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Palestine Journal of Mathematics
Vol. 11(4)(2022) , 319–324 © Palestine Polytechnic University-PPU 2022
ON PRINCIPALLY µ- LIFTING MODULES
Enas Mustafa Kamil
Communicated by Jawad Abuhlail
MSC 2010 Classifications: Primary 16D10; Secondary 16D80.
Keywords and phrases: µ-hollow modules, principally µ-lifting modules.
Abstract Let Rbe an arbitrary ring with identity and let Mbe a left R- module. In this pa-
per, I introduce a class of modules analogous to that of principally lifting modules. The module
Mis called principally µ-lifting module if for every cyclic submodule Aof M, there exists a
submodule Dof Asuch that M=D⊕D0, D0≤Mand A∩D0<µD0. I also introduce a gener-
alization of µ-hollow modules, namely, a module Mis said to be principally µ-hollow module if
every proper cyclic submodule of Mis µ-small in M. I prove some results of principally lifting
modules can be extended to principally µlifting modules for this general
1 Introduction
Through this paper, Rdenotes an associative ring with identity, and all modules are unital left
R- modules. A submodule Aof Mis called µ-small in Mif, for every 2submodule Kof M,
the equality M=A+Ksuch that M
Kis cosingular implies M=K. Clearly that every small
submodule is µ- small in M. A module Mis called µ-hollow module if every proper submodule
of Mis µ-small in M, see [1]. A submodule Pis a µ-supplement of Nin Mif M=P+N
and P∩Nis µ-small in P, while Mis called µsupplemented if every submodule of Mhas a
µ-supplement in M, see [2]. A module Mis called µ-lifting module if for every submodule A
of M, there exists a submodule Dof Asuch that M=D⊕D0, D0≤Mand A∩D0<µD0,
see [3]. In this paper, we define principally µ-hollow modules as a generalization of µ-hollow
modules. Also as generalization of µ-lifting modules , I introduce the concept of principally
µ-lifting module as follows: An R-module Mis called principally µ-lifting module if for every
cyclic submodule Aof M, there exists a submodule Dof Asuch that M=D⊕D0, D0≤M
and A∩D0µD0.
This paper is organized as follows: Section two is devoted to define principally µ−hollow
modules and the basic properties of these modules that will be used in the sequel.
In section three, the notion of principally µ-lifting modules is introduced. It is shown that
every µ-lifting module is principally µ-lifting. I give an example to show that the reverse impli-
cation need not hold in general.
In what follows, by Z, Q and Z
nZ we denote, respectively, integers, rational numbers, and the
ring of integers and the Z- module of integers modulo n.
2 Principally µ-hollow modules.
In this section, I introduce the concept of principally µ−0 hollow modules and investigate the
basic properties of these modules.
Definition 2.1. A nonzero R-module Mis called principally µ-hollow module if every proper
cyclic submodule of Mis µ-small in M.
1- Z4as Z-module is principally µ-hollow module.
2- Z6as Z-module is not principally µ-hollow, since {0,3}and {0,2,4}are not µsmall in Z6.
320 Enas Mustafa Kamil
3- Zas Z-module is not principally µ-hollow because nZ is not µ-small in Zfor each positive
integer n.
4- Every simple module is a principally µ-hollow. For example Z2as Z-module.
5- It is clear that every µ-hollow module is principally µ-hollow. But the converse is not true
in general. For example: since every cyclic submodule of Qas Z- module is small in Q
and hence µ-small, then Qis principally µ-hollow but Qis not µhollow since Qis not
supplemented Z-module.
Proposition 2.2. A nonzero epimorphic image of principally µ-hollow is principally µ-hollow.
Proof. Let f:M→M0be a nonzero epimorphism and let Mbe a principally µ-hollow module,
we have to show that M0is principally µ-hollow, let Abe a proper cyclic submodule of M0,
then f−1(A)is a proper cyclic submodule of M, if f−1(A) = M, then A=M0, which is a
contradiction. Since Mis principally µ-hollow, then f−1(A)µM, and hence AµM0.
Corollary 2.3. Let Mbe a principally µ-hollow and let Abe a submodule of M. Then M
Ais
principally µ-hollow
Proof. Let π:M→M
Abe the natural epimorphism. Since Mis principally µ-hollow, then by
previous proposition M
Ais principally µ-hollow.
Note In general, the converse of previous corollary is not true as the following example shows:
Consider Zas Z-module, note that Z
4Z∼
=Z4is principally µ-hollow, but Zis not principally
µ-hollow.
The following proposition gives a condition for the converse of corollary (2.4).
Proposition 2.4. If Ais µ-small submodule of Mand M
Ais principally µ-hollow, then Mis
principally µ-hollow.
