On n-Absorbing Submodules

حول المقاسات الجزئية الممتصة من النوع ن

Presented by

Osama Abed El-Karim Naji

Supervised by

Mohammed M. Al-Ashker

Professor of Mathematics

Arwa E. Ashour

Associate Professor of Mathematics

A thesis submitted in partial fulfillment

of the requirements for the degree of

Master of Mathematics

July/2016

زةــغ – تــالميــــــت اإلســـــــــبمعـالج

العليبشئون البحث العلمي والدراسبث

ـلـــــــــــــــــومـعـالت ــــــــــــــــــــليـك

بثـــــيـــــبضــــــريير ــــتــســبجـــم

The Islamic University–Gaza

Research and Postgraduate Affairs

Faculty of Science

Master of Mathematics

i

إقــــــــــــــرار

أنا الموقع أدناه مقدم الرسالة التي تحمل العنوان:

On n-Absorbing Submodules

حول المقاسات الجزئية الممتصة من النوع ن

أقر بأن ما اشتممت عميو ىذه الرسالة إنما ىو نتاج جيدي الخاص، باستثناء ما تمت اإلشارة إليو حيثما ورد، وأن

ككل أو أي جزء منيا لم يقدم من قبل االخرين لنيل درجة أو لقب عممي أو بحثي لدى أي مؤسسة ىذه الرسالة

تعميمية أو بحثية أخرى.

Declaration

I understand the nature of plagiarism, and I am aware of the University’s policy on

this.

The work provided in this thesis, unless otherwise referenced, is the researcher's own

work, and has not been submitted by others elsewhere for any other degree or

qualification.

:Osama Abed El-Karim Naji Student's name أسامة عبد الكريم ناجي اسم الطالب:

:Osama Naji Signature أســامــة ناجي التوقيع:

:Date 2016\6\26 2016\6\26 التاريخ:

ii

Abstract

Let R be a commutative ring with a nonzero identity and M be a unitary

R-module. Chin Pi. Lu studied prime submodules. Many authors have

investigated some generalizations of prime ideals and submodules

to 2-absorbing ( n-absorbing ) ideals and submodules.

In this thesis we prove several results concerning 2-absorbing

submodules and extended some of them to n-absorbing submodules. We

also investigate the sufficient and necessary conditions for a submodule

N to be (primary - classical - almost) 2-absorbing submodule of M.

iii

الملخص

، باي لو تشن درس Rمقاسا أحاديا عمى Mحمقة ابدالية بيا محايد غير صفري، وليكن Rلتكـن

الباحثين قاموا بتقصي الحمقات المثالية األولية . والعديد من Mالمقاسات الجزئية األولية لـ

والمقاسات الجزئية األولية وعمموىا إلى حمقات مثالية ثنائية االمتصاص )أو نونية االمتصاص(،

وكذلك المقاسات الجزئية.

في ىذه الرسالة نبرىن عدة نتائج نحاول من خالليا الحصول عمى الشرط الكافي والالزم لممقاس

ن أولي ثنائي االمتصاص أو تقميدي ثنائي االمتصاص أو أنو يكاد يكون ثنائي الجزئي ليكو

االمتصاص.

iv

To My parents

My brothers and My sisters

v

Acknowledgments

First I would like to express my sincere thanks to my God. Then I am

grateful to my father Abed El-Karim Nagi for his encouragement,

guidance, and support from the initial to final level in my academic steps,

and my lovely mother for subsidization, love, assistance and patience.

Without them this work would never have come to existence.

I would like to thank prof. Mohammed Al-Ashker and Dr. Arwa

Ashour, my supervisors, for their many suggestions and constant support

during this research.

I am extremely and sincerely thankful to all the staff members of the

Mathematics department and all my teachers who taught me to come to

this stage of learning.

I would like to express my sincere thanks to my sisters, brothers and my

family whose love, care and sacrifice enabled me to reach this level of

learning.

Contents

Abstract ii

Acknowledgements v

Introduction 1

1 Basic Concepts 4

1.1 Rings and Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Modules and Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Prime and Almost Prime Submodules . . . . . . . . . . . . . . . . . . . . . . . 16

1.4 Primary Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 2-Absorbing Submodules 21

2.1 Definition and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Weakly 2-Absorbing Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3 Classes of 2-Absorbing submodules . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.3.1 2-Absorbing Primary Submodules . . . . . . . . . . . . . . . . . . . . . 36

2.3.2 Classical 2-Absorbing Submodules . . . . . . . . . . . . . . . . . . . . . 42

2.3.3 Almost 2-Absorbing Submodules . . . . . . . . . . . . . . . . . . . . . . 49

2.3.4 Almost 2-Absorbing Primary Submodules . . . . . . . . . . . . . . . . . 58

3 n-Absorbing Submodules 66

3.1 n-Absorbing Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.2 n-Absorbing Compactly Packed Modules . . . . . . . . . . . . . . . . . . . . . . 75

Conclusion 79

vi

Introduction

Among the most famous concepts of ring theory is ” prime ideals ”. In the late

decades many authors in Algebra focussed on generalizing the concept of ” prime

ideals ”. For example primary, primal, almost prime, 2-absorbing, n-absorbing

ideals,...etc where studied. Also these previous concepts were generalized to

submodules instead of ideals.

Let R be a commutative ring with a nonzero identity and M be a unitary

R-module. Prime submodules play an important role in the theory of modules

over commutative rings. A proper submodule N of M is prime if it satisfies the

property that for a ∈ R and m ∈ M , am ∈ N implies that m ∈ N or aM ⊆ N

(equivalently a ∈ (N : M)).

The concept of 2-absorbing ideals - which is a generalization of prime ide-

als of a commutative ring was introduced in [11] by Badawi. He define a

proper ideal I of a commutative ring R to be a 2-absorbing ideal if abc ∈ I

for a, b, c ∈ R, then ab ∈ I or ac ∈ I or bc ∈ I. Later many properties of this

ideal were studied in [6] by Anderson and Badawi. Darani and Soheilina in [19]

extended the notion of 2-absorbing ideals to 2-absorbing submodules which is

a generalization of prime submodules of an R-module M . They define a proper

submodule N of an R-module M to be a 2-absorbing submodule if whenever

1

a, b ∈ R, m ∈M with abm ∈ N , then am ∈ N or bm ∈ N or ab ∈ (N : M).

Darani and Soheilina in [20] have introduced the concept of n-absorbing

submodules and transferred several results parallel to n-absorbing ideals in a

commutative ring - which was introduced by Anderson and Badawi in [6]. A

proper submodule N of M is said to be an n-absorbing submodule if whenever

a1...anm ∈ N for some a1, ..., an ∈ R and m ∈M , then either a1...an ∈ (N : M)

or there are n− 1 of the a′is whose product with m belongs to N .

In this thesis we are concerned with the properties of n-absorbing submod-

ules, especially when n = 2, and studied some concepts related to 2-absorbing

submodules such as : weakly 2-absorbing, classical 2-absorbing, 2-absorbing pri-

mary, almost 2-absorbing and almost 2-absorbing primary submodules. These

are considered as generalizations to the concepts related to prime submodules.

Among the main results that we obtained, is characterization of some classes

of 2-absorbing submodules. We proved the following results: :

• In cyclic modules, a submodule is a 2-absorbing if and only if it is a classical

2-absorbing. [Corollary 2.3.29].

• In finitely generated faithful multiplication modules, a submodule is an

almost 2-absorbing (resp. almost 2-absorbing primary) if and only if it’s

residual is an almost 2-absorbing ideal (resp. almost 2-absorbing primary

ideal). [Theorem 2.3.49 (resp. Theorem 2.3.67)].

• If a, b ∈ (N : M) and 0ab = 0 (zero submodule), then N is a 2-absorbing

submodule if and only if N is a weakly 2-absorbing. [Corollary 2.2.9].

• In any module we give characterization of 2-absorbing, n-absorbing,

2-absorbing primary, classical 2-absorbing, almost 2-absorbing and almost

2-absorbing primary submodules. [Theorem 2.1.16, Theorem 3.1.10,

2

Theorem 2.3.12, Theorem 2.3.26, Theorem 2.3.39 and Theorem 2.3.64,

respectively].

These are in addition to many other results through the thesis.

The thesis consists of three chapters. In chapter one we recall the definitions

of prime, primary and almost prime submodules. we also recall some basic

definitions and results from ring and module theory that we need in our work.

In the second chapter, we investigate the concept of 2-absorbing sub-

modules and enumerate most of our main results. Where we study

some classes of submodules that are related to 2-absorbing submodules:

weakly 2-absorbing , 2-absorbing primary , classical 2-absorbing , almost

2-absorbing and almost 2-absorbing primary submodules.

In chapter three, we look at the n-absorbing submodules and transfer some

results that we got on 2-absorbing submodules to suit the concept of n-absorbing

submodules. Also in this chapter we studied few results on n-absorbing com-

pactly packed modules. Information about n-absorbing compactly packed mod-

ules is rare, specially when we notice that the available results are general on

compactly packed modules, and do not have something special for n-absorbness.

We assume throughout this thesis that all rings are commutative rings with

identity and all modules will be unitary.

3

Chapter 1

Basic Concepts

In this chapter we recall some basic definitions and results that we need through-

out our thesis.

1.1 Rings and Ideals

Definition 1.1.1. [26] A ring is a nonempty set R together with two binary

operations, addition (denoted by a+ b) and multiplication (denoted by ab), such

that for all a, b, c in R:

1. a+ b = b+ a.

2. (a+ b) + c = a+ (b+ c).

3. There is an additive identity 0. That is, there is an element 0 in R such

that a+ 0 = a for all a in R.

4. There is an element −a in R such that a+ (−a) = 0.

4

5. (ab)c = a(bc).

6. a(b+ c) = ab+ ac and (a+ b)c = ac+ bc.

If in addition: ab = ba for all a, b ∈ R, then R is said to be a commutative

ring. If R contains an element 1 such that 1a = a1 = a for all a ∈ R, then R is

said to be a ring with identity (or sometimes: a ring with unity).

Definition 1.1.2. [26] A non-empty subset S of a ring R is a subring of R if S

itself is a ring under the same operations of addition and multiplication defined

on R.

Definition 1.1.3. [26] A subring A of a ring R is called an ideal of R if for

every r ∈ R and every a ∈ A both ra and ar are in A.

Example 1.1.4. {0} and R are ideals of any ring R.

Proposition 1.1.5. [26] A nonempty subset A of a ring R is an ideal of R if

(i) a− b ∈ A whenever a, b ∈ A, and

(ii) ra ∈ A whenever a ∈ A and r ∈ R.

Example 1.1.6. For any positive integer number n, nZ is an ideal of the ring

Z.

Definition 1.1.7. [26] Let X be a subset of a ring R. If X = {a1, a2, ..., an} then

the ideal < X >= {r1a1 + r2a2 + ...+ rnan|ri ∈ R} is called the ideal generated

by X. If X consists of a single element, say a, then < X >=< a > is called a

principal ideal.

Example 1.1.8. In the ring Z of integer numbers, nZ is a principal ideal for

any positive integer n.

5

Definition 1.1.9. [26] A principal ideal ring is a ring in which every ideal is

principal.

Remark 1.1.10. [18] If A and B are ideals of a ring R, the product of A and B

is the ideal defined as:

AB = {a1b1 + a2b2 + ...+ anbn : ai ∈ A, bi ∈ B, n a positive integer}

Proposition 1.1.11. [34] Let A, B, and C be ideals of a ring R such that

A = B ∪ C, then A = B or A = C.

Proof. Suppose that A 6= B. Since A = B ∪ C, then ∃a ∈ A − B such that

a ∈ C. Note that A = (A− B) ∪ (A ∩ B). Now, let x be any arbitrary element

in A. If x ∈ A − B, then x ∈ C. Let x ∈ A ∩ B, then x − a ∈ A − B, since A

and B are ideals. Thus x − a ∈ C. Hence (x − a) + a = x ∈ C, since C is an

ideal. Thus A = C.

Definition 1.1.12. [18] A proper ideal A of a ring R is a maximal ideal of R

if whenever I is an ideal of R such that A ⊆ I ⊆ R, then A = I or I = R.

Example 1.1.13. The ideal 3Z is a maximal in Z but 4Z is not maximal, since

4Z ( 2Z ( Z.

Definition 1.1.14. [18] An ideal A of a ring R is said to be minimal ideal if

{0} and A are the only ideals of R contained in A.

Definition 1.1.15. [36] A proper ideal P of a ring R is a prime ideal if for

any a, b ∈ R, ab ∈ P implies that either a ∈ P or b ∈ P .

Example 1.1.16. (1) In the ring Z[X] of all polynomials with integer coeffi-

cients, the ideal generated by 2 and X is a prime ideal. It consists of all those

6

polynomials whose constant coefficient is even.

(2) In the ring Z, the zero ideal and pZ are prime ideals where p is prime integer.

Proposition 1.1.17. [31] An ideal P in a ring R is a prime ideal if and only if

it satisfies the following property: If A and B are ideals in R such that AB ⊆ P ,

then A ⊆ P or B ⊆ P .

Definition 1.1.18. [26] Let I be an ideal of a ring R, the radical of I, denoted

by√I, is the ideal

√I =

⋂P , where the intersection runs over all prime ideals of

R containing I. Equivalently,√I = {r ∈ R : rn ∈ I, for some integer n > 0}.

An ideal I is said to be a radical ideal if√I = I.

Remark 1.1.19. [18] Every prime ideal is a radical ideal.

Example 1.1.20. In Z12, < 6 > is a radical ideal but is not prime, since

2.3 ∈< 6 >, 2 /∈< 6 >, 3 /∈< 6 >.

Definition 1.1.21. [26] A proper ideal P of a ring R is a primary ideal if for

any a, b ∈ R such that ab ∈ P , either a ∈ P or bn ∈ P for some positive integer

n.

Example 1.1.22. [26] For any prime integer p and any positive integer n ≥ 2,

pZ is a prime ideal in Z while pnZ is a primary ideal in Z which is not prime .

Proposition 1.1.23. [26] If Q is a primary ideal in a ring R. Then√Q is a

prime ideal in R.

Proof. Since Q is a proper ideal in R, then 1 /∈ Q and hence 1 /∈√Q, so

√Q is

proper ideal in R. Let ab ∈√Q and a /∈

√Q, then (ab)n ∈ Q for some positive

integer n, and hence anbn ∈ Q. Since a /∈√Q, an /∈ Q. Since Q is primary,

there is a positive integer k such that (bn)k ∈ Q, hence b ∈√Q. Therefore

√Q

7

is a prime ideal.

Remark 1.1.24. The converse of Proposition 1.1.23 is not true. Consider the

ideal I = (y2, xy) ⊂ Z2[x, y], we have√I = (y) is prime but I is not primary

since xy ∈ I and y /∈ I, xn /∈ I for any n.

Definition 1.1.25. [5] A proper ideal I of a ringR is called weakly prime ideal

if for any a, b ∈ R such that 0 6= ab ∈ I then a ∈ I or b ∈ I.

Definition 1.1.26. [17] A proper ideal I of a ringR is called almost prime ideal

if for any a, b ∈ R such that ab ∈ I − I2 then a ∈ I or b ∈ I.

Remark 1.1.27. From the definitions, any prime ideal is weakly prime and any

weakly prime ideal is almost prime.

Example 1.1.28. (1) In Z6, {0} is weakly prime ideal but is not prime, since

2.3 ∈ {0} and 2 /∈ {0}, 3 /∈ |0|.

(2) In Z12, < 4 > is almost prime since < 4 >2=< 4 >, but it is not weakly

prime since 0 6= 2.2 ∈< 4 > and 2 /∈< 4 >.

Definition 1.1.29. [31] A nonempty subset S of a ring R is said to be multi-

plicatively closed if 1 ∈ S and for a, b ∈ S, ab ∈ S.

Definition 1.1.30. [31] An ideal A of a ring R is called a multiplication ideal

if for every ideal B ⊆ A there exists an ideal C such that B = AC.

Definition 1.1.31. [31] A ring R is called a multiplication ring if all its ideals

are multiplication ideals.

Example 1.1.32. The ring of integer numbers Z is a multiplication ring.

Definition 1.1.33. [18] Let S be a nonempty subset of a ring R, then

ann(S) = {a ∈ R : aS = 0} is an ideal of R called the annihilator of S.

8

Example 1.1.34. In Z, let S = 2Z, then ann(S) = 0.

Definition 1.1.35. [27] A ring R is called a faithful ring if it has no nonzero

annihilator. Thus, if R is a faithful ring, aR 6= 0 for every nonzero a in R.

Example 1.1.36. The ring of integer numbers Z is a faithful ring.

Definition 1.1.37. [18] Let A and B be ideals of a ring R, then the set

(B : A) = {a ∈ R : aA ⊆ B} is an ideal of R.

Definition 1.1.38. [18] A ring R is said to be a local ring if it has a unique

maximal ideal.

Example 1.1.39. Zp is a local ring, where p is prime integer.

1.2 Modules and Submodules

Definition 1.2.1. [18] Let R be a ring. A (left) R-module is an additive

abelian group M together with a function R ×M → M (the image of (r,m)

being denoted by rm) such that for all r, s ∈ R and m,m1,m2 ∈M :

(i) r(m1 +m2) = rm1 + rm2.

(ii) (r + s)m = rm+ sm.

(iii) r(sm) = (rs)m.

If in addition 1m = m for all m ∈M (1 is the identity element of R), then M is

said to be a unitary left R−module.

A right R−module is defined similarly via a function M × R → M denoted

(m, r) 7→ mr and satisfies the obvious analogues of (i)− (iii).

9

Since we only deal with commutative rings in this thesis, then every leftR−module

M can be given the structure of a right R−module by defining mr = rm for

r ∈ R,m ∈M . From now on, every module M is assumed to be both a left and

a right module with mr = rm for r ∈ R,m ∈M .

Definition 1.2.2. [18] Let R be a ring, M an R−module and N a nonempty

subset of M . N is a submodule of M provided that N is an additive subgroup

of M and rm ∈ N for all r ∈ R and m ∈ N .

Definition 1.2.3. [18] A submodule that is not the entire module is called a

proper submodule.

Example 1.2.4. [18]

(1) For any ring R, we can consider R as an R-module, and the ideals in the

ring R as submodules of the module R.

(2) If I is an ideal of R and N is a submodule of M , then

IN = {r1m1 + ...+ rnmn : ri ∈ I, mi ∈ N, n a positive integer}

is a submodule of M .

