Math. J. Okayama Univ. 52 (2010), 123–131

SOME PROPERTIES OF EF-EXTENDING RINGSTRUONG CONG QUYNH and LE VAN THUYETAbstract. In [16], Thuyet and Wisbauer considered the extending
property for the class of (essentially) finitely generated submodules. A
module M is called ef-extending if every closed submodule which contains essentially a finitely generated submodule is a direct summand of
M . A ring R is called right ef-extending if RR is an ef-extending module. We show that a ring R is right ef-extending and the R-dual of every
simple left R-module is simple if and only if R is semiperfect right continuous with Sl = Sl ≤e RR . We also prove that a ring R is a QF-ring(ω)
if and only if R is left Kasch and RR is ef-extending if and only if R isright AGP-injective satisfying DCC on right (or left) annihilators and(R ⊕ R)R is ef-extending.

1. Introduction and Definitions
Throughout the paper, R represents an associative ring with identity
1 6= 0 and all modules are unitary R-modules. We write MR (resp., R M ) toindicate that M is a right (resp., left) R-module. We also write J (resp., Zr ,Sr ) for the Jacobson radical (resp., the right singular ideal, the right socleof R) and E(MR ) for the injective hull of MR . If X is a subset of R, theright (resp., left) annihilator of X in R is denoted by rR (X) (resp., lR (X))
or simply r(X) (resp., l(X)) if no confusion appears. If N is a submoduleof M (resp., proper submodule) we denote by N ≤ M (resp., N < M ).
Moreover, we write N ≤e M , N ≤⊕ M and N ≤max M to indicate that N
is an essential submodule, a direct summand and a maximal submodule ofM , respectively. A module M is called uniform if M 6= 0 and every non-zero
submodule of M is essential in M . It is called that a module M has finite
uniform dimension if M does not contain an infinite direct sum of non-zero
submodules. Let M, N be R-modules. M is said to be N -injective if, for any
submodule H of N , every R-homomorphism f : H −→ M can be extended
to an R-homomorphism f¯ : N −→ M . Let M be any module. A submodule
K of M is closed (in M ), if K ≤e L ≤ M , then K = L. A ring R is calledright AGP-injective if for each 0 6= a ∈ R, there exists n ∈ IN such thatan 6= 0 and lr(an ) = Ran ⊕ Xa with Xa ≤ R R (see [17]). A ring R is called
QF if it is right (or left) artinian and right (or left) self-injective. A ring
R is said to be right PF if RR is an injective cogenerator in the category
Mathematics Subject Classification. 16D50, 16D70, 16D80.
Key words and phrases. ef-extending rings, extending (or CS) rings, PF rings, QF
rings.
123

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T. C. QUYNH AND L. V. THUYET

of right R-modules. A ring R is called semiregular if R/J is von Neumannregular and idempotents lift modulo J. A ring R is called right Kasch if
every simple right R-module is embeded in R.
We consider the following conditions on a module MR :C1: Every submodule of M is essential in a direct summand of M.
C2: Every submodule of M that is isomorphic to a direct summand of
M is itself a direct summand of M.
C3: M1 ⊕ M2 is a direct summand of M for any two direct summand
M1 , M2 of M with M1 ∩ M2 = 0.
Module MR is called extending (or CS) (resp., continuous) if it satisfies
C1 (resp., both C1 and C2). R is called right extending (resp., continuous)
if RR is an extending module (resp., continuous). Module MR is called
quasi-continuous if it satisfies C1 and C3. A module M is called uniformextending if every uniform submodule is essential in a direct summand ofM.
Recently, the theory of extending modules has been developed. Some
results for extending modules contribute to plentiful theory of ring andmodule, particularly decomposition into direct sum of indecomposable (oruniform) modules and application to theory of QF-rings. In [11], Oshiro
considered a ring R such that every projective R-module is an extending
module (i.e., R is co-H) and in [12] he proved that rings with this property are (left) Artinian. Also in [3, 7] it is proved that R(I) is an extending
module for any index set I if and only if every projective R-module is anextending module.
In [3], Dung, Huynh, Smith and Wisbauer studied QF-rings via extending
modules. In [6], Pardo and Asensio proved that R is right PF if and only
if R is right cogenerator right extending. This result generalizes Osofsky’s
result in [13]. In [18], Yousif proved that R is right PF if and only if R ⊕ R is
extending as a right R-module and the R-dual of every simple left R-module
is simple.
There are some interesting generalization of extending modules. For example, in [15] Smith and Tercan studied weak extending module and C11
module and in [16], Thuyet and Wisbauer considered the extending property for the class of (essentially) finitely generated submodules and defined
ef-extending module. A module M is called ef-extending if every closed submodule which contains essentially a finitely generated submodule is a direct
summand of M . A ring R is called right ef-extending if RR is an ef-extending
module.
(ω)
In this paper, we prove that R is QF if and only if R is left Kasch and RR
is ef-extending. Moreover, we prove that R is QF if and only if R is rightSOME PROPERTIES OF EF-EXTENDING RINGS