Proof. Let m∈Mand assume that Rm +N=Mfor some submodule Nof Mwith M
N
cosingular. Then R(m+A) = Rm+A
Ais a cyclic submodule of M
Aand M
A=(Rm+A)
A+N+A
A
and M
A+Nis cosingular. Since M
Ais principally µ-hollow, then M=N+A. But Ais µ-small
submodule of M, therefore M=N. Thus, Mis principally µ-hollow.
Proposition 2.5. Every direct summand of µ-hollow is µ-hollow.
Proof. Clear from Prop. (2.3).
A direct sum of µ-hollow modules need not be µ-hollow as the following example shows.
The Z-modules Z2and Z8are µ-hollow, but Z8⊕Z2is not µ-hollow module.
Let Mbe an R-module. Recall that a submodule Aof Mis called a fully invariant if
g(A)≤A, for every g∈End(M)and Mis called duo module if every submodule of Mis
fully invariant. See [4].
Now, we give conditions under which the direct sum of µ-hollow modules is a µhollow.
Proposition 2.6. Let M=M1⊕M2be a duo module , then Mis principally µ-hollow if and
only if M1and M2are principally µ-hollow, provided that A∩Miis proper cyclic submodule of
Mi, i =1,2,∀A⊂M.
Proof. (⇒)Clear by Prop. (2.6).
(⇐) Let mR be a proper submodule of M. Since Mis a duo module, then mR = (mR∩
M1)⊕(mR ∩M2). Hence each of mR ∩M1, mR∩M2is a proper submodule of M1mR µM.
Thus Mis µ-hollow.
Recall that an R-module Mis called distributive if for all A, B and C≤M , A ∩(B+C)
= (A∩B)+(A∩C). See [5].
By similar argument, one can easily prove the following proposition.
ON PRINCIPALLY µ- LIFTING MODULES 321
Proposition 2.7. Let M1and M2be R-modules and let M=M1⊕M2be a distributive module.
Then Mis principally µ-hollow if and only if M1and M2are principally µhollow, provided that
A∩Miis proper cyclic submodule of Mi, i =1,2,∀A⊂M.
3 Principally µ−lifting modules
Motivated by the generalization of µ-lifting modules, I introduce principally µ−lifting modules.
This section is devoted to investigating some properties of this class of modules.
Definition 3.1. An R-module Mis called principally µ-lifting module if for every cyclic sub-
module Aof M, there exists a submodule Dof Asuch that M=D⊕D0, D0≤Mand
A∩D0<<µD0
(1) Every principally µ-hollow module is principally µ-lifting module. The converse is not
true in general, for example Z6as Z- module is principally µ-hollow but not principally
µ-hollow module.
(2) Every semisimple module is principally µ-lifting module. The converse is not true in gen-
eral, for example Z4as Z- module.
(3) It is clear that every µ-lifting module is principally µ-lifting. The converse is not true in
general, for example: Consider Qas Z- module. Since Qis principally µhollow, then Q
is principally µ-lifting Z-module which is not µ-lifting.
(4) Let pbe a prime integer, and nany positive integer Then the Z-module M=Z
Zpnis a
principally µ-lifting module.
(5) Zas Z-module is not principally µ-lifting.
(6) Z8⊕Z2as Z-module is not principally µ-lifting, since it is not principally lifting cosingular
module.
(7) Principally µ-lifting modules are closed under isomorphisms.
The following proposition gives a condition under which principally µ-hollow and principally
µ-lifting modules are equivalent.
Proposition 3.2. Let Mbe an indecomposable module. Then Mis principally µhollow if and
only if Mprincipally is µ-lifting.
Proof. Let Mbe an indecomposable principally µ-lifting and let Abe a propoer cyclic submod-
ule of M, there exists a submodule Dof Asuch that M=D⊕D0, D0≤Mand A∩D0<<µD0.
But Mis indecomposable, therefore D=0 and D0=M, hence we conclude that AµM.
Thus Mis principally µ-hollow. The converse is obvious.
The following propositions give characterizations of µ-lifting modules.
Proposition 3.3. Let Mbe an R-module. Then Mis principally µ-lifting if and only if for every
cyclic submodule Aof M, there exists a submodule Dof Asuch that M=D⊕D0, D0≤Mand
A∩D0<µM.
Proof. Clear.
Proposition 3.4. Let Mbe an R-module. The following statements are equivalent.
(1) Mis principally µ-lifting module.
(2) Every cyclic submodule Aof Mcan be written as A=D⊕S, where Dis a direct summand
of Mand SµM.
