EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 3, 2019, 1187-1198
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On C-co-epi-retractable modules

Abdoul Djibril Diallo1, Papa Cheikhou Diop2,∗, Mamadou Barry1

1 Département de Mathématiques et Informatique, Faculté des Sciences et Techniques,
Université Cheikh Anta Diop, Dakar, Sénégal
2 Département de Mathématiques , UFR Sciences et Technologies, Université de Thiès,
Thiès, Sénégal

Abstract. In this paper, we introduce the notion of c-co-epi-retractable modules. An R-moduleM
is called c-co-epi-retractable if it contains a copy of its factor module by a complement submodule.
The ring R is called c-co-pri if RR is c-co-epi-retractable. Conditions are found under which, a c-co-
epi-retractable module is extending, retractable, semi-simple, quasi-injective, injective and simple.
Also, we investigate when c-co-epi-retractable modules have finite uniform dimension. Finally,
right SI-rings, semi-simple artinian rings and quasi-Frobenius rings are characterized in termes of
c-co-epi-retractable modules.

2010 Mathematics Subject Classifications: 13B10,13C05,13C13

Key Words and Phrases: co-epi-retractable modules, c-co-epi-retactable modules, extending
modules, Rickart modules

1. Introduction

Throughout all rings are associative with identity and all modules are unitary right
module. In [10], Ghorbani introduced the co-epi-retractable modules. An R-module M
is called co-epi-retractable if it contains any of its factor modules. A ring R is called
co-pri if RR is a co-epi-retractable module. It is was shown in [10], that a ring R is co-
pri iff its right ideals is the right annihilator of an element of R. Also co-pi-retractable
modules have been investigated by Mostafanasab [15]. He studied the simplicity and the
semi-simplicity of co-pi-retractable modules. Recall that a module M is called extending
if every complement submodule is a direct summand. Motivated by the definition of a
co-epi-retractable module and the definition of a extending module, we say that a module
is c-co-epi-retractable if it contains a copy of its factor modules by a complement submod-
ule. Every co-epi-retractable module and every extending module is c-co-epi-retractable.
In particular uniform modules and semi-simple modules are c-co-epi-retractable. In this

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3461

Email addresses: cheikpapa@yahoo.fr (P. C. Diop),
dialloabdoulaziz58@yahoo.fr (A. D. Diallo ), mansabadion1@hotmail.com (M. Barry )

http://www.ejpam.com 1187 c© 2019 EJPAM All rights reserved.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1188

paper, we investigate when c-co-epi-retractable modules are extending, continuous, quasi-
continuous, semi-simple, retractable, quasi-injective, injective and simple. Also, we prove
under certain conditions that a c-co-epi-retractable module has finite uniform dimension.
Finally, we characterize some well-known rings with the help of c-co-epi-retratacble mod-
ules.
Our paper is structured as follows:
In the second section, we give preliminary definitions and results which we will use through-
out this paper.
In the third section, we define the c-co-epi-retractable modules. Our aim in this section
is to work on the concept of c-co-epi-retractable modules. We show, among others, the
following results.
(1) For an R-module with regular endomorphism ring, the properties, c-co-epi-retractable,
extending, continuous and quasi-continuous are all equivalent.
(2) If M is a c-co-epi-retractable module with Udim(M) = n ≥ 2, then M is retractable
and for any 0 6= C ⊆c M , M/C is uniform.
(3) Let R be a right self-injective ring and M be a self-hereditary R-module. Then M is
c-co-epi-retractable iff it is finitely generated semi-simple injective.
(4) Let M be a c-co-epi-retractable R-module such that S satisfies DCC for cyclic right
ideals. If for any finitely generated right ideal I ⊆ S, r(KerI) = I then M has finite
uniform dimension.
(5) The following conditions are euivalent for a right SI-ring R:

(a) R
(N)
R is extending.

(b) Every R-module is c-co-epi-retractable.
(c) Every R-module is extending.