125

AGP-injective satisfying DCC on right (or left) annihilators and (R ⊕ R)R
is ef-extending.
2. Main Results.
From the definition of ef-extending module and ring, we have:i) A right extending ring is right ef-extending. But the converse is nottrue in general.
Example. Let K be a division ring and K V a left K-vector space with infinite
dimension. Take S = End(K V ), then it right ef-extending but not right
extending.
ii) Every finitely generated submodule of an ef-extending module M isessential in a direct summand of M .
Lemma 2.1 ([16]). Every direct summand of an ef-extending module is
ef-extending.
It is well-known M is an extending module if and only if every closed
submodule of M is a direct summand (see [3]).
Lemma 2.2. Let M be a module such that Soc(M ) is finitely generated andessential in M . Then M is an extending module if and only if M is an
ef-extending module.
Proof. Assume that M is ef-extending. Let N be a closed submodule of
M . We have Soc(N ) ≤ Soc(M ), and so Soc(N ) is finitely generated byhypothesis.
On the other hand, Soc(N ) = N ∩ Soc(M ). For every x ∈ N such thatxR ∩ Soc(N ) = 0, xR ∩ (N ∩ Soc(M )) = 0 or xR ∩ Soc(M ) = 0. Since
Soc(M ) ≤e M , xR = 0. It follows that x = 0. So Soc(N ) ≤e N. Hence N isa closed submodule which contains essentially a finitely generated submodule. Thus N is a direct summand of M .
Corollary 2.3. Let R be a left perfect ring. Then R is right extending if
and only if R is right ef-extending.
If R M is a left R-module, recall that the R-dual M ∗ = HomR (R M, R) of
M is a right R-module via (f r)(m) = f (rm) for all r ∈ R, f ∈ M ∗ , andm ∈ M.
Proposition 2.4. The following statements are equivalent for a ring R.
(1) R is right ef-extending and the R-dual of every simple left R-module
is simple.
(2) R is semiperfect right continuous ring with Sr = Sl ≤e RR .

126

T. C. QUYNH AND L. V. THUYETProof. (1) ⇒ (2). By [10, Theorem 4.8], we have R is semiperfect, left Kaschwith Sr = Sl ≤e RR . Since R is right ef-extending, ei R is ef-extending for
all i = 1, 2, . . . , n. It follows that ei R is uniform (because ei R is indecomposable) and so Soc(ei R) is simple and essential in ei R for all i = 1, 2, . . . , n.
Therefore Sr is finitely generated and hence R is right extending by Lemma
2.2. Thus R is right continuous by [10, Theorem 4.10].
(2) ⇒ (1) is clear.

