/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 8, No. 3, 2015, 417-430
ISSN 1307-5543 – www.ejpam.com

Classical 2-Absorbing Submodules of Modules over

Commutative Rings

Hojjat Mostafanasab1,!, Ünsal Tekir2 and Kürşat Hakan Oral3

1 Department of Mathematics and Applications, University of Mohaghegh Ardabili, P. O. Box 179,
Ardabil, Iran
2 Department of Mathematics, Marmara University, Ziverbey, Goztepe, Istanbul 34722, Turkey
3 Department of Mathematics, Yildiz Technical University, Davutpasa Campus, Esenler, Istanbul,
Turkey

Abstract. In this article, all rings are commutative with nonzero identity. Let M be an R-module.
A proper submodule N of M is called a classical prime submodule, if for each m " M and elements
a, b " R, abm " N implies that am " N or bm " N . We introduce the concept of "classical 2-absorbing
submodules" as a generalization of "classical prime submodules". We say that a proper submodule N
of M is a classical 2-absorbing submodule if whenever a, b, c " R and m " M with abcm " N , then
abm " N or acm " N or bcm " N .
2010 Mathematics Subject Classifications: 13A15, 13C99, 13F05

Key Words and Phrases: Classical prime submodule, Classical 2-absorbing submodule

1. Introduction

Throughout this paper, we assume that all rings are commutative with 1 #= 0. Let R be
a commutative ring and M be an R-module. A proper submodule N of M is said to be a
prime submodule, if for each element a " R and m " M , am " N implies that m " N or
a " (N :R M) = {r " R | rM $ N}. A proper submodule N of 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 .
This notion of classical prime submodules has been extensively studied by Behboodi in [9, 10]
(see also, [11], in which, the notion of “weakly prime submodules” is investigated). For more
information on weakly prime submodules, the reader is referred to [3, 4, 12].

Badawi gave a generalization of prime ideals in [5] and said such ideals 2-absorbing ideals.
A proper ideal I of R is a 2-absorbing ideal of R if whenever a, b, c " R and abc " I , then
ab " I or ac " I or bc " I . He proved that I is a 2-absorbing ideal of R if and only if

!Corresponding author.

Email addresses: h.mostafanasab@gmail.com (H. Mostafanasab), utekir@marmara.edu.tr (Ü. Tekir),
khoral@yildiz.edu.tr (K. Hakan Oral)

http://www.ejpam.com 417 c% 2015 EJPAM All rights reserved.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 418

whenever I1, I2, I3 are ideals of R with I1 I2 I3 $ I , then I1 I2 $ I or I1 I3 $ I or I2 I3 $ I .
Anderson and Badawi [2] generalized the notion of 2-absorbing ideals to n-absorbing ideals.
A proper ideal I of R is called an n-absorbing (resp. a strongly n-absorbing) ideal if whenever
x1 · · · xn+1 " I for x1, . . . , xn+1 " R (resp. I1 . . . In+1 $ I for ideals I1, . . . , In+1 of R), then
there are n of the xi ’s (resp. n of the Ii ’s) whose product is in I . The reader is referred to
[6–8] for more concepts related to 2-absorbing ideals. Yousefian Darani and Soheilnia in [13]
extended 2-absorbing ideals to 2-absorbing submodules. A proper submodule N of M is called
a 2-absorbing submodule of M if whenever abm " N for some a, b " R and m " M , then
am " N or bm " N or ab "

!

N :R M
"

. Generally, a proper submodule N of M is called an
n-absorbing submodule if whenever a1 . . . anm " N for a1, . . . an " R and m " M , then either
a1 . . . an " (N :R M) or there are n& 1 of ai ’s whose product with m is in N , see [14]. Several
authors investigated properties of 2-absorbing submodules, for example [15].

In this paper we introduce the definition of classical 2-absorbing submodules. A proper sub-
module N of an R-module M is called classical 2-absorbing submodule if whenever a, b, c " R

and m " M with abcm " N , then abm " N or acm " N or bcm " N . Clearly, every classi-
cal prime submodule is a classical 2-absorbing submodule. We show that every Noetherian
R-module M contains a finite number of minimal classical 2-absorbing submodules (Theorem
3). Further, we give the relationship between classical 2-absorbing submodules, classical prime
submodules and 2-absorbing submodules (Proposition 2, Proposition 7). Moreover, we charac-
terize classical 2-absorbing submodules in (Theorem 2, Theorem 4). In (Theorem 7, Theorem
8) we investigate classical 2-absorbing submodules of a finite direct product of modules.

2. Characterizations of Classical 2-Absorbing Submodules

First of all we give a module which has no classical 2-absorbing submodule.

