IHJPAS. 36(2)2023 

354 
This work is licensed under a Creative Commons Attribution 4.0 International License. 

 

   

 

 

 

 

 

 

 

 
 

 

 

 

Abstract 

Let ℋ be a module over a commutative ring 𝑅 with identity. In this paper we intoduce the 
concept of Strongly Pseudo Nearly Semi-2-Absorbing submodule, where a  proper submodule ℱ 
of an 𝑅-module ℋ is said to be Strongly Pseudo Nearly Semi-2-Absorbing submodule of ℋ  if 
whenever 𝑢2ϰ ∈ ℱ, for 𝑢 ∈ 𝑅, 𝜘 ∈ ℋ implies that either 𝑢𝜘 ∈ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2 ∈
[ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ):𝑅 ℋ], this concept is a generalization of 2_Absorbing submodule, semi 2-
Absorbing submodule, and strong form of (Nearly–2–Absorbing, Pseudo_2_Absorbing, and 

Nearly Semi–2–Absorbing) submodules. Several properties characterizations, and examples 

concerning this new notion are given. We study the relation between Strongly Pseudo Nearly 

Semei-2-Absorbing submodule and (2_Absorbing, Nearly_2_Absorbing, Pseudo_2_Absorbing, 

and Nearly Semi–2–Absorbing) submodules and the converse of this relation is true under certain 

condition. Also, we introduced many characterizations of  Strongly Pseudo Nearly Semei-2-

Absorbing submodules in some types of modules. 

 

Keywords:2_Absorbing.submodules,.Pseudo_2_Absorbing.submodules,.Nearly_2_Absorbingsu

bmodules, Nearly Semei-2-Absorbing submodules, STPNS-2-Absorbing submodules. 

 

1. Introduction 

 Throughout this paper, all rings are assumed commutative with identity and all R-modules are 

left unitary. According Darani and Soheiline in 2011, we introduce a concept 2-Absorbing 

submodule. Many researchers have generalized the concept of 2_Absorbing submodules in 

different way. Recently, in 2018 Reem and Shwkea introduced the concept of Nearly–2–

Absorbing submodules as new generalization of 2_Absorbing submodules. Also, in 2019 Haibat 

and Omer introduced the concept Pseudo_2_Absorbing submodules as new generalization of 

2_Absorbing submodules. In 1967 Goodearl made a generalization of the concept of 2_Absorbing 

which is semi-2-Absorbing submodule. Also, in 2019 Akram and Haibat, Haibat and Omer made 

a generalization of the concept of 2_Absorbing an illusion Nearly semi-2-Absorbing and Pseudo 

semi-2-Absorbing. Our goal in this paper was studied the concept Strongly Pseudo Nearly Semei-

2-Absorbing submodules as new generalization of 2_Absorbing submodule, strong form of both 

Nearly–2–Absorbing, Pseudo_2_Absorbing, Nearly Semi–2–Absorbing and Pseudo semi-2-

Absorbing submodules are given Several properties and characterizations of this concepts in some 

doi.org/10.30526/36.2.3021 

Article history: Received 17 September 2022, Accepted 6 November 2022, Published in April 2023. 

 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

 

Strongly Pseudo Nearly Semei-2-Absorbing Submodule(𝐈) 
 

Mohmad E. Dahash 
Department of Mathematics College of 

Computer Science and Mathematics 
Tikrit University / Iraq. 

mohmad.e.dahash35391@st.tu.edu.iq 

 

Haibat K. Mohammadali 
Department of Mathematics College of 

Computer Science and Mathematics 

Tikrit University / Iraq. 
H.mohammadali@tu.edu.iq 

 

 
 

https://creativecommons.org/licenses/by/4.0/
mailto:mohmad.e.dahash35391@st.tu.edu.iq
mailto:H.mohammadali@tu.edu.iq
mailto:H.mohammadali@tu.edu.iq


IHJPAS. 36(2)2023 

355 
 

types of modules. 

 

2.  Basic Properties 

In this section we introduce the basic properties of the concept Strongly Pseudo Nearly Semei-

2-Absorbing submodules. 

Definition 2.1 [1]. 

Let ℋ be an 𝑅-module and ℱ ⊂ ℋ is called 2-absorbing if whenever 𝑢𝑣ℎ ∈ ℱ , for  u, 𝑣 ∈ 𝑅, ℎ ∈
ℋ, then either 𝑢ℎ ∈ ℱ or 𝑣ℎ ∈ ℱ or  𝑢𝑣 ∈ [ℱ ℋ𝑅

: ]. 

Definition 2.2 [2]. 

𝑠𝑜𝑐(ℋ) is the intersection of all essential submodule of ℋand a nonzero submodule ℚ of ℋ is a 
essential in ℋ if ℚ ∩ ℂ ≠ (0) for any nonzero submodule ℂ of ℋ. 

Definition 2.3 [3]. 

𝐽(ℋ) is the intersection of all maximal submodule of ℋ and a proper submodule ℬ of an R._module 
ℋ is called maximal submodule if whenever ℬ ⊆ ℒ ⊆ ℋ for some submodule ℒ of  ℋ then either 
ℬ = ℒ or ℒ = ℋ. 

 Definition 2.4 [4]. 

Let ℋ be an 𝑅-module and ℱ ⊂ ℋ is called Nearly-2-Absorbing submodule if whenever 𝑢𝑣𝑚 ∈
ℱ , for 𝑢, 𝑣 ∈ 𝑅, ℎ ∈ ℋ, implies that either 𝑢ℎ ∈ ℱ + 𝐽(ℋ) or 𝑣ℎ ∈ ℱ + 𝐽(ℋ) or 𝑢𝑣 ∈
[ℱ + 𝐽(ℋ)𝑅 : ℋ]. 

Definition 2.5 [5]. 

Let ℋ be an 𝑅-module and ℱ ⊂ ℋ is called Pseudo-2-Absorbing submodule if whenever 𝑢𝑣𝑚 ∈
ℱ , for 𝑢, 𝑣 ∈ 𝑅, ℎ ∈ ℋ, implies that either 𝑢ℎ ∈ ℱ + 𝑆𝑜𝑐(ℋ) or 𝑣ℎ ∈ ℱ + 𝑆𝑜𝑐(ℋ) or 𝑢𝑣 ∈
[ℱ + 𝑆𝑜𝑐(ℋ)𝑅 : ℋ]. 

Definition 2.6 [2]. 

Let ℋ be an 𝑅-module and ℱ ⊂ ℋ is called semi-2-absorbing if whenever 𝑢2𝑦 ∈ ℱ , for𝑢 ∈ 𝑅, 
𝑦 ∈ ℋ implies that 𝑢𝑦 ∈ ℱ or  𝑢2 ∈ [ℱ :𝑅 ℋ ]. 

Definition 2.7 [6]. 

Let ℋ be an 𝑅-module and ℱ ⊂ ℋ is called Nearly semi-2 absorbing submodule if whenever 
a2y ∈ N, for a ∈ R, y ∈ M implies that ay ∈ ℱ + J(ℋ) or a2 ∈ [ℱ + J(ℋ) ℋR

: ]. 

Definition 2.8 [5]. 

A proper submodule ℱ of an R-module ℋ  is called Pseudo semi-2 absorbing submodule if 
whenever a2y ∈ N, for a ∈ R, y ∈ M implies that ay ∈ ℱ + soc(ℋ) or a2 ∈ [ℱ + soc(ℋ) ℋR

: ]. 

Lemma 2.9 [ 7, Lemma. ( 2.3.15)]. 
Let  ℒ  , ℚ and  ℬ  be  submodule  of   an  𝑅_module  ℋ with   ℚ  ℬ  . Then  ( ℒ  +  ℚ)  ∩  ℬ   =
 ( ℒ  ∩  ℬ  )  +  ℚ =  ( ℒ  ∩  ℬ  )  +  ( ℚ ∩  ℬ  ). 

