On n -Absorbing Submodules À¸Â ¦À WnY ¦W Q±^ ¦©SjS ¦¾Âa › thesis › 119577.pdfOn n...
Transcript of On n -Absorbing Submodules À¸Â ¦À WnY ¦W Q±^ ¦©SjS ¦¾Âa › thesis › 119577.pdfOn n...
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
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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.
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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.
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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
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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.
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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).
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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
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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 .
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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.
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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
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