Example 1. Let p be a fixed prime integer and !0 = !' {0}. Then

E
!

p
"

:=

#

! " "/# | != r

pn
+# for some r " # and n " !0

$

is a nonzero submodule of the #-module "/#. For each t " !0, set

Gt :=

#

! " "/# | != r

pt
+# for some r " #

$

.

Notice that for each t " !0, Gt is a submodule of E
!

p
"

generated by 1
pt +# for each t " !0. Each

proper submodule of E
!

p
"

is equal to Gi for some i " !0 (see, [17, Example 7.10]). However, no

Gt is a classical 2-absorbing submodule of E
!

p
"

. Indeed, 1
pt+3 +# " E
!

p
"

. Then

p3
%

1
pt+3 +#
&

= 1
pt +# " Gt but p2

%
1

pt+3 +#
&

= 1
pt+1 +# /" Gt.

Theorem 1. Let f : M ( M ) be an epimorphism of R-modules.

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

submodule of M.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 419

(ii) 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. (i) Since f is epimorphism, f &1(N )) is a proper submodule of M . Let a, b, c " R and
m " M such that abcm " f &1(N )). Then abc f (m) " N ). Hence ab f (m) " N ) or ac f (m) " N )

or bc f (m) " N ), and thus abm " f &1(N )) or acm " f &1(N )) or bcm " f &1(N )). So, f &1(N ))
is a classical 2-absorbing submodule of M .
(ii) 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 ).

As an immediate consequence of Theorem 1 we have the following corollary.

Corollary 1. Let M be an R-module and L $ N be submodules of M. Then N is a classical

2-absorbing submodule of M if and only if N/L is a classical 2-absorbing submodule of M/L.

Proposition 1. 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 for some a, b, c " R and m " M , 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.

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

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

(ii) N is a classical prime submodule of M if and only if N is a 2-absorbing submodule of M

and (N :R M) is a prime ideal of R.

Proof. (i) 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.
(ii) It is evident that if N is classical prime, then it is 2-absorbing. Also, [3, Lemma 2.1]

implies that (N :R M) is a prime ideal of R. Assume that N is a 2-absorbing submodule of M

and (N :R M) is a prime ideal of R. Let abm " N for some a, b " R and m " M such that neither
am " N nor bm " N . Then ab " (N :R M) and so either a " (N :R M) or b " (N :R M).This
contradiction shows that N is classical prime.

he following example shows that the converse of Proposition 2(i) is not true.

Example 2. Let R = # and M = #p

'

#q

'

" where p, q are two distinct prime integers. One

can easily see that the zero submodule of M is a classical 2-absorbing submodule. Notice that

pq(1,1,0) = (0,0,0), but p(1,1,0) #= (0,0,0), q(1,1,0) #= (0,0,0) and pq(1,1,1) #= 0. So

the zero submodule of M is not 2-absorbing. Also, part (ii) of Proposition 2 shows that the

zero submodule is not a classical prime submodule. Hence the two concepts of classical prime

submodules and of classical 2-absorbing submodules are different in general.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 420

Let M be an R-module and N a submodule of M . For every a " R, {m " M | am " N} is
denoted by (N :R a). It is easy to see that (N :M a) is a submodule of M containing N .

Theorem 2. Let M be an R-module and N be a proper submodule of M. The following conditions

are equivalent:

(i) N is classical 2-absorbing;

(ii) For every a, b, c " R, (N :M abc) = (N :M ab)' (N :M ac)' (N :M bc);

(iii) For every a, b " R and m " M with abm /" N, (N :R abm) = (N :R am)' (N :R bm);

(iv) For every a, b " R and m " M with abm /" N, (N :R abm) = (N :R am) or

(N :R abm) = (N :R bm);

(v) For every a, b " R and every ideal I of R and m " M with abIm $ N, either abm " N or

aIm $ N or bIm $ N;

(vi) For every a " R and every ideal I of R and m " M with aIm #$ N, (N :R aIm) = (N :R am)

or (N :R aIm) = (N :R Im);

(vii) For every a " R and every ideals I , J of R and m " M with aIJm $ N, either aIm $ N or

aJm $ N or IJm $ N;

(viii) For every ideals I , J of R and m " M with IJm #$ N, (N :R IJm) = (N :R Im) or

(N :R IJm) = (N :R Jm);

(ix) For every ideals I , J , K of R and m " M with IJKm $ N, either IJm $ N or IKm $ N or

JKm $ N;

(x) For every m " M\N, (N :R m) is a 2-absorbing ideal of R.

Proof. (i)+ (ii) Suppose that N is a classical 2-absorbing submodule of M . Let
m "
!

N :M abc
"

. Then abcm " N . Hence abm " N or acm " N or bcm " N . Therefore
m "
!

N :M ab
"

or m "
!

N :M ac
"

or m "
!