(3) For every cyclic submodule Aof M, there exists a direct summand Dof Msuch that D≤A
and D≤µce Ain M.
322 Enas Mustafa Kamil
Proof. (1)⇒(2)Suppose that Mis a principally µ-lifting module and let Abe cyclic a sub-
module of M, then there exists a submodule Dof Asuch that M=D⊕D0, D0≤Mand
A∩D0µM, by proposition (3.4). Now, A=A∩M=A∩(D⊕D0)=D⊕(A∩D0). Thus
we get the result.
(2) ⇒(3) Let Abe a cyclic submodule of M. By (2), A=D⊕S, where Dis a direct summand
of Mand S <<µM. We have to show that A
DµM
D, let M
D=A
D+U
D,M
Uis cosingular, then
M
D=D+S
D+U
Dand hence M=D+S+U=S+U. But S <<µM, therefore M=U.
(3) ⇒(1) Let Abe a cyclic submodule of M. By (3), there exists a direct summand Dof M
such that D≤Aand D≤µce Ain M. We want to show that A∩D0<µD0, let D0=
(A∩D0)+U, D0
Uis cosingular. Since M=D+D0=D+(A∩D0)+U, then M
D=
D+(A∩D0)+U
D=D+(A∩D0)
D+U+D
D. Since D≤D+(A∩D0)≤Aand D≤µce Ain M,
then D≤µce D+(A∩D0)in M, by [, proposition (2.5)] and M
U+D=D+D0
U+D=(D+U)+D0
U+D∼
=
D0
D0∩(U+D)=D0
Uwhich is cosingular, hence M
D=U+D
D, implies that M=U+Dand clearly
that U∩D=0, then M=U⊕D, that is U=D0. Thus Mis a principally µ-lifting module.
Theorem 3.5. Let Mbe an R-module. The following statements are equivalent.
(1) Mis principally µ-lifting module.
(2) Every cyclic submodule Aof Mhas a µ-supplement Bin Msuch that A∩Bis a direct
summand of A.
Proof. (1) ⇒(2) Let Mbe principally µ-lifting module and let Abe a cyclic submodule of M.
By proposition (3.5), there exists a direct summand Dof Msuch that D≤Aand D≤uce Ain
M. Now, A=A∩M=A∩(D⊕D0)=D⊕(A∩D0). Since D≤A, then M=A+D0and
A∩D0<µD0. Hence D0is µ-supplement of Aand A∩D0is a direct summand of A.
(2) ⇒(1) Let Abe a cyclic submodule of M. By (2) Ahas a µ-supplement Bin Msuch that
A∩Bis a direct summand of A. Then M=A+B , A ∩B <µBand A= (A∩B)⊕Y,Y≤A.
Since M=A+B= (A∩B) + Y+B=Y+Band A∩B∩Y=B∩Y={0}, then M=B⊕
Y. Thus Mis principally µ-lifting module.
The following proposition gives another characterization of µ-lifting module.
Proposition 3.6. Let Mbe an R-module. Then Mis principally µ-lifting module if and only if
for every cyclic submodule Aof M, there exists an idempotent f∈End(M)such that f(M)≤A
and (I−f)(A)µ(I−f)(M).
Proof. (⇒)Assume that Mis a principally µ-lifting module and let Abe a cyclic submodule
of M. By characterization (3.6) Ahas a µ-supplement Bin Msuch that A∩Bis a direct
summand of A, then M=A+B, A ∩B <<µBand A= (A∩B)⊕X, X ≤A. Note
M=A+B= (A∩B) + X+B=X+Band A∩B∩X=B∩X={0}, implies that
M=B⊕X. Now define the following map f:M→X, it is clear that fis an idempotent and
f(M)≤A. It is sufficient to prove that (I−f)(A)µ(I−f)(M). One can easily show that
(I−f)(A) = A∩(I−f)(M) = A∩BµB= (I−f)(M).
(⇐)Let Abe a cyclic submodule of M. By our assumption, there exists an idempotent f∈
End(M)such that f(M)≤Aand (I−f)(A)µ(I−f)(M), clearly that M=f(M)⊕(I−
f)(M)and A∩(I−f)(M)=(I−f)(A)µ(I−f)(M). Thus Mis principally µ−lifting.
Proposition 3.7. Any direct summand of principally µ-lifting is principally µ-lifting.
Proof. Let M=M1⊕M2be a principally µ-lifting and let Abe a cyclic submodule of M1,
then A=D⊕S, where Dis a direct summand of Mand S <<µM, by characterization (3.5).
Since Dis a direct summand of Mcontained in M1, then Dis a direct summand of M1and
S <µM, S ≤M1and M1is a direct summand of M, then S <<µM1. Thus M1is µ-lifting.