For an R-module M , S = EndR(M) denotes the endomorphism ring of M . For φ ∈ S,
Kerφ and Imφ stand for kernel and image of φ, respectively. The notations N ≤ M ,
N ≤e M and N ≤⊕ M mean that N is a submodule of M , an essential submodule and a
direct summand of M , respectively. Also E(M) denotes the injective hull of M .

2. Preliminaries

In this section, we are going to give preliminary definitions and results which we
will use throughout this paper.

Definition 1. 1. An R-module M is called CS module if every complement submodule of
M is a direct summand.
2. An R-module M is called continuous if it is a CS module and satisfies the following
condition: (C2) Every submodule of M that is isomorphic to a direct summand M is itself
a direct summand of M .
3. An R-module M is called quasi-continuous if it is a CS module and satisfies the
following condition: (C3) If N and K are direct summands of M with N ∩K = {0}, then
N ⊕K is a direct summand of M .



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1189

Definition 2. Let M be an R-module, put Z(M) = {m ∈ M : annR(m) ≤e R}. M
is called nonsingular if Z(M) = {0}, and singular if Z(M) = M . The Goldie torsion
submodule Z2(M) of M is defined by Z(M/Z(M)) = Z2(M)/Z(M). M is Z2-torsion if,
Z2(M) = M .

3. C-co-epi-retractable modules and some applications

Definition 3. An R-module M is called c-co-epi-retractable if, for every complement
submodule N of M , there exists a monomorphism f : M/N −→ M . The ring R is called
c-co-pri if RR is c-co-epi-retractable.

Remark 1. Clearly, every co-epi-retractable module is c-co-epi-retractable while the con-
verse is not true. For example Q as Z-module is c-co-epi-retractable but it is not co-epi-
retractable.

Lemma 1. The following statements are equivalent for an R-module M :
(1) M is a c-co-epi-retractable module.
(2) There exists ϕ ∈ EndR(M) such that Kerϕ = N for any nonzero N ⊆c M .

Proposition 1. Let M be a c-co-epi-retractable R-module. Then a fully invariant com-
plement submodule of a M is also c-co-epi-retractable.

Proof.
Let M be a c-co-epi-retractable module and N ⊆c M with N fully invariant. Let K ⊆c N .
Then, K ⊆c M . So, there is an endomorphism f : M −→M such that K = Kerf . Then
f |N : N −→ N and K = Ker(f |N ). Therefore, N is a c-co-epi-retractable module.

Corollary 1. Every fully invariant direct summand of a c-co-epi-retractable module is
c-co-epi-retractable.

Proposition 2. Let M be an R-module with S = EndR(M) regular.
Then the following conditions are equivalent:
(1) M is a c-co-epi-retractable module.
(2) M is an extending module.

Proof.
(1) ⇒ (2) Suppose M is a c-co-epi-retractable module and K a complement submodule
of M . Then, there is 0 6= g ∈ EndR(M) such that Kerg = K. By our assumption,
Kerg = K is a direct summand of M . Therefore, M is an extending module.
(2)⇒ (1) is obvious.

Corollary 2. A ring R is regular c-co-pri if and only if it is right nonsingular right
continuous.

Proposition 3. A ring R is c-co-pri if and only if every complement right ideal of R is
the right annihilator of an element of R.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1190

Proof.
Let I be a right complement ideal of R. If R is c-co-pri, there is a monomorphism
f : R/I −→ R. Set x = f(1 + I), then I = r(x), where r(x) denotes the right annihilator
of x. On the other hand, if I = r(x) is a right complement ideal of R for an element
x ∈ R, then R/I ∼= xR.

Let M be an R-module. The left annihilator of N ≤ M in S = EndR(M) is denoted
by LS(N) = {φ ∈ S : φN = {0}} and the right annihilator of a left ideal I of S is
rM (I) = {m ∈M : Im = {0}}

Recall that an R-module is called Baer if, for all N ≤M , LS(N) = Se, with e2 = e ∈ S.
Equivalenly, M is Baer if, for all ideal I ≤S S, rM (I) = eM with e2 = e ∈ S. An R-module
M is called Rickart if any endomorphism of M has a direct summand kernel. A module
M is called K-nonsingular if, ∀ϕ ∈ End(M),
Kerϕ ≤e M implies ϕ = 0.