N :M bc
"

. Consequently,

!

N :M abc
"

=
!

N :M ab
"

'
!

N :M ac
"

'
!

N :M bc
"

.

(ii)+ (iii) Let abm /" N for some a, b " R and m " M . Assume that x " (N :R abm). Then
abxm " N , and so m " (N :M abx). Since abm /" N , m /" (N :M ab). Thus by part (i),
m " (N :M ax) or m " (N :M bx), whence x " (N :R am) or x " (N :R bm). Therefore
(N :R abm) = (N :R am)' (N :R bm).
(iii)+ (iv) By the fact that if an ideal (a subgroup) is the union of two ideals (two subgroups),
then it is equal to one of them.
(iv)+ (v) Let for some a, b " R, an ideal I of R and m " M , abIm $ N . Hence I $ (N :R abm).
If abm " N , then we are done. Assume that abm /" N . Therefore by part (iv) we have that
I $ (N :R am) or I $ (N :R bm), i.e., aIm $ N or bIm $ N .
(v)+ (vi)+ (vii)+ (viii)+ (i x) Have proofs similar to that of the previous implications.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 421

(i x)+ (i) Is trivial.
(i x), (x) Straightforward.

Corollary 2. Let R be a ring and I be a proper ideal of R.

(i) RI is a classical 2-absorbing submodule of R if and only if I is a 2-absorbing ideal of R.

(ii) Every proper ideal of R is 2-absorbing if and only if for every R-module M and every proper

submodule N of M, N is a classical 2-absorbing submodule of M.

Proof. (i) Let I be a classical 2-absorbing submodule of R. Then by Theorem 2, (I :R 1) = I

is a 2-absorbing ideal of R. For the converse see part (i) of Proposition 2.
(ii) Assume that every proper ideal of R is 2-absorbing. Let N be a proper submodule of an

R-module M . Since for every m " M\N , (N :R m) is a proper ideal of R, then it is a 2-absorbing
ideal of R. Hence by Theorem 2, N is a classical 2-absorbing submodule of M . We have the
converse immediately by part (i).

Proposition 3. Let M be an R-module and
(

Ki | i " I
)

be a chain of classical 2-absorbing sub-

modules of M. Then *i"I Ki is a classical 2-absorbing submodule of M.

Proof. Suppose that abcm " *i"I Ki for some a, b, c " R and m " M . Assume that
abm /" *i"I Ki and acm /" *i"I Ki . Then there are t, l " I where abm /" Kt and acm /" Kl .
Hence, for every Ks $ Kt and every Kd $ Kl we have that abm /" Ks and acm /" Kd . Thus, for
every submodule Kh such that Kh $ Kt and Kh $ Kl we get bcm " Kh. Hence bcm " *i"I Ki .

A classical 2-absorbing submodule of M is called minimal, if for any classical 2-absorbing
submodule K of M such that K $ N , then K = N . Let L be a classical 2-absorbing submodule
of M . Set

!=
(

K | K is a classical 2-absorbing submodule of M and K $ L
)

.

If
(

Ki : i " I
)

is any chain in !, then *i"I Ki is in !, by Proposition 3. By Zorn’s Lemma, !
contains a minimal member which is clearly a minimal classical 2-absorbing submodule of M .
Thus, every classical 2 -absorbing submodule of M contains a minimal classical 2-absorbing
submodule of M . If M is a finitely generated, then it is clear that M contains a minimal classical
2-absorbing submodule.

Theorem 3. Let M be a Noetherian R-module. Then M contains a finite number of minimal

classical 2-absorbing submodules.

Proof. Suppose that the result is false. Let ! denote the collection of proper submodules
N of M such that the module M/N has an infinite number of minimal classical 2-absorbing
submodules. Since 0 " ! we get ! #= $. Therefore ! has a maximal member T , since M is a
Noetherian R-module. It is clear that T is not a classical 2-absorbing submodule. Therefore,
there exists an element m " M\T and ideals I , J , K in R such that I JKm $ T but I Jm #$ T ,
IKm #$ T and JKm #$ T . The maximality of T implies that M/ (T + I Jm), M/ (T + IKm)



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 422

and M/ (T + JKm) have only finitely many minimal classical 2-absorbing submodules. Sup-
pose P/T be a minimal classical 2-absorbing submodule of M/T . So I JKm $ T $ P, which
implies that I Jm $ P or IKm $ P or JKm $ P. Thus P/ (T + I Jm) is a minimal classical
2-absorbing submodule of M/ (T + I Jm) or P/ (T + IKm) is a minimal classical 2-absorbing
submodule of M/ (T + IKm) or P/ (T + JKm) is a minimal classical 2-absorbing submodule
of M/ (T + JKm). Thus, there are only a finite number of possibilities for the submodule P.
This is a contradiction.