Definition 1.2.5. [18] Let R be a ring, M an R-module and m ∈ M , the

cyclic submodule generated by m is a submodule of M having the form

Rm = {rm : r ∈ R}.

Definition 1.2.6. [18] An R-module M is said to be finitely generated if

there is a finite subset {x1, ..., xn} of M such that M = Rx1 + ...+Rxn. In this

case we say that M is called a module generated by x1, ..., xn.

Definition 1.2.7. [18] A mapping f from R-module M to R-module N is a

homomorphism if f(x+ y) = f(x) + f(y) and f(ax) = af(x), for all x, y ∈M

10

and a ∈ R. A homomorphism f : M → N is an epimorphism if it maps M

onto N .

Definition 1.2.8. [18] If N is a submodule of an R-module M , then M/N

together with the operations: (x+N) + (y+N) = (x+ y) +N and a(x+N)

= ax+N for x, y ∈M and a ∈ R is called the factor module.

Definition 1.2.9. [31] The mapping φ from M into the factor module M/N

defined by φ(x) = x+N is an epimorphism, called the canonical homomorphism

from M onto M/N .

Definition 1.2.10. [18] Let M and N be an R-modules and let f : M → N be

a homomorphism, then Ker(f) = {x : x ∈ M and f(x) = 0} is a submodule of

M called the kernel of f .

Definition 1.2.11. [31] Let M be an R-module. A proper submodule N of M

is said to be irreducible if N is not the intersection of two submodules of M

that properly contain it.

Definition 1.2.12. [18] Let R be a ring and let M be an R-module. The

annihilator of M in R is the ideal

ann(M) = {r ∈ R : rm = 0 for all m ∈M}

.

Definition 1.2.13. [18] Let A be an ideal of the ring R and let M be an R-

module. The annihilator of A in M is the submodule

ann(A) = {m ∈M : rm = 0 for all r ∈ R}

.

11

Definition 1.2.14. [31] Let N be a submodule of an R-module M . The residual

of N by M, denoted (N : M), is the ideal (N : M) = {r ∈ R : rM ⊆ N}.

If x ∈M , then the ideal (N : x) is defined by (N : x) = {r ∈ R : rx ∈ N}.

Definition 1.2.15. [31] Let M be an R-module, A be an ideal of R and N

be a submodule of M . The residual of N by A, denoted by (N : A), is the

submodule (N : A) = {x : x ∈ M and Ax ⊆ N}. If A consists of one element,

say a, then (N : a) = Na = {x : x ∈M and ax ∈ N}.

Proposition 1.2.16. [31] Let K,L and N be submodules of R-module M and

let A and B be ideals of R. Then:

(1) (L ∩N : M) = (L : M) ∩ (N : M).

(2) A ⊆ B implies (N : A) ⊇ (N : B).

(3) ((N : A) : B) = (N : AB).

(4) (L ∩N : A) = (L : A) ∩ (N : A).

(5) L ⊆ N implies that (L : A) ⊆ (N : A) and (L : K) ⊆ (N : K).

Definition 1.2.17. [25] An R-module M is called a multiplication module

if every submodule N of M is of the form IM , for some ideal I of R

Proposition 1.2.18. [25] Let M be a multiplication R-module and N be a sub-

module of M , then N = (N : M)M .

Proof. Let N be a submodule of a multiplication module M , then there exists

I of R such that N = IM . Since I ⊆ (N : M) and N = IM ⊆ (N : M)M ⊆ N

so that N = (N : M)M .

Definition 1.2.19. [36] Let R be a ring and M be an R-module. The set of

12

zero divisors of M , denoted by Zd(M) is defined by

Zd(M) = {r ∈ R : for some x ∈M and x 6= 0, rx = 0}

Proposition 1.2.20. [10] Let M be an R-module and N be a proper submodule

of M . Then Zd(M/N) =⋃

x∈M−N(N : x)

Definition 1.2.21. [26] M is called a faithful R-module if (0 : M) = 0.

Definition 1.2.22. [33] An R-module M is called a cancellation module if

whenever AM = BM with A and B are ideals of R implies A = B.

Proposition 1.2.23. [39] Let M be a finitely generated faithful multiplication

module, then M is a cancellation.

Lemma 1.2.24. [29] Let N be a submodule of a finitely generated faithful mul-

tiplication R-module M . Then, we have (IN : M) = I(N : M) for every ideal I

of R.

Proof. Since M is a multiplication R-module, then

I(N : M)M = IN = (IN : M)M

then the result follows because M is a cancellation.

Lemma 1.2.25. [3] Let N be a submodule of a faithful multiplication R-module

M and I be a finitely generated faithful multiplication ideal of R. Then,

(1) N = (IN : I).

(2) If N ⊆ IM , then (JN : I) = J(N : I) for any ideal J of R.

(3) (N : I) = ((N : M) : I)M = (N : IM)M .

13

Lemma 1.2.26. [14] Let M be an R-module and let N be a proper submodule

of M . Then (N/((N : M)N) : M/((N : M)N)) = (N : M).

Proof. First we prove that (N/((N : M)N) : M/((N : M)N)) ⊆ (N : M). Let

x ∈ (N/((N : M)N) : M/((N : M)N)), then

x(M/((N : M)N)) ⊆ N/((N : M)N).

Suppose x /∈ (N : M), so xM * N which implies that

xM + ((N : M)N) * N + ((N : M)N),

and hence x(M + ((N : M)N)) * N + ((N : M)N), which is a contradiction.

Now we prove the other inclusion, Let x ∈ (N : M), then xM ⊆ N and so

x(M/((N : M)N) ⊆ N/((N : M)N), hence

x ∈ N/((N : M)N) : M/((N : M)N)).

Lemma 1.2.27. [18] Let R be a ring and M be an R-module. Let S be a

multiplicatively closed set in R. Let T be the set of all pairs (x, s), where x ∈M

and s ∈ S. Define a relation on T by (x, s) ∼ (x′, s′), if and only if there exists

t ∈ S such that t(sx′ − s′x) = 0. Then ∼ is an equivalence relation on T .

Definition 1.2.28. [18] For (x, s) ∈ T which is defined in Lemma 1.2.27, denote

the equivalence class of ∼ which contains (x, s) by xs. Let S−1M denote the set

of all equivalence classes of T with respect to this relation. We can make S−1M

into an R-module by setting xs

+ yt

= tx+syst

and axs

= axs

, where x, y ∈M , t, s ∈ S

and a ∈ R. The module S−1M is called the module of fractions of M with

respect to S (or quotient module of M).

14

Definition 1.2.29. [18] Since R may be considered as an R-module we can form

the quotient ring S−1R. An element of S−1R has the form as

, where a ∈ R and

s ∈ S. We can make S−1R into a ring by setting asbt

= abst

, where a, b ∈ R and

s, t ∈ S. The ring S−1R is called the ring of fractions of R with respect to

S (a quotient ring of R).

Definition 1.2.30. [30] Let N be a submodule of an R-module M , and let S

be multiplicatively closed subset of R. An S-component of N is denoted by NS

and defined as NS = {m : m ∈M and sm ∈ N for some s ∈ S}.

Definition 1.2.31. [18] A proper submodule N of M is said to be a maximal

submodule of M if whenever N ′ is a submodule of M such that N ⊆ N ′ ⊆ M ,

either N = N ′ or N ′ = M .

Definition 1.2.32. A module M is called a local module if it has a unique

maximal submodule.

Definition 1.2.33. [36] An R-module M is said to satisfy the ascending chain

condition (ACC) on submodules (or to be Noetherian) if for every chain

N1 ⊆ N2 ⊆ N3 ⊆ ... of submodules of M , there is an integer n such that Ni = Nn

for all i ≥ n.

Definition 1.2.34. [36] An R-module M is said to satisfy the descending chain

condition (DCC) on submodules (or to be Artinian) if for every chain

N1 ⊇ N2 ⊇ N3 ⊇ ... of submodules of M , there is an integer n such that Ni = Nn

for all i ≥ n.

Definition 1.2.35. [36] An R-module M is said to satisfy the maximal condition

on submodules if every nonempty collection of submodules of M contains a

maximal element (with respect to set theoretic inclusion).

15

Theorem 1.2.36. [36] Let M be an R-module, then the following statements

are equivalent:

(i) M satisfies the (ACC) on submodules.

(ii) M satisfies the maximal condition on submodules.

(iii) Every submodule of M is finitely generated.

1.3 Prime and Almost Prime Submodules

Definition 1.3.1. [32] Let M be an R-module. A proper submodule N of M

is said to be a prime submodule if whenever rm ∈ N for r ∈ R and m ∈ M

implies that either m ∈ N or rM ⊆ N (equivalently r ∈ (N : M)).

Proposition 1.3.2. [32] If N is a prime submodule of an R-module M , then

(N : M) is prime ideal in R.

Proof. (N : M) is a proper ideal, since 1 /∈ (N : M). Let ab ∈ (N : M) and

b /∈ (N : M). Then bM * N , that is there exists m ∈ M with bm /∈ N . But

a(bm) = (ab)m ∈ N and N is prime, therefore aM ⊆ N . Thus a ∈ (N : M).

Definition 1.3.3. [7] Let M be an R-module and N be a proper submodule

of M . N is called a weakly prime submodule of M if, whenever r ∈ R and

m ∈M such that 0 6= rm ∈ N , then either m ∈ N or r ∈ (N : M).

Remark 1.3.4. [7]

1. Every prime submodule of a module is a weakly prime submodule. How-

ever, the converse may not hold, for example the {0} is weakly prime in the

16

Z-module Z4 (by definition), but is not prime since 2.2 ∈ {0} and 2 /∈ {0}.

2. If N is a weakly prime submodule of an R-module M , then (N : M) is not

weakly prime ideal of R in general. For example, let M denote the cyclic

Z-module Z8. Take N = {0}. Certainly N is a weakly prime submodule

of M , but (N : M) = 8Z is not a weakly prime ideal of Z. Note that

0 6= 2.4 ∈ (N : M), while neither 2 ∈ (N : M) nor 4 ∈ (N : M).

Proposition 1.3.5. [7] Let R be a ring, M a faithful cyclic R-module, and N

a weakly prime submodule of M . Then (N : M) is a weakly prime ideal of R.

Proof. Assume that M = Rx, where x ∈ M , and let 0 6= ab ∈ (N : M)

with a /∈ (N : M). Then there exists r ∈ R such that rax = a(rx) /∈ N , so

ax /∈ N . As 0 6= abM ⊆ N , it follows that 0 6= abx ∈ N (for if abx = 0, then

ab ∈ (0 : x) ⊆ (0 : M) = 0, a contradiction), so 0 6= abx = bax ∈ N implies

b ∈ (N : M) since N is a weakly prime submodule of M . Thus (N : M) is

weakly prime ideal.

Definition 1.3.6. [29] Let M be an R-module. A proper submodule N of M

is called an almost prime submodule of M if, whenever r ∈ R and m ∈ M

such that rm ∈ N − (N : M)N , then either m ∈ N or r ∈ (N : M).

Remark 1.3.7. [14] Any weakly prime submodule is almost prime. However, the

converse need not necessarily be true. For example, we consider the Z-module

M = Z24 and the proper submodule N of M generated by 8, < 8 >= {8, 16, 0}.

Then clearly (N : M)N = 8Z < 8 >= N . Hence N is almost prime. On the

other hand, 0 6= 4.4 ∈ N with 4 /∈ N and 4 /∈ (N : M) and so N is not weakly

prime.

17

Theorem 1.3.8. [29] Let M be an R-module and N be a proper submodule of

M . Then, N is almost prime in M if and only if N/(N : M)N is weakly prime

in M/(N : M)N .

Proof. Suppose that N is almost prime in M . Let r ∈ R and m ∈M , such that

0 6= r(m + (N : M)N) ∈ N/(N : M)N . Then, rm ∈ N − (N : M)N and so

either m ∈ N or r ∈ (N : M). Hence, either m + (N : M)N ∈ N/(N : M)N

or r ∈ (N : M) = (N/(N : M)N : M/(N : M)N) and so N/(N : M)N

is weakly prime in M/(N : M)N . Conversely, assume that N/(N : M)N is

weakly prime in M/(N : M)N and let r ∈ R and m ∈ M such that

rm ∈ N − (N : M)N . Then, 0 6= r(m + (N : M)N) ∈ N/(N : M)N and

hence either m + (N : M)N ∈ N/(N : M)N and so m ∈ N

r ∈ (N/(N : M)N : M/(N : M)N) = (N : M).

1.4 Primary Submodules

Definition 1.4.1. [36] Let M be an R-module. A proper submodule N of M

is said to be a primary submodule if rm ∈ N for r ∈ R and m ∈ M implies

that either m ∈ N or rnM ⊆ N for some positive integer n.

Directly from the definition every prime submodule is primary. The converse

need not be true, (see Example 1.1.22).

Proposition 1.4.2. [31] If N is a primary submodule of an R-module M , then

(N : M) is a primary ideal in R, and hence√

(N : M) is prime ideal in R.

18

Proof. Let ab ∈√

(N : M), where a, b ∈ R, then for some positive integer n we

have anbnM = (ab)nM ⊆ N . If b /∈√

(N : M) then bnx /∈ N for some x ∈ M .

Since anbnx ∈ N but bnx /∈ N we have ankM ⊆ N for some positive integer k.

Thus a ∈√

(N : M).

Definition 1.4.3. [8] A proper submodule N of an R-module M is said to be

a weakly primary submodule if whenever 0 6= rm ∈ N , for some r ∈ R,

m ∈M , then m ∈ N or rnM ⊆ N for some n ∈ N .

Theorem 1.4.4. [8] Let M be an R-module, and N a proper submodule of M .

Then the following statements are equivalent:

(1) N is a weakly primary submodule of M .

(2) For m ∈M −N ,√

(N : Rm) =√

(N : M) ∪ (0 : Rm).

(3) For m ∈M −N ,√

(N : Rm) =√

(N : M) or√

(N : Rm) = (0 : Rm).

Proof. (1) ⇒ (2) Let H =√

(N : M) ∪ (0 : Rm). Let a ∈√

(N : Rm) where

m ∈ M − N . Then akm ∈ N for some positive integer k. If akm 6= 0, then

ak ∈ (N : M) since N is weakly primary, hence a ∈√

(N : M). If akm = 0, then

assume that s is the smallest integer with asm = 0. If s = 1, then a ∈ (0 : Rm).

Otherwise, a ∈√

(N : M), so√

(N : Rm) ⊆√

(N : M) ∪ (0 : Rm) = H.

For the other inclusion assume that b ∈ H. Clearly, if b ∈ (0 : Rm), then

b ∈ (N : Rm) ⊆√

(N : Rm). If b ∈√

(N : M), then bt ∈ (N : M) ⊆ (N : Rm)

for some positive integer t, so b ∈√

(N : Rm).

19

(2)⇒ (3) Since if an ideal is a union of two ideals, then it is equal to one of

them. (See Proposition 1.1.11).

(3) ⇒ (1) Suppose that 0 6= rm ∈ N with r ∈ R and m ∈ M − N . Then

r ∈ (N : Rm) ⊆√

(N : Rm) and r /∈ (0 : Rm). It follows from (3) that

r ∈√

(N : Rm) =√

(N : M), as required.

20

Chapter 2

2-Absorbing Submodules

The concept of 2-absorbing ideals was introduced by Badawi in [11] as a gen-

eralization of prime ideals. Darani and Soheilinia generalized this concept to

submodules in [19] to be a generalization of prime submodules. In this chap-

ter we investigate the concept of 2-absorbing submodules and study some of its

classes.

2.1 Definition and Properties

Definition 2.1.1. [11] A proper ideal I of R is said to be a 2-absorbing ideal

if whenever a, b, c ∈ R with abc ∈ I then ab ∈ I or ac ∈ I or bc ∈ I .

Remark 2.1.2. [11] Every prime ideal is a 2-absorbing ideal.

Definition 2.1.3. [19] A proper submodule N of an R-module M is said to be

a 2-absorbing submodule if whenever a, b ∈ R and m ∈ M with abm ∈ N

then ab ∈ (N : M) or am ∈ N or bm ∈ N .

21

Example 2.1.4. 1. For the Z-module Z8 , N = {0, 4} is a 2-absorbing sub-

module of Z8 but is not prime . To see this, let a, b ∈ Z, m ∈ Z8 be such

that abm ∈ N = {0, 4}. We have (N : Z8) = 4Z.

Case 1: If abm ≡ 0 (mod 8), then 8|abm. If a ≡ 0 (mod 8), then

ab ∈ 4Z = (N : Z8) and am = 0 ∈ N . Assume that a 6≡ 0 (mod 8),

b 6≡ 0 (mod 8) and m 6= 0 then we have the following cases:

i. 2|a , 2|b, and m is even in Z8. In which case ab ∈ 4Z.

ii. 4|a, which implies that ab ∈ 4Z. (Similar argument for b in place of a

can be done).

Case 2: abm ≡ 4 (mod 8). Hence 4|abm. If 4|a, we are done. Suppose

that 4 - a, 4 - b and m 6= 0 in Z8. Then we have the cases:

i. 2|a and 2|b, hence ab ∈ 4Z.

ii. 2|a, 2 - b and m is even, then we have am ≡ 0 (mod 8) if a or m is

divisible by 4 , or am ≡ 4 (mod 8) if neither a nor m is divisible by 4.

iii. 2 - a, 2 - b and m = 4 and that gives am ≡ 4 (mod 8).

Thus N is 2-absorbing submodule of Z8. N is not prime because 2.2 ∈ N

but 2 /∈ N and 2 /∈ 4Z.

2. Let R be a ring, then N = {(k, ..., k) ∈ Rn : k ∈ R} is 2-absorbing Z-

submodule of Rn .

Example 2.1.5. [38]

1. On the Z-module Z , nZ is a 2-absorbing submodule if n = 0 , p or pq ,

where p, q are prime integers. To see this, if n = 0 or n = p, where p is

prime integer, then nZ is prime submodule and hence it is 2-absorbing (See

Proposition 2.1.6). If n = pq where p, q are prime integer. Let xyz ∈ pqZ

22

then there exists r ∈ Z such that ayz = pqr, then p|x so x = pt for some

t ∈ Z. Now xyz = ptyz = pqr then tyz = qr so either q|t or q|y or q|z.

If q|t then pq|x and we are done. If q|y then there exists u ∈ Z such that

y = qu, so we have xy = (pt)(qu) = pq(tu) ∈ pqZ. Similarly if q|z, then

xz ∈ pqZ.