Theorem 2.5. The following statements are equivalent for a ring R.
(1) R is right PF.
(2) R ⊕ R is ef-extending as a right R-module and the R-dual of every
simple left R-module is simple.
Proof. (1) ⇒ (2) is clear.
(2) ⇒ (1). Assume that R ⊕ R is ef-extending as a right R-module andthe R-dual of every simple left R-module is simple. Then R is semiperfect
right continuous ring with Sr = Sl ≤e RR by Proposition 2.4. Since R issemiperfect, right ef-extending and Sr ≤e RR , then Sr is finitely generated.
Hence Soc(R⊕R)R is finitely generated and essential in (R⊕R)R . Therefore(R ⊕ R)R is extending by Lemma 2.2. We have J = Zr (since R is rightcontinuous) and R is semiregular (since R is semiperfect), then (R ⊕ R)R
satisfies the C2 by [10, Example 7.18], and so (R ⊕ R)R is continuous. Thus
R is right self-injective by [10, Theorem 1.35].
Corollary 2.6 ([18], Theorem 2). The following statements are equivalent
for a ring R.
(1) R is right PF.
(2) R ⊕ R is extending as a right R-module and the R-dual of every
simple left R-module is simple.
(ω)

We write RR to indicate a countable direct sum of copies of the right
R-module RR .
Theorem 2.7. The following statements are equivalent for a ring R.
(1) R is QF.
(ω)(2) R is left Kasch and RR is ef-extending.
Proof. We prove (2) ⇒ (1). Let T be a maximal left ideal of R. Since R
is left Kasch, r(T ) 6= 0. There exists 0 6= a ∈ r(T ) or T ≤ l(a) whichyields T = l(a) by maximality of T and so r(T ) = rl(a). Since R is rightef-extending, aR ≤e eR for some e2 = e ∈ R. On the other hand, aR ≤rl(a) ≤ eR and then rl(a) ≤e eR. Hence r(T ) ≤e eR. It implies that R
is semiperfect by [10, Lemma 4.1]. Thus R = e1 R ⊕ · · · ⊕ en R, where{ei }ni=1 is the complete set of orthogonal local idempotents. For every i 6= j

SOME PROPERTIES OF EF-EXTENDING RINGS

127(i, j ∈ {1, 2, · · · , n}), let f : ei R −→ ej R be a monomorphism. We haveei R ∼= f (ei R) ≤ ej R. Since R satisfies the right C2 (because R is left
Kasch), then f (ei R) is a direct summand of ej R or f (ei R) = ej R (because
ej R is indecomposable). Hence f is an isomorphism. Since R is right efextending, every uniform right ideal of R is essential
in direct summand ofL
ei R is ei0 R-injective
RR . Therefore for every i0 ∈ {1, 2, · · · , n},{1,2,··· ,n}\{i0 }by [3, Corollary 8.9]. Since ei R is also ef-extending, indecomposable and so
ei R is quasi-continuous. By [9, Theorem 2.13], R is right quasi-continuous.
Thus R is right continuous.
By Utumi’s Theorem (see [10, Theorem 1.26]), J = Zr . By [10, Example
7.18], (R⊕R)R satisfies the C2 and so (R⊕R)R is continuous. Thus R is rightself-injective by [10, Theorem 1.35]. It implies that ei R is injective for every
L
(ω)
εi R
i = 1, 2, . . . , n. On the other hand, RR = (e1 R ⊕ · · · ⊕ en R)(ω) =is uniform-extending, for some countable set ω ′ and εω′i ∈ {e1 , e2 , · · · , en } for(ω)′
each i ∈ ω . By [3, Corollary 8.10], RR is injective. By a well-known resultof Faith ([4]), R has ACC on right annihilators and hence R is QF.

Corollary 2.8 ([18], Theorem 3). The following statements are equivalent
for a ring R.
(1) R is QF.
(ω)(2) R is left Kasch and RR is extending.
Corollary 2.9. The following statements are equivalent for a ring R.
(1) R is right PF.
(2) R is left, right Kasch and (R ⊕ R)R is ef-extending.
Proof. (1) ⇒ (2). Assume that R is right PF. Then R is left, right Kaschand R is right self-injective by [5, Theorem 2.8]. Hence (R⊕R)R is extending
module [10, Theorem 1.35].
(2) ⇒ (1). In the proof of Theorem 2.7, we showed that R is right PF. 
Lemma 2.10. Assume that RR = e1 R ⊕ e2 R ⊕ · · · ⊕ en R, where each ei R
is uniform for all i = 1, 2, . . . , n. If every monomorphism RR −→ RR is an
epimorphism, then R is semiperfect.
Proof. By [10, Lemma 4.26].