We recall from [5] that if I is a 2-absorbing ideal of a ring R, then either
-

I = P where P

is a prime ideal of R or
-

I = P1 * P2 where P1, P2 are the only distinct minimal prime ideals
of I .

Corollary 3. Let N be a classical 2-absorbing submodule of an R-module M. Suppose that

m " M\N and
*

(N :R m) = P where P is a prime ideal of R and (N :R m) #= P. Then for each

x "
*

(N :R m)\(N :R m), (N :R xm) is a prime ideal of R containing P. Furthermore, either

(N :R xm) $ (N :R ym) or (N :R ym) $ (N :R xm) for every x , y "
*

(N :R m)\(N :R m).

Proof. By Theorem 2 and [5, Theorem 2.5].

Corollary 4. Let N be a classical 2-absorbing submodule of an R-module M. Suppose that

m " M\N and
*

(N :R m) = P1 * P2 where P1 and P2 are the only nonzero distinct prime ideals

of R that are minimal over (N :R m). Then for each x "
*

(N :R m)\(N :R m), (N :R xm)

is a prime ideal of R containing P1 and P2. Furthermore, either (N :R xm) $ (N :R ym) or

(N :R ym) $ (N :R xm) for every x , y "
*

(N :R m)\(N :R m).

Proof. By Theorem 2 and [5, Theorem 2.6].

An R-module M is called a multiplication module if every submodule N of M has the form
I M for some ideal I of R. Let N and K be submodules of a multiplication R-module M with
N = I1M and K = I2M for some ideals I1 and I2 of R. The product of N and K denoted by NK

is defined by NK = I1 I2M . Then by [1, Theorem 3.4], the product of N and K is independent
of presentations of N and K .

Proposition 4. Let M be a multiplication R-module and N be a proper submodule of M. The

following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M;

(ii) If N1N2N3m $ N for some submodules N1, N2, N3 of M and m " M, then either N1N2m $ N

or N1N3m $ N or N2N3m $ N.

Proof. (i) + (ii) Let N1N2N3m $ N for some submodules N1, N2, N3 of M and m " M .
Since M is multiplication, there are ideals I1, I2, I3 of R such that N1 = I1M , N2 = I2M and
N3 = I3M . Therefore I1 I2 I3m $ N , and so either I1 I2m $ N or I1 I3m $ N or I2 I3m $ N .
Hence N1N2m $ N or N1N3m $ N or N2N3m $ N .
(ii) + (i) Suppose that I1 I2 I3m $ N for some ideals I1, I2, I3 of R and some m " M . It is
sufficient to set N1 := I1M , N2 := I2M and N3 = I3M in part (ii).



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 423

In [16], Quartararo et al. said that a commutative ring R is a u-ring provided R has the
property that an ideal contained in a finite union of ideals must be contained in one of those
ideals; and a um-ring is a ring R with the property that an R-module which is equal to a finite
union of submodules must be equal to one of them. They show that every Bézout ring is a
u-ring. Moreover, they proved that every Prüfer domain is a u-domain. Also, any ring which
contains an infinite field as a subring is a u-ring, [17, Exercise 3.63].

Theorem 4. Let R be a um-ring, M be an R-module and N be a proper submodule of M. The

following conditions are equivalent:

(i) N is classical 2-absorbing;

(ii) For every a, b, c " R, (N :M abc) = (N :M ab) or (N :M abc) = (N :M ac) or

(N :M abc) = (N :M bc);

(iii) For every a, b, c " R and every submodule L of M, abcL $ N implies that abL $ N or

acL $ N or bcL $ N;

(iv) For every a, b " R and every submodule L of M with abL #$ N, (N :R abL) = (N :R aL) or

(N :R abL) = (N :R bL);

(v) For every a, b " R, every ideal I of R and every submodule L of M, abI L $ N implies that

abL $ N or aI L $ N or bI L $ N;

(vi) For every a " R, every ideal I of R and every submodule L of M with

aI L #$ N, (N :R aI L) = (N :R aL) or (N :R aI L) = (N :R I L);

(vii) For every a " R, every ideals I , J of R and every submodule L of M, aIJ L $ N implies that

aI L $ N or aJ L $ N or IJ L $ N;

(viii) For every ideals I , J of R and every submodule L of M with IJ L #$ N, (N :R IJ L) = (N :R I L)

or (N :R IJ L) = (N :R J L);

(ix) For every ideals I , J , K of R and every submodule L of M, IJK L $ N implies that IJ L $ N

or IK L $ N or JK L $ N;

(x) For every submodule L of M not contained in N, (N :R L) is a 2-absorbing ideal of R.

Proof. Similar to the proof of Theorem 2.