2. Some modules do not have any 2-absorbing submodules . Let Q/Z be a

Z-module and let p be a fixed prime integer , then

Z(p∞) = {α ∈ Q/Z : α = r/pn + Z for some r ∈ Z and n ≥ 0}

is a nonzero submodule of Q/Z .

Let Gt = {α ∈ Q/Z : α = r/pt+Z for some r ∈ Z } for all t ≥ 0 . Now

Gt is not 2-absorbing submodule of Z(p∞) , because p2(1/pt+2 + Z) ∈ Gt

while p(1/pt+2 + Z) /∈ Gt and p2 /∈ (Gt : Z(p∞)) = 0. Since each proper

submodule of Z(p∞) is equal to Gt for some t ≥ 0 , so Z(p∞) does not have

any 2-absorbing submodule .

Proposition 2.1.6. [38] If either N is a prime submodule of M or N is the

intersection of two prime submodules of M , then N is a 2-absorbing submodule

of M .

Proof. Let N be prime submodule of M and let a, b ∈ R and m ∈ M with

abm ∈ N then ab ∈ (N : M) or m ∈ N , which implies that ab ∈ (N : M) or

am ∈ N or bm ∈ N . Let N1 and N2 be two prime submodules of M , we have

to show that N1∩N2 is 2-absorbing submodule of M . Assume that a, b ∈ R and

m ∈ M with abm ∈ N thus abm ∈ Ni for i = 1, 2. Now abm ∈ Ni implies that

a ∈ (Ni : M) or b ∈ (Ni : M) or m ∈ Ni for i = 1, 2 by Proposition 1.3.2 since

Ni is prime. If a ∈ (N1 : M) and a ∈ (N2 : M) then a ∈ (N1 ∩ N2 : M) and so

23

ab ∈ (N1∩N2 : M) . If a ∈ (N1 : M) and b ∈ (N2 : M), then ab ∈ (N1∩N2 : M).

If a ∈ (N1 : M) and m ∈ N2 , then am ∈ N1 ∩ N2. We do the same in other

cases.

Remark 2.1.7. [38]

1. The intersection of three nonzero prime submodules is not necessarily 2-

absorbing.

For example 2Z , 3Z and 5Z are prime Z-modules of Z , but 2Z∩ 3Z∩

5Z = 30Z which is not 2-absorbing submodule of Z , since 2.3.5 ∈ 30Z but

2.5 /∈ 30Z and 3.5 /∈ 30Z and 2.3 /∈ (30Z : Z) = 30Z .

2. The intersection of two nonzero 2-absorbing submodules is not necessarily

2-absorbing.

For example 4Z ∩ 3Z = 12Z in Z-module Z which is not 2-absorbing

submodule since 2.2.3 ∈ 12Z but 2.3 /∈ 12Z and 2.2 /∈ (12Z : Z) = 12Z.

3. The intersection of prime submodule and a 2-absorbing submodule is not

necessarily 2-absorbing. For example ( see the previous example).

Proposition 2.1.8. [20] Let N be a 2-absorbing submodule of an R-module M

. For every elements a, b ∈ R and every submodule K of M , abK ⊆ N implies

that ab ∈ (N : M) or aK ⊆ N or bK ⊆ N .

Proof. Assume that ab /∈ (N : M), aK * N and bK * N . Then ax /∈ N and

by /∈ N for some x, y ∈ K. As abx, aby ∈ N we have ay ∈ N and bx ∈ N .

Now it follows from ab(x + y) ∈ N that either a(x + y) ∈ N or b(x + y) ∈ N .

Consequently, either by ∈ N or ax ∈ N which are contradictions.

24

Theorem 2.1.9. [38] If N is a 2-absorbing submodule of M then (N : M) is a

2-absorbing ideal in R.

Proof. Let a, b, c ∈ R , abc ∈ (N : M) with ac /∈ (N : M) and bc /∈ (N : M).

We show that ab ∈ (N : M). There are m1,m2 ∈ M such that acm1 /∈ N

and bcm2 /∈ N , but ab(cm1 + cm2) ∈ N . By assumption , N is a 2-absorbing

submodule , so that a(cm1 + cm2) ∈ N or b(cm1 + cm2) ∈ N or ab ∈ (N : M)

. If ab ∈ (N : M) we are done. If a(cm1 + cm2) ∈ N , then acm2 /∈ N , since if

acm2 ∈ N then acm1 ∈ N which is contradiction. Now acm2 /∈ N and bcm2 /∈ N

while abcm2 ∈ N , thus ab ∈ (N : M). With the same argument, we can show

that if b(cm1 + cm2) ∈ N , then ab ∈ (N : M) . This complete the proof.

Remark 2.1.10. [38] In general, the converse of Theorem 2.1.9 is not true . From

Example 2.1.5(2) we have (Gt : Z(p∞)) = 0 is a 2-absorbing ideal of Z for all

t ≥ 0 , but Gt is not 2-absorbing submodule of Z(p∞) .

Theorem 2.1.11. [19] Let M be a cyclic R-module and N be a submodule of

M . If (N : M) is a 2-absorbing ideal of R then N is 2-absorbing submodule of

M .

Proof. Let M = Rm for some m ∈M and assume that (N : M) is a 2-absorbing

ideal in R and let abx ∈ N for some a, b ∈ R and x ∈M , then there exists c ∈ R

such that abx = abcm ∈ N , then abc ∈ (N : m) = (N : M) then ab ∈ (N : M)

or ac ∈ (N : M) or bc ∈ (N : M) . Therefore ab ∈ (N : M) or ax ∈ N or

bx ∈ N . Hence N is 2-absorbing .

Proposition 2.1.12. [38] If N is 2-absorbing submodule of M then (N : m) is

2-absorbing ideal in R for all m ∈M −N .

25

Proof. For m ∈M −N , (N : m) is a proper ideal of R. Assume that a, b, c ∈ R

and abc ∈ (N : m). Thus abcm = a(bc)m ∈ N , since N is a 2-absorbing

submodule of M , then am ∈ N or bcm ∈ N or abc ∈ (N : M). If am ∈ N or

bcm ∈ N , then a ∈ (N : m) or bc ∈ (N : m) and we are done. If abc ∈ (N : M)

then the assertion follows by Theorem 2.1.9.

Proposition 2.1.13. Let N be a 2-absorbing submodule of M . If the set of all

zero divisors of M/N , Zd(M/N), forms an ideal in R, then it is a 2-absorbing

ideal.

Proof. Let a, b, c ∈ R with abc ∈ Zd(M/N) , then by Proposition 1.2.20, abc ∈

(N : m′) for some m′ ∈ M − N . But N is 2-absorbing submodule of M and

m′ ∈ M −N then by Proposition 2.1.12 (N : m′) is 2-absorbing ideal of R. So

ab ∈ (N : m′) or bc ∈ (N : m′) or ac ∈ (N : m′) , then ab ∈⋃

x∈M−N(N : x)

or bc ∈⋃

x∈M−N(N : x) or ac ∈

⋃x∈M−N

(N : x). Hence ab ∈ Zd(M/N) or bc ∈

Zd(M/N) or ac ∈ Zd(M/N) , Thus Zd(M/N) is 2-absorbing ideal of R.

Remark 2.1.14. The set of all zero divisors may not be an ideal. For example,

consider the Z-module M = Z6, we have 2, 3 ∈ Zd(M) but 2 + 3 /∈ Zd(M).

Theorem 2.1.15. [37] Let N be a 2-absorbing submodule of M . Then (N : M) is

a prime ideal of R if and only if (N : m) is a prime ideal of R for all m ∈M−N .

Proof. Assume that a, b ∈ R , m ∈ M − N and ab ∈ (N : m). Then abm ∈ N .

We have am ∈ N or bm ∈ N or ab ∈ (N : M) since N is a 2-absorbing

submodule of M . If am ∈ N or bm ∈ N we are done. If ab ∈ (N : M), then

by assumption either a ∈ (N : M) or b ∈ (N : M). Thus either a ∈ (N : m) or

b ∈ (N : m). So (N : m) is a prime ideal. Conversely, suppose that ab ∈ (N : M)

26

for some a, b ∈ R and assume that there exist m,m′ ∈ M such that am /∈ N

and bm′ /∈ N . Since abm, abm′ ∈ N it follows that bm ∈ N and am′ ∈ N since

(N : m) and (N : m′) are prime ideals of R. If m + m′ ∈ N , then am ∈ N

which is a contradiction. Thus m + m′ /∈ N . Now by ab(m + m′) ∈ N we have

a(m+m′) ∈ N or b(m+m′) ∈ N since (N : m+m′) is prime ideal of R which

is a contradiction. Thus aM ⊆ N or bM ⊆ N which implies that (N : M) is

prime.

Theorem 2.1.16. Let M be an R-module and N be a proper submodule of M .

then the following are equivalent :

(1) N is 2-absorbing submodule of M .

(2) For a, b ∈ R such that ab /∈ (N : M) , Nab = Na ∪ Nb and so Nab = Na or

Nab = Nb.

Proof. (1) ⇒ (2) Assume that ab /∈ (N : M) and let m ∈ Nab ,then abm ∈ N ,

since N is 2-absorbing then am ∈ N or bm ∈ N , which implies that m ∈ Na or

m ∈ Nb thus m ∈ Na ∪ Nb. Let x ∈ Na ∪ Nb ,then ax ∈ N or bx ∈ N , hence

abx ∈ N so x ∈ Nab.

(2)⇒ (1) Let a, b ∈ R and m ∈ M with abm ∈ N . Assume that ab /∈ (N : M),

then m ∈ Nab = Na or m ∈ Nab = Nb, which implies that am ∈ N or bm ∈ N .

Thus N is 2-absorbing submodule of M .

The following example shows that if N is not a 2-absorbing submodule of M ,

then the second statement in the previous theorem does not hold.

Example 2.1.17. Let M = Z be a module over itself, and let N = 8Z, N is not

a 2-absorbing submodule of M and N2.2 = 2Z = N4 6= N2 = 4Z.

27

Theorem 2.1.18. Let N be a 2-absorbing submodule of an R-module M . Let

y ∈M and r ∈ R− (N : y). If (N : M) is a prime ideal of R then

(N : y) = (N : ry).

Proof. Let a ∈ (N : ry), then ray ∈ N . Since N is 2-absorbing then ay ∈ N

or ry ∈ N or ra ∈ (N : M). If ay ∈ N then a ∈ (N : y) .If ry ∈ N then then

r ∈ (N : y) which contradicts the hypothesis. If ra ∈ (N : M) then r ∈ (N : M)

or a ∈ (N : M). if r ∈ (N : M) then r ∈ (N : y) which again contradicts the

hypothesis. If a ∈ (N : M) then a ∈ (N : y).

For the reverse inclusion, let a ∈ (N : y) then ay ∈ N , then ray ∈ N hence

a ∈ (N : ry), which completes the proof.

Proposition 2.1.19. Let N be a 2-absorbing submodule of an R-module M . Let

y ∈M and a, b ∈ R such that ab /∈ (N : M) . Then

(N : aby) = (N : ay) ∪ (N : by).

Proof. Let r ∈ (N : aby) then raby = ab(ry) ∈ N . Since N is 2-absorbing then

ary ∈ N or bry ∈ N or ab ∈ (N : M). By hypothesis we have ab /∈ (N : M).

then r ∈ (N : ay) or r ∈ (N : by), hence r ∈ (N : ay) ∪ (N : by). For the

reverse inclusion, let r ∈ (N : ay) ∪ (N : by). If r ∈ (N : ay) then ray ∈ N then

raby ∈ N , hence r ∈ (N : aby). Similarly if r ∈ (N : by), then r ∈ (N : aby).

Theorem 2.1.20. [21] Let f : M →M ′ be an epimorphism of R-modules.

(1) If N ′ is a 2-absorbing submodule of M ′ then f−1(N ′) is a 2-absorbing sub-

module of M .

(2) If N is a 2-absorbing submodule of M containing ker(f) then f(N) is a

2-absorbing submodule of M ′.

28

Proof. (1)Let a, b ∈ R and m ∈M such that abm ∈ f−1(N ′) then abf(m) ∈ N ′,

but N ′ is 2-absorbing submodule of M ′, so ab ∈ (N ′ : M ′) or af(m) ∈ N ′ or

bf(m) ∈ N ′. If ab ∈ (N ′ : M ′) then abM ′ ⊆ N ′, then abM ⊆ f−1(N ′), so

ab ∈ (f−1(N ′) : M). If af(m) ∈ N ′ then f(am) ∈ N ′ then am ∈ f−1(N ′).

Similarly if bf(m) ∈ N ′, then bm ∈ f−1(N ′). Thus ab ∈ (f−1(N ′) : M) or

am ∈ f−1(N ′) or bm ∈ f−1(N ′) and hence f−1(N ′) is 2-absorbing submodule of

M .

(2) Let a, b ∈ R, m′ ∈ M ′ and abm′ ∈ f(N). Then there exists n ∈ N such

that abm′ = f(n). Since f is an epimorphism therefore for some m ∈ M we

have f(m) = m′. Thus abf(m) = f(n). This implies f(abm − n) = 0 implies

abm−n ∈ ker(f) ⊆ N . So abm ∈ N . Now, since N is a 2-absorbing submodule,

therefore am ∈ N or bm ∈ N or ab ∈ (N : M). Thus am′ ∈ f(N) or bm′ ∈ f(N)

or abM ⊆ N implies abf(M) ⊆ f(N) so ab ∈ (f(N) : M ′). Hence f(N) is a

2-absorbing submodule ofM ′.

Corollary 2.1.21. Let f : M → M ′ be an isomorphism of R-modules, then N

is a 2-absorbing submodule of M iff f(N) is a 2-absorbing submodule of M ′

Proof. By Theorem 2.1.20 and the fact that ker(f) = {0} ⊆ N .

Remark 2.1.22.

1. If f is not onto may cause that f−1(N) is not a proper submodule of M . For

example consider the homomorphism f : Z4 → Z4 defined by f(0) = f(2) = 0 ,

f(1) = f(3) = 2, let N ′ = {0, 2}, then f−1(N ′) = Z4.

2. Ignoring the condition ” ker(f) ⊆ N ” may cause that f(N) is not a proper

submodule of M ′. Consider the epimorphism f : Z12 → Z4 defined by

29

f(x) = x (mod 4), ker(f) = {0, 4, 8}, let N = {0, 3, 6, 9} then f(N) = Z4.

Theorem 2.1.23. [19] Let N,K be R-submodules of M with K ⊆ N . Then N is

a 2-absorbing submodule of M if and only if N/K is a 2-absorbing R-submodule

of M/K.

Proof. Suppose that N is a 2-absorbing submodule of M and let a, b ∈ R and

m ∈ M be such that ab(m + K) ∈ N/K. Then abm ∈ N and N is 2-absorbing

gives ab ∈ (N : M) or am ∈ N or bm ∈ N . Therefore ab ∈ (N/K : M/K) or

a(m + K) ∈ N/K or b(m + K) ∈ N/K, that is N/K is 2-absorbing submodule

of M/K. Conversely, assume that N/K is a 2-absorbing submodule of M/K.

Suppose that a, b ∈ R and m ∈ M are such that abm ∈ N . Then we have

ab(m + K) ∈ N/K. Therefor ab ∈ (N/K : M/K) or a(m + K) ∈ N/K or

b(m + K) ∈ N/K since N/K is 2-absorbing submodule in M/K. Therefore

ab ∈ (N : M) or am ∈ N or bm ∈ N . This implies that N is a 2-absorbing

submodule of M .

Proposition 2.1.24. [38] Suppose S is a multiplicatively closed subset of R and

S−1M is the module of fraction of M . Then the following statements hold.

(1)If N is a 2-absorbing submodule of M , then S−1N is a 2-absorbing submodule

of the S−1R-module S−1M .

(2)If S−1N is a 2-absorbing submodule of S−1M such that Zd(M/N) ∩ S = φ,

then N is a 2-absorbing submodule of M .

Proof. (1) Assume that a, b ∈ R, s, t, l ∈ S, m ∈ M and abmstl∈ S−1N . Then

there exists s′ ∈ S such that s′abm ∈ N . By assumption, N is a 2-absorbing

submodule of M , thus s′am ∈ N or bm ∈ N or s′ab ∈ (N : M). If s′am ∈ N ,

30

then s′ams′sl

= amsl∈ S−1N , and if bm ∈ N , then bm

tl∈ S−1N and we are done.

Now assume that s′ab ∈ (N : M). Thus, s′abs′st

= abst∈ S−1(N : M).Hence,

abst∈ (S−1N :S−1R S−1M). Therefore, S−1N is a 2-absorbing submodule of

S−1M .

(2) Assume that a, b ∈ R, m ∈M and abm ∈ N . Thus, abm1∈ S−1N . Hence,

am1∈ S−1N or bm

1∈ S−1N or ab

1∈ (S−1N :S−1R S−1M). If am

1∈ S−1N , then

there exists s ∈ S such that sam ∈ N . Thus, am ∈ N , since S ∩Zd(M/N) = φ.

If bm1∈ S−1N , we do the same. If ab

1∈ (S−1N :S−1R S

−1M) then

abS−1M ⊆ S−1N . Now we have to show that abM ⊆ N . Assume that m′ ⊆M ,

thus abm′

1∈ abS−1M ⊆ S−1N so that there exists t ∈ S such that tabm′ ∈ N .

Hence, abm′ ∈ N , since S ∩ Zd(M/N) = φ. Therefore, abM ⊆ N and N is a

2-absorbing submodule of M .

Theorem 2.1.25. [28] Let R = R1 × R2 and each Ri is a commutative

ring with identity. Let Mi be an Ri-module and M = M1 × M2

with (r1, r2)(m1,m2) = (r1m1, r2m2), be an R-module, where ri ∈ Ri, mi ∈Mi.

Then we have:

(1) If N1 is a 2-absorbing submodule of M1, then N1 × M2 is a 2-absorbing

submodule of M .

(2) If N2 is a 2-absorbing submodule of M2, then M1 × N2 is a 2-absorbing

submodule of M .

Proof. (1) Suppose that N1 is a 2-absorbing submodule of M1 and let(a1, a2)

and(b1, b2) be elements in R and (x, y) ∈M such that

31

(a1, a2)(b1, b2)(x, y) ∈ N1 ×M2.

Then(a1b1x, a2b2y) ∈ N1 ×M2. Therefore a1b1x ∈ N1. Since N1 is 2-absorbing

submodule of M1 then either a1x ∈ N1 or b1x ∈ N1 or a1b1 ∈ (N1 : M1). If

a1x ∈ N1 then (a1, a2)(x, y) ∈ N1×M2. If b1x ∈ N1 then (b1, b2)(x, y) ∈ N1×M2.

