 2 cases to be considered :

(i) abm /∈ (rM : M)β(rM),

(ii) abm ∈ (rM : M)β(rM).



REFERENCES 334

First, we consider Case (i). Since rM is a ϕ1-β-absorbing submodule of M , ab + ab ∈
(rM : M) or a(m + m) ∈ rM or b(m + m) ∈ rM . Next, Case (ii) is considered. Since
abm ∈ rM and rbm ∈ rM , (ab + rb)m ∈ rM . If (ab + rb)m /∈ (rM : M)β(rM), then
(a + r)b + (a + r)b ∈ (rM : M) or (a + r)(m + m) ∈ rM or b(m + m) ∈ rM . Since
rb + rb ∈ (rM : M) and r(m + m) ∈ rM , ab + ab ∈ (rM : M) or a(m + m) ∈ rM or
b(m + m) ∈ rM . Assume that (ab + rb)m ∈ (rM : M)β(rM). Since β(rM) = rβ(M),
(ab + rb)m and abm are elements of r(rM : M)β(M). This implies that rbm ∈ r(rM :
M)β(M). Thus rbm = ry for some y ∈ (rM : M)β(M). So bm − y ∈ {0}r ⊆ β(rM).
Therefore bm = (bm−y)+y ∈ (rM : M)β(rM)+{0}r ⊆ β(rM). We have b(m+m) ∈ rM .
Therefore rM is a β-absorbing submodule of M .

Acknowledgements

This work was supported by School of Science, King Mongkut’s Institute of Technology
Ladkrabang under Grant no.2565-02-05-001.

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