Proposition 5. Let R be a um-ring and N be a proper submodule of an R-module M. Then N

is a classical 2-absorbing submodule of M if and only if N is a 3-absorbing submodule of M and

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

Proof. It is trivial that if N is classical 2-absorbing, then it is 3-absorbing. Also, Theorem 4
implies that (N :R M) is a 2-absorbing ideal of R. Now, assume that N is a 3-absorbing submod-
ule of M and (N :R M) is a 2-absorbing ideal of R. Let a1a2a3m " N for some a1, a2, a3 " R and
m " M such that neither a1a2m " N nor a1a3m " N nor a2a3m " N . Then a1a2a3 " (N :R M)



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 424

and so either a1a2 " (N :R M) or a1a3 " (N :R M) or a2a3 " (N :R M). This contradiction
shows that N is classical 2-absorbing.

Proposition 6. Let M be an R-module and N be a classical 2-absorbing submodule of M. The

following conditions hold:

(i) For every a, b, c " R and m " M, (N :R abcm) = (N :R abm)' (N :R acm)' (N :R bcm);

(ii) If R is a u-ring, then for every a, b, c " R and m " M, (N :R abcm) = (N :R abm) or

(N :R abcm) = (N :R acm) or (N :R abcm) = (N :R bcm).

Proof. (i) Let a, b, c " R and m " M . Suppose that r " (N :R abcm). Then abc(rm) " N .
So, either ab(rm) " N or ac(rm) " N or bc(rm) " N . Therefore, either r " (N :R abm) or
r " (N :R acm) or r " (N :R bcm). Consequently
(N :R abcm) = (N :R abm)' (N :R acm)' (N :R bcm).
(ii) Use part (i).

Proposition 7. Let R be a um-ring, M be a multiplication R-module and N be a proper submodule

of M. The following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M;

(ii) If N1N2N3N4 $ N for some submodules N1, N2, N3, N4 of M, then either N1N2N4 $ N or

N1N3N4 $ N or N2N3N4 $ N;

(iii) If N1N2N3 $ N for some submodules N1, N2, N3 of M, then either N1N2 $ N or N1N3 $ N

or N2N3 $ N;

(iv) N is a 2-absorbing submodule of M;

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

Proof. (i)+ (ii) Let N1N2N3N4 $ N for some submodules N1, N2, N3, N4 of M . Since M

is multiplication, there are ideals I1, I2, I3 of R such that N1 = I1M , N2 = I2M and N3 = I3M .
Therefore I1 I2 I3N4 $ N , and so I1 I2N4 $ N or I1 I3N4 $ N or I2 I3N4 $ N . Thus by Theorem 4,
either N1N2N4 $ N or N1N3N4 $ N or N2N3N4 $ N .
(ii)+ (iii) Is easy.
(iii)+ (iv) Suppose that I1 I2K $ N for some ideals I1, I2 of R and some submodule K of M .
It is sufficient to set N1 := I1M , N2 := I2M and N3 = K in part (iii).
(iv)+ (i) By part (i) of Proposition 2.
(iv)+ (v) By [15, Theorem 2.3].
(v)+ (iv) Let I1 I2K $ N for some ideals I1, I2 of R and some submodule K of M . Since M is
multiplication, then there is an ideal I3 of R such that K = I3M . Hence I1 I2 I3 $ (N :R M)which
implies that either I1 I2 $ (N :R M) or I1 I3 $ (N :R M) or I2 I3 $ (N :R M). If I1 I2 $ (N :R M),
then we are done. So, suppose that I1 I3 $ (N :R M). Thus I1 I3M = I1K $ N . Similarly if
I2 I3 $ (N :R M), then we have I2K $ N .



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 425

Definition 1. Let R be a um-ring, M be an R-module and S be a subset of M\{0}. If for all

ideals I , J , Q of R and all submodules K, L of M, (K + I J L)* S #= . and (K + IQL)* S #= . and

(K + JQL) * S #= . implies (K + I JQL) * S #= ., then the subset S is called classical 2-absorbing

m-closed.

Proposition 8. Let R be a um-ring, M be R-module and N a submodule of M. Then N is a

classical 2-absorbing submodule if and only if M\N is a classical 2-absorbing m-closed.