Lemma 2. ([16], Lemma 2.14)
Any K-nonsingular extending module is Baer.

In the two following results, we show that for a c-co-epi-retractable R-module or a
module with c-co-pri endomorphism ring the properties Rickart and Baer are equivalent.

Proposition 4. Let M be a c-co-epi-retractable R-module. Then M is Rickart if and only
if M is Baer.

Proof.
Suppose M is Rickart. Since, M is c-co-epi-retractable, it is also extending by Proposition
2. Now, suppose Kerf ≤e M for some f ∈ EndR(M). The property of Rickart implies
that Kerf ≤⊕ M , and so Kerf = M . Consequently, f = 0. Therefore, according to
Lemma 2, M is Baer. The converse implication is clear.

Corollary 3. Let R be a c-co-pri ring. Then R is Baer if and only if R is right Rickart.

Proposition 5. Let M be an R-module for which S is c-co-pri. Then the following
statements are equivalent:
(1) M is Baer.
(2) M is Rickart.
(3) S is right Rickart.

Proof.
(1)⇒ (2) This is clear.
(2)⇒ (3) Follows from Proposition 2.2.1 in [13].
(3)⇒ (1) Let N be a submodule of M . Since S is right Rickart, it is also right nonsingu-
lar. Thus, LS(N) is a complement right ideal in S. Because S is c-co-pri, it follows from
Proposition 2 that S is right extending. Therefore, LS(N) = S(1−e) for some e = e2 ∈ S,



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1191

and hence M is Baer.

Recall that an R-module N is said to subgenerated by M if N is isomorphic to a
submodule of an M -generated module, i.e N is a kernel of a morphism between M -
generated modules.
We denote by σ[M ], the full subcategory of mod-R whose objects are all R-modules
subgenerated by M .
Recall that an R-module M is self-hereditary if every submodule of M is projective in
σ[M ].

Theorem 1. Let R be a right self-injective ring and M be a self-hereditary R-module.
Then the following conditions are equivalent:
(1) M is c-co-epi-retractable.
(2) M is extending.
(3) M is continuous.
(4) M is finitely generated semi-simple injective.

Proof.
(1) ⇒ (2) Suppose M is c-co-epi-retractable. Thus for any complement submodule C of
M , there exists a submodule N of M such that M/C ∼= N . Consequently, the property
of self-hereditary implies that C is a direct summand of M . Hence M is extending.
(2)⇒ (3) Suppose M is extending. Thus, according to Theorem 10.5 in [7], M is nonsingu-
lar and has finite uniform dimension. Hence, there exists uniform independent submodules
U1, ...., Un of M such that V = U1 ⊕ U2 ⊕ ... ⊕ Un is an essential submodule of M . Set
0 6= ui ∈ Ui, 1 ≤ i ≤ n. Then, Ui = uiR. It is easy to see that M = V . Therefore, M is
finitely generated semi-simple injective. This means that M is continuous.
(3)⇒ (4) Follows from an argument similar to the one in (2)⇒ (3).
(4)⇔ (1) It is easy to see.

Corollary 4. A right self-injective right hereditary ring is semi-simple artinian.

Recall that a module M is said to be retractable if for any 0 6= N ≤ M , there exists
a nonzero homomorphism form M to N . A module M has finite uniform dimension n
(written Udim(M) = n) if there is an essential submodule V ≤e M that is a direct sum
of n uniform submodules.

Remark 2. A c-co-epi-retractable module with finite uniform dimension need not be re-
tractable. In fact, the Z-module Q is c-co-epi-retractable with finite uniform dimension
but it is not retractable. Clearly, a c-co-epi-retractable need not to have a finite uniform
dimension. For example extending modules are c-co-epi-retractable which need not have
finite uniform dimension.