If a1b1 ∈ (N1 : M1), then (a1, a2)(b1, b2) ∈ (N1 ×M2 : M). Thus N1 ×M2 is a

2-absorbing submodule of M .

(2) The proof is quite similar to (1).

2.2 Weakly 2-Absorbing Submodules

Definition 2.2.1. [12] A proper ideal I of a ring R is said to be weakly 2-

absorbing ideal if whenever a, b, c ∈ R with 0 6= abc ∈ I then ab ∈ I or ac ∈ I or

bc ∈ I.

Definition 2.2.2. [19] A proper submodule N of R-module M is said to

be 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 .

Remark 2.2.3. [19] From the definition, every 2-absorbing submodule is weakly

2-absorbing but the converse does not necessarily hold. For example consider

the case where R = Z, M = Z/30Z and N = 0. Then 2.3.(5 + 30Z) = 0 ∈ N

while 2.3 /∈ (N : M), 2.(5 + 30Z) /∈ N and 3.(5 + 30Z) /∈ N . Therefore N is not

2-absorbing while it is weakly 2-absorbing.

Example 2.2.4. Consider the Z-module Z[x]. The submodule < x, 2 > is a

weakly 2-absorbing submodule of Z[x] with (< x, 2 >: Z[x]) = 2Z.

32

Remark 2.2.5. [28] If N is a weakly 2-absorbing submodule which is not 2-

absorbing. Then the ideal (N : M) is not a weakly 2-absorbing ideal of R

generally. For example let M denote the cyclic Z-module Z/27Z and N = 0. So,

N is a weakly 2-absorbing submodule of M , but (N : M) = 27Z is not a weakly

2-absorbing ideal of R since 0 6= 3.3.3 ∈ 27Z but 3.3 /∈ 27Z.

Theorem 2.2.6. [22] Let N be a weakly 2-absorbing submodule of a faithful

cyclic module M over the ring R. Then (N : M) is a weakly 2-absorbing ideal

of R.

Proof. Let N be a weakly 2-absorbing submodule of M . Since M is a cyclic

module we assume M = Rm for some m ∈ M . Let a, b, c ∈ R such that

0 6= abc ∈ (N : M). Suppose ab /∈ (N : M) and bc /∈ (N : M). Then we

have to show that ac ∈ (N : M). As ab /∈ (N : M) and bc /∈ (N : M) then

abm /∈ N and bcm /∈ N . If abcm = 0 then abc ∈ (0 : m) = (0 : M) which is

a contradiction since M is a faithful module. Thus 0 6= abcm = ac(bm) ∈ N .

Since N is a weakly 2-absorbing submodule with a(bm) /∈ N and c(bm) /∈ N ,

therefore ac ∈ (N : M). Thus (N : M) is weakly 2-absorbing ideal of R.

Theorem 2.2.7. [21] Let x ∈M and a ∈ R. Then the following hold:

(1) if ann(a) ⊆ aM , then the submodule aM is2-absorbing if and only if aM is

weakly 2-absorbing.

(2) if ann(x) ⊆ (Rx : M), then the submodule Rx is 2-absorbing if and only if

Rx is weakly 2absorbing.

Proof. (1) Let aM be a weakly 2-absorbing submodule of M and

suppose r, s ∈ R and m ∈ M such that rsm ∈ aM . If 0 6= rsm then

33

since aM is weakly 2-absorbing therefore we have rm ∈ aM or sm ∈ aM or

rs ∈ (aM : M) which implies that aM is 2-absorbing. Therefore we may assume

that rsm = 0 . Clearly, r(s + a)m = rsm + ram ∈ aM .If r(s + a)m 6= 0 then

we have(s + a)m ∈ aM or rm ∈ aM or r(s + a) ∈ (aM : M). Since am ∈ aM

and ra ∈ (aM : M) therefore rm ∈ aM or sm ∈ aM or rs ∈ (aM : M), and we

are done. Now suppose that r(s + a)m = 0. Then since rsm = 0 then we have

arm = 0 and so rm ∈ ann(a) ⊆ aM . Thus rm ∈ aM and consequently aM is

2-absorbing submodule of M . By Remark 2.2.3 if aM is 2-absorbing submodule

of M , then it is weakly 2-absorbing submodule of M .

(2)By Remark 2.2.3, it is enough to prove that if Rx is weakly 2-absorbing sub-

module of M then Rx is 2-absorbing submodule of M . So let Rx be a weakly

2-absorbing submodule of M and suppose r, s ∈ R and m ∈M with rsm ∈ Rx.

Since Rx is weakly 2-absorbing submodule, we may assume rsm = 0, oth-

erwise Rx is 2 -absorbing. Now rs(x + m) ∈ Rx. If rs(x + m) 6= 0 then

we have rs ∈ (Rx : M) or r(x + m) ∈ Rx or s(x + m) ∈ Rx, as Rx is

a weakly 2-absorbing submodule. Hence rs ∈ (Rx : M) or rm ∈ Rx or

sm ∈ Rx. Now let rs(x + m) = 0. Then rsm = 0 implies rsx = 0. Hence

rs ∈ ann(x) ⊆ (Rx : M).Thus Rx is 2-absorbing.

Theorem 2.2.8. [22] Let N be a proper submodule of an R-module M . Then N

is a weakly 2-absorbing submodule of M if and only if for any r1r2 ∈ R−(N : M),

we have Nr1r2 = 0r1r2 ∪Nr1 ∪Nr2.

Proof. Let N be a weakly 2-absorbing submodule of M . Let m ∈ Nr1r2 . Then

r1r2m ∈ N . If 0 6= r1r2m ∈ N for any r1r2 ∈ R− (N : M), then either r1m ∈ N

or r2m ∈ N which implies m ∈ Nr1 or m ∈ Nr2 . If r1r2m = 0, then m ∈ 0r1r2 .

34

Hence,Nr1r2 ⊆ 0r1r2 ∪ Nr1 ∪ Nr2 , the other inclusion is trivial. Conversely, let

r1, r2 ∈ R and m ∈ M such that 0 6= r1r2m ∈ N and suppose r1r2 /∈ (N : M).

Then by the given hypothesis m ∈ Nr1 or m ∈ Nr2 . This implies r1m ∈ N or

r2m ∈ N . Hence, N is weakly a 2-absorbing submodule of M .

Corollary 2.2.9. Let N be a proper submodule of an R-module M . If

for a, b ∈ R− (N : M), 0ab is the zero submodule of M , then N is a 2-absorbing

submodule if and only if N is a weakly 2-absorbing.

Proof. Directly from Theorem 2.1.16 and Theorem 2.2.8 .

Theorem 2.2.10. [21] Let R = R1×R2 and each Ri is a commutative ring with

identity. Let Mi be an Ri-module for each i = 1, 2 and let M = M1 ×M2 with

(r1, r2)(m1,m2) = (r1m1, r2m2), be an R-module, where ri ∈ Ri, mi ∈Mi. Then

the following are equivalent :

(1) N is a 2-absorbing R1-submodule of M1.

(2) N ×M2 is a 2-absorbing R-submodule of M .

(3) N ×M2 is a weakly 2-absorbing R-submodule of M .

Proof. (1)⇒ (2) From Theorem 2.1.25

(2)⇒ (3) From Remark 2.2.3.

(3) ⇒ (1) Let a, b ∈ R1, x ∈ M1 such that abx ∈ N . Then for each

0 6= y ∈M2, we have(0, 0) 6= (a, 1)(b, 1)(x, y) ∈ N×M2. But N×M2 is a weakly

2-absorbing R-submodule of M , therefore we have either(a, 1)(x, y) ∈ N ×M2

or(b, 1)(x, y) ∈ N ×M2 or(a, 1)(b, 1) ∈ (N ×M2 : M),that is either ax ∈ N or

bx ∈ N or ab ∈ (N : M1). This shows that N is a 2-absorbing submodule of

M1.

35

2.3 Classes of 2-Absorbing submodules

In this section we study some known classes of 2-absorbing submodules and

introduce some new classes.

2.3.1 2-Absorbing Primary Submodules

Definition 2.3.1. [13] A proper ideal I of R is said to be a 2-absorbing primary

ideal of R if whenever a, b, c ∈ R with abc ∈ I, then ab ∈ I or ac ∈√I or

bc ∈√I.

Definition 2.3.2. [23] Let M be an R-module and N be a proper submodule

of M . Then N is said to be a 2-absorbing primary submodule of M if whenever

a, b ∈ R and m ∈ M with abm ∈ N , then ab ∈√

(N : M) or am ∈ N or

bm ∈ N .

Remark 2.3.3. [23] From the definitions of 2-absorbing and 2-absorbing primary

submodules we have that every 2-absorbing submodule is a 2-absorbing primary

submodule but the converse need not be true, as shown in the following example.

Example 2.3.4. Consider R = Z and an R-module M = Z54.

Take the submodule N = {0 , 27} of M . Then (N : M) = 27Z and√(N : M) = {a ∈ R : anM ⊆ N} = 3Z. Now, 3.3.3 ∈ N but 3.3 /∈ N and

3.3 /∈ (N : M). Therefore, N is not a 2-absorbing submodule of M but it is a

2-absorbing primary submodule of M , since 3.3 ∈√

(N : M).

Remark 2.3.5. A 2-absorbing primary submodule need not be primary.

Example 2.3.6. Consider a submodule N = 10Z of a Z-module Z. Then 10Z is

36

a 2-absorbing primary submodule. But it is not a primary submodule as 2.5 ∈ N

but neither 2 ∈ N nor 5 ∈√

(N : Z) = 10Z.

Theorem 2.3.7. [23] Let N be a 2-absorbing primary submodule of an R-module

M . Then (N : M) is a 2-absorbing primary ideal of R.

Proof. Let abc ∈ (N : M) for some a, b, c ∈ R. Let ab /∈ (N : M)

and bc /∈√

(N : M). This implies ab /∈ (N : M) and bc /∈ (N : M). So, there

exist m1,m2 ∈ M such that abm1 /∈ N and bcm2 /∈ N but ac(bm1 + bm2) ∈ N .

SinceN is a 2-absorbing primary submodule, then we have either ac ∈√

(N : M)

or a(bm1 +bm2) ∈ N or c(bm1 +bm2) ∈ N . If ac ∈√

(N : M), then we are done.

If a(bm1 + bm2) ∈ N , then abm2 /∈ N . Consider abcm2 = ac(bm2) ∈ N . Since

N is a 2-absorbing primary submodule and abm2 /∈ N , bcm2 /∈ N , therefore

ac ∈√

(N : M). Similarly, if c(bm1 + bm2) ∈ N , then we have cbm1 /∈ N .

Consider abcm1 = ac(bm1) ∈ N . Since N is a 2-absorbing primary submodule

and abm1 /∈ N , bcm1 /∈ N , therefore ac ∈√

(N : M). This implies, in each case,

(N : M) is a 2-absorbing primary ideal of R.

Remark 2.3.8. The converse of the above theorem is not true. If (N : M)

is 2-absorbing primary ideal, then N may not be 2-absorbing primary. Let us

consider the Z-moduleM = Z×Z andN = (0, 6)Z be the submodule ofM . Then

(N : M) = 0 but N is not a 2-absorbing primary module. Since 2.3.(0, 1) ∈ N

but neither 2.(0, 1) ∈ N nor 3.(0, 1) ∈ N also 2.3 /∈√

(N : M).

Proposition 2.3.9. Let N be a 2-absorbing primary submodule of an R-module

M . If (N : M) is a radical ideal then N is a 2-absorbing submodule.

Proof. suppose that abm ∈ N for a, b ∈ R and m ∈M , then am ∈ N or bm ∈ N

37

or ab ∈√

(N : M) = (N : M), implying that N is a 2-absorbing submodule.

Corollary 2.3.10. Let N be a 2-absorbing primary submodule of an R-module

M . If (N : M) is a prime ideal in R then N is a 2-absorbing submodule.

Proof. Since every prime ideal is radical (Remark 1.1.19) , then by Proposition 2.3.9

N is a 2-absorbing submodule .

Theorem 2.3.11. Let N be a 2-absorbing primary submodule of an R-module

M . Then (N : m) is a 2-absorbing primary ideal of R for all m ∈M −N .

Proof. For m ∈M −N , (N : m) is a proper ideal of R. Assume that a, b, c ∈ R

and abc ∈ (N : m). Thus abcm = a(bc)m ∈ N , N is a 2-absorbing primary

submodule of M , so that am ∈ N or bcm ∈ N or abc ∈√

(N : M). If am ∈ N or

bcm ∈ N , then a ∈ (N : m) or bc ∈ (N : m) and we are done. If abc ∈√

(N : M)

then anbncn = (abc)n ∈ (N : M) for some positive integer n. Since N is 2-

absorbing primary submodule then by Theorem 2.3.7, (N : M) is 2-absorbing

primary ideal, hence anbn = (ab)n ∈ (N : M) or bncn = (bc)n ∈√

(N : M) or

ancn = (ac)n ∈√

(N : M). If (ab)n ∈ (N : M) then (ab)n ∈ (N : m) and hence

ab ∈√

(N : m). If (bc)n ∈√

(N : M) then (bc)nn′ ∈ (N : M) ⊆ (N : m) for

some positive integer n′, hence bc ∈√

(N : m). Similarly if (ac)n ∈√

(N : M),

then ac ∈√

(N : m).

38

Theorem 2.3.12. Let M be an R-module and N be a proper submodule of M .

Then the following are equivalent :

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

(2) For a, b ∈ R such that ab /∈√

(N : M) , Nab = Na ∪Nb and so Nab = Na or

Nab = Nb.

Proof. (1)⇒ (2) Assume that ab /∈√

(N : M) and let m ∈ Nab ,then abm ∈ N .

Since N is 2-absorbing primary then am ∈ N or bm ∈ N , then m ∈ Na or

m ∈ Nb and thus m ∈ Na∪Nb. Let x ∈ Na∪Nb ,then ax ∈ N or bx ∈ N , hence

abx ∈ N so x ∈ Nab.

(2)⇒ (1) Let a, b ∈ R and m ∈M with abm ∈ N . Assume that ab /∈√

(N : M),

then m ∈ Nab = Na or m ∈ Nab = Nb, which implies that am ∈ N or bm ∈ N .

Thus N is a 2-absorbing primary submodule of M .

Theorem 2.3.13. Let N be a 2-absorbing primary submodule of an R-module

M . Let y ∈ M and r ∈ R − (N : y). If (N : M) is a prime ideal of R then√(N : y) =

√(N : ry).

Proof. Let a ∈√

(N : ry) then anry ∈ N for some positive integer n. Then

any ∈ N or ry ∈ N or ran ∈√

(N : M). If any ∈ N then a ∈√

(N : y) and

we are done. If ry ∈ N then r ∈ (N : y) which contradicts the hypothesis. If

ran ∈√

(N : M) then rn′ann

′= (ran)n

′ ∈ (N : M) for some positive integer

n′. Since (N : M) is prime then either rn′ ∈ (N : M) or ann

′ ∈ (N : M), If

rn′ ∈ (N : M) then r ∈ (N : M) ⊆ (N : y) because (N : M) is prime which

contradicts the hypothesis. If ann′ ∈ (N : M) ⊆ (N : y) then a ∈

√(N : y). For

the reverse inclusion, let a ∈√

(N : y) then any ∈ N for some positive integer

39

n, implies that rany ∈ N and hence a ∈√

(N : ry).

Theorem 2.3.14. [23] Let f : M →M ′ be an epimorphism of R-modules.

(1) If N ′ is a 2-absorbing primary submodule of M ′ then f−1(N ′) is a 2-absorbing

primary submodule of M .

(2) If N is a 2-absorbing primary submodule of M containing ker(f) then f(N)

is a 2-absorbing primary submodule of M ′.

Proof. (1) Let a, b ∈ R and m ∈M such that abm ∈ f−1(N ′) then abf(m) ∈ N ′.

But N ′ is a 2-absorbing submodule of M ′, so ab ∈√

(N ′ : M ′) or af(m) ∈ N ′

or bf(m) ∈ N ′. If ab ∈√

(N ′ : M ′) then (ab)nM ′ ⊆ N ′ for some positive integer

n, then (ab)nM ⊆ f−1(N ′), so ab ∈√

(f−1(N ′) : M). If af(m) ∈ N ′ then

f(am) ∈ N ′ then am ∈ f−1(N ′). Similarly bf(m) ∈ N ′ implies bm ∈ f−1(N ′).

Thus ab ∈√

(f−1(N ′) : M) or am ∈ f−1(N ′) or bm ∈ f−1(N ′) and hence

f−1(N ′) is a 2-absorbing primary submodule of M .

(2) Let a, b ∈ R and y ∈ M ′ such that aby ∈ f(N). Then there exists n ∈ N

such that aby = f(n). Since f is an epimorphism therefore for some m ∈ M

we have f(m) = y. Thus abf(m) = f(n). This implies f(abm − n) = 0 which

gives abm − n ∈ kerf ⊆ N . So abm ∈ N . Since N is a 2-absorbing primary

submodule of M , am ∈ N or bm ∈ N or ab ∈√

(N : M). This gives ay ∈ f(N)

or by ∈ f(N) or ab ∈√

(f(N) : M ′). Therefore f(N) is a 2-absorbing primary

submodule of M .

Theorem 2.3.15. [23] Let N be a submodule of an R-module M and let K

be any submodule of M contained in N . Then N/K is a 2-absorbing primary

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

40

Proof. Let N be a 2-absorbing primary submodule of M . Consider the canonical

map f : N → N/K. Then (by Theorem 2.3.14(2))N/K is a 2-absorbing primary

submodule of M/K . Conversely, let abm ∈ N for a, b ∈ R and m ∈ M . Then

(abm + K) ∈ N/K which implies that ab(m + K) ∈ N/K . Since N/K is a

2-absorbing primary submodule, either a(m + K) ∈ N/K or b(m + K) ∈ N/K

or ab ∈√

(N/K : M/K) . This implies, either am ∈ N or bm ∈ N

or ab ∈√

(N : M). Hence N is a 2-absorbing primary submodule of M .

Theorem 2.3.16. Let R1, R2 be rings and R = R1 × R2. Let Mi be an

Ri-module ∀i ∈ {1, 2} and M = M1 ×M2 with (r1, r2)(m1,m2) = (r1m1, r2m2),

be an R-module, where ri ∈ Ri, mi ∈Mi ∀i ∈ {1, 2}. Then we have:

(1) If N1 is a 2-absorbing primary submodule of M1, then N1 ×M2 is

a 2-absorbing primary submodule of M .