Lemma 2.10 [ 7, Example .(12) (c)]. 
It  well – known  that  an 𝑅 _ module  ℋ  is a  semi simple   if  and  only  if  for  each  submodule  

ℱ of  ℋ  , 𝑠𝑜𝑐 (
ℋ

ℱ 
) =

𝑠𝑜𝑐(ℋ)+ℱ

ℱ
 . 

Lemma 2.11 [ 7, Example(12), P. 239]. 

Let ℱ be a submodule of a semi simple 𝑅 _ module  ℋ then 𝐽 (
ℋ

ℱ 
) =

𝐽(ℋ)+ℱ

ℱ
 .  

 



IHJPAS. 36(2)2023 

356 
 

Definition 2.12 [8]. 

An element 𝑎 ∈ 𝑅 is an idempotent element if 𝑎2 = 𝑎. 𝑎 = 𝑎. And if every element 𝑎 of a ring 𝑅 
is an idempotent, then 𝑅 is a Boolean ring. 

Lemma 2.13 [ 9, Theorem(2.2)] 
If 𝑅 is a Boolean ring, then 𝑅 is a regular ring. 

Definition 2.14 [10]. 

An 𝑅-module H is called regular module if every submodule of H is a pure. 

Lemma 2.15 [10]. 

If  ℋ is a regular R-module, then 𝐽( ℋ) = 0. 

Definition 2.16 [2] 

An R_module  ℋ is  said  to  be  a semi simple , if  every  submodule  of  ℋ is a direct  summand  
of  ℋ, that is  if  ℱ is  a submodule  of  ℋ , then  ℋ = ℱ⨁𝐾  for  some  submodule  𝐾 of  ℋ. 

Lemma 2.17 [ 7, proposition . (9.14) (c)] 
If  ℋ is a semi-simple R-module, then J( ℋ) = 0. 

Lemma 2.18[ 5, remark (1.2)]. 
It is clear that every 2-Absorbing submodule of an 𝑅-module ℋ is Pseudo 2-Absorbing submodule. 

Lemma 2.19 [ 4, proposition (2.8)]. 
Let 𝐴 be a Nearly 2-Absorbing submodule of an 𝑅-module ℋ with 𝐽(ℋ) ⊆ 𝐴. Then 𝐴 is 2-
Absorbing submodule. 

Lemma 2.20 [ 6, proposition (2.11)]. 
Let ℋ be an 𝑅-module and 𝐴 is a proper submodule of ℋ with 𝑠𝑜𝑐(ℋ) ⊆ 𝐴. Then 𝐴 is Semi 2-
Absorbing of ℋ if and only if 𝐴 Nearly Semi 2-Absorbing of ℋ. 

3.  The Results  

In this section we introduce the definition of Strongly Pseudo Nearly Semi-2-Absorbing 

submodule. Example, characterizations, some basic properties of this concept are given: 

Definition 3.1 

     A proper submodule ℱ of an 𝑅-module ℋ is said to be Strongly Pseudo Nearly Semi-2-
Absorbing submodule of ℋ (for short STPNS) if whenever 𝑢2ϰ ∈ ℱ, for 𝑢 ∈ 𝑅, 𝜘 ∈ ℋ implies 
that either 𝑢𝜘 ∈ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2 ∈ [ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ):𝑅 ℋ]. 

Remarks and Examples 3.2 

1. Let ℋ = 𝑍36 , 𝑅 = 𝑍 and the submodule ℱ = 〈4̅〉 is STPNS-2-Absorbing submodule of ℋ since 
𝑠𝑜𝑐(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 ∩ 〈6̅〉 ∩ 𝑍36 = 〈6̅〉 and 𝐽(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢 ∈ 𝑍 and 

ϰ ∈ 𝑍36 such that 𝑢
2ϰ ∈ 〈4̅〉, implies that either  𝑢ϰ ∈ ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36)) = 〈2̅〉 +

(〈6̅〉 ∩ 〈6̅〉) = 〈2̅〉 𝑢2 ∈ [ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36))𝑅
: 𝑍36] =  2𝑍. That is 2

2. 4̅ ∈ 〈2̅〉, implies that 

2. 4̅ = 8̅ ∈ 〈2̅〉 or 22 = 4 ∈ [ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36))𝑅
: 𝑍36] = 2𝑍.  

2. Every 2-Absorbing submodule is STPNS-2-Absorbing, but the converse is not true. 

Proof 

Assume that ℱ is 2-Absorbing submodule of ℋ and 𝑢2𝑚 ∈ ℱ, for 𝑢 ∈ 𝑅, 𝑚 ∈ ℋ, that is 𝑢 . 𝑢. 𝑚 ∈
ℱ. Since ℱ is 2-Absorbing submodule, then either  𝑢𝑚 ∈ ℱ ⊆ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) or 𝑢. 𝑢ℋ =



IHJPAS. 36(2)2023 

357 
 

𝑢2ℋ ⊆ ℱ ⊆ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))implies that 𝑢2 ∈ [ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))𝑅 : ℋ]. Hence ℱ is 
STPNS-2-Absorbing submodule of ℋ. 
For the converse consider the following example: 

Let ℋ = 𝑍36 , 𝑅 = 𝑍 and the submodule ℱ = 〈12̅̅̅̅ 〉 is STPNS_2_Absorbing submodule of ℋ since 
𝑠𝑜𝑐(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 ∩ 〈6̅〉 ∩ 𝑍36 = 〈6̅〉 and 𝐽(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢 ∈ 𝑍 and 

ϰ ∈ 𝑍36 such that 𝑢
2ϰ ∈ 〈12̅̅̅̅ 〉, implies that either 𝑢ϰ ∈ ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36)) = 〈12̅̅̅̅ 〉 +

(〈6̅〉 ∩ 〈6̅〉) = 〈6̅〉 𝑢2 ∈ [ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36))𝑅
: 𝑍36] =  6𝑍. That is 2

2. 3̅ ∈ 〈12̅̅̅̅ 〉, implies that 

2. 3̅ = 6̅ ∈ ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36)) = 6̅ ∈ 〈12̅̅̅̅ 〉 + (〈6̅〉 ∩ 〈6̅〉) = 〈6̅〉. But ℱ = 〈12̅̅̅̅ 〉 is not 

2_Absorbing submodule of ℋ, since 2.3. 2̅ ∈ 〈12̅̅̅̅ 〉, but 2. 2̅ ∉ 〈12̅̅̅̅ 〉 and 3. 2̅ ∉ 〈12̅̅̅̅ 〉 and 2.3 ∉
[〈12̅̅̅̅ 〉 𝑍36𝑅

: ] = 12𝑍. 

3. Every STPNS-2-Absorbing  submodule is Nearly-2-Absorbing, but the converse is not true. 

Proof  

Suppose that ℱ is STPNS-2-Absorbing submodule of ℋand 𝑢2𝑚 ∈ ℱ, for 𝑢 ∈ 𝑅, 𝑚 ∈ ℋ. Since 
ℱ is STPNS-2-Absorbing submodule, then 𝑢𝑚 ∈ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ⊆ ℱ + 𝐽(ℋ). Hence ℱ 
is Nearly-2-Absorbing submodule of ℋ. 
For the converse consider the following example: 

Let ℋ = 𝑍48 , 𝑅 = 𝑍 and the submodule ℱ = 〈24̅̅̅̅ 〉 is Nearly-2-Absorbing submodule of 𝑀 since 
𝑠𝑜𝑐(𝑍48) = 〈2̅〉 ∩ 〈4̅〉 ∩ 〈8̅〉 = 〈8̅〉 and 𝐽(𝑍48) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢, 𝑣 ∈ 𝑍 and 𝑚 ∈

𝑍48 such that 𝑢𝑣𝑚 ∈ 〈24̅̅̅̅ 〉, implies that either  𝑢𝑚 ∈ ℱ + (𝐽(𝑍48)) = 〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉 or  

𝑣𝑚 ∈ ℱ + (𝐽(𝑍48)) = 〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉. That is 2.4. 3̅ ∈ 〈24〉, implies that 2. 3̅ = 6̅ ∈ ℱ +

(𝐽(𝑍48)) = 〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉 and 4. 3̅ = 12̅̅̅̅ ∈ ℱ + (𝐽(𝑍48)) = 〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉. But ℱ =
〈24̅̅̅̅ 〉 is not STPNS-2-Absorbing submodule of ℋ, since 22. 6̅ ∈ 〈24̅̅̅̅ 〉, but 2. 6̅ ∉ 〈24〉 +

(𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48)) = 〈24̅̅̅̅ 〉 and  2
2 ∉ [〈24̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48))𝑅

: 𝑍48] =  24𝑍. 