Proposition 6. If M is a c-co-epi-retractable R-module with Udim(M) = n ≥ 2, then
the following assertions are verified:
(1) M is retractable.
(2) For every 0 6= C ⊆c M , M/C is uniform.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1192

Proof.
(1) Let 0 6= N ≤M . Since Udim(N) <∞, N contains a uniform submodule, say U . After
replacing U by an essential closure, we may assume that U is a complement submodule
of M . By the c-co-epi-retractable condition on M , there exists a monomorphism f :
M/U −→M . Consider the inclusion map i : U −→M . Thus, f = ij is a monomorphism
where j : M/U −→ U . Consequently, j is a monomorphism. Now, consider the inclusion
map i1 : U −→ N . So, i1jπ : M −→ N is a nonzero homomorphism where π : M −→M/U
is the natural surjection. Therefore, M is retractable.
(2) Since Udim(M) = n ≥ 2, there exist complements submodules Ci ⊆c M(1 ≤ i ≤ n)
such that each M/Ci is uniform and C1∩ ...∩Cn = 0. Thus, there exists a monomorphism
f : M −→ ⊕n

iM/Ci. Let 0 6= C ⊆c M . Since M is c-co-epi-retractable, there exists a
monomorphism g : M/C −→M . Hence, h = fg : M/C −→ ⊕n

iM/Ci is a monomorphism.
Consider the inclusion map i : M/Ci −→ ⊕n

iM/Ci. Thus, h = ii1 is a monomorphism
where i1 : M/C −→ M/Ci. Therefore, i1 is a monomorphism. It follows that M/C is
uniform.

Corollary 5. An R-module M is simple if and only if M is Artinian c-co-epi-retractable
and every endomorphisme of M is a monomorphism.

Let N and M be R-modules and S = EndR(M). We denote by N the set of R-
submodules of N and by H the set of S-submodules of HomR(N,M)S .
For X ∈ H we put:
Ker(X) = ∩{Kerg|g ∈ X} ∈ N .

In the next result, we investigate when c-co-epi-retractable R-modules have finite uni-
form dimension.

Proposition 7. Let M be a c-co-epi-retractable R-module such that S satisfies DCC for
cyclic right ideals. If for any finitely generated right ideal I ⊆ S, r(KerI) = I, then has
finite uniform dimension.

Proof.
In view of Proposition 6.30 in [12], we need to show that the complements in M satisfy
ACC. Now, let C1 ⊆ C2 ⊆ .... be an ascending chain of complement submodules of M .
By the c-co-epi-retractable condition on M , there is fi ∈ S such that each Ci is of the
form Kerfi = KerfiS. With applying r(−) to this chain, we see that f1S ⊇ f2S ⊇ .....
By our assumption, there is some n such that fiS = fnS for all i ≥ n. Hence, Ci = Cn

for all i ≥ n. Therefore, M has finite uniform dimension.

Recall that an R-module M is said to be have the summand sum property (SSP , for
short) if, the sum of any two direct summands of M is again a direct summand of M .

Corollary 6. Let M be a quasi-injective R-module such that R⊕M has the SSP . Assume
that S satisfies DCC for cyclic right ideals. Then M is finitely generated semi-simple.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1193

Proof.
Since R⊕M has the SSP , we infer from Proposition 3.4 in [9] that every cyclic submodule
of M is a direct summand of M . Since M is quasi-injective, r(KerI) = I for any finitely
generated right ideal I ⊆ S by ([19], 28.1). But M is c-co-epi-retractable. Then, according
to Proposition 7, M has finite uniform dimension. Therefore, M is finitely generated semi-
simple.

Corollary 7. If M is a quasi-injective R-module such that S satisfies DCC for cyclic
right ideals, then M is a finite direct sum of uniform submodules.