(2) If N2 is a 2-absorbing primary submodule of M2, then M1 ×N2 is

a 2-absorbing primary submodule of M .

Proof. (1) Suppose that N1 is a 2-absorbing primary submodule of M1

and let (a1, a2), (b1, b2) ∈ R and (x, y) ∈ M such that

(a1, a2)(b1, b2)(x, y) ∈ N1 × M2. Then (a1b1x, a2b2y) ∈ N1 × M2. Therefore

a1b1x ∈ N1. Since N1 is 2-absorbing primary submodule of M1 then ei-

ther a1x ∈ N1 or b1x ∈ N1 or a1b1 ∈√

(N1 : M1). If a1x ∈ N1

then (a1, a2)(x, y) ∈ N1×M2. Similarly b1x ∈ N implies (b1, b2)(x, y) ∈ N1×M2.

If a1b1 ∈√

(N1 : M1), then an1bn1 = (a1b1)n ∈ (N1 : M1) for some positive in-

teger n. Then (a1n, a2

n)(b1n, b2

n) = [(a1, a2)(b1, b2)]n ∈ (N1 ×M2 : M), hence

(a1, a2)(b1, b2) ∈√

(N1 ×M2 : M). Thus N1 ×M2 is a 2-absorbing.

(2) Similar to (1).

41

2.3.2 Classical 2-Absorbing Submodules

Definition 2.3.17. [16] A proper submodule N of R-module M is called a

classical prime submodule, if for each m ∈ M and a, b ∈ R, abm ∈ N implies

that am ∈ N or bm ∈ N .

Definition 2.3.18. [35] A proper submodule N of R-module M is called a clas-

sical 2-absorbing submodule if whenever a, b, c ∈ R and m ∈ M with

abcm ∈ N , then abm ∈ N or acm ∈ N or bcm ∈ N .

Example 2.3.19. (1) Let R = Z and M = R×R.

The submodule N = {(k, k) : k ∈ R} is a classical 2-absorbing submodule of M .

(2) Let R = Z and M = Z3 ⊕ Q ⊕ Z. The submodule N = 0 ⊕ {0} ⊕ Z is a

classical 2-absorbing submodule of M . To see this, let a, b, c, z ∈ Z, w ∈ Q and

x ∈ Z3 such that abc(x, w, z) ∈ N . Hence abcx = 0 and abcw = 0.

If abcz 6= 0 then w = 0. We have 3|abcx, then 3|ab or 3|cx, if

3|ab then ab(x, w, z) = (abx, 0, abz) = (0, 0, abz) ∈ N . Similarly if

3|cx, then c(x, w, z) = (cx, 0, cz) = (0, 0, cz) ∈ N . Now if abcz = 0, then one

of a, b, c, z is zero, firstly we take one of the scalars is zero, say a, then

a(x, w, z) = (0, 0, 0) ∈ N and hence ab(x, w, z) ∈ N . if a, b, c 6= 0 and z = 0,

since abcw = 0 then w = 0 ( this was a previous case). If a, b, c 6= 0, z = 0 and

w 6= 0 then abcw 6= 0 so abc(x, w, z) /∈ N , a contradiction. Thus N is a classical

2-absorbing submodule of M .

Example 2.3.20. [35] Some module do not have any classical 2-absorbing sub-

module . Let p be a fixed prime integer and N0 = N ∪ {0}.

Then E(p) = {α ∈ Q/Z : α = rpn

+Z for some r ∈ Z and n ∈ N0} is a nonzero

submodule of the Z-module Q/Z. For each t ∈ N0, set

42

Gt = {α ∈ Q/Z : α = rpt

+ Z for some r ∈ Z}. Notice that for each t ∈ N0,

Gt is a submodule of E(p) generated by 1pt

+ Z. Each proper submodule of E(p)

is equal to Gi for some i ∈ Z0. However, no Gt is a classical 2-absorbing sub-

module of E(p). Indeed, 1pt+3 + Z ∈ E(p). Then p3( 1

pt+3 + Z) = 1pt

+ Z ∈ Gt but

p2( 1pt+3 + Z) = 1

pt+1 + Z /∈ Gt.

Proposition 2.3.21. [35] Let M be an R-module and N1, N2 be classical prime

submodules of M . Then N1 ∩N2 is a classical 2-absorbing submodule of M .

Proof. Let a, b, c ∈ R and m ∈ M with abcm ∈ N1 ∩N2. Since N1 is a classical

prime submodule, then we may assume that am ∈ N1. Likewise, assume that

bm ∈ N2. Hence abm ∈ N1 ∩N2 which implies N1 ∩N2 is a classical 2-absorbing

submodule.

Remark 2.3.22. The intersection of two classical 2-absorbing submodules need

not to be a classical 2-absorbing. For example in Z-module Z, 3Z and 4Z are

classical 2-absorbing submodules, but 3Z∩4Z = 12Z is not a classical 2-absorbing

submodule since 2.2.3.1 ∈ 12Z but 2.2.1 /∈ 12Z and 2.3.1 /∈ 12Z.

Theorem 2.3.23. [35] Let f : M →M ′ be an epimorphism of R-module.

(1) If N ′ is a classical 2-absorbing submodule of M ′ then f−1(N ′) is a classical

2-absorbing submodule of M .

(2) If N is a classical 2-absorbing submodule of M containing ker(f) then f(N)

is a classical 2-absorbing submodule of M ′.

Proof. (1) Let a, b, c ∈ R and m ∈ M such that abcm ∈ f−1(N ′). Then

abcf(m) ∈ N ′. Hence abf(m) ∈ N ′ or acf(m) ∈ N ′ or bcf(m) ∈ N ′, and

thus abm ∈ f−1(N ′) or acm ∈ f−1(N ′) or bcm ∈ f−1(N ′). So, f−1(N ′) is a

43

classical 2-absorbing submodule of M .

(2) Let a, b, c ∈ R and m′ ∈M ′ be such that abcm′ ∈ f(N). By assumption there

exists m ∈M such that m′ = f(m) and so f(abcm) ∈ f(N). Since Ker(f) ⊆ N ,

we have abcm ∈ N . It implies that abm ∈ N or acm ∈ N or bcm ∈ N . Hence

abm′ ∈ f(N) or acm′ ∈ f(N) or bcm′ ∈ f(N). Consequently f(N) is a classical

2-absorbing submodule of M ′.

Proposition 2.3.24. [35] Let N be a proper submodule of an R-module M .

(1) If N is a 2-absorbing submodule of M , then N is a classical 2-absorbing

submodule of M .

(2) If N is a 2-absorbing submodule of M and (N : M) is a prime ideal of R,

then N is a classical prime submodule of M .

Proof. (1) Assume that N is a 2-absorbing submodule of M . Let a, b, c ∈ R

and m ∈ M such that abcm ∈ N . Therefore either acm ∈ N or bcm ∈ N or

ab ∈ (N : M). The first two cases lead us to the claim. In the third case we

have that abm ∈ N . Consequently N is a classical 2-absorbing submodule.

(2) Assume that N is a 2-absorbing submodule of M and (N : M) is a prime

ideal of R. Let abm ∈ N for some a, b ∈ R and m ∈M , then am ∈ N or bm ∈ N

or ab ∈ (N : M). If am ∈ N or bm ∈ N we are done. Assume that ab ∈ (N : M)

this implies that a ∈ (N : M) or b ∈ (N : M) and so am ∈ N or bm ∈ N . In

each case, N is a classical prime submodule of M .

Remark 2.3.25. The following example shows that the converse of Proposition

2.3.24(1) is not true. LetR = Z andM = Z3⊕Z5⊕Z. The zero submodule of

M is a classical 2-absorbing submodule , but is not 2-absorbing

since 3.5(1, 1, 0) = (0, 0, 0), but 3(1, 1, 0) 6= (0, 0, 0), 5(1, 1, 0) 6= (0, 0, 0) and

44

3.5 /∈ (0 : Z3 ⊕ Z5 ⊕ Z) = 0.

Theorem 2.3.26. Let M be an R-module and N be a proper submodule of M .

Then N is a classical 2-absorbing submodule of M if and only if (N : m) is a

2-absorbing ideal of R for every m ∈M −N .

Proof. (⇒) (N : m) is proper since m ∈ M − N . Let rsl ∈ (N : m) for some

r, s, l ∈ R then rslm ∈ N . Since N is a classical 2-absorbing submodule then

rsm ∈ N or rlm ∈ N or slm ∈ N , hence rs ∈ (N : m) or rl ∈ (N : m) or

sl ∈ (N : m).

(⇐) let a, b, c ∈ R and m ∈M with abcm ∈ N . If m ∈ N we are done. Assume

that m /∈ N then abc ∈ (N : m). Since (N : m) is a 2-absorbing ideal then

ab ∈ (N : m) or ac ∈ (N : m) or bc ∈ (N : m), and hence abm ∈ N or acm ∈ N

or bcm ∈ N .

Theorem 2.3.27. Let M be a cyclic R-module and N be a submodule of M . If

N is a classical 2-absorbing submodule, then N is a 2-absorbing submodule of

M .

Proof. Let M = Rm for some m ∈M . Suppose that abx ∈ N for some a, b ∈ R

and x ∈ M . Then there exists an element c ∈ R such that x = cm. Therefore

abx = abcm ∈ N and since N is a classical 2-absorbing submodule then abm ∈ N

or acm ∈ N or bcm ∈ N , and hence ab ∈ (N : m) = (N : M) or ax ∈ N or

bx ∈ N .

Remark 2.3.28. If we delete the condition ”cyclic” in Theorem 2.3.27, the the-

orem does not remain true. To see this let M = Z3 ×Z5 ×Z be a Z-module, M

is not cyclic, and let N = {0}, by Remark 2.3.25, N is a classical 2-absorbing

45

submodule but not a 2-absorbing.

Corollary 2.3.29. Let M be a cyclic R-module and N be a submodule of M .

Then N is a classical 2-absorbing submodule if and only if N is a 2-absorbing

submodule of M .

Proof. Directly from Proposition 2.3.24 and Theorem 2.3.27

Proposition 2.3.30. [35] Let M be an R-module and N be a classical 2-absorbing

submodule of M . Then for every a, b, c ∈ R and m ∈M ,

(N : abcm) = (N : abm) ∪ (N : acm) ∪ (N : bcm).

Proof. It is clear that N : abm) ∪ (N : acm) ∪ (N : bcm) ⊆ (N : abcm). For the

other inclusion, let a, b, c ∈ R and m ∈ M . Suppose that r ∈ (N : abcm). Then

abc(rm) ∈ N . So, either ab(rm) ∈ N or ac(rm) ∈ N or bc(rm) ∈ N . Therefore,

either r ∈ (N : abm) or r ∈ (N : acm) or r ∈ (N : bcm). Consequently

(N : abcm) = (N : abm) ∪ (N : acm) ∪ (N : bcm).

Proposition 2.3.31. [35] Let N be a submodule of an R-module M and S be a

multiplicatively closed subset of R. If S−1N is a classical 2-absorbing submodule

of S−1R-module S−1M such that Zd(M/N) ∩ S = φ, then N is a classical 2-

absorbing submodule of M .

Proof. Assume that S−1N is a classical 2-absorbing submodule of S−1M and

Zd(M/N) ∩ S = φ. Let a, b, c ∈ R and m ∈ M such that abcm ∈ N . Then

a1b1c1m1∈ S−1N . Therefore a

1b1m1∈ S−1N or a

1c1m1∈ S−1N or b

1c1m1∈ S−1N .

If a1b1m1∈ S−1N , then there exists u ∈ S such that uabm ∈ N .

But Zd(M/N) ∩ S = φ, hence abm ∈ N . Similarly, if b1c1m1∈ S−1N , then

46

bcm ∈ N and if a1c1m1∈ S−1N then acm ∈ N . Consequently N is a classical

2-absorbing submodule of M .

Theorem 2.3.32. Let M be an R-module and N be a proper irreducible sub-

module of M , such that Nr = Nr3 ∀r ∈ R then N is a classical 2-absorbing

submodule of M .

Proof. Let r1, r2, r3 ∈ R and m ∈ N with r1r2r3m ∈ N , and assume that

r1r2m /∈ N , r1r3m /∈ N and r2r3m /∈ N . We have

N ⊆ (N +Rr1r2m) ∩ (N +Rr1r3m) ∩ (N +Rr2r3m).

Let n ∈ (N + Rr1r2m) ∩ (N + Rr1r3m) ∩ (N + Rr2r3m), then

n = n1 + s1r1r2m = n2 + s2r1r3m = n3 + s3r2r3m where n1, n2, n3 ∈ N and

s1, s2, s3 ∈ R, then

r12n = r1

2n1 + s1r13r2m = r1

2n2 + s2r13r3m = r1

2n3 + s3r12r2r3m , since

r12n3, s3r1

2r2r3m ∈ N so s1r13r2m ∈ N implies s1r2m ∈ Nr13 but Nr13 = Nr1 .

Therefore s1r1r2m ∈ N and so n ∈ N .

Hence (N + Rr1r2m) ∩ (N + Rr1r3m) ∩ (N + Rr2r3m) ⊆ N consequently

(N + Rr1r2m) ∩ (N + Rr1r3m) ∩ (N + Rr2r3m) = N , a contradiction because

N is an irreducible. Hence N is a classical 2-absorbing submodule of M .

Theorem 2.3.33. Let M be an R-module and N be a classical 2-absorbing

submodule of M such that (N : y) is a prime ideal of R for y ∈ M − N . For

x ∈M if (N : x)−⋃

xi∈M−N(N : xi) 6= φ then N = (N +Rx)∩

⋂xi∈M−N

(N +Rxi).

Proof. Suppose that N is a classical 2-absorbing submodule of M . Let

ab ∈ (N : x) −⋃

xi∈M−N(N : xi) where a, b ∈ R, then abx ∈ N and abxi /∈ N for

every xi ∈M −N . It is Clear that N ⊆ (N +Rx) ∩⋂

xi∈M−N(N +Rxi). For the

47

reverse inclusion, let n ∈ (N+Rx)∩⋂

xi∈M−N(N+Rxi) then n = n′+r′x = ni+rixi

for every xi ∈M −N , where n′, ni ∈ N and r′, ri ∈ R.

Now abn = abn′+ abr′x = abni + abrixi and abr′x, abn′, abni ∈ N so abrixi ∈ N .

Since N is a classical 2-absorbing submodule and abxi /∈ N then arixi ∈ N

or brixi ∈ N . If arixi ∈ N then ari ∈ (N : xi), if xi ∈ N then rixi ∈ N ,

assume that xi /∈ N then (N : xi) is a prime, and hence either a ∈ (N : xi)

or ri ∈ (N : xi). If a ∈ (N : xi) we get abxi ∈ N which is a contradiction.

So ri ∈ (N : xi) and hence rixi ∈ N . Similarly if brixi ∈ N we get rixi ∈ N .

Thus we have n = ni + rixi ∈ N so (N + Rx) ∩⋂

xi∈M−N(N + Rxi) ⊆ N . Hence

N = (N +Rx) ∩⋂

xi∈M−N(N +Rxi).

Corollary 2.3.34. Let M be an R-module and N be a classical 2-absorbing

submodule of M such that (N : y) is a prime ideal of R for y ∈ M − N . For

x ∈M −N if (N : x)−⋃

xi∈M−N(N : xi) 6= φ then N is not irreducible.

Proof. By Theorem 2.3.33, N = (N+Rx)∩⋂

xi∈M−N(N+Rxi). Since x ∈M−N

we haveN ⊂ (N+Rx) andN ⊂⋂

xi∈M−N(N+Rxi). ThusN is not irreducible.

48

2.3.3 Almost 2-Absorbing Submodules

In this section we aim to investigate and study some properties of almost

2-absorbing submodules. M. Bataineh gives definitions of almost 2-absorbing

ideal and submodule as follows :

Definition 2.3.35. [15] A proper ideal I of a ring R is called an almost

2-absorbing ideal if a, b, c ∈ R with abc ∈ I − I2 implies that ab ∈ I or ac ∈ I,

or bc ∈ I.

Definition 2.3.36. [15] A proper submodule N of anR-module M is called an

almost 2-absorbing submodule of M if, whenever a, b ∈ R and m ∈M such that

abm ∈ N − (N : M)N , implies that ab ∈ (N : M) or am ∈ N , or bm ∈ N .

We get the following results :

Remark 2.3.37. It is clear that, any 2-absorbing submodule is weakly 2-absorbing

and any weakly 2-absorbing submodule is almost 2-absorbing.

Example 2.3.38. Let R = Z , M = Z48 and let N =< 16 >.

Then (N : M)N = 16Z < 16 > = < 16 > and hence N is almost 2-absorbing

submodule, but N is not a 2-absorbing since 2.2.4 ∈< 16 > but 2.4 /∈< 16 > and

2.2 /∈ (N : M).

Theorem 2.3.39. Let M be an R-module and N be a proper submodule of M .

The following are equivalent :

(1) N is an almost 2-absorbing submodule.

(2) For a, b ∈ R such that ab /∈ (N : M), Nab = Na ∪Nb ∪ [(N : M)N ]ab.

Proof. (1)⇒ (2) Let N be an almost 2-absorbing submodule of M , and assume

49

that ab /∈ (N : M), let m ∈ Nab then abm ∈ N . If abm /∈ (N : M)N then

am ∈ N or bm ∈ N and hence m ∈ Na or m ∈ Nb. If abm ∈ (N : M)N then

m ∈ [(N : M)N ]ab. The other containment holds for any submodule.

(2) ⇒ (1) Let a, b ∈ R and m ∈ M with abm ∈ N − (N : M)N . Assume that

ab /∈ (N : M) then m ∈ Nab = Na ∪ Nb ∪ [(N : M)N ]ab, but abm /∈ (N : M)N

then m ∈ Na or m ∈ Nb, thus am ∈ N or bm ∈ N .

Proposition 2.3.40. Let M be an R-module and N be a proper submodule of

M , then N is an almost 2-absorbing submodule in M if and only if for any

a, b ∈ R and submodule K of M such that abK−{0} ⊆ N − (N : M)N , we have

ab ∈ (N : M) or aK ⊆ N or bK ⊆ N .

Proof. (⇒) Assume that ab /∈ (N : M) with the assumption

abK ⊆ [N−(N : M)N ]∪{0}. We have to show that either aK ⊆ N or bK ⊆ N .