Proof. Suppose that N is a classical 2-absorbing submodule of M and I , J , Q are ideals
of R and K , L are submodules of M such that (K + I J L) * S #= . and (K + IQL) * S #= . and
(K + JQL) * S #= . where S = M\N . Assume that (K + I JQL) * S = .. Then K + I JQL $ N

and so K $ N and I JQL $ N . Since N is a classical 2-absorbing submodule, we get I J L $ N

or IQL $ N or JQL $ N . If I J L $ N , then we get (K + I J L) * S = ., since K $ N . This is a
contradiction. By the other cases we get similar contradictions. Now for the converse suppose
that S = M\N is a classical 2-absorbing m-closed and assume that I JQL $ N for some ideals
I , J , Q of R and submodule L of M . Then we get for submodule K = (0), K + I JQL $ N .
Thus (K + I JQL) * S = .. Since S is a classical 2-absorbing m-closed, (K + I J L) * S = . or
(K + IQL) * S = . or (K + JQL) * S = .. Hence I J L $ N or IQL $ N or JQL $ N . So N is a
classical 2-absorbing submodule.

Proposition 9. Let R be a um-ring, M be an R-module, N a submodule of M and S = M\N. The

following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M;

(ii) S is a classical 2-absorbing m-closed;

(iii) For every ideals I , J , Q of R and every submodule L of M, if I J L * S #= . and IQL * S #= .
and JQL * S #= ., then IJQL * S #= .;

(iv) For every ideals I , J , Q of R and every m " M, if I Jm * S #= . and IQm * S #= . and

JQm* S #= ., then IJQm* S #= ..
Proof. It follows from the previous Proposition, Theorem 2 and Theorem 4.

Theorem 5. Let R be a um-ring, M be an R-module and S be a classical 2-absorbing m-closed.

Then the set of all submodules of M which are disjoint from S has at least one maximal element.

Any such maximal element is a classical 2-absorbing submodule.

Proof. Let " = {N | N is a submodule of M and N * S = .}. Then (0) " " #= .. Since "
is partially ordered by using Zorn’s Lemma we get at least a maximal element of ", say P,
with property P * S = .. Now we will show that P is classical 2-absorbing. Suppose that
I JQL $ P for ideals I , J , Q of R and submodule L of M . Assume that I J L #$ P or IQL #$ P or
JQL #$ P. Then by the maximality of P we get (I J L + P) * S #= . and (IQL + P) * S #= . and
(JQL + P) * S #= .. Since S is a classical 2-absorbing m-closed we have (I JQL + P) * S #= ..
Hence P * S #= ., which is a contradiction. Thus P is a classical 2-absorbing submodule of
M .



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 426

Theorem 6. Let R be a um-ring and M be an R-module.

(i) If F is a flat R-module and N is a classical 2-absorbing submodule of M such that
F / N #= F /M , then F / N is a classical 2-absorbing submodule of F /M .

(ii) Suppose that F is a faithfully flat R-module. Then N is a classical 2-absorbing submodule
of M if and only if F / N is a classical 2-absorbing submodule of F /M .

Proof. (i) Let a, b, c " R. Then we get by Theorem 4,
!

N :M abc
"

=
!

N :M ab
"

or
!

N :M abc
"

=
!

N :M ac
"

or
!

N :M abc
"

=
!

N :M bc
"

. Assume that
!

N :M abc
"

=
!

N :M ab
"

.
Then by [4, Lemma 3.2],
!

F / N :F/M abc
"

= F /
!

N :M abc
"

= F /
!

N :M ab
"

=
!

F / N :F/M ab
"

.

Again Theorem 4 implies that F / N is a classical 2-absorbing submodule of F /M .
(ii) Let N be a classical 2-absorbing submodule of M and assume that F / N = F / M .

Then 0 ( F / N
$( F / M ( 0 is an exact sequence. Since F is a faithfully flat module,

0( N
$( M ( 0 is an exact sequence. So N = M , which is a contradiction. So

F / N #= F /M . Then F / N is a classical 2-absorbing submodule by (1). Now for conversely,
let F / N be a classical 2-absorbing submodule of F / M . We have F / N #= F / M and so
N #= M . Let a, b, c " R. Then

!

F / N :F/M abc
"

=
!

F / N :F/M ab
"

or
!

F / N :F/M abc
"

=
!

F / N :F/M ac
"

or
!

F / N :F/M abc
"

=
!

F / N :F/M bc
"

by Theorem 4.
Assume that
!

F / N :F/M abc
"

=
!

F / N :F/M ab
"

. Hence

F /
!

N :M ab
"

=
!

F / N :F/M ab
"

=
!

F / N :F/M abc
"

= F /
!

N :M abc
"

.

So 0 ( F /
!

N :M ab
" $( F /
!

N :M abc
"

( 0 is an exact sequence. Since F is a faith-

fully flat module, 0(
!

N :M ab
" $(
!

N :M abc
"

( 0 is an exact sequence which implies that
!

N :M ab
"

=
!

N :M abc
"

. Consequently N is a classical 2-absorbing submodule of M by The-
orem 4.

Corollary 5. Let R be a um-ring, M be an R-module and X be an indeterminate. If N is a classical

2-absorbing submodule of M, then N[X ] is a classical 2-absorbing submodule of M[X ].