Proof.
Suppose M is quasi-injective such that S satisfies DCC for cyclic right ideals. Thus by
([19], 28.1), r(KerI) = I for any finitely generated right ideal I ⊆ S. Therefore, according
to Proposition 7, M is a finite direct sum of uniform submodules.

Corollary 8. Let R be a right self-injective ring and M a nonsingular R-module such
that S satisfies DCC for cyclic right ideals. Then the following conditions are equivalent.
(1) M is quasi-injective.
(2) M is semi-simple injective.

Recall that an R-module M is called compressible if for every nonzero submodule N
of M there is a monomorphism f : M −→ N .

Proposition 8. An R-module is simple if and only if it is compressible c-co-epi-retractable
and contains a maximal complement submodule.

Proof.
The necessity is clear. Conversely, assume that M is compressible c-co-epi-retractable and
contains a maximal complement submodule C. Then there is a submodule N of M such
that M/C ∼= N . Hence, N is simple. By the compressible condition on M , there is a
monomorphism f : M −→ N . Thus, M is isomorphic to a submodule of M . As f 6= 0,
M = N , and so M is simple.

Corollary 9. An R-module is simple if and only if it is compressible finitely generayed
c-co-epi-retractable.

Recall that an R-module M is cohereditary if every factor module of M is injective.
Now, let us introduce the following notion.

Definition 4. An R-module module is called c-cohereditary if M/C is injective for each
nonzero proper complement submodule C of M . The ring R is called c-cohereditary if RR

is c-cohereditary.

Proposition 9. A c-cohereditary c-co-epi-retractable R-module M is injective. Moreover,
M is a direct sum of a nonsingular module and an injective module.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1194

Proof.
SupposeM is c-co-epi-retractable. It is well known that Z2(M) is a complement submodule
of M . By the c-co-epi-retractable condition on M , there exists a nonzero endomorphism
f of M such that Kerf = Z2(M). Hence, M/Kerf ∼= Imf . By our assumption, Imf is
injective, and so a direct summand of M . Thus, there exists a submodule K of M such
that M = Imf ⊕ K. By hypthesis again, M/Imf ∼= K is injective. Therefore, M is
injective. The last part is clear since Imf is nonsingular.

Corollary 10. A c-co-pri right c-cohereditary ring is right self-injective.

Corollary 11. A right extending right c-cohereditary ring is right self-injective.

Corollary 12. Any extending c-cohereditary R-module is injective

We end this section with some applications of c-co-epi-retractable modules regarding
the characterization of right SI, semi-simple artinian and quasi-Frobenius rings. Recall
that a ring R is said to be right SI if every singular R-module is injective.

Lemma 3. (([17], Lemma 3.1) and ([11], Theorem 3))
If R is a right SI-ring, then R is right nonsingular right hereditary and every singular
R-module is semi-simple.

Lemma 4. ([4], Corollary 3.2)
Let M be an R-module having C3-condition. If M = M1 ⊕M2 and f : M1 −→ M2 is a
monomorphism, then Imf ≤⊕ M2.

Theorem 2. The following conditions are equivalent for a ring R.
(1) R is a right SI-ring.
(2) Every Z2-torsion c-co-epi-retractable R-module is injective.
(3) Every Goldie-torsion R-module has C3-condition.

Proof.
(1)⇔ (2) Follows from Theorem 3 in [11].
The implication (1)⇒ (3) is clear.
(3) ⇒ (1) Let M be a cyclic Z2-torsion R-module. It is clear that M ⊕ E(M) is Goldie-
torsion and has the C3-condition by (4). Consider the inclusion map i : M −→ E(M).
Hence, i(M) = M ≤⊕ E(M) by Lemma 4. It follows that M is injective. This means
that every cyclic singular R-module is injective. Therefore, according to ([7], 17.4), R is a
right SI-ring.

The following lemmas are crucial in the establishement of the next theorem.