By hypothesis and Theorem 2.3.39, K ⊆ Nab = Na ∪ Nb ∪ [(N : M)N ]ab, the

case K ⊆ [(N : M)N ]ab is not possible by hypothesis. Therefore K ⊆ Na ∪Nb.

It is enough to show that either K ⊆ Na or K ⊆ Nb. Assume not, so there

exist t1, t2 ∈ K such that t1 /∈ Na and t2 /∈ Nb, then t1 ∈ Nb and t2 ∈ Na.

But t1+t2 ∈ Na∪Nb, if t1+t2 ∈ Na then as t2 ∈ Na we have t1 = t1+t2−t2 ∈ Na,

a contradiction. If t1 + t2 ∈ Nb, then t2 ∈ Nb another contradiction. So, we must

have that either K ⊆ Na or K ⊆ Nb.

(⇐) Suppose that abm ∈ N − (N : M)N for a, b ∈ R and m ∈ M . Then,

ab(m) − {0} ⊆ N − (N : M)N and so ab ∈ (N : M) or a(m) ⊆ N or

b(m) ⊆ N . Therefore, ab ∈ (N : M) or am ∈ N or bm ∈ N , thus N is almost

2-absorbing.

Lemma 2.3.41. Let I be an ideal of R and N be an almost 2-absorbing sub-

50

module of M . If a ∈ R, m ∈M and Iam−{0} ⊆ N − (N : M)N , then am ∈ N

or Im ⊆ N or Ia ⊆ (N : M).

Proof. Let am /∈ N and Ia * (N : M). Then there exists b ∈ I such that

ba /∈ (N : M). Now, bam ∈ N − (N : M)N implies that bm ∈ N , since N

is an almost 2-absorbing submodule of M . We have to show that Im ⊆ N .

Let c be an arbitrary element in I. Thus (b + c)am ∈ N − (N : M)N . Hence,

either (b + c)m ∈ N or (b + c)a ∈ (N : M). If (b + c)m ∈ N , then by bm ∈ N

it follows that cm ∈ N . If (b + c)a ∈ (N : M), then ca /∈ (N : M), but

cam ∈ N − (N : M)N . Thus cm ∈ N . Hence, we conclude that Im ⊆ N .

Lemma 2.3.42. Let I, J be ideals of R and N be an almost 2-absorbing sub-

module of M . If m ∈ M and IJm − {0} ⊆ N − (N : M)N , then Im ⊆ N or

Jm ⊆ N or IJ ⊆ (N : M).

Proof. Let Im * N and Jm * N (so I * (N : m) and J * (N : m)).

We have to show that IJ ⊆ (N : M). Assume that c ∈ I and d ∈ J .

By assumption there exists a ∈ I − (N : m) such that am /∈ N but

aJm − {0} ⊆ N − (N : M)N . So by Lemma 2.3.41 shows that aJ ⊆ (N : M)

and so (I − (N : m))J ⊆ (N : M). Similarly there exists b ∈ J − (N : m)

such that Ib ⊆ (N : M) and also I(J − (N : m)) ⊆ (N : M). Thus we have

ab ∈ (N : M), ad ∈ (N : M) and cb ∈ (N : M). By a + c ∈ I and b + d ∈ J

it follows that (a + c)(b + d)m ∈ N − (N : M)N . Therefore, (a + c)m ∈ N or

(b+ d)m ∈ N or (a+ c)(b+ d) ∈ (N : M). If (a+ c)m ∈ N , then cm /∈ N hence,

c ∈ I− (N : m) which implies that cd ∈ (N : M). Similarly by (b+d)m ∈ N , we

get cd ∈ (N : M). If (a+ c)(b+ d) ∈ (N : M), then ab+ ad+ cb+ cd ∈ (N : M)

51

and so cd ∈ (N : M). Therefore, IJ ⊆ (N : M).

Theorem 2.3.43. Let N be a proper submodule of M . The following statements

are equivalent:

(1) N is an almost 2-absorbing submodule of M .

(2) If IJL− {0} ⊆ N − (N : M)N for some ideals I, J of R and a submodule

L of M , then IL ⊆ N or JL ⊆ N or IJ ⊆ (N : M).

Proof. (1) ⇒ (2) Let N be an almost 2-absorbing submodule of M . Assume

that IL * N , JL * N and IJ * (N : M), then there exist m, m′ ∈ L such

that Im * N and Jm′ * N . Thus by Lemma 2.3.42, Jm ⊆ N and Im′ ⊆ N .

Since IJ(m + m′) − {0} ⊆ N − (N : M)N we have either I(m + m′) ⊆ N or

J(m + m′) ⊆ N . If I(m + m′) ⊆ N , then Im ⊆ N which is a contradiction.

Similarly, if J(m + m′) ⊆ N we get Jm′ ⊆ N which a gain is a contradiction.

Therefore IL ⊆ N or JL ⊆ N or IJ ⊆ (N : M).

(2) ⇒ (1) Let a, b ∈ R and m ∈ M such that abm ∈ N − (N : M)N . Then

(a)(b)(m) − {0} ⊆ N − (N : M)N then by assumption we get (a)(m) ⊆ N or

(b)(m) ⊆ N or (a)(b) ⊆ (N : M), hence am ∈ N or bm ∈ N or ab ∈ (N : M).

Theorem 2.3.44. Let M be an R-module and N be a proper submodule of M ,

then N is almost 2-absorbing submodule in M if and only if N/(N : M)N is

weakly 2-absorbing submodule in M/(N : M)N .

Proof. (⇒) Suppose N is almost 2-absorbing submodule in M . Let

a, b ∈ R, m ∈ M such that 0 6= ab(m + (N : M)N) ∈ N/(N : M)N .

Then abm ∈ N − (N : M)N , and so am ∈ N or bm ∈ N or ab ∈ (N : M) since

N is almost 2-absorbing. So ab ∈ (N/(N : M)N : M/(N : M)N) = (N : M)

52

(by Lemma 1.2.26) or a(m + (N : M)N) ∈ N/(N : M)N or

b(m + (N : M)N) ∈ N/(N : M)N . Hence N/(N : M)N is weakly 2-absorbing

in M/(N : M)N .

(⇐) Assume that N/(N : M)N is weakly 2-absorbing submodule in

M/(N : M)N . Let a, b ∈ R, m ∈ M such that abm ∈ N − (N : M)N . So

0 6= ab(m+ (N : M)N) ∈ N/(N : M)N .

Then we have ab ∈ (N/(N : M)N : M/(N : M)N) = (N : M)

or a(m + (N : M)N ∈ N/(N : M)N or b(m + (N : M)N) ∈ N/(N : M)N .

That is, am ∈ N or bm ∈ N or ab ∈ (N : M). Hence N is almost 2-absorbing

submodule in M .

Theorem 2.3.45. Let N,K be R-submodules of M with K ⊆ N . If N is

an almost 2-absorbing submodule of M then N/K is an almost 2-absorbing R-

submodule of M/K.

Proof. Let a, b ∈ R and m + K ∈ M/K such that

ab(m + K) ∈ (N/K) − (N/K : M/K)N/K. Since (N : M) = (N/K : M/K)

then, abm + K ∈ N/K − (N : M)N/K and so abm ∈ N − (N : M)N . As N is

almost 2-absorbing in M , then am ∈ N or bm ∈ N or ab ∈ (N : M). Therefore,

a(m+K) ∈ N/K or b(m+K) ∈ N/K or ab ∈ (N/K : M/K), and hence N/K

is almost 2-absorbing in M/K.

Remark 2.3.46. The converse of above theorem is not be true in general. For

example, for any non almost 2-absorbing submodule N of an R-module M , we

have 0 = N/N is a weakly 2-absorbing (and so almost 2-absorbing) submodule

of M/N .

53

Proposition 2.3.47. Let N be an almost 2-absorbing submodule of R-module

M . If S is a multiplicatively closed subset of R, then S−1N is almost 2-absorbing

submodule in R-module S−1M .

Proof. Let a, b ∈ R, s ∈ S and m ∈ M such that

ab(ms

) ∈ S−1N − (S−1N : S−1M)S−1N . Then, abms∈ S−1N − S−1((N : M)N).

Indeed, if abms∈ S−1((N : M)N), then there is t ∈ S such that

abms

= r1n1+r2n2+...+rknk

t= r1

n1

t+ r2

n2

t+ ... + rk

nk

t, where ri ∈ (N : M) and

ni ∈ N , i = 1, 2, , k. Thus, abms∈ (N : M)(S−1N) ⊆ (S−1N : S−1M)S−1N ,

which is a contradiction. As abms∈ S−1N , there is t ∈ S, such that

tabm ∈ N − (N : M)N , since N is almost 2-absorbing then tam ∈ N or

tbm ∈ N or ab ∈ (N : M) ⊆ (S−1N : S−1M) and hence tamts

= ams∈ S−1N or

tbmts

= bms∈ S−1N or ab ∈ (S−1N : S−1M).

Proposition 2.3.48. Let Q be a submodule of an R-module M , N be any R-

module. If Q ⊕ N is an almost 2-absorbing submodule of M ⊕ N then Q is an

almost 2-absorbing submodule of M .

Proof. Suppose Q ⊕ N is an almost 2-absorbing submodule in M ⊕ N .

Let a, b ∈ R, m ∈ M such that abm ∈ Q − (Q : M)Q. Then we get

ab(m, 0) ∈ (Q ⊕ N) − (Q ⊕ N : M ⊕ N)(Q ⊕ N). Since Q ⊕ N is almost

2-absorbing, then ab ∈ (Q⊕N : M⊕N) or a(m, 0) ∈ Q⊕N or b(m, 0) ∈ Q⊕N ,

that is, am ∈ Q or bm ∈ Q or ab ∈ (Q : M). Hence, Q is almost 2-absorbing

submodule in M .

54

In the remaining part of this section, we appeal to Lemma 1.2.24, Lemma 1.2.25

and Proposition 1.2.23. For this, we consider only finitely generated faithful

multiplication R-module. If M is a multiplication R-module and N = IM ,

K = JM are two submodules of M , then the product NK of N and K is

defined as NK = (IM)(JM) = (IJ)M (See [4]).

Theorem 2.3.49. Let M be a finitely generated faithful multiplication R-module

and N be a proper submodule of M . The following are equivalent :

(1) N is almost 2-absorbing submodule in M .

(2) (N : M) is almost 2-absorbing ideal in R.

(3) N = QM for some almost 2-absorbing ideal Q of R.

Proof. (1)⇒ (2) Suppose N is almost 2-absorbing submodule and let a, b, c ∈ R

such that abc ∈ (N : M) − (N : M)2. Then, abcM − {0} ⊆ N − (N : M)N .

Indeed, if abcM ⊆ (N : M)N , then by Lemma 1.2.24, abc ∈ ((N : M)N : M)

= (N : M)2, a contradiction. Now, since N is almost 2-absorbing submodule

then by Proportion 2.3.40 implies that ab ∈ (N : M) or acM ⊆ N or bcM ⊆ N

(and so ac ∈ (N : M) or bc ∈ (N : M)). Hence, (N : M) is almost 2-absorbing

ideal in R.

(2) ⇒ (1) Let a, b ∈ R and m ∈ M , such that abm ∈ N − (N : M)N . Then,

ab((m) : M) ⊆ ((abm) : M) ⊆ (N : M). Moreover, ab((m) : M) * (N : M)2

because otherwise, if ab((m) : M) ⊆ (N : M)2 = ((N : M)N : M), then,

ab(m) = ab((m) : M)M ⊆ (N : M)N , a contradiction. As (N : M) is almost

2-absorbing ideal in R, then, ab ∈ (N : M) or a((m) : M) ⊆ (N : M) or

b((m) : M) ⊆ (N : M) (by Proposition 2.3.40). In the second case, we obtain

(am) ⊆ a(m) = a((m) : M)M ⊆ (N : M)M = N and so am ∈ N , similarly we

55

have bm ∈ N . Thus N is an almost 2-absorbing submodule in M .

(2)⇔ (3) We choose Q = (N : M).

Proposition 2.3.50. Let M be a local multiplication R-module with a unique

maximal submodule Q and (Q : M)Q = 0, then any proper submodule of M is

almost 2-absorbing if and only if it is weakly 2-absorbing .

Proof. (⇒) For any proper submodule N of M , N ( Q, (N : M)N = 0, because

(Q : M)Q = 0. Whenever a, b ∈ R, m ∈M such that abm ∈ N − (N : M)N we

have 0 6= abm ∈ N . So if N is almost 2-absorbing, then it is weakly 2-absorbing

in M .

(⇐) It is trivial, because any weakly 2-absorbing submodule is almost 2-absorbing.

Theorem 2.3.51. Let N be a submodule of a faithful multiplication R-module

M and I be a finitely generated faithful multiplication ideal of R. Then, N is an

almost 2-absorbing submodule of IM if and only if (N : I) is almost 2-absorbing

in M .

Proof. Suppose that N is almost 2-absorbing submodule in IM .

Let a, b ∈ R and m ∈ M , such that abm ∈ (N : I) − ((N : I) : M)(N : I).

Then, abIm − {0} ⊆ N − (N : IM)N . In fact, if abIm ⊆ (N : IM)N , then

by Lemma 1.2.25, abm ∈ ((N : IM)N : I) = (N : IM)(N : I)

= ((N : I) : M)(N : I), a contradiction. As N is almost 2-absorbing

submodule in IM , then by Proposition 2.3.40, with K = Im we have, aIm ⊆ N

or bIm ⊆ N or ab ∈ (N : IM). If aIm ⊆ N or bIm ⊆ N , then, am ∈ (N : I)

or bm ∈ (N : I). Suppose ab ∈ (N : IM), so that abIM ⊆ N . Then again

56

by Lemma 1.2.25, abM = ab(IM : I) ⊆ (abIM : I) ⊆ (N : I), and so,

ab ∈ ((N : I) : M). Therefore, (N : I) is almost 2-absorbing submodule in M .

Conversely, suppose that (N : I) is almost 2-absorbing submodule in M . Let

a, b ∈ R and K be a submodule of IM such that abK −{0} ⊆ N − (N : IM)N .

Then, ab(K : I) ⊆ (abK : I) ⊆ (N : I).

Moreover, if ab(K : I) ⊆ ((N : I) : M)(N : I) = (N : IM)(N : I), then,

abK = ab(IK : I) = ab(K : I)I ⊆ (N : IM)(N : I)I = (N : IM)N , a

contradiction. As (N : I) is almost 2-absorbing submodule in M , then

ab ∈ ((N : I) : M) = (N : IM) or a(K : I) ⊆ (N : I) or b(K : I) ⊆ (N : I),

which implies that aK = a(K : I)I ⊆ a(N : I)I = aN ⊆ N or

bK = b(K : I)I ⊆ b(N : I)I ⊆ N . Hence, N is almost 2-absorbing submodule

in IM .

Theorem 2.3.52. Let M be a finitely generated faithful multiplication R-module

and P be a proper submodule of M , then P is almost 2-absorbing submodule

in M if and only if whenever N , K and L are submodules of M such that

NKL− {0} ⊆ P − (P : M)P , we have NK ⊆ P or NL ⊆ P or KL ⊆ P .

Proof. (⇒) Suppose P is almost 2-absorbing submodule inM . So by Theorem 2.3.49,

(P : M) is almost 2-absorbing ideal in R. We have N = (N : M)M ,

K = (K : M)M and L = (L : M)M . Then NKL = (N : M)(K : M)(L : M)M .

SupposeNKL−{0} ⊆ P−(P : M)P , butNK * P , NL * P and KL * P .

Then (N : M)(K : M) * (P : M) , (N : M)(L : M) * (P : M)

and (K : M)(L : M) * (P : M). As (P : M) is almost 2-absorbing in

R and by Theorem 2.3.43, so (N : M)(K : M)(L : M) * (P : M) or

(N : M)(K : M)(L : M) ⊆ (P : M)2. If (N : M)(K : M)(L : M) * (P : M),

57

then NKL = (N : M)(K : M)(L : M)M * (P : M)M = P , which is a contra-

diction. If (N : M)(K : M)(L : M) ⊆ (P : M)2, then

NKL = (N : M)(K : M)(L : M)M ⊆ (P : M)2M = (P : M)P , which is a

contradiction. Therefore, NK ⊆ P or NL ⊆ P or KL ⊆ P .

(⇐) To prove that P is almost 2-absorbing submodule in M , by Theorem 2.3.49

it is enough to prove that (P : M) is almost 2-absorbing ideal in R. Let a, b, c ∈ R

such that abc ∈ (P : M) − (P : M)2, then abcM − {0} ⊆ P − (P : M)P . Take

aM = N , bM = K and cM = L, we get NKL − {0} ⊆ P − (P : M)P . By

assumption, NK ⊆ P or NL ⊆ P or KL ⊆ P ; that is, abM ⊆ P or acM ⊆ P or

bcM ⊆ P . Hence ab ∈ (P : M) or ac ∈ (P : M) or bc ∈ (P : M). Thus (P : M)

is almost 2-absorbing ideal in R; that is, P is almost 2-absorbing submodule in

M .

2.3.4 Almost 2-Absorbing Primary Submodules

In [15] was introduced the concept of almost 2-absorbing submodule as a

generalization of 2-absorbing submodule [19]. In this section we introduce and

study the concept of almost 2-absorbing primary submodules .

Definition 2.3.53. A proper ideal I ofR is called an almost 2-absorbing primary

ideal if a, b, c ∈ R with abc ∈ I − I2 implies that ab ∈ I or ac ∈√I, or bc ∈

√I.

Definition 2.3.54. A proper submodule N of an R-module M is called an

almost 2-absorbing submodule of M if, whenever a, b ∈ R and m ∈M such that

abm ∈ N − (N : M)N , implies that ab ∈√

(N : M) or am ∈ N , or bm ∈ N .

Definition 2.3.55. Let M be an R-module and N be a proper submodule of M .

Then N is said to be a weakly 2-absorbing primary submodule of M if whenever

58

a, b ∈ R and m ∈ M with 0 6= abm ∈ N , then ab ∈√

(N : M) or am ∈ N or

bm ∈ N .

Remark 2.3.56. It is clear that, any 2-absorbing primary submodule is weakly

2-absorbing primary and any weakly 2-absorbing primary submodule is almost

2-absorbing primary. However, the converses are not necessarily true.

Example 2.3.57. (1) Consider the Z-module Z30, N = {0} is a weakly

2-absorbing primary submodule but is not 2-absorbing primary, because

0 = 2.3.5 ∈ N , but 2.5 /∈ N , 3.5 /∈ N and 2.3 /∈√

(N : M) = {0}.