4. Every STPNS-2-Absorbing  submodule is Pseudo-2-Absorbing, but the converse is not true. 

Proof  

Suppose that ℱ is STPNS-2-Absorbing submodule of ℋand 𝑢2𝑚 ∈ ℱ, for 𝑢 ∈ 𝑅, 𝑚 ∈ ℋ. Since 
ℱ is STPNS-2-Absorbing submodule, then 𝑢𝑚 ∈ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ⊆ ℱ + 𝑠𝑜𝑐(ℋ). Hence 
ℱ is pseudo-2-Absorbing submodule of ℋ. 
For the converse consider the following example: 

Let ℋ = 𝑍48 , 𝑅 = 𝑍 and the submodule ℱ = 〈12̅̅̅̅ 〉 is pseudo-2-Absorbing submodule of 𝑀 since 
 𝑠𝑜𝑐(𝑍48) = 〈2̅〉 ∩ 〈4̅〉 ∩ 〈8̅〉 = 〈8̅〉 and 𝐽(𝑍48) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢, 𝑣 ∈ 𝑍 and 𝑚 ∈

𝑍48 such that 𝑢𝑣𝑚 ∈ 〈12̅̅̅̅ 〉, implies that either 𝑢𝑚 ∈ ℱ + (𝑠𝑜𝑐(𝑍48)) = 〈12̅̅̅̅ 〉 + (〈8̅〉) = 〈4̅〉 

or 𝑣𝑚 ∈ ℱ + (𝑠𝑜𝑐(𝑍48)) = 〈12̅̅̅̅ 〉 + (〈8̅〉) = 〈4̅〉or 𝑢𝑣 ∈ [ℱ + (𝑠𝑜𝑐(𝑍48))𝑅
: 𝑍48] =  4𝑍. That is 

2.2. 3̅ ∈ 〈12̅̅̅̅ 〉, implies that 2.2 ∈ [ℱ + (𝑠𝑜𝑐(𝑍48))𝑅
: 𝑍48] =  4𝑍 . But ℱ = 〈12̅̅̅̅ 〉 is not STPNS-2-

Absorbing submodule of ℋ, since 22. 3̅ ∈ 〈12̅̅̅̅ 〉, but 2. 3̅ ∉ 〈12̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48)) = 〈12̅̅̅̅ 〉 

and 22 ∉ [〈12̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48))𝑅
: 𝑍48] = 12𝑍. 

5. Every semi-2-Absorbing submodule is STPNS-2-Absorbing, but the converse is not true. 

Proof  

Let ℱ be a semi-2-Absorbing submodule of ℋ, and 𝑢2ϰ ∈ ℱ for 𝑢 ∈ 𝑅, ϰ ∈ ℋ. Since ℱ  is a semi-
2-Absorbing, then either 𝑢ϰ ∈ ℱ ⊆ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2ℋ ⊆ ℱ ⊆ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ). 



IHJPAS. 36(2)2023 

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That is either 𝑢ϰ ∈ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2 ∈ [ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) :𝑅 ℋ]. Hence ℱ is 
STPNS-2-absorbing submodule of ℋ. 
For the converse consider the following example: 

Let ℋ = 𝑍36 , 𝑅 = 𝑍 and the submodule ℱ = 〈12̅̅̅̅ 〉 is STPNS_2_Absorbing submodule of ℋ since 
𝑠𝑜𝑐(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 ∩ 〈6̅〉 ∩ 𝑍36 = 〈6̅〉 and 𝐽(𝑍36) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢 ∈ 𝑍 and 

ϰ ∈ 𝑍36 such that 𝑢
2ϰ ∈ 〈12̅̅̅̅ 〉, implies that either 𝑢ϰ ∈ ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36)) = 〈12̅̅̅̅ 〉 +

(〈6̅〉 ∩ 〈6̅〉) = 〈6̅〉 or 𝑢2 ∈ [ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36))𝑅
: 𝑍36] =  6𝑍. That is 2

2. 3̅ ∈ 〈12̅̅̅̅ 〉, implies 

that 2. 3̅ = 6̅ ∈ ℱ + (𝐽(𝑍36) ∩ 𝑠𝑜𝑐(𝑍36)) = 6̅ ∈ 〈12̅̅̅̅ 〉 + (〈6̅〉 ∩ 〈6̅〉) = 〈6̅〉. But ℱ = 〈12̅̅̅̅ 〉 is not 

semi-2_Absorbing submodule of ℋ, since 22. 3̅ ∈ 〈12̅̅̅̅ 〉, but 2. 3̅ ∉ 〈12̅̅̅̅ 〉 and 22 ∉ [〈12̅̅̅̅ 〉 𝑍36𝑅
: ] =

12𝑍. 

6. Every STPNS-2-Absorbing  submodule is Nearly semi-2-Absorbing but, the converse is not 

true. 

Proof  

Clear. 

For the converse consider the following example: 

Let ℋ = 𝑍48 , 𝑅 = 𝑍 and the submodule ℱ = 〈24̅̅̅̅ 〉 is Nearly semi-2-Absorbing submodule of 𝑀 
since 𝑠𝑜𝑐(𝑍48) = 〈2̅〉 ∩ 〈4̅〉 ∩ 〈8̅〉 = 〈8̅〉 and 𝐽(𝑍48) = 〈2̅〉 ∩ 〈3̅〉 = 〈6̅〉 that is for all 𝑢 ∈ 𝑍 and 

𝑚 ∈ 𝑍48 such that 𝑢
2𝑚 ∈ 〈24̅̅̅̅ 〉, implies that either  𝑢𝑚 ∈ ℱ + (𝐽(𝑍48)) = 〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉 or  

𝑢2 ∈ [〈24̅̅̅̅ 〉 + 𝐽(𝑍48)𝑅 : 𝑍48] = 4𝑍. That is 2
2. 6̅ ∈ 〈24̅̅̅̅ 〉, implies that 2. 6̅ = 12̅̅̅̅ ∈ ℱ + (𝐽(𝑍48)) =

〈24̅̅̅̅ 〉 + (〈6̅〉) = 〈6̅〉 and 22 ∈ [〈24̅̅̅̅ 〉 + 𝐽(𝑍48)𝑅 : 𝑍48] = 4𝑍. But ℱ = 〈24̅̅̅̅ 〉 is not STPNS-2-

Absorbing submodule of ℋ, since 22. 6̅ ∈ 〈24̅̅̅̅ 〉, but 2. 6̅ ∉ 〈24〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48)) = 〈24̅̅̅̅ 〉 

and  22 ∉ [〈24̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48))𝑅
: 𝑍48] =  24𝑍. 

Proposition 3.3 

        Let ℱ be a proper submodule of an 𝑅-module ℋ. Then ℱ is a STPNS-2-Absorbing submodule 
of ℋ if and only if Ι2ℒ ⊆ ℱ for Ι is an ideal of 𝑅 and ℒ is a submodule of ℋ implies that either 
Ιℒ ⊆ ℱ + (J(ℋ) ∩ soc(ℋ)) or Ι2 ⊆ [ℱ + (J(ℋ) ∩ soc(ℋ)):𝑅 ℋ]. 