Lemma 5. ([7], Corollary 11.4)

Let R be ring such that R
(A)
R is extending, then the following statements hold true:

(1) Every nonsingular R-module is extending.
(2) Every nonsingular R-module is projective.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1195

Lemma 6. ([7], 7.11)
An R-module M is extending if and only if M = Z2(M)⊕M ′, for some submodule M ′ of
M , such that M ′ and Z2(M) are both extending and Z2(M) is M ′-injective.

Now, we are able to prove the following result.

Theorem 3. For a right SI-ring R, the following conditions are equivalent:

(1) R
(N)
R is extending.

(2) R
(N)
R is c-co-epi-retractable.

(3) Every R-module is extending.
(4) Every R-module is c-co-epi-retractable.

Proof.
(1)⇒ (2) It is easy to see.

(2)⇒ (3) Suppose R
(N)
R is c-co-epi-retracatble. Hence, for every complement right ideal I

of R(N), there exists a right ideal J of R(N) such that R(N)/I ∼= J . Then it follows from
Lemma 3 that R(N) is an extending R-module. Thus, by ([17], Propositions 3.4 and 3.9),

R is right Noetherian. So, according to Corollary 11.12 in [7], R
(A)
R is extending for any

index set A. Now, let M be any R-module. Thus, by Lemma 5, M = Z2(M) ⊕ N for
some submodule N of M and clearly N is nonsingular. In view of Lemma 5 again, N is
an extending module. Moroever, since R is right nonsingular, Z2(M) = Z(M) is singular.
Hence, Z2(M) is injective. Therefore, according to Lemma 6, M is extending, as desired.
(3)⇒ (4) is clear.

(4) ⇒ (1) By (4), R
(N)
R is c-co-epi-retracatble. But R is right hereditary. Thus, by the

proof of (2)⇒ (3), R
(N)
R is extending.

Corollary 13. The the following conditions are equivalent for a ring R:
(1) R is semi-simple artinian.

(2) R is a regular right SI-ring and R
(N)
R is c-co-epi-retractable.

(3) R is a regular right SI-ring and R
(N)
R is extending.

(4) R is right SI-ring and R
(N)
R is continuous.

(5) R is right SI-ring and R
(N)
R is quasi-continuous.

Proof.
(1)⇒ (2)⇒ (3)⇒ (4)⇒ (5) are clear.

(5)⇒ (1) Assume that R is a right SI-ring such that R
(N)
R is quasi-continuous. In particu-

lar, R
(N)
R is extending. Then, by Theorem 3 every R-module is extending. So by ([7], 13.5),

R is an Artinian serial ring. Thus, by ([6], Proposition 6.1 (4)), R
(N)
R is quasi-injective,

and hence RR is quasi-injective. Consequently, R is right self-injective by ([12], Remark
6.71(2B)). Since R is right Artinian, RR has finite uniform dimension. Because R is right
SI, it is right nonsingular by Lemma 3. Now, RR is nonsingular extending and has finite
uniform dimension. Thus, RR is a finite direct sum of uniform submodules. But R is
right self-injective. Then as in the proof of (2) ⇒ (3) in theorem 1, RR is semi-simple.
Therefore, R is semi-simple arinian.



A. D. Diallo, P. C. Diop , M. Barry / Eur. J. Pure Appl. Math, 12 (3) (2019), 1187-1198 1196

Corollary 14. If R is a right SI-ring such that R
(N)
R is c-co-epi-retractable, then all

R-modules with a regular endomorphism ring are quasi-injective.

Proof.
Let M be an R-module with a regular endomorphism ring. Thus, by Theorem 3, M is
extending. It follows that M is quasi-continuous. On the other hand, R is Artinian serial
by ([7], 13.5). Hence by ([6], Proposition 6.1 (4)), M is quasi-injective.

Note that along the lines of the proof of the above Theorem we have shown that if R
is a right SI-ring such that R(N) is right extending, then R(A) is right extending for any
index set A.