Proof. Assume that N is a classical 2-absorbing submodule of M . Notice that R[X ] is a
flat R-module. So by Theorem 6, R[X ] / N 0 N[X ] is a classical 2-absorbing submodule of
R[X ]/M 0 M[X ].

For an R-module M , the set of zero-divisors of M is denoted by ZR(M).

Proposition 10. Let M be an R-module, N be a submodule and S be a multiplicative subset of R.

(i) If N is a classical 2-absorbing submodule of M such that
!

N :R M
"

* S = ., then S&1N is a

classical 2-absorbing submodule of S&1M.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 427

(ii) If S&1N is a classical 2-absorbing submodule of S&1M such that ZR(M/N)*S = ., then N

is a classical 2-absorbing submodule of M.

Proof. (i) Let N be a classical 2-absorbing submodule of M and
!

N :R M
"

*S = .. Suppose
that a1

s1

a2
s2

a3
s3

m
s4
" S&1N . Then there exist n " N and s " S such that a1

s1

a2
s2

a3
s3

m
s4
= n

s . Therefore
there exists an s) " S such that s)sa1a2a3m= s)s1s2s3s4n " N . So a1a2a3 (s

!m) " N for s! = s)s.
Since N is a classical 2-absorbing submodule we get a1a2 (s

!m) " N or a1a3 (s
!m) " N or

a2a3 (s
!m) " N . Thus a1a2m

s1s2s4
=

a1a2(s
!m)

s1s2s4s! " S&1N or a1a3m
s1s3s4
" S&1N or a2a3m

s2s3s4
" S&1N .

(ii) Assume that S&1N is a classical 2-absorbing submodule of S&1M and ZR(M/N)*S = ..
Let a, b, c " R and m " M such that abcm " N . Then a

1
b
1

c
1

m
1 " S&1N . Therefore a

1
b
1

m
1 " S&1N

or a
1

c
1

m
1 " S&1N or b

1
c
1

m
1 " S&1N . We may assume that a

1
b
1

m
1 " S&1N . So there exists u " S

such that uabm " N . But ZR(M/N)* S = ., whence abm " N . Consequently N is a classical
2-absorbing submodule of M .

Let Ri be a commutative ring with identity and Mi be an Ri-module, for i = 1,2. Let
R = R1 1 R2. Then M = M1 1 M2 is an R-module and each submodule of M is in the form of
N = N1 1 N2 for some submodules N1 of M1 and N2 of M2.

Theorem 7. Let R = R1 1 R2 be a decomposable ring and M = M1 1M2 be an R-module where

M1 is an R1-module and M2 is an R2-module. Suppose that N = N1 1 N2 is a proper submodule

of M. Then the following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M;

(ii) Either N1 = M1 and N2 is a classical 2-absorbing submodule of M2 or N2 = M2 and N1 is

a classical 2-absorbing submodule of M1 or N1, N2 are classical prime submodules of M1,

M2, respectively.

Proof. (i)+ (ii) Suppose that N is a classical 2-absorbing submodule of M such that
N2 = M2. From our hypothesis, N is proper, so N1 #= M1. Set M ) = M

{0}1M2
. Hence N ) = N

{0}1M2

is a classical 2-absorbing submodule of M ) by Corollary 1. Also observe that M ) 2= M1 and
N ) 2= N1. Thus N1 is a classical 2-absorbing submodule of M1. Suppose that N1 #= M1 and
N2 #= M2. We show that N1 is a classical prime submodule of M1. Since N2 #= M2, there exists
m2 " M2\N2. Let abm1 " N1 for some a, b " R1 and m1 " M1. Thus

(a, 1)(b, 1)(1,0)(m1, m2) = (abm1, 0) " N = N1 1 N2.

So either (a, 1)(1,0)(m1, m2) = (am1, 0) " N or (b, 1)(1,0)(m1, m2) = (bm1, 0) " N . Hence
either am1 " N1 or bm1 " N1 which shows that N1 is a classical prime submodule of M1.
Similarly we can show that N2 is a classical prime submodule of M2.
(ii) + (i) Suppose that N = N1 1 M2 where N1 is a classical 2-absorbing (resp. classical
prime) submodule of M1. Then it is clear that N is a classical 2-absorbing (resp. classical
prime) submodule of M . Now, assume that N = N1 1 N2 where N1 and N2 are classical prime
submodules of M1 and M2, respectively. Hence (N1 1 M2) * (M1 1 N2) = N1 1 N2 = N is a
classical 2-absorbing submodule of M , by Proposition 1.