A ring R is called I-finite if R contains no infinite sets of orthogonal
idempotents (see [10]).
Proposition 2.11. The following statements are equivalent for a ring R.
(i) R is a semiperfect, right continuous ring.

128

T. C. QUYNH AND L. V. THUYET(ii) R is a right ef-extending ring, Zr = J, and R has DCC on principal
projective right ideals.
(iii) R is a right ef-extending, right C2 ring, and R has DCC on principal
projective right ideals.
(iv) R is a right ef-extending, right C2 and I-finite ring.
Proof. (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear.
(iv) ⇒ (i). Since R is I-finite, write 1 = u1 + · · · + un , where the ui are
orthogonal primitive idempotents. Hence R = u1 R ⊕ · · · ⊕ un R, where eachuk R is uniform because it is an ef-extending module. On the other hand, R
satisfies C2, every monomorphism RR −→ RR is epimorphism. Thus R issemiperfect by Lemma 2.10. By a similar proof of Theorem 2.7, R is rightcontinuous.

Corollary 2.12. The following statements on a ring R are equivalent:(1) R is a semiperfect right self-injective ring.
(2) (R ⊕ R)R is ef-extending, Zr = J and R has DCC on principal
projective right ideals.
Proof. (1) ⇒ (2) is clear.
(2) ⇒ (1). By Proposition 2.11, R is semiperfect. Thus by a similar proofof Theorem 2.7, R is right self-injective.

Next, we consider properties of a ring which has DCC on right (left)annihilators such that (R ⊕ R)R is ef-extending.
Theorem 2.13. The following statements are equivalent for a ring R;(1) R is QF.
(2) R is right AGP-injective satisfying DCC on right annihilators and(R ⊕ R)R is ef-extending.
(3) R is right AGP-injective satisfying DCC on left annihilators and(R ⊕ R)R is ef-extending.
Proof. (2) ⇒ (1). Assume that there is monomorphism f : R −→ R which
not epimorphism. Let a = f (1). Then r(an ) = 0, ∀n ≥ 1. Assume thataR 6= R. Since R is right AGP-injective, there exist a positive integer m ≥ 1
and X1 ≤ R R such that am 6= 0 and lr(am ) = Ram ⊕ X1 . It implies thatR = Ram ⊕ X1 (since r(am ) = 0) and so Ram = Re for some e2 = e ∈ R.
Then
0 = r(am ) = r(Ram ) = r(Re) = r(e) = (1 − e)R,and hence e = 1 or Ram = R. It implies that R = Ra, i.e., ba = 1
for some b ∈ R. If ab 6= 1, then by [8, Example 21.26], there are someeij = ai bj −ai+1 bj+1 ∈ R, i, j ∈ IN such that eij ekl = δjk eil for all i, j, k ∈ IN ,