Theorem 4. The following conditions are equivalent for a ring R with R = R/Z2(RR).
(1) R is semi-simple.
(2) Every c-co-epi-retractable R-module is R-injective.
(3) Every nonsingular R-module is quasi-injective.
(4) Every nonsingular R-module is quasi-continuous.
(5) Every nonsingular R-module has C3-condition.
(6) Every submodule of a nonsingular R-module is a C3-module.
(7) Every submodule of R⊕R is a C3-module.

Proof.
The implication (1)⇔ (2) follows from a similar proof to ([1], Theorem 4.5).
The implication (1)⇒ (3) is clear by ([3], Theorem 3.2).
Implications (3)⇒ (4)⇒ (5)⇒ (6)⇒ (7) are easy to see.
(7)⇒ (1) By ([3], Theorem 3.2), we need to show that R is semi-simple. Let I be a right
ideal of R. Thus I ⊕ R, being a submodule of R ⊕ R is a C3-module by (7). Now, let
i : I −→ R be the inclusion map. By Lemma 4, I is a direct summand of R. Therfore, R
is semi-simple.

Corollary 15. The following conditions are equivalent for a ring R with
R/Z2(R) = R.
(1) R is quasi-Frobenius.
(2) Every c-co-epi-retractable R-module is R-injective and Z2(RR) is an Artinian injective
R-module.
(3) Every c-co-epi-retractable R-module is R-injective and Z2(RR) is a Noetherian injective
R-module.

Proof.
(1)⇒ (2) Since R is quasi-Frobenius, it is right continuous. Hence, R is a continuous R-
module. Thus, RR = Z2(RR)⊕R′ for a continuous R-module R′. It follows that R is right
nonsingular right continuous. Consequently, R is regular. Since R is right Noetherian,
R, also, is right Noetherian. Thus, the property of regular implies that R is semi-simple.
Thus, in view of Theorem 4, every c-co-epi-retractable R-module is R-injective. The last
part is clear since RR is injective and Artinian.



REFERENCES 1197

(2) ⇒ (1) Assume that every c-co-epi-retractable R-module is R-injective and Z2(RR)
is an Artinian injective ring. Since every c-co-epi-retractable R-module is R-injective, we
infer from theorem 4 that R is a semi-simple ring. Thus, R is semi-simple as an R-module.
Since R is a nonsingular R-module, R is a projective R-module. So, Z2(RR) ≤⊕ R, say
R = Z2(RR) ⊕ R′ where R′ is semi-simple ring. By our assumption, R is right Artinian
right self-injective. Consequently, R is quasi-Frobenius.
Similarly, (3) is equivalent to (1).

Proposition 10. The following statements are equivalent for a ring R.
(1) R is an Artinian serial ring with J2(R) = 0.
(2) Every submodule of a co-c-epi-retractable R-module is extending.
(3) Every submodule of an extending R-module is extending.

Proof.
(1)⇒ (2) follows from ([7], 13.5).
(2)⇒ (1) Let M be any R-module. Then M⊕E(M), being a submodule of E(M)⊕E(M)
is extending by (2). In view of Proposition 2.7 in [14], M is CS. Therefore, R is a Artinian
serial ring with J2 = 0 by ([7], 13.5).
Similarly, (1) and (3) are equivalent.

Proposition 11. The following conditions are equivalent for a ring R.
(1) R is semi-simple artinian.
(2) Every c-co-epi-retractable R-module is semi-simple.
(3) Every c-co-epi-retractable R-module is injective.
(4) Every submodule of a c-co-epi-retractable R-module is quasi-continuous.

Proof.
(1)⇒ (2) This is clear.
(2)⇒ (1) Let M be any R-module. By (2), E(M) is semi-simple, and hence M = E(M).
Therefore, R is semi-simple artinian.
(1)⇒ (4) is clear.
(4)⇒ (1) Let M be any R-module. Then M⊕E(M), being a submodule of E(M)⊕E(M)
is quasi-continuous by (2). Consequently, M ⊕E(M) has C3-condition. By Lemma 4, M
injective and so R is semi-simple artinian.
(1)⇔ (3) follows from Corollary 2 in [11].

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