H. Mostafanasab, Ü. Tekir and K. Hakan Oral / Eur. J. Pure Appl. Math, 8 (2015), 417-430 428

Lemma 1. Let R= R11R21 · · ·1Rn be a decomposable ring and M = M11M21 · · ·1Mn be an

R-module where for every 13 i 3 n, Mi is an Ri-module, respectively. A proper submodule N of M

is a classical prime submodule of M if and only if N = 1n
i=1Ni such that for some k " {1,2, . . . , n},

Nk is a classical prime submodule of Mk, and Ni = Mi for every i " {1,2, . . . , n}\{k}.
Proof. (+) Let N be a classical prime submodule of M . We know N = 1n

i=1Ni where for
every 13 i 3 n, Ni is a submodule of Mi , respectively. Assume that Nr is a proper submodule
of Mr and Ns is a proper submodule of Ms for some 1 3 r < s 3 n. So, there are mr " Mr\Nr

and ms " Ms\Ns. Since

(0, . . . , 0,

r-th
+,-.

1Rr
, 0, . . . , 0)(0, . . . , 0,

s-th
+,-.

1Rs
, 0, . . . , 0)(0, . . . , 0,

r-th
+,-.

mr , 0, . . . , 0,
s-th
+,-.

ms , 0, . . . , 0)

=(0, . . . , 0) " N ,

then either

(0, . . . , 0,

r-th
+,-.

1Rr
, 0, . . . , 0)(0, . . . , 0,

r-th
+,-.

mr , 0, . . . , 0,
s-th
+,-.

ms , 0, . . . , 0)

=(0, . . . , 0,
r-th
+,-.

mr , 0, . . . , 0) " N

or

(0, . . . , 0,

s-th
+,-.

1Rs
, 0, . . . , 0)(0, . . . , 0,

r-th
+,-.

mr , 0, . . . , 0,
s-th
+,-.

ms , 0, . . . , 0)

=(0, . . . , 0,
s-th
+,-.

ms , 0, . . . , 0) " N ,

which is a contradiction. Hence exactly one of the Ni ’s is proper, say Nk. Now, we show that
Nk is a classical prime submodule of Mk. Let abmk " Nk for some a, b " Rk and mk " Mk.
Therefore

(0, . . . , 0,
k-th
+,-.

a , 0, . . . , 0)(0, . . . , 0,

k-th
+,-.

b , 0, . . . , 0)(0, . . . , 0,
k-th
+,-.

mk , 0, . . . , 0)

=(0, . . . , 0,

k-th
+ ,- .

abmk, 0, . . . , 0) " N ,

and so

(0, . . . , 0,
k-th
+,-.

a , 0, . . . , 0)(0, . . . , 0,
k-th
+,-.

mk , 0, . . . , 0) = (0, . . . , 0,
k-th
+,-.

amk , 0, . . . , 0) " N

or

(0, . . . , 0,

k-th
+,-.

b , 0, . . . , 0)(0, . . . , 0,
k-th
+,-.

mk , 0, . . . , 0) = (0, . . . , 0,

k-th
+,-.

bmk , 0, . . . , 0) " N .

Thus amk " Nk or bmk " Nk which implies that Nk is a classical prime submodule of Mk.
(4) Is easy.



REFERENCES 429

Theorem 8. Let R= R1 1 R2 1 · · ·1 Rn (23 n<5) be a decomposable ring and

M = M1 1 M2 1 · · · 1 Mn be an R-module where for every 1 3 i 3 n, Mi is an Ri-module,

respectively. For a proper submodule N of M the following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M;

(ii) Either N = 1n
t=1Nt such that for some k " {1,2, . . . , n}, Nk is a classical 2-absorbing

submodule of Mk, and Nt = Mt for every t " {1,2, . . . , n}\{k} or N = 1n
t=1Nt such that for

some k, m " {1,2, . . . , n}, Nk is a classical prime submodule of Mk, Nm is a classical prime

submodule of Mm, and Nt = Mt for every t " {1,2, . . . , n}\{k, m}.

Proof. We argue induction on n. For n = 2 the result holds by Theorem 7. Then let
33 n<5 and suppose that the result is valid when K = M1 1 · · ·1Mn&1. We show that the
result holds when M = K 1Mn. By Theorem 7, N is a classical 2-absorbing submodule of M if
and only if either N = L1Mn for some classical 2-absorbing submodule L of K or N = K 1 Ln

for some classical 2-absorbing submodule Ln of Mn or N = L 1 Ln for some classical prime
submodule L of K and some classical prime submodule Ln of Mn. Notice that by Lemma 1, a
proper submodule L of K is a classical prime submodule of K if and only if L = 1n&1

t=1Nt such
that for some k " {1,2, . . . , n& 1}, Nk is a classical prime submodule of Mk, and Nt = Mt for
every t " {1,2, . . . , n& 1}\{k}. Consequently we reach the claim.

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