Next, we give some various conditions under which the quotient of µ-lifting module is µ-
lifting.
Proposition 3.8. Let Mbe a principally µ-lifting R-module and let Abe a cyclic submodule of
M. Then M
Ais principally µ-lifting in each of the following cases.
ON PRINCIPALLY µ- LIFTING MODULES 323
(1) For every direct summand Dof M, D+A
Ais a direct summand of M
A.
(2) Mis distributive module.
Proof. (1) Suppose that Mis principally µ-lifting R-module and let X
Abe a cyclic submodule
of M
A, then Xis cyclic submodule of M, hence there exists D≤Xsuch that M=D⊕
D0, D0≤Mand D≤µce Xin M. By hypothesis, D+A
Ais a direct summand of M
A. Then
D+A
A≤µce X
Ain M
A. Thus M
Ais principally µ-lifting.
(2) Suppose that Mis distributive module, we use (1) to show that M
Ais principally µ-lifting.
Let Dbe a direct summand of M, M =D⊕D0, D0≤M, then M
A=D+D0
A=D+A
A+D0+A
A
and D+A
A∩D0+A
A=(D+A)∩D0+[(D+A)∩A]
A=(A∩D0)+(D∩A)+A
A=A. Hence D+A
Ais a direct
summand of M
A. So, by (1) Mis principally µ-lifting module.
Lemma 3.9. [23, lemma5−4]: Let Mbe an R-module, if M=M1⊕M2, then M
A=A+M1
A⊕
A+M2
A, for every fully invariant submodule Aof M.
Proposition 3.10. Let Mbe a principally µ-lifting module if Ais a fully invariant submodule of
M, then M
Ais principally µ-lifting module.
Proof. Let X
Abe a cyclic submodule of M
A. Since Mis principally µ-lifting, there exists a
submodule Dof Xsuch that D≤µce Xin Mand M=D⊕D0, D0≤M. By lemma (3.10)
we have M
A=D+A
A⊕D0+A
A, let f:M
D→M
D+Abe a map defined by f(m+D) = m+D+
A, ∀m∈M, it is clear that fis an epimorphism. Now, since D≤µce Xin M, X
DµM
Dand
fX
D<µfM
D, by [1] which implies that X
D+AµM
D+A, hence D+A≤µce Xin Mand
hence D+A
A≤µce X
Ain M
A, by [1]. Thus M
Ais principally µ−lifting module.
Remark 3.11. A direct sum of principally µ-lifting modules need not be principally µ−lifting
module as the following example shows.
Let M=Z8⊕Z2as Z-module. It is clear that Z8and Z2are principally µ-lifting Zmodules,
but Mis not principally µ-lifting module.
Proposition 3.12. Let M=M1⊕M2be a duo module such that M1and M2are principally
µ-lifting modules, then Mis principally µ-lifting.
Proof. Let M=M1⊕M2be a duo module and let mR be a submodule of M, then mR is a
fully invariant submodule of M. Hence mR =mR ∩M=mR ∩(M1⊕M2)=(mR ∩M1)
⊕(mR ∩M2). Since M1and M2are principally µ-lifting modules, then mR ∩M1=A1⊕A2
and mR ∩M2=A3⊕A4, where A1and A3are direct summands of M1and M2respectively
and A2, A4are µ-small submodules of M1and M2respectively, by Prop. (3.5). It is clear that
A1⊕A3is a direct summand of Mand A2⊕A4is µ-small submodule of M. Thus Mis
principally µ-lifting.
By the same argument, one can easily prove the following proposition.
Proposition 3.13. Let M=M1⊕M2be a distributive module such that M1and M2are princi-
pally µ-lifting modules, then Mis principally µ-lifting.
We end this section by the following diagram.
324 Enas Mustafa Kamil
References
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[2] E. M. Kamil, and W. Khalid, 2018. On µ-supplemented and cofinitely µ- supplemented, Sci. Int. (Lahore),
: 567-572.
[3] E. M. Kamil, and W. Khalid, 2019. On µ- lifting modules Iraqi Journal of Science 60 359-365.
[4] C. Ozcan, A. Harmanci and P. F. Smith, " Duo Modules", Glasgow Math. Journal Trust, 48, 533-545,
(2006).
[5] V. Erdogdu , 1987. Distributive Modules ,Canada Math. Bull, 30, pp:248-254.
Author information
Enas Mustafa Kamil, College of pharmacy, AL-KitabaUniversity, Kirkuk , Iraq,.
E-mail:
Enas.M.Kamil@uoalkitab.edu.iq
Received: November 12, 2021.
Accepted: January 16, 2022.