SOME PROPERTIES OF EF-EXTENDING RINGS

129

where δjk are the Kronecker deltas. Notice eij 6= 0 for all i, j ∈ IN , byconstruction. Set ei = eii . Then ei ej = δij ei , ∀i, j ∈ IN . Therefore we haver(e1 ) > r({e1 , e2 }) > · · · ,this is a contradiction because R has DCC on right annihilators. Henceab = 1 and so aR = R. This is a contradiction to our assumption. Thus f
is an epimorphism.
On the other hand, R is I-finite, there exists an orthogonal set of primitive
idempotents {ei }ni=1 such that RR = e1 R⊕e2 R⊕...⊕en R. Since R is right efextending, ei R is ef-extending and so ei R is uniform for every i = 1, 2, . . . , n.
Thus R is semiperfect by Lemma 2.10. But R is right AGP-injective, J = Zr .
Therefore R is right C2 by [10, Example 7.18]. Then R is QF by the laterpart of the proof in Theorem 2.7.
(3) ⇒ (1). Assume that there is monomorphism f : R −→ R which
not epimorphism. The same argument of (2) ⇒ (1), there exists a set oforthogonal idempotents {ei ∈ R|ei 6= 0, i ∈ IN }. Therefore we havel(e1 ) > l({e1 , e2 }) > · · · ,this is a contradiction because R has DCC on left annihilators. Thus R is
QF by the later part of the proof in (2) ⇒ (1).
Theorem 2.14. The following statements are equivalent.
(1) R is QF.
(2) (R ⊕ R)R is ef-extending, R is right C2 and satisfies ACC on leftannihilators.
(3) (R ⊕ R)R is ef-extending, R is right C2 and satisfies ACC on rightannihilators.
Proof. (1) ⇒ (2), (3) is clear.
(2) ⇒ (1). Assume that there is monomorphism f : R −→ R which not
epimorphism. Let a = f (1). If aR 6= R, then an R > an+1 R for all n ≥ 1(because r(a) = 0). Moreover, an R ∼= R for all n ≥ 1. Then for every= aR ∼2n ≥ 1, there exists 0 6= en = en ∈ R such that an R = en R because R is rightC2. Hence we have a strict ascending chainl(e1 ) < l(e2 ) < · · · ,this is a contradiction. Thus aR = R or f is epimorphism. It implies that R
is semiperfect by Lemma 2.10. Then R is QF by the later part of the proof
in Theorem 2.7.
(3) ⇒ (1). By (2) ⇒ (1), for every monomorphism f : R −→ R which is
not epimorphism and for every n ≥ 1, there exists 0 6= e2n = en ∈ R such

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T. C. QUYNH AND L. V. THUYET

that an R = en R with a = f (1). Hence we have
e1 R > e2 R > · · · ,this is a contradiction by [10, Lemma B.6]. Thus R is QF by the same
argument of (2) ⇒ (1).
From this theorem, we have the following proposition.
Proposition 2.15. The following statements are equivalent for a ring R.
(1) R is QF.
(2) (R ⊕ R)R is ef-extending, R satisfies ACC on right annihilators andSl ≤e RR .
Proof. We prove (2) ⇒ (1). By [14, Theorem 2.9 ], R is semiprimary.
Therefore R is left Kasch by [10, Lemma 4.2]. Thus R is QF.
In [1], the authors proved that R is QF if and only if R is right Artinian,(R ⊕ R)R or R (R ⊕ R) is extending and Sr ≤ Sl . In this paper, we extend
this result in case (R ⊕ R)R or R (R ⊕ R) is ef-extending.
Theorem 2.16. Assume that R has ACC on right annihilators with Sr ≤e
RR . Then the following statements are equivalent.
(1) R is QF.
(2) (a) (R ⊕ R)R or R (R ⊕ R) is ef-extending.
(b) Sr ≤ Sl .
Proof. (1) ⇒ (2) is clear.
(2) ⇒ (1). Since R has ACC on right annihilators with Sl ≤e RR , R issemiprimary by [14, Theorem 2.9]. Hence R is left Kasch by [10, Lemma
4.2]. If (R ⊕ R)R is ef-extending, then R is QF by Proposition 2.15.
If R (R⊕R) is ef-extending, then R is left ef-extending. Therefore Soc(Re)is simple for every local idempotent e ∈ R. It implies that Sr e is simple orzero. Moreover, Sr is essential in RR and so Soc(eR) 6= 0. It implies that R
is right Kasch by [10, Theorem 3.12]. Therefore R is left C2. Thus R is QF
by Theorem 2.14.
Acknowledgment. The authors would like to thank the referee for his/her
many valuable suggestions and comments.
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Truong Cong Quynh
Department of Mathematics
Danang University
VietName-mail address: tcquynh@dce.udn.vn
Le Van Thuyet
Department of Mathematics
Hue University
VietName-mail address: lvthuyethue@gmail.com(Received May 18, 2008 )(Revised August 13, 2008 )