(2) Let N be any submodule that is not 2-absorbing primary of R-module M such

that (N : M)N = N , then N is almost 2-absorbing primary but not 2-absorbing

primary.

Remark 2.3.58. Since (N : M) ⊆√

(N : M) for any submodule N of an R-

module M , then any almost 2-absorbing submodule is an almost 2-absorbing

primary submodule of M .

Proposition 2.3.59. Let N be a submodule of an R-module M , and (N : M)

be a radical ideal in R, then N is almost 2-absorbing primary if and only if N

is almost 2-absorbing submodule.

Proof. (⇒) Suppose N is almost 2-absorbing primary submodule, let a, b ∈ R,

m ∈ M with abm ∈ N − (N : M)N , then am ∈ N or bm ∈ N or

ab ∈√

(N : M) = (N : M) ( since (N : M) is radical ), hence N is almost

2-absorbing submodule.

(⇐) It is trivial, by Remark 2.3.58.

Corollary 2.3.60. Let N be a submodule of an R-module M , and (N : M) be

59

a prime ideal in R, then N is almost 2-absorbing primary if and only if N is

almost 2-absorbing submodule.

Proof. By Remark 1.1.19, (N : M) is a radical ideal in R.

Theorem 2.3.61. Let N,K be R-submodules of M with K ⊆ N . If N is an

almost 2-absorbing primary submodule of M then N/K is an almost 2-absorbing

primary R-submodule of M/K.

Proof. Let a , b ∈ R and m ∈ M such that

ab(m + K) ∈ (N/K) − (N/K : M/K)N/K. Since (N : M) = (N/K : M/K)

then, abm + K ∈ N/K − (N : M)N/K and so abm ∈ N − (N : M)N . As N is

almost 2-absorbing primary in M , then am ∈ N or bm ∈ N or ab ∈√

(N : M).

Therefore, a(m+K) ∈ N/K or b(m+K) ∈ N/K or ab ∈√

(N/K : M/K), and

hence N/K is almost 2-absorbing primary in M/K.

Proposition 2.3.62. Let N be an almost 2-absorbing primary submodule of R-

module M . If S is a multiplicatively closed subset of R, then S−1N is almost

2-absorbing primary submodule in R-module S−1M .

Proof. Let a , b ∈ R, s ∈ S and m ∈ M such that

ab(ms

) ∈ S−1N − (S−1N : S−1M)S−1N . Then, abms∈ S−1N − S−1((N : M)N).

Indeed, if abms∈ S−1((N : M)N), then there is t ∈ S such that

abms

= r1n1+r2n2+...+rknk

t= r1

n1

t+ r2

n2

t+ ... + rk

nk

t, where ri ∈ (N : M) and

ni ∈ N , i = 1, 2, ..., k. Thus, abms∈ (N : M)(S−1N) ⊆ (S−1N : S−1M)S−1N ,

which is a contradiction. As abms∈ S−1N , there is t ∈ S,

such that ab(tm) ∈ N(N : M)N , since N is almost 2-absorbing primary then

60

a(tm) ∈ N or b(tm) ∈ N or ab ∈√

(N : M) ⊆√

(S−1N : S−1M) and hence

tamts

= ams∈ S−1N or tbm

ts= bm

s∈ S−1N or ab ∈

√(S−1N : S−1M).

Proposition 2.3.63. Let Q be a submodule of R-module M , N be an any R-

module. If Q ⊕ N is an almost 2-absorbing primary submodule of M ⊕ N then

Q is an almost 2-absorbing primary submodule of M .

Proof. Suppose Q ⊕ N is almost 2-absorbing primary submod-

ule in M ⊕ N . Let a, b ∈ R, m ∈ M such that abm ∈ Q − (Q : M)Q.

Then we get ab(m, 0) ∈ (Q ⊕N) − (Q ⊕N : M ⊕N)(Q ⊕N). Since Q ⊕N is

almost 2-absorbing primary, then ab ∈√

(Q⊕N : M ⊕N) or a(m, 0) ∈ Q⊕N

or b(m, 0) ∈ Q⊕N , that is, am ∈ Q or bm ∈ Q or ab ∈√

(Q : M). Hence, Q is

almost 2-absorbing primary submodule in M .

Theorem 2.3.64. Let M be an R-module and N be a proper submodule of M .

The following are equivalent :

(1) N is an almost 2-absorbing primary submodule.

(2) For a, b ∈ R such that ab /∈√

(N : M), Nab = Na ∪Nb ∪ [(N : M)N ]ab.

Proof. (1) ⇒ (2) Let N be an almost 2-absorbing primary submodule, and

assume that ab /∈√

(N : M), let m ∈ Nab then abm ∈ N . If abm /∈ (N : M)N

then am ∈ N or bm ∈ N and hence m ∈ Na or m ∈ Nb. If abm ∈ (N : M)N

then m ∈ [(N : M)N ]ab. The other containment holds for any submodule.

(2) ⇒ (1) Let a, b ∈ R and m ∈ M with abm ∈ N − (N : M)N . Assume that

ab /∈√

(N : M) then m ∈ Nab = Na ∪Nb ∪ [(N : M)N ]ab, but abm /∈ (N : M)N

then m ∈ Na or m ∈ Nb, thus am ∈ N or bm ∈ N .

61

Proposition 2.3.65. Let M be an R-module and N be a proper submodule of

M , then N is an almost 2-absorbing primary submodule in M if and only if for

any a, b ∈ R and submodule K of M such that abK −{0} ⊆ N − (N : M)N , we

have ab ∈√

(N : M) or aK ⊆ N or bK ⊆ N .

Proof. (⇒) Assume that ab /∈√

(N : M), then by Theorem 2.3.64

K ⊆ Nab = Na ∪ Nb ∪ [(N : M)N ]ab, but abK * (N : M)N then K ⊆ Na or

K ⊆ Nb and hence aK ⊆ N or bK ⊆ N .

(⇐) Suppose that abm ∈ N − (N : M)N for a, b ∈ R and m ∈ M . Then,

ab(m)−{0} ⊆ N−(N : M)N and so ab ∈√

(N : M) or a(m) ⊆ N or b(m) ⊆ N .

Therefore, ab ∈√

(N : M) or am ∈ N or bm ∈ N , thus N is almost 2-absorbing

primary.

Theorem 2.3.66. Let M be an R-module and N be a proper submodule

of M , then N is almost 2-absorbing primary submodule in M if and only

if N/(N : M)N is weakly 2-absorbing primary submodule in M/(N : M)N .

Proof. (⇒) Suppose N is almost 2-absorbing primary submodule in M .

Let a, b ∈ R, m ∈ M such that 0 6= ab(m + (N : M)N) ∈ N/(N : M)N . Then

abm ∈ N − (N : M)N , and so am ∈ N or bm ∈ N or ab ∈√

(N : M) since

N is almost 2-absorbing primary. So ab ∈√

(N/(N : M)N : M/(N : M)N)

=√

(N : M) (by Lemma 1.2.26) or a(m + (N : M)N) ∈ N/(N : M)N or

b(m + (N : M)N) ∈ N/(N : M)N . Hence N/(N : M)N is weakly 2-absorbing

primary in M/(N : M)N .

(⇐) Assume N/(N : M)N is weakly 2-absorbing primary in M/(N : M)N .

Let a , b ∈ R, m ∈ M such that abm ∈ N − (N : M)N . So

62

0 6= ab ( m + ( N : M ) N ) ∈ N/ ( N : M ) N . Then we have

ab ∈√

(N/ (N : M) N : M/ (N : M) N ) =√

(N : M) or

a(m + (N : M)N) ∈ N/(N : M)N or b(m + (N : M)N) ∈ N/(N : M)N . That

is, am ∈ N or bm ∈ N or ab ∈√

(N : M). Hence N is almost 2-absorbing

primary submodule in M .

Theorem 2.3.67. Let M be a finitely generated faithful multiplication R-module

and N be a proper submodule of M . The following are equivalent :

(1) N is almost 2-absorbing primary submodule in M .

(2) (N : M) is almost 2-absorbing primary ideal in R.

(3) N = QM for some almost 2-absorbing primary ideal Q of R.

Proof. (1) ⇒ (2) Suppose N is almost 2-absorbing primary submod-

ule and let a, b, c ∈ R such that abc ∈ (N : M) − (N : M)2. Then,

abcM − {0} ⊆ N − (N : M)N . Indeed, if ab(cM) ⊆ (N : M)N , then by

Lemma 1.2.24, abc ∈ ((N : M)N : M) = (N : M)2, a contradiction. Now,

since N is almost 2-absorbing primary submodule then by Proportion 2.3.65

we have ab ∈√

(N : M) or acM ⊆ N or bcM ⊆ N (and so ac ∈ (N : M) or

bc ∈ (N : M)). Hence, (N : M) is almost 2-absorbing primary ideal in R.

(2) ⇒ (1) Let a, b ∈ R and m ∈ M , such that abm ∈ N − (N : M)N . Then,

ab((m) : M) ⊆ ((abm) : M) ⊆ (N : M). Moreover, ab((m) : M) * (N : M)2

because otherwise, if ab((m) : M) ⊆ (N : M)2 ⊆ ((N : M)N : M), then,

ab(m) = ab((m) : M)M ⊆ (N : M)N , a contradiction. As (N : M) is almost

2-absorbing primary ideal in R, then, ab ∈√

(N : M) or a((m) : M) ⊆ (N : M)

or b((m) : M) ⊆ (N : M) (by Proposition 2.3.65). In the second case, we obtain

(am) ⊆ a(m) = a((m) : M)M ⊆ (N : M)M = N and so am ∈ N , similarly we

63

have bm ∈ N . Thus N is an almost 2-absorbing primary submodule in M .

(2)⇔ (3) We choose Q = (N : M).

Proposition 2.3.68. Let M be a local multiplication R-module with a unique

maximal submodule Q and (Q : M)Q = 0, then any proper submodule of M is

almost 2-absorbing primary if and only if it is weakly 2-absorbing primary .

Proof. (⇒) For any proper submodule N of M , N ( Q, (N : M)N = 0, because

(Q : M)Q = 0. Whenever a, b ∈ R, m ∈ M such that abm ∈ N − (N : M)N

we have 0 6= abm ∈ N . So if N is almost 2-absorbing primary, then it is weakly

2-absorbing primary in M .

(⇐) It is trivial, by Remark 2.3.56.

Theorem 2.3.69. Let N be a submodule of a faithful multiplication R-module

M and I be a finitely generated faithful multiplication ideal of R. Then, N is an

almost 2-absorbing primary submodule of IM if and only if (N : I) is an almost

2-absorbing primary in M .

Proof. Suppose that N is almost 2-absorbing primary submodule in IM .

Let a, b ∈ R and m ∈ M , such that abm ∈ (N : I) − ((N : I) : M)(N : I).

Then, abIm − {0} ⊆ N − (N : IM)N . In fact, if abIm ⊆ (N : IM)N ,

then by Lemma 1.2.25, abm ∈ ((N : IM)N : I) = (N : IM)(N : I)

= ((N : I) : M)(N : I), a contradiction. As N is almost 2-absorbing primary

submodule in IM , then by Proposition 2.3.65, with K = Im we have aIm ⊆ N

or bIm ⊆ N or ab ∈√

(N : IM). If aIm ⊆ N or bIm ⊆ N , then, am ∈ (N : I)

or bm ∈ (N : I). Suppose ab ∈√

(N : IM) then ab ∈√

((N : I) : M), because

(N : IM) = ((N : I) : M). Therefore (N : I) is almost 2-absorbing primary

64

submodule of M .

Conversely, suppose that (N : I) is almost 2-absorbing primary submod-

ule in M . Let a, b ∈ R and K be a submodule of IM such that

abK − {0} ⊆ N − (N : IM)N . Then, ab(K : I) ⊆ (abK : I) ⊆ (N : I).

Moreover, if ab(K : I) ⊆ ((N : I) : M)(N : I) = (N : IM)(N : I), then,

abK = ab(IK : I) = ab(K : I)I ⊆ (N : IM)(N : I)I = (N : IM)N , a con-

tradiction. As (N : I) is almost 2-absorbing primary submodule in M , then

ab ∈√

((N : I) : M) =√

(N : IM) or a(K : I) ⊆ (N : I) or b(K : I) ⊆ (N : I),

which implies that aK = a(K : I)I ⊆ a(N : I)I ⊆ aN ⊆ N or

bK = b(K : I)I ⊆ b(N : I)I ⊆ N . Hence, N is almost 2-absorbing primary

submodule in IM .

65

Chapter 3

n-Absorbing Submodules

In this chapter we extend the definition of 2-absorbing to n-absorbing submod-

ules, where n is any positive integer .

3.1 n-Absorbing Submodules

Definition 3.1.1. [6] A proper ideal I of a ring R is said to be an n-absorbing

ideal if whenever a1...an+1 ∈ I for a1, ..., an+1 ∈ R then there are n of a′is whose

product is in I.

Definition 3.1.2. [20] A proper submodule N of an R-module M is called an

n-absorbing submodule if whenever a1...anm ∈ N for a1, ..., an ∈ R and m ∈M ,

then either a1...an ∈ (N : M) or there are n− 1 of a′is whose product with m is

in N .

Example 3.1.3. (1) Let R = Z and M = Rm . The submodule N = {(k, ..., k) :

k ∈ R} is an n-absorbing submodule of M .

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(2) On Z-module Z , nZ is n-absorbing submodule if n = 0, p or p1p2...pn , where

p′is are prime integers .

Proposition 3.1.4. [20] Let M be a cyclic multiplication R-module. Then N is

an n-absorbing submodule of M if and only if (N : M) is an n-absorbing ideal

of R.

Proof. Let M be a cyclic R-module generated by m ∈ M . Let N be an n-

absorbing submodule of M . Assume that a1...an+1 ∈ (N : M). For every

1 ≤ i ≤ n let ai be the element of R which is obtained by eliminating ai from

a1...an. Assume that aian+1 /∈ (N : M) for every 1 ≤ i ≤ n. Then as M = Rm,

we have aian+1m /∈ N . So it is follows from (a1...an)(an+1m) ∈ N and the fact

that N is n-absorbing that a1...an ∈ (N : M), that is, (N : M) is n-absorbing .

Conversely, assume that (N : M) is an n-absorbing ideal of R. Let a1, ..., an ∈ R

and x ∈ M such that a1...anx ∈ N . Then there exists an+1 ∈ R such that

x = an+1m. Thus a1...an+1m ∈ N . So a1...anan+1 ∈ (N : m) = (N : M). But

(N : M) is an n-absorbing ideal in R, so there are n of the a′is whose product is

in (N : M). This implies that either a1...an ∈ (N : M) or there are n− 1 of a′is

whose product with x is in N , that is, N is n-absorbing.

Remark 3.1.5. In general, Proportion 3.1.4 is not true. For example let

Q/Z be a Z-module and let p be a fixed prime integer , then

Z(p∞) = {α ∈ Q/Z : α = r/pn + Z for some r ∈ Z and n ≥ 0}

is a nonzero submodule of Q/Z .

Let Gt = {α ∈ Q/Z : α = r/pt + Z for some r ∈ Z } for all t ≥ 0 . Now Gt

is not n-absorbing submodule of Z(p∞) , because pn(1/pt+n + Z) ∈ Gt while

pn−1(1/pt+n + Z) /∈ Gt and pn /∈ (Gt : Z(p∞)) = 0. But (Gt : Z(p∞)) = 0 is an

67

n-absorbing ideal in Z.

Proposition 3.1.6. If N is an n-absorbing submodule of an R-module M then

(N : m) is an n-absorbing ideal in R for all m ∈M −N .

Proof. For m ∈ M −N , (N : m) is a proper ideal of R. Assume that

a1...an+1 ∈ (N : m) for a1, ..., an+1 ∈ R. Then a1...an+1m = a1...an(an+1m) ∈ N .

Since N is an n-absorbing submodule then a1...an ∈ (N : M) ⊆ (N : m) or there

are n − 1 of the a′is , 1 ≤ i ≤ n whose product with an+1m in N , the latter

case means that there are n− 1 of the a′is, 1 ≤ i ≤ n whose product with an+1

belongs to (N : m). Thus (N : m) is an n-absorbing ideal in R.

Proposition 3.1.7. Let N be an n-absorbing submodule of M . If the set of all

zero divisors of M/N , Zd(M/N), forms an ideal in R, then it is an n-absorbing

ideal of R.

Proof. Let a1...an+1 ∈ Zd(M/N) for a1, ..., an+1 ∈ R, then by Proposition 1.2.20,

a1...an+1 ∈ (N : m′) for some m′ ∈M −N . Since N is an n-absorbing submod-

ule then (N : m′) is an n-absorbing ideal of R. So there are n of a′is whose

product belongs to (N : m′) and hence there are n of a′is whose product belongs

to Zd(M/N).

Theorem 3.1.8. [20] If Nj is an nj-absorbing submodule of M for every

1 ≤ j ≤ k, then N1 ∩ ... ∩Nk is an n-absorbing submodule of M where

n = n1 + ... + nk. In particular, if N1, ..., Nn are prime submodule of M , then

N1 ∩ ... ∩Nn is an n-absorbing submodule of M .

Proof. Let a1, ..., an ∈ R and m ∈M with a1...anm ∈ N1∩ ...∩Nk = N such that

68

no subset consisting of n−1 of the a′is can satisfy the condition that the product

of its elements with m belong to N . As a1...anm ∈ N1∩ ...∩Nk, so a1...anm ∈ Nj

for every 1 ≤ j ≤ k. Therefore a1...an ∈ (Nj : M) for every 1 ≤ j ≤ k since

Nj is assumed to be an nj-absorbing submodule of M and nj ≤ n. Therefore

a1...an ∈k⋂i=1

(Nj : M) = (N : M),that is, N is n-absorbing.

Theorem 3.1.9. [22] Let N be an n-absorbing submodule of M . Then Nr is an

n-absorbing submodule of M containing N for all r ∈ R− (N : M).

Proof. Let a1...anm ∈ Nr for some a1, ..., an ∈ R and m ∈ M . Then

a1a2...an(rm) ∈ N so, either a1...an ∈ (N : M) or there are n−1 of the a′is whose

product with rm is in N . It is Clear that, if a1...an ∈ (N : M) ⊆ (Nr : M), then

a1...an ∈ (Nr : M) and hence we are done. In other case, there are n− 1 of ais

whose product with rm is in N , we have that there is a product of (n − 1) of

the a′is with m is in Nr . Thus Nr is an n-absorbing submodule of M .

