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Generalizations of Semiregular RingsGuangshi Xiao a & Wenting Tong aa Department of Mathematics, Nanjing University, Nanjing, ChinaPublished online: 03 Sep 2006.

To cite this article: Guangshi Xiao & Wenting Tong (2005): Generalizations of Semiregular Rings ,Communications in Algebra, 33:10, 3447-3465

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Communications in Algebra®, 33: 3447–3465, 2005Copyright © Taylor & Francis, Inc.ISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/AGB-00927870500232984

GENERALIZATIONS OF SEMIREGULAR RINGS#

Guangshi Xiao and Wenting TongDepartment of Mathematics, Nanjing University, Nanjing, China

In this article, we call a ring R right generalized semiregular if for any a ∈ R thereexist two left ideals P�L of R such that lr�a� = P ⊕ L, where P ⊆ Ra and Ra ∩L is small in R. The class of generalized semiregular rings contains all semiregularrings and all AP-injective rings. Some properties of these rings are studied and someresults about semiregular rings and AP-injective rings are extended. In addition, wecall a ring R semi-�-regular if for any a ∈ R there exist a positive integer n ande2 = e ∈ anR such that �1− e�an ∈ J�R�, the Jacobson radical of R. It is shown thata ring R is semi-�-regular if and only if R/J�R� is �-regular and idempotents can belifted modulo J�R�.

Key Words: Exchange rings; AP-injective rings; Semi-�-regular rings; Generalized semiregular rings.

2000 Mathematics Subject Classification: 16D50; 16E40; 16D40; 16A30.

INTRODUCTION

Let R be a ring and M a right R-module. M is called right principallyinjective (P-injective) (Nicholson and Yousif, 1995) if every R-homomorphism froma principal right ideal aR to M extends to one from RR to M . M is said to be almostprincipally injective (or AP-injective for short) (Page and Zhou, 1998) if, for anya ∈ R, there exists an S-submodule Xa of M such that lM�rR�a�� = Ma⊕ Xa as leftS-modules, where S = End�M�. R is called a right P-injective (resp. AP-injective)ring if RR is P-injective (resp. AP-injective). From Page and Zhou (1998), we knowthat P-injective rings are AP-injective, but the converse is not true (see Page andZhou, 1998, Example 1.5).

A ring R is called semiregular (Nicholson, 1976) if, for any a ∈ R, there existsan idempotent g ∈ Ra such that a�1− g� ∈ J�R�.

In Section 1, we call a ring R right generalized semiregular if, for any a ∈ R,there exist two left ideals P�L of R such that lr�a� = P ⊕ L, where P ⊆ Ra andRa ∩ L is small in R. We prove that these rings are non-trivial generalizations ofboth right AP-injective rings and semiregular rings. If the n× n matrix ring Mn�R�

over a ring R is right generalized semiregular, then so is R (see Theorem 1.15).

Received September 17, 2003; Accepted May 23, 2004#Communicated by Alberto Facchini.Address correspondence to Guangshi Xiao, Department of Mathematics, Nanjing University,

Nanjing 210093, China; E-mail: xgs01cn@yahoo.com.cn

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The notion of semiregular rings is left-right symmetric, but we do not know whetherthis is true for generalized semiregular rings.

In Section 2, we study the relations between right generalized semiregular ringsand right self-injective rings. An example is given to answer the question raised byCamillo and Yu (1994) in the negative. If R is a right MI and right generalizedsemiregular ring with J�R� ⊆ Z�RR�, then R is right self-injective (see Theorem 2.7).In particular, if R is a right MI ring, then R is a right generalized semiregular ringwith J�R� ⊆ Z�RR� and maximum condition on right annihilators if and only if Ris quasi-Frobeniusean. A ring R is right self-injective if and only if R is a rightgeneralized semiregular ring and RR is weakly injective. From Nicholson and Yousif(1995), we know that if R is a right P-injective ring, then RR satisfies C2. We provethat if R is a right generalized semiregular ring with J�R� ⊆ Z�RR�, then RR alsosatisfies C2.

In Section 3, the regularity of right generalized semiregular rings ischaracterized. If R is a semiprimitive and right generalized semiregular ring, then thecenter C�R� of R is von Neumann regular. If R is a right generalized semiregular ringwith J�R� ⊆ Z�RR� which has a classical right quotient ring Q, then Q is stronglyregular if and only if R is reduced. According to Xiao et al. (2003, Theorem 2.3), weknow that a ring R is von Neumann regular if and only if R is a right AP-injectiveand right PP ring. We generalize this result and prove that a ring R is von Neumannregular if and only if R is a right generalized semiregular ring with J�R� ⊆ Z�RR�and for any 0 �= a ∈ R, there exists a positive integer m with am �= 0 such that amRis projective (see Theorem 3.12).

In Section 4, a ring R is called semi-�-regular if, for any a ∈ R, there exista positive integer n and e2 = e ∈ anR such that an − ean ∈ J�R�. We prove that aring R is semi-�-regular if and only if R/J�R� is �-regular and idempotents can belifted modulo J�R�. Then semiregular rings and �-regular rings are semi-�-regular,and semi-�-regular rings are exchange rings. Two examples show that semi-�-regularrings need not be semiregular or �-regular and exchange rings need not be semi-�-regular. We also prove that, if M is a projective left R-module, then J�E� = �� ∈ E ��M is small in M�, and E is semi-�-regular if and only if for every � ∈ E there existsa positive integer n such that �nM lies over a summand of M , where E = End�M�.

Throughout this article, all rings R considered are associative with unit, andall R-modules are unitary. For any non-empty subset X of R, r�X� (resp. l�X�)is reserved for the right (resp. left) annihilator of X in R. If X = a, we usuallyabbreviate it to r�a� (resp. l�a�). The symbols J�R�, Z�RR�, Z�RR�, Soc�RR�, andSoc�RR� will stand for the