Proof   

(⟹) Suppose that ℱ is STPNS-2-Absorbing submodule of an 𝑅-module ℋ, and Ι2ℒ ⊆ ℱ, for 𝛪 is 
an ideal of 𝑅 and ℒ is a submodule of ℋ and 𝐼2 ⊈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ].To prove that Ιℒ ⊆
ℱ + J(ℋ) ∩ soc(ℋ). Let ϰ ∈ 𝛪ℒ, implies that ϰ = 𝑢1ϰ1 + 𝑢2ϰ2 + ⋯ + 𝑢𝑛 ϰ𝑛 for 𝑢𝑖 ∈ 𝛪, ϰ𝑖 ∈ 𝐿 
for all 𝑖 = 1,2,3, … , 𝑛, thus 𝑢𝑖

2ϰ𝑖 ∈ Ι
2ℒ ⊆ ℱ, for 𝑢𝑖

2 ∉ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ], but ℱ is 
STPN-2-Absorbing of ℋ and 𝑢𝑖

2 ∉ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ], implies that 𝑢𝑖 ϰ𝑖 ∈ ℱ + J(ℋ) ∩
soc(ℋ) for 𝑖 = 1,2,3, … , 𝑛. Hence ϰ ∈ ℱ + J(ℋ) ∩ soc(ℋ) it follows that 𝛪ℒ ⊆ ℱ + J(ℋ) ∩
soc(ℋ). 

(⟸) Suppose that 𝑢2ϰ ∈ ℱfor 𝑢 ∈ 𝑅, ϰ ∈ ℋ, implies that (𝑢2)(ϰ) ⊆ ℱ, by hypothesis we have 
either (𝑢)(ϰ) ⊆ ℱ + J(ℋ) ∩ soc(ℋ) or (𝑢2) ⊆  [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. That is 𝑢ϰ ∈ ℱ +
J(ℋ) ∩ soc(ℋ) or 𝑢2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. Hence ℱ is STPNS-2-Absorbing submodule 
of ℋ. 

 As a direct consequence of Proposition 3.3 we have the following corollaries. 

Corollary 3.4 

        Let ℱ be a proper submodule of an 𝑅-module ℋ. Then ℱ  is STPNS-2-Absorbing submodule 
of ℋ if and only if u2ℒ ⊆ ℱ for 𝑢 ∈ 𝑅 and ℒ is a submodule of ℋ implies that either 𝑢ℒ ⊆ ℱ +
J(ℋ) ∩ soc(ℋ)  or u2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. 



IHJPAS. 36(2)2023 

359 
 

Corollary 3.5 

        Let ℱ be a proper submodule of an 𝑅-module ℋ. Then ℱ is STPNS-2-Absorbing submodule 
of ℋif and only if Ι2ℋ ⊆ ℱ for I is an ideal of 𝑅, implies that either Iℋ ⊆ ℱ + J(ℋ) ∩
soc(ℋ)   or 𝐼2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. 

Proposition 3.6 

         Let  ℋ be an 𝑅-module and ℱ be a proper submodule of ℋ. Then ℱ + J(ℋ) ∩ soc(ℋ) is 
STPNS-2-Absorbing submodule of ℋ if and only if [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢

2ϰ] ⊆ [ℱ + J(ℋ) ∩
soc(ℋ):𝑅 𝑟ϰ] for each ϰ ∈ ℋ or 𝑢

2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. 

Proof 

(⟹)  let 𝜔 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢
2ϰ] and 𝑢2 ∉ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ], since 𝜔 ∈

[ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢
2ϰ], then 𝑢2𝜔ϰ ∈ ℱ + J(ℋ) ∩ soc(ℋ). But ℱ + J(ℋ) ∩ soc(ℋ) is 

STPNS-2-Absorbing submodule of ℋ and 𝑢2 ∉ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ], then 𝑢𝜔ϰ ∈

(ℱ + J(ℋ) ∩ soc(ℋ)) + J(ℋ) ∩ soc(ℋ) = ℱ + J(ℋ) ∩ soc(ℋ).That is 𝜔 ∈ [ℱ + J(ℋ) ∩

soc(ℋ):𝑅 𝑢ϰ] .Thus  [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢
2ϰ] ⊆ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ]. It is clear that 

[ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ] ⊆ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢
2ϰ].   Hence[ℱ + J(ℋ) ∩

soc(ℋ):𝑅 𝑢
2ϰ] = [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ].  

(⟸) Let 𝑢2ϰ ∈ ℱ + J(ℋ) ∩ soc(ℋ), for 𝑢 ∈ 𝑅, ϰ ∈ ℋ, then by hypothesis [ℱ + J(ℋ) ∩
soc(ℋ):𝑅 𝑢

2ϰ] = [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ] or 𝑢
2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. If [ℱ +

J(ℋ) ∩ soc(ℋ):𝑅 𝑢
2ϰ] = [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ] and 𝑢

2ϰ ∈ ℱ + J(ℋ) ∩ soc(ℋ)  then 
[ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢

2ϰ] = 𝑅, it follows that [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 𝑢ϰ] = 𝑅, hence 𝑢ϰ ∈
ℱ + J(ℋ) ∩ soc(ℋ) ⊆ ℱ + J(ℋ) ∩ soc(ℋ) + J(ℋ) ∩ soc(ℋ), so 𝑢ϰ ∈ ℱ + J(ℋ) ∩ soc(ℋ) +
J(ℋ) ∩ soc(ℋ) or  𝑢2ℋ ⊆ ℱ + J(ℋ) ∩ soc(ℋ) + J(ℋ) ∩ soc(ℋ). That is ℱ + J(ℋ) ∩ soc(ℋ) 
is STPN-2-Absorbing submodule of ℋ. 

Proposition 3.7 

        Let ℋ be an 𝑅-module and ℱ be a proper submodule of ℋ with J(ℋ) ∩ soc(ℋ) ⊆ ℱ. Then 
ℱ is STPNS-2-Absorbing submodule of ℋ if and only if [ℱ:ℋ 𝑢

2] ⊆ [ℱ:ℋ 𝑢] or 𝑢
2 ∈

[ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. 

Proof 

(⟹) Suppose that ℱ is STPNS-2-Absorbing submodule of ℋ, let ϰ ∈ [ℱ:𝑅 𝑢
2] and  𝑢2 ∉

[ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. Since ϰ ∈ [ℱ:𝑅 𝑢
2], implies that 𝑢2ϰ ∈ ℱ, but ℱ is STPNS-2-

Absorbing submodule of ℋ and 𝑢2 ∉ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ], implies that 𝑢ϰ ∈ ℱ + J(ℋ) ∩
soc(ℋ). Since J(ℋ) ∩ soc(ℋ) ⊆ ℱ, then ℱ + J(ℋ) ∩ soc(ℋ) = ℱ, implies that 𝑢ϰ ∈ ℱ, hence 
ϰ ∈ [ℱ:ℋ 𝑢], thus [ℱ:ℋ 𝑢

2] ⊆ [ℱ:ℋ 𝑢]. Clear [ℱ:ℋ 𝑢] ⊆ [ℱ:ℋ 𝑢
2], hence [ℱ:ℋ 𝑢

2] = [ℱ:ℋ 𝑢]. 

(⟸) Let 𝑢2ϰ ∈ ℱfor 𝑢 ∈ 𝑅, ϰ ∈ ℋ, then ϰ ∈ [ℱ:ℋ 𝑢
2] = [ℱ:ℋ 𝑢], by hypothesis, implies that 

ϰ ∈ [ℱ:ℋ 𝑢], then 𝑢ϰ ∈ ℱ or 𝑢
2 ∈ [ℱ + J(ℋ) ∩ soc(ℋ):𝑅 ℋ]. Hence ℱ is STPNS-2-Absorbing 

submodule of ℋ. 

Remark 3.8 
The intersection of two STPNS-2-Absorbing submodule of an 𝑅-module ℋ need not to be STPNS-
2-Absorbing submodule of ℋ. The following example explains that: 

Consider the Z_module 𝑍48 and the submodules ℒ=〈3̅〉 and ℚ=〈4̅〉 are STPNS_2_Absorbing 
submodules of the Z_module 𝑍48 (because 〈3̅〉 and 〈4̅〉 are 2-Absorbing of 𝑍48 ), but ℒ ∩ ℚ = 〈12̅̅̅̅ 〉 

is not STPNS_2_Absorbing, since 22. 3̅ ∈ 〈12̅̅̅̅ 〉, but 2. 3̅ ∉ 〈12̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48)) = 〈12̅̅̅̅ 〉 

and  22 ∉ [〈12̅̅̅̅ 〉 + (𝐽(𝑍48) ∩ 𝑠𝑜𝑐(𝑍48))𝑅
: 𝑍48]  = 12𝑍. 



IHJPAS. 36(2)2023 

360 
 

The inverse of above remark satisfy under certain condition.   

Proposition 3.9 

 Let ℒ and ℱ be a proper submodules of an 𝑅-module ℋ, with 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) ⊆ ℒ or 𝐽(ℋ) ∩
𝑠𝑜𝑐(ℋ) ⊆ ℱ . If ℒ and ℱ are STPNS-2-Absorbing submodules of ℋ, then ℒ ∩ ℱ  is STPNS-2-
Absorbing submodule of ℋ. 

Proof 

  Assume that  𝑢2ϰ ∈  ℒ ∩ ℱ   for 𝑢 ∈ 𝑅, ϰ ∈ ℋ, implies that 𝑢2ϰ ∈  ℒ and 𝑢2ϰ ∈  ℱ  . Since both 
ℒ and ℱ are STPNS-2-Absorbing submodule of ℋ, then either 𝑢ϰ ∈ ℒ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 
𝑢2ℋ ⊆ ℒ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) and ϰ ∈ ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2ℋ ⊆ ℱ + 𝐽(ℋ) ∩
𝑠𝑜𝑐(ℋ)Thus either 𝑢ϰ ∈ (ℒ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ∩ (ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) or 𝑢2ℋ ⊆ (ℒ +
𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ∩ (ℱ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)). If 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) ⊆ ℱ , then ℱ + 𝐽(ℋ) ∩
𝑠𝑜𝑐(ℋ) = ℱ, hence either  𝑢ϰ ∈ (ℒ + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ∩ ℱ or 𝑢2ℋ ⊆ (ℒ + 𝐽(ℋ) ∩
𝑠𝑜𝑐(ℋ)) ∩ ℱ. Thus by lemma 2.9we have either 𝑢ϰ ∈ (ℒ ∩ ℱ) + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) or 𝑢2ℋ ⊆
(ℒ ∩ ℱ) + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ). Therefore ℒ ∩ ℱ is STPNS-2-Absorbing submodule of ℋ. In similar 
way ℒ ∩ ℱ is STPNS-2-Absorbing submodule of  ℋ if 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) ⊆ ℒ. 

Proposition 3.10 

Let  ℱ  be STPNS_2_Absorbing  submodule of  an R_ module  ℋ and  ℒ is a submodule  of  ℋ  

with  ℒ ⊆ ℱ,   then  
ℱ

ℒ
 is  STPNS_2_Absorbing  submodule  of  an R_ module   

ℋ

ℒ
 . 

Proof  

Let  ℱ  be STPNS_2_Absorbing  submodule of  ℋ,  and  𝑢2(𝑒 + ℒ) = 𝑢2𝑒 + ℒ ∈
ℱ

ℒ
  for 𝑢 𝑅 and 

𝑒 + ℒ ∈
ℋ

ℒ
  , 𝑒  ℋ , implies  that  𝑢2𝑒 ∈ ℱ. Since ℱ is STPNS_2_Absorbing  submodule of  ℋ ,  

then either  𝑢𝑒 ∈ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))  or  𝑢2 ∈ [ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))𝑅 : ℋ]. That  is  either  
𝑢𝑒 ∈ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))  or 𝑢2ℋ ⊆ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)). It  follows  that  either  𝑢(𝑒 +

ℒ) ∈
ℱ+(𝐽(ℋ)∩𝑠𝑜𝑐(ℋ))

ℒ
    or   𝑢2(

ℋ

ℒ
) ⊆

ℱ+(𝐽(ℋ)∩𝑠𝑜𝑐(ℋ))

ℒ
 . That is either  𝑢(𝑒 + ℒ) ∈

ℱ

ℒ
+

ℱ+(𝐽(ℋ)∩𝑠𝑜𝑐(ℋ))

ℒ
⊆

ℱ

ℒ
+ (𝐽 (

ℋ

ℒ
) ∩ 𝑠𝑜𝑐 (

ℋ

ℒ
) or 𝑢2 (

ℋ

ℒ
) ⊆

ℱ+(𝐽(ℋ)∩𝑠𝑜𝑐(ℋ))

ℒ
⊆

ℱ

ℒ
+ (𝐽 (

ℋ

ℒ
) ∩ 𝑠𝑜𝑐 (

ℋ

ℒ
). 

That  is  either  𝑢(𝑒 + ℒ) ⊆
ℱ

ℒ
+ (𝐽 (

ℋ

ℒ
) ∩ 𝑠𝑜𝑐 (

ℋ

ℒ
) or  𝑢2 ∈ [

ℱ

ℒ
+ (𝐽 (

ℋ

ℒ
) ∩ 𝑠𝑜𝑐 (

ℋ

ℒ
))𝑅:

ℋ

ℒ
].  Hence  

 
ℱ

ℒ
   is STPNS_ 2_ Absorbing   submodule  of  an  R_ module   

ℋ

ℒ
 . 

Proposition 3.11  

Let  ℋ be  an 𝑅_module ,  and  ℱ  is  a proper  submodule  of   ℋ with   𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) ⊆ ℱ . 
Then  ℱ is STPNS_ 2_ Absorbing   submodule  of   ℋ if  and  only  if  [ℱℋ : I] is  
STPNS_ 2_ Absorbing   submodule  of   ℋ for each  ideal  I  of  R . 
Proof  

(  )  Let  𝑢2ℒ ⊆ [ℱℋ : I]  for  u  R  , ℒ is  a submodule  of  ℋ , then  𝑢
2(Iℒ) ⊆ ℱ. Since  ℱ  is 

STPNS_ 2_ Absorbing   submodule  of   ℋ then   by  proposition (3.3)  either 𝑢(Iℒ) ⊆ ℱ +
(𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) or  𝑢2 ∈ [ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ))𝑅: ℋ]. But  𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ) ⊆ ℱ,  implies  

that ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) = ℱ .Thus  either  𝑢(Iℒ) ⊆ ℱ  or   𝑢2 ∈ [ℱ𝑅 : ℋ].That  is   either   

𝑢ℒ ⊆ [ℱℋ : I] or 𝑢
2ℋ ⊆ ℱ ⊆ [ℱℋ : I]   . It  follows  that  either  𝑢ℒ ⊆ [ℱℋ : I] ⊆ [ℱℋ : I] + 𝐽(ℋ) ∩

𝑠𝑜𝑐(ℋ) or 𝑢2ℋ ⊆ [ℱℋ : I] ⊆ [ℱℋ : I] + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ). That is either  𝑢ℒ ⊆ [ℱℋ : I] + 𝐽(ℋ) ∩
𝑠𝑜𝑐(ℋ) or 𝑢2 ∈ [[ℱℋ : I] + 𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)𝑅 : ℋ]. Hence [ℱℋ : I] is STPNS_ 2_ Absorbing   
submodule  of   ℋ . 

(  )  Suppose  that  [ℱℋ : I] is STPNS_ 2_ Absorbing   submodule  of   ℋ for  every  non_zero 
ideal I of  R . Put   I = R   we  get [ℱℋ : R] = ℱ is STPNS_ 2_ Absorbing   submodule  of   ℋ. 



IHJPAS. 36(2)2023 

361 
 

4.  The Relations of STPNS-2-Absorbing Submodules with 2-Absorbing Submodules And 

Other Form of Submodules 

In this section we introduce the relations Of STPNS-2-Absorbing submodules with 2-Absorbing 

submodules and other form of submodules.  

The converse of Remarks and Examples 3.2 (2) is true under certain conditions where given the 

following propositions. 

 

 

Proposition 4.1 

Let ℋ be an  Ɍ_module over a Boolean ring 𝑅 with either 𝐽( ℋ) = 0 or 𝑠𝑜𝑐( ℋ) = 0, and 𝐴 is a 
proper submodule of ℋ.  Then 𝛢 is STPNS_2_Absorbing of ℋ if and only if 𝛢 is 2_Absorbing  
submodule of ℋ. 

Proof 

(⇒) Let ɑ𝑏𝑦 ∈ 𝐴 for ɑ, 𝑏 ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝑅 is Boolean ring, then (ɑ𝑏)2𝑦 ∈ 𝐴 with   (ɑ𝑏)2 =
(𝑎𝑏) ∉ [𝐴 + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)):Ɍ ℋ] and 𝑏𝑦 ∉ 𝐴 + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) or ɑ𝑦 ∈ 𝐴 + (𝐽( ℋ) ∩
𝑠𝑜𝑐( ℋ)). Since 𝐽( ℋ) = 0, then 𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ) = 0 so 𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) = 𝐴 if 
𝑠𝑜𝑐( ℋ) = 0, then ( ℋ) ∩ 𝑠𝑜𝑐( ℋ) = 0 so 𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) = 𝐴. In both case we get 
(ɑ𝑏)2 = (𝑎𝑏) ∉ [𝐴:Ɍ ℋ] and  𝑏𝑦 ∉ 𝐴, to prove that ɑ𝑦 ∈ 𝐴. Since  𝛢 is a STPNS_ 2_Absorbing 
submodule of ℋ and (ɑ𝑏)2 ∉ [𝐴 + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)):Ɍ ℋ], implies that (ɑ𝑏)𝑦 ∈ 𝐴 + (𝐽( ℋ) ∩
𝑠𝑜𝑐( ℋ)) that is  ɑ(𝑏𝑦) ∈ 𝐴 + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)). So 𝑅 is Boolean ring then ɑ(𝑏𝑦) = ɑ2(𝑏𝑦) ∈
𝐴 + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)). Now by Lemma 2.13, we have 𝑎 = ɑ2𝑏. Thus ɑ2(𝑏𝑦) = 𝑎𝑦 ∈ 𝐴 +
(𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)), so 𝐴  + 𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ) = 𝐴 it follows that 𝑎𝑦 ∈ 𝐴. Hence  𝛢 is 
2_Absorbing  submodule of ℋ.   

(⇐)  Direct. 

Proposition 4.2 

Let ℋ be an Ɍ_module over a Boolean ring 𝑅, 𝐴 is a proper submodule of ℋ. Then 𝛢 is 
STPNS_2_Absorbing of ℋ if and only if 𝛢 is 2_Absorbing  submodule of ℋ. 

Proof 

(⇒) Since 𝑅 is a Boolean ring, then by Lemma 2.13 and 2.15 𝐽( ℋ) = 0, so  (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) =
(0) ∩ 𝑠𝑜𝑐( ℋ) = (0). Thus the proof follows as in proposition 4.1. 

(⇐)  Direct. 

The converse of Remarks and Examples 3.2 (3) is true under certain conditions where given the 

following propositions. 

 

Proposition 4.3  

Let ℋ be an Ɍ_module, and 𝛢 ⊂ ℋ with 𝑠𝑜𝑐( ℋ) =  ℋ. Then 𝛢 is Nearly_2_Absorbing if and 
only if 𝛢 STPNS_2_Absorbing submodule of ℋ.  

 

Proof 

(⇒) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ, that is ɑ. 𝑎. 𝑦 ∈ 𝐴 with 𝑎. 𝑎 ∉ [𝐴 + 𝐽( ℋ):Ɍ ℋ]. Since 𝐴 is 
Nearly-2-Absorbing and 𝑎. 𝑎 ∉ [𝐴 + 𝐽( ℋ):Ɍ ℋ], then ɑ𝑦 ∈ 𝐴 + 𝐽( ℋ) . But 𝐽( ℋ) ⊆ ℋ, so 
𝐽( ℋ) ∩ ℋ = 𝐽( ℋ), that is 𝑎𝑦 ∈ 𝛢  + 𝐽( ℋ) ∩ ℋ. Since 𝑠𝑜𝑐( ℋ) =  ℋ it follows that 𝑎𝑦 ∈
𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)).  Thus 𝛢 is STPNS_2_Absorbing submodule of ℋ.  

(⇐)  Direct. 



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362 
 

The following corollary is a direct result of Proposition 4.3. 

 

 

Corollary 4.4 

Let ℋ be a semi-simple, and 𝛢 ⊂ ℋ . Then 𝛢 is Nearly_2_Absorbing if and only if 𝛢 is 
STPNS_2_Absorbing submodule of ℋ. 

Proposition 4.5 

Let ℋ be an Ɍ_module over a Boolean ring 𝑅, and 𝛢 ⊂ ℋ . Then the following are Valente: 

1. 𝛢  is 2_Absorbing submodule of ℋ. 

2. 𝛢  is STPNS_2_Absorbing submodule of ℋ. 

3. 𝛢  is Nearly_2_Absorbing submodule of ℋ. 

Proof  

(1) ⇔ (2) See  Proposition 4.2. 

(2) ⇔ (3) See Corollary 4.4. 

The converse of Remarks and Examples 3.2 (4) is true under certain conditions where given the 

following propositions.  

Proposition 4.6 

Let ℋ be an Ɍ_module, and 𝛢 ⊂ ℋ  with 𝐽( ℋ) =  ℋ. Then 𝛢 is Pseudo_2_Absorbing if and only 
if 𝛢 is STPNS_2_Absorbing submodule of ℋ. 

Proof 

(⇒) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ, that is ɑ. 𝑎. 𝑦 ∈ 𝐴 with 𝑎. 𝑎 ∉ [𝐴 + 𝐽( ℋ):Ɍ ℋ]. Since 𝐴 is 
Pseudo-2-Absorbing and 𝑎. 𝑎 ∉ [𝐴 + 𝑠𝑜𝑐( ℋ):Ɍ ℋ], then ɑ𝑦 ∈ 𝐴 + 𝑠𝑜𝑐( ℋ). But 𝑠𝑜𝑐( ℋ) ⊆
ℋ, so ℋ ∩ 𝑠𝑜𝑐( ℋ) = 𝑠𝑜𝑐(ℋ), that is either 𝑎𝑦 ∈ 𝛢 + ℋ ∩ 𝑠𝑜𝑐( ℋ). But 𝐽( ℋ) =  ℋ, it 
follows that 𝑎𝑦 ∈ 𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)). Thus 𝛢 is STPNS_2_Absorbing submodule of ℋ.  

(⇐)  Direct. 

Proposition 4.7 

Let ℋ be an  Ɍ_module, and 𝛢 ⊂ ℋ  with 𝑠𝑜𝑐( ℋ) ⊆ 𝛢 and 𝐽( ℋ) ⊆ 𝛢 . Then the following are 
Valente: 

1 𝛢  is 2_Absorbing submodule of ℋ. 

2. 𝛢  is Pseudo_2_Absorbing submodule of ℋ. 

3. 𝛢  is STPNS_2_Absorbing submodule of ℋ. 

4. 𝛢  is Nearly_2_Absorbing submodule of ℋ. 

Proof  

(1) ⇒ (2) Direct by Lemma 2.19. 

(2) ⇒ (3) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ, that is ɑ. 𝑎. 𝑦 ∈ 𝐴 with 𝑎. 𝑎 ∉ [𝐴 + 𝑠𝑜𝑐( ℋ):Ɍ ℋ]. 
Since 𝐴 is Pseudo-2-Absorbing and 𝑎. 𝑎 ∉ [𝐴 + 𝑠𝑜𝑐( ℋ):Ɍ ℋ], then ɑ𝑦 ∈ 𝐴 + 𝑠𝑜𝑐( ℋ). But 
𝑠𝑜𝑐( ℋ) ⊆ 𝛢 then 𝛢  + 𝑠𝑜𝑐( ℋ) = 𝛢. Thus either 𝑎𝑦 ∈ 𝐴 ⊆ 𝛢  + (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ))  . That is  
𝛢 is STPNS_2_Absorbing submodule of ℋ. 

(3) ⇒ (4) Direct by Remarks and Examples 3.2 (3). 

(4) ⇒ (1) Direct by Lemma 2.19. 



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363 
 

The converse of Remarks and Examples 3.2 (5) is true under certain conditions where given the 

following propositions. 
Proposition 4.8 

Let ℋ be a semi-simple Ɍ_module, 𝐴 is a proper submodule of ℋ . Then 𝛢 is 
STPNS_2_Absorbing of ℋ if and only if 𝛢 is Semi 2_Absorbing  submodule of ℋ. 

Proof 

(⇒) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝐴 is STPNS_2_ Absorbing submodule of ℋ, then 𝑎𝑦 ∈
𝛢  + (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ)) or ɑ2 ∈ [𝛢  + (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ)):Ɍ ℋ]. Since ℋ is semi-simple, then 
by Lemma 2.17 𝐽( ℋ) = 0, so  (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) = (0) ∩ 𝑠𝑜𝑐( ℋ) = (0). Thus either 𝑎𝑦 ∈
𝐴  or ɑ2 ∈ [𝛢 ∶Ɍ ℋ] . Hence 𝛢 is a Semi 2_Absorbing submodule of ℋ.   

(⇐)  Direct. 

The converse of Remarks and Examples 3.2 (6) is true under certain conditions where given the 

following propositions. 
Proposition 4.9 

Let ℋ be an Ɍ_module, and 𝛢 ⊂ ℋ with 𝑠𝑜𝑐( ℋ) =  ℋ. Then 𝛢 is Nearly Semi _2_Absorbing if 
and only if 𝛢 STPNS_2_Absorbing submodule of ℋ.  

Proof 

(⇒) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝐴 is Nearly Semi _2_ Absorbing submodule of ℋ, then 
𝑎𝑦 ∈ 𝛢  + 𝐽(ℋ) or ɑ2 ∈ [𝛢  + 𝐽(ℋ):Ɍ ℋ]. But 𝐽( ℋ) ⊆ ℋ, so 𝐽( ℋ) ∩ ℋ = 𝐽( ℋ), that is 
either 𝑎𝑦 ∈ 𝐴  + 𝐽( ℋ) ∩ ℋ or ɑ2 ∈ [𝐴  + 𝐽( ℋ) ∩ ℋ ∶Ɍ ℋ]. Since 𝑠𝑜𝑐( ℋ) =  ℋ it follows 
that either 𝑎𝑦 ∈ 𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) or ɑ2 ∈ [𝛢  + (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ)):Ɍ ℋ]. Thus 𝛢 is 
STPNS_2_Absorbing submodule of ℋ.  

(⇐)  Direct. 

Proposition 4.10 

Let ℋ be an  Ɍ_module over a Boolean ring 𝑅 , and 𝛢 is a proper submodule of  ℋ with 𝐽(ℋ) ⊆
𝐴 . Then the following are statement: 

1 𝛢  is STPNS_2_Absorbing submodule of ℋ. 

2. 𝛢  is Nearly Semi _2_Absorbing submodule of ℋ. 

3. 𝛢  is Semi 2_Absorbing submodule of ℋ. 

4. 𝛢  is 2_Absorbing submodule of ℋ. 

Proof  

(1) ⇒ (2) Direct by Remarks and Examples 3.2 (6). 

(2) ⇒ (3) Direct by Lemma 2.20. 

(3) ⇒ (4) Let ɑ𝑏𝑦 ∈ 𝐴 for ɑ, 𝑏 ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝑅 is Boolean ring, then (ɑ𝑏)2𝑦 ∈ 𝐴 with   
(ɑ𝑏)2 = (𝑎𝑏) ∉ [𝐴:Ɍ ℋ] and 𝑏𝑦 ∉ 𝐴, to prove that ɑ𝑦 ∈ 𝐴. Since  𝛢 is a Semi_ 2_Absorbing 
submodule of ℋ and (ɑ𝑏)2 ∉ [𝐴:Ɍ ℋ], implies that (ɑ𝑏)𝑦 ∈ 𝐴 that is  ɑ(𝑏𝑦) ∈ 𝐴.So 𝑅 is Boolean 
ring then ɑ(𝑏𝑦) = ɑ2(𝑏𝑦) ∈ 𝐴. Now by Lemma 2.13 we have 𝑎 = ɑ2𝑏. Thus ɑ2(𝑏𝑦) = 𝑎𝑦 ∈ 𝐴. 
Hence  𝛢 is 2_Absorbing  submodule of ℋ. 

(4) ⇒ (1) Direct by Remarks and Examples 3.2 (2). 

The following proposition we establish the relation between STPNS_2_Absorbing submodules 

and Pseudo semi 2_Absorbing submodules. 

 



IHJPAS. 36(2)2023 

364 
 

Proposition 4.11 

Let ℋ be an 𝑅-module and  𝐴 ⊂ ℋ. If  𝐴 is STPNS_2_Absorbing submodule of  an 𝑅-module ℋ 
then 𝐴  is Pseudo semi 2_Absorbing submodule. 

Proof  

Suppose that ℱ is STPNS-2-Absorbing submodule of ℋ, and  let 𝑢2𝑚 ∈ ℱ, for 𝑢 ∈ 𝑅, 𝑚 ∈ ℋ. 
Since ℱ is STPNS-2-Absorbing submodule, then either 𝑢𝑚 ∈ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ⊆ ℱ +
𝑠𝑜𝑐(ℋ) or  𝑢2ℋ ⊆ ℱ + (𝐽(ℋ) ∩ 𝑠𝑜𝑐(ℋ)) ⊆ ℱ + 𝑠𝑜𝑐(ℋ). Hence ℱ is Pseudo semi-2-
Absorbing submodule of ℋ. 

The converse of Proposition 4.11 is true under certain conditions where given the following 

propositions. 

Proposition 4.12 

Let ℋ be an Ɍ_module, and 𝛢 ⊂ ℋ  with 𝐽( ℋ) =  ℋ. Then 𝛢 is Pseudo Semi _2_Absorbing if 
and only if 𝛢 is STPNS_2_Absorbing submodule of ℋ. 
Proof 

(⇒) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝐴 is Pseudo Semi _2_ Absorbing submodule of ℋ, then 
𝑎𝑦 ∈ 𝛢  + 𝑠𝑜𝑐(ℋ) or ɑ2 ∈ [𝛢  + 𝑠𝑜𝑐(ℋ):Ɍ ℋ]. But 𝑠𝑜𝑐( ℋ) ⊆ ℋ then 𝑠𝑜𝑐( ℋ) ∩ ℋ =
𝑠𝑜𝑐( ℋ) that is either 𝑎𝑦 ∈ 𝛢 + 𝑠𝑜𝑐( ℋ) ∩ ℋ or ɑ2 ∈ [𝛢 + 𝑠𝑜𝑐( ℋ) ∩ ℋ:Ɍ ℋ]. Since 𝐽( ℋ) =
 ℋ, it follows that either 𝑎𝑦 ∈ 𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)) or ɑ2 ∈ [𝐴  + (𝐽( ℋ) ∩ 𝑠𝑜𝑐( ℋ)):Ɍ ℋ]. 
Thus 𝛢 is STPNS_2_Absorbing submodule of ℋ.  
(⇐)  Direct by Proposition 4.11. 
 

Proposition 4.13 

Let ℋ be an Ɍ_module over a Boolean ring 𝑅, and 𝛢 a proper submodule of ℋ with 𝑠𝑜𝑐( ℋ) ⊆
𝐽( ℋ) and 𝐽( ℋ) ⊆ 𝐴 . Then the following are statement: 

1. 𝛢  is 2_Absorbing submodule of ℋ. 
2. 𝛢  is Pseudo_2_Absorbing submodule of ℋ. 
3. 𝛢  is Pseudo Semi _2_Absorbing submodule of ℋ. 
4. 𝛢  is STPNS_2_Absorbing submodule of ℋ. 
5. 𝛢  is Nearly_2_Absorbing submodule of ℋ. 
6. 𝛢  is Nearly Semi _2_Absorbing submodule of ℋ. 
7. 𝛢  is Semi 2_Absorbing submodule of ℋ. 

 
Proof  

(1) ⇒ (2) Direct by Lemma 2.18. 
(2) ⇒ (3) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ, that is ɑ. 𝑎. 𝑦 ∈ 𝐴 with 𝑎. 𝑎 ∉ [𝐴 + 𝑠𝑜𝑐( ℋ):Ɍ ℋ]. 
Since 𝐴 is Pseudo-2-Absorbing and 𝑎. 𝑎 ∉ [𝐴 + 𝑠𝑜𝑐( ℋ):Ɍ ℋ], then ɑ𝑦 ∈ 𝐴 + 𝑠𝑜𝑐( ℋ). But 
𝑠𝑜𝑐( ℋ) ⊆ 𝐽( ℋ) then (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ)) = 𝑠𝑜𝑐( ℋ). Thus we have either 𝑎𝑦 ∈ 𝛢  + (𝐽(ℋ) ∩
𝑠𝑜𝑐( ℋ)). That is  𝛢 is STPNS-2-Absorbing submodule of ℋ. 
(3) ⇒ (4) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝐴 is Pseudo Semi_2_Absorbing submodule of ℋ, 
then 𝑎𝑦 ∈ 𝛢  + 𝑠𝑜𝑐(ℋ) or ɑ2 ∈ [𝛢  + 𝑠𝑜𝑐(ℋ):Ɍ ℋ]. But 𝑠𝑜𝑐( ℋ) ⊆ 𝐽( ℋ),  then 𝑠𝑜𝑐( ℋ) ∩
𝐽( ℋ) = 𝑠𝑜𝑐( ℋ), so either 𝑎𝑦 ∈ 𝐴  + (𝐽(ℋ) ∩ 𝑠𝑜𝑐( ℋ)) or ɑ2 ∈ [𝐴  + (𝐽(ℋ) ∩
𝑠𝑜𝑐( ℋ)):Ɍ ℋ] . That is 𝛢 is STPNS_2_Absorbing submodule of ℋ. 
(4) ⇒ (5) Direct by Remarks and Examples 3.2 (3). 
(5) ⇒ (6) Let ɑ2𝑦 ∈ 𝐴 for ɑ ∈ Ɍ, 𝑦 ∈ ℋ, that is ɑ. 𝑎. 𝑦 ∈ 𝐴. Since 𝐴 is Nearly-2-Absorbing, then 
either 𝑎. 𝑎 ∈ [𝐴 + 𝐽( ℋ):Ɍ ℋ] or  ɑ𝑦 ∈ 𝐴 + 𝐽( ℋ). Hence 𝛢  is Nearly Semi-2-Absorbing 
submodule of ℋ. 
(6) ⇒ (7) Direct by Lemma 2.20 



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365 
 

(7) ⇒ (1) Let ɑ𝑏𝑦 ∈ 𝐴 for ɑ, 𝑏 ∈ Ɍ, 𝑦 ∈ ℋ. Since 𝑅 is Boolean ring, then (ɑ𝑏)2𝑦 ∈ 𝐴 with   
(ɑ𝑏)2 = (𝑎𝑏) ∉ [𝐴:Ɍ ℋ] and 𝑏𝑦 ∉ 𝐴, to prove that ɑ𝑦 ∈ 𝐴. Since  𝛢 is a Semi-2-Absorbing 
submodule of ℋ and (ɑ𝑏)2 ∉ [𝐴:Ɍ ℋ], implies that (ɑ𝑏)𝑦 ∈ 𝐴 that is  ɑ(𝑏𝑦) ∈ 𝐴.So 𝑅 is Boolean 
ring then ɑ(𝑏𝑦) = ɑ2(𝑏𝑦) ∈ 𝐴. Now by Lemma 2.13 we have 𝑎 = ɑ2𝑏. Thus ɑ2(𝑏𝑦) = 𝑎𝑦 ∈ 𝐴. 
Hence  𝛢 is 2-Absorbing  submodule of ℋ. 
 

5. Conclusion 

 

We will present the most important propositions in this research: 

 

. Let ℋ be an Ɍ-module over a Boolean ring 𝑅, and 𝛢 ⊂ ℋ. Then the following are statement: 

1. 𝛢  is 2-Absorbing submodule of ℋ. 

2. 𝛢  is STPNS-2-Absorbing submodule of ℋ. 

3. 𝛢  is Nearly-2-Absorbing submodule of ℋ. 

 

. Let ℋ be an  Ɍ-module, and 𝛢 ⊂ ℋ  with 𝑠𝑜𝑐( ℋ) ⊆ 𝛢 and 𝐽( ℋ) ⊆ 𝛢 . Then the following are 
statement: 

1 𝛢  is 2-Absorbing submodule of ℋ. 

2. 𝛢  is Pseudo-2-Absorbing submodule of ℋ. 

3. 𝛢  is STPNS-2-Absorbing submodule of ℋ. 

4. 𝛢  is Nearly-2-Absorbing submodule of ℋ. 

. Let ℋ be an  Ɍ-module over a Boolean ring 𝑅 , and 𝛢 is a proper submodule of  ℋ with 𝐽(ℋ) ⊆
𝐴 . Then the following are statement: 

1 𝛢  is STPNS-2-Absorbing submodule of ℋ. 

2. 𝛢  is Nearly Semi -2-Absorbing submodule of ℋ. 

3. 𝛢  is Semi 2_Absorbing submodule of ℋ. 

4. 𝛢  is 2-Absorbing submodule of ℋ. 

. Let ℋ be an Ɍ-module over a Boolean ring 𝑅, and 𝛢 a proper submodule of ℋ with 𝑠𝑜𝑐( ℋ) ⊆
𝐽( ℋ) and 𝐽( ℋ) ⊆ 𝐴 . Then the following are Valente: 

1. 𝛢  is 2-Absorbing submodule of ℋ. 
2. 𝛢  is Pseudo-2-Absorbing submodule of ℋ. 
3. 𝛢  is Pseudo Semi -2-Absorbing submodule of ℋ. 
4. 𝛢  is STPNS-2-Absorbing submodule of ℋ. 
5. 𝛢  is Nearly-2-Absorbing submodule of ℋ. 
6. 𝛢  is Nearly Semi -2-Absorbing submodule of ℋ. 
7. 𝛢  is Semi 2-Absorbing submodule of ℋ. 

 

 

References 

1. Darani, A.Y ; Soheilniai. F. 2-Absorbing and Weakly 2-AbsorbingSubmodules, Tahi Journal. 

Math, 2011,(9), 577-584. 

2.Goodearl, K. R. Ring Theory; Marcel Dekker, Inc. New York and Basel. 1976 , 206. 

3.Dauns, J. Prime Modules. Journal  Re. Angew, Math, 1978,(2),156-181. 

4. Reem,  T. Shwkea,  M. Nearly 2-Absorbing Submodules and Related Concept, Tikrit Journal 

for Pure Sci, 2018, (2),(3), 215-221. 



IHJPAS. 36(2)2023 

366 
 

5. Haibat,  K. Omar,  A. Pseudo 2-Absorbing and Pseudo Semi 2-Absorbing Submodules, AIP 

Conference Proceedings, 2019, 2096,020006(2019),1-9. 

6. Haibat,  K. Akram,  S. Nearly Semi-2-Absorbing Submodules, Italian Journal of Pure and App. 

Math, 2019, (41),620-627. 

7.  Kash, F. Modules and Rings.  London  Math. Soc. Monographs New York , Academic  Press , 

1982, (370) . 

8. Fraleigh, J. B. A First Course in Abstract Algebra, (New York: Addison- Wesley Publishing 

Company), 2003. 

9. Wardayani, A. ;Kharismawati, I.; Sihwaningrum, I. Regular Rings and Their Properties, IOP  

Conference Series, 2020, 1494, 012020,1-4. 

10. Mahmood, S. Y. Regular Modules, M. Sc. Thesis, university of Baghdad,1993.