Theorem 3.1.10. Let N be a submodule of an R-module M . The following are

equivalent :

(1) N is an n-absorbing submodule.

(2) For a1, ..., an ∈ R such that a1...an /∈ (N : M), Na1...an =n⋃i=1

Nai , where

ai = a1...ai−1ai+1...an.

Proof. (1) ⇒ (2) Let m ∈ Na1...an and assume that a1...an /∈ (N : M), then

a1...anm ∈ N . Since N is an n-absorbing submodule then there are n− 1 of a′is,

1 ≤ i ≤ n, such that aim ∈ N , ai = a1...ai−1ai+1...an, hence m ∈ Nai . For the

other containment, let m ∈n⋃i=1

Nai , then ajm ∈ N for some j ∈ {1, ..., n}, then

aj ajm = a1...anm ∈ N so m ∈ Na1...an .

69

(2) ⇐ (1) Let a1, ..., an ∈ R and m ∈ M such that a1...anm ∈ N , assume

that a1...an /∈ (N : M), then m ∈ Na1...an =n⋃i=1

Nai then m ∈ Naj for some

j ∈ {1, ..., n}, implies ajm = a1...aj−1aj+1...anm ∈ N . Thus N is an n-absorbing

submodule.

Proposition 3.1.11. Let N be an n-absorbing submodule of an R-module M ,

y ∈M and a1, ..., an ∈ R. If a1...an /∈ (N : M) then

(N : a1...any) =n⋃i=1

(N : aiy)

where ai = a1...ai−1ai+1...an.

Proof. Let r ∈ (N : a1...any) then ra1...any = a1...an(ry) ∈ N . Since N is

an n-absorbing submodule and a1...an /∈ (N : M) then ai(ry) ∈ N , where

ai = a1...ai−1ai+1...an, for some i , hence r ∈ (N : aiy). For the reverse

inclusion, let r ∈n⋃i=1

(N : aiy), then r ∈ (N : ajy) for some j ∈ {1, ..., n}. Then

raj ajy = ra1...any ∈ N implies r ∈ (N : a1...any).

Definition 3.1.12. Let N be a proper submodule of an R-module M . The

positive integer n is said to be the index of N if N is an n-absorbing submodule

of M but is not (n− 1)-absorbing. And we write ind(N) = n.

Remark 3.1.13. If ind(N) = n, then N is a k-absorbing for all k ≥ n.

The following theorem and its corollary are attempts to make a reduction of

ind(N) (in fact a reduction of the index of some submodule containing N).

Theorem 3.1.14. [22] Let N be an n-absorbing submodule of M with n ≥ 2 and

(N : M) ⊂√

(N : M) . Suppose that r ∈√

(N : M)− (N : M) and let t ≥ 2 be

the least positive integer such that rt ∈ (N : M). Then Nrt−1 = (N : rt−1) is an

(n− t+ 1)-absorbing submodule of M containing N .

70

Proof. Choose 2 ≤ t ≤ n. Then n − t + 1 ≥ 1. Let a1...an−t+1m ∈ (N : rt−1)

for some a1, ..., an−t+1 ∈ R and m ∈ M . Since rt−1a1...an−t+1m ∈ N

and N is an n-absorbing submodule of M , therefore either rt−1aim ∈ N

or rt−2a1...an−t+1m ∈ N or rt−1a1...an−t+1 ∈ (N : M) where

ai = a1a2...ai−1ai+1...an−t+1 for all 1 ≤ i ≤ n − t + 1. If rt−1aim ∈ N or

rt−1a1...an−t+1 ∈ (N : M), then we are done. Hence assume that rt−1aim /∈ N

and rt−1a1...an−t+1 /∈ (N : M). Since N is an n-absorbing submodule of M ,

therefore rt−2a1...an−t+1m ∈ N . Now rt ∈ (N : M) and rt−1a1...an−t+1m ∈ N

imply that rrt−2a1...an−t(an−t+1 + r)m ∈ N . Again, Since N is an

n-absorbing and rt−1aim /∈ N for any 1 ≤ i ≤ (n − t + 1) and

rrt−2a1...an−t(an−t+1 + r) /∈ (N : M) (as rt ∈ (N : M)), we must have

rt−2a1...an−t(an−t+1 + r)m = rt−2a1...an−t+1m + rt−1a1...an−tm ∈ N . As

rt−2a1...an−t+1m ∈ N , we have rt−1a1...an−tm ∈ N , a contradiction, since we

assumed that the product of rt−1 with any n − t of the ais with m is not in

N . Thus rt−1aim ∈ N or rt−1a1...an−t+1 ∈ (N : M), and hence Nrt−1 is an

(n− t+ 1)-absorbing submodule of M .

Corollary 3.1.15. [22] Let N be an n-absorbing submodule of M with n ≥ 2

and (N : M) ⊂√

(N : M) . Suppose that r ∈√

(N : M) − (N : M) and

rn ∈ (N : M), but rn−1 /∈ (N : M). Then Nrn−1 is a prime submodule of M

containing N .

Proof. Clearly, Nrn−1 is an (n − n + 1)-absorbing submodule of M containing

N by Theorem 3.1.14, and thus Nrn−1 is a prime submodule of M containing

N .

71

Now we give a necessary and sufficient condition for capabelity of reducing

(by 1) the index of the residual (N : M) of the proper submodule N of M .

Theorem 3.1.16. Let N be an n-absorbing submodule of an R-module

M . Then (N : M) is an (n − 1)-absorbing ideal of R if and only if

(N : m) is an (n− 1)-absorbing ideal of R for all m ∈M −N .

Proof. (⇒) Let a1, ..., an ∈ R, m ∈ M − N and a1...an ∈ (N : m).

Then a1...anm ∈ N . Since N is an n-absorbing submodule of M ,

then a1...an ∈ (N : M) or there are n − 1 of the a′is whose product with m

is in N . If a1...an ∈ (N : M) then by assumption there are n − 1 of the a′is,

1 ≤ i ≤ n, whose product belongs to (N : M) and hence there are n− 1 of the

a′is, 1 ≤ i ≤ n, whose product belongs to (N : m). In the other case, if there are

n− 1 of the a′is whose product with m is in N , and hence there are n− 1 of the

a′is, 1 ≤ i ≤ n, whose product belongs to (N : m) and we are done.

(⇐) Suppose that a1...an ∈ (N : M) for some a1, ..., an ∈ R and assume that

for every i, 1 ≤ i ≤ n, there exists mi ∈ M such that aimi /∈ N , where

ai = a1...ai−1ai+1...an. By a1...anmi ∈ N it follows that ajmi ∈ N , where

j 6= i and aj = a1...aj−1aj+1...an, since (N : mi) is (n − 1)-absorbing ideal. If∑ni=1mi ∈ N , then ajmj ∈ N since ajmi ∈ N , ∀i 6= j, which is a contradiction.

Thus∑n

i=1mi /∈ N . Now by a1...an∑n

i=1mi ∈ N we have

a1...an ∈ (N :∑n

i=1mi), then there are n − 1 of the a′is whose product is in

(N :∑n

i=1 mi), hence there are n − 1 of the a′is whose product with∑n

i=1mi

belongs to N , then we must have akmk ∈ N , for some k ∈ {1, ..., n}, which is a

contradiction. Thus there are n−1 of the a′is whose product with M is contained

in N . Therefore (N : M) is (n− 1)-absorbing ideal of R.

72

Theorem 3.1.17. Let f : M →M ′ be an epimorphism of R-modules.

(1) If N ′ is an n-absorbing submodule of M ′ then f−1(N ′) is an n-absorbing

submodule of M .

(2) If N is an n-absorbing submodule of M containing ker(f) then f(N) is an

n-absorbing submodule of M ′.

Proof. (1) Let a1, ..., an ∈ R and m ∈ M such that a1...anm ∈ f−1(N ′) then

a1...anf(m) ∈ N ′, but N ′ is n-absorbing submodule of M ′, so a1...an ∈ (N ′ : M ′)

or aif(m) ∈ N ′, where ai = a1...ai−1ai+1...an. If a1...an ∈ (N ′ : M ′) then

a1...anM′ ⊆ N ′, then a1...anM ⊆ f−1(N ′), so a1...an ∈ (f−1(N ′) : M). If

aif(m) ∈ N ′ then f(aim) ∈ N ′ so aim ∈ f−1(N ′). Thus f−1(N ′) is an

n-absorbing submodule of M .

(2) Let a1, ..., an ∈ R, m′ ∈ M ′ and a1...anm′ ∈ f(N). Then there exists t ∈ N

such that a1...anm′ = f(t). Since f is an epimorphism therefore for some m ∈M

we have f(m) = m′. Thus a1...anf(m) = f(t). This implies f(a1...anm− t) = 0,

so a1...anm − t ∈ ker(f) ⊆ N . Thus a1...anm ∈ N . Now, since N is an

n-absorbing therefore aim ∈ N or a1...an ∈ (N : M). Thus aim′ ∈ f(N) or

a1...an ∈ (f(N) : M ′). Hence f(N) is an n-absorbing submodule of M ′.

Theorem 3.1.18. [22] Let N and K be submodules of M such that K ⊆ N .

Then N is an n-absorbing submodule of M if and only if N/K is an n-absorbing

submodule of M/K.

Proof. Suppose N is an n-absorbing submodule of M . Let

a1a2...an(m + K) ∈ N/K for some a1, a2, ..., an ∈ R and m ∈ M . Then

a1a2...anm ∈ N . Since N is an n-absorbing submodule, therefore ei-

73

ther a1a2a3...an ∈ (N : M) or aim ∈ N where ai = a1a2...ai−1ai+1...an for some

1 ≤ i ≤ n. Therefore a1a2...an ∈ (N/K : M/K) or ai(m + K) ∈ N/K. Hence

N/K is an n-absorbing submodule of M/K. Conversely, assume that N/K is an

n-absorbing submodule of M/K and a1a2...anm ∈ N for some a1, a2, ..., an ∈ R

and m ∈M . Since N/K is an n-absorbing submodule of M/K, therefore we have

a1a2...an(m+K) ∈ N/K implies a1a2...an ∈ (N/K : M/K) or ai(m+K) ∈ N/K.

Therefore a1a2...an ∈ (N : M) or aim ∈ N . This implies that N is an

n-absorbing submodule of M .

Theorem 3.1.19. Suppose S is a multiplicatively closed subset of R and S−1M

is the module of fraction of M . Then the following statements hold.

(1) If N is an n-absorbing submodule of M , then S−1N is an n-absorbing

submodule of S−1M .

(2)If S−1N is an n-absorbing submodule of S−1M such that Zd(M/N)∩ S = φ,

then N is an n-absorbing submodule of M .

Proof. (1) Assume that a1, ..., an ∈ R, s1, ..., sn, l ∈ S, m ∈ M and

a1...anms1...snl

∈ S−1N . Then there exists s′ ∈ S such that s′a1...anm =

a1...an(s′m) ∈ N . By assumption, N is an n-absorbing submodule of M , thus

a1...an ∈ (N : M) or ais′m ∈ N , where ai = a1...ai−1ai+1...an for some 1 ≤ i ≤ n.

If ais′m ∈ N then ais

′ms1...si−1si+1...sns′l

= aimsil∈ S−1N , and if a1...an ∈ (N : M) then

a1...ans1...sn

∈ S−1(N : M) ⊆ (S−1N : S−1M) . Therefore, S−1N is an n-absorbing

submodule of S−1M .

(2) Let a1, ..., an ∈ R and m ∈ M be such that a1...anm ∈ N . Then

a1...anm1

∈ S−1N . Since S−1N is an n-absorbing submodule of S−1M , either

a1...an1∈ (S−1N :S−1R S−1M) or aim

1∈ S−1N , where ai = a1...ai−1aa+1..an for

74

some 1 ≤ i ≤ n. Therefore, there exists s ∈ S such that saim ∈ N . This

implies aim ∈ N , since S ∩ Zd(M/N) = φ. Now, consider the case when

a1...an1∈ (S−1N :S−1R S

−1M), then a1...anS−1M ⊆ S−1N . Now we have to show

a1...anM ⊆ N . Assume that m′ ∈ M , then a1...anm′

1∈ a1...anS

−1M ⊆ S−1N ,

so there exists t ∈ S such that ta1...anm ∈ N . Since S ∩ Zd(M/N) = φ, then

a1...anm′ ∈ N , therefor a1...anM ⊆ N . Hence N is an n-absorbing submodule

of M .

3.2 n-Absorbing Compactly Packed Modules

The concept of compactly packed modules was introduced by Al-Ani in [1]

and it was generalized to primary compactly packed modules by El-Atrash and

Ashour in [24] . AlAshker, Ashour and Abu Mallouh generalized this concept to

primal compactly packed modules in [2]. In this section we give a new general-

ization, called n-absorbing compactly packed modules, and study some results

concerning it.

Definition 3.2.1. [1] A proper submodule N of an R-module M is compactly

packed if for each family {Pα}α∈M of prime submodules of M with N ⊆⋃α∈M

Pα

, N ⊆ Pβ for some β ∈M . A module M is compactly packed if every proper

submodule is compactly packed .

Definition 3.2.2. A proper submodule N of an R-module M is an n-absorbing

compactly packed if for each family {Pα}α∈M of n-absorbing submodules of M

with N ⊆⋃α∈M

Pα, N ⊆ Pβ for some β ∈M . A module M is an n-absorbing

compactly packed if every proper submodule is n-absorbing compactly packed .

75

Since every prime submodule is n-absorbing and every m-absorbing

submodule is n-absorbing for n ≥ m, where n and m are positive integers, we

have the following property

Proposition 3.2.3. (1) Every n-absorbing compactly packed module is com-

pactly packed.

(2) Every n-absorbing compactly packed module is m-absorbing compactly packed,

∀m ≤ n, where m and n are positive integers.

Proposition 3.2.4. Let f : M →M ′ be an epimorphism R-module.

(1) If M is an n-absorbing compactly packed module then so is M ′.

(2) If M ′ is an n-absorbing compactly packed module and every n-absorbing sub-

module of M containing ker(f), then M is an n-absorbing compactly packed.

Proof. (1) Let M be an n-absorbing compactly packed module and N ′ be a

proper submodule of M ′ suppose that N ′ ⊆⋃α∈∆

Pα, where Pα is an n-absorbing

submodule of M ′ for each α ∈ ∆. Since f is an epimorphism R-module,

f−1(N ′) ⊆ f−1(⋃α∈∆

Pα). Thus f−1(N ′) ⊆⋃α∈∆

f−1(Pα). Since Pα

is an n-absorbing submodule for each α ∈ ∆, by Theorem 3.1.17 f−1(Pα) is

an n-absorbing submodule of M for each α ∈ ∆. But M is an n-absorbing com-

pactly packed, thus there exists β ∈ ∆ such that f−1(N ′) ⊆ f−1(Pβ). There-

fore N ′ ⊆ Pβ for some β ∈ ∆ and hence N ′ is an n-absorbing compactly packed

submodule of M . Thus M ′ is an n-absorbing compactly packed module.

(2) Suppose that M ′ is an n-absorbing compactly packed and ker(f) contained

in each n-absorbing submodule of M . Let N ⊆⋃α∈∆

Pα where N is a sub-

module of M and Pα is an n-absorbing submodule of M for each α ∈ ∆ so

f(N) ⊆ f(⋃α∈∆

Pα), thus f(N) ⊆⋃α∈∆

f(Pα). But ker(f) ⊆ Pα for each α ∈ ∆.

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Therefore by Theorem 3.1.17 f(Pα) is an n-absorbing submodule of M ′ for each

α ∈ ∆. Since M ′ is an n-absorbing compactly packed module, f(N) ⊆ f(Pβ)

for some β ∈ ∆. Thus for any x ∈ N , f(x) ∈ f(N) ⊆ f(Pβ), so ∃b ∈ Pβ such

that f(x) = f(b), so we have f(x − b) = 0, that is x − b ∈ ker(f) ⊆ Pβ, so we

have x − b ∈ Pβ. Since b ∈ Pβ we have x ∈ Pβ. Therefore N ⊆ Pβ and hence

N is an n-absorbing compactly packed. Thus M is an n-absorbing compactly

packed.

Since n-absorbing compactly packed modules are special modules of ” com-

pactly packed ” modules then any property of compactly packed modules will

be inherited by n-absorbing compactly packed modules. Therefore many results

on compactly packed modules that were proved in [1] hold automatically on

n-absorbing compactly packed modules, we mention for example :

Proposition 3.2.5. Let M be an R-module. If every proper submodule of M is

cyclic then M is n-absorbing compactly packed.

Theorem 3.2.6. If M is an n-absorbing compactly packed module which has

at least one maximal submodule then M satisfies the (ACC) on n-absorbing

submodules.

Since every finitely generated or multiplication R-module has a proper

maximal submodule (see [9]), the following corollary holds.

Corollary 3.2.7. Let M be an n-absorbing compactly packed R-module. If M

is finitely generated or a multiplication R-module then M satisfies the (ACC)

on n-absorbing submodules.

77

At the end of this section, some questions arise about n-absorbing submodules

( or ideals )and their relations with (ACC) or (DCC) on submodules

• If M satisfies (ACC) on submodules and N is a proper submodule of M ,

must N be an n-absorbing submodule of M (especially when we know that

(N : r) ⊆ (N : r2) ⊆ (N : r3) ⊆ ... for any r ∈ R).

• If R satisfies (ACC) on ideals and N be an n-absorbing submodule (es-

pecially we know that (N : m) ⊆ (N : 2m) ⊆ ... for any m ∈ M), must

(N : rm) be an n-absorbing ideal of R for some positive integer r.

78

Conclusion

In this thesis, the residual (N : M) or (N : m) or (N : r) play a central role in

its uses for proving many theorems.

Still we have a crucial question : ” If N is an n-absorbing submodule of the

R-module M , must (N : M) be an n-absorbing ideal in R ?” .

If the answer can be ”YES”, then the results of 2-absorbing submodules would

hold on n-absorbing submodules for every positive integer n.

At the end of this thesis, the following diagram shows the relations between

submodules .

79

Bibliography

[1] Al-Ani, Z. Compactly Packed Modules and Coprimely Packed

Modules, M.sc. Thesies, college of science, university of Baghdad, (1998).

[2] AL-Ashker, M.; Ashour, A. and Abu Mallouh, A. on primal compactly

packed modules, Palestine Journal of Mathematics Vol. 3(Spec 1), pp.

481-488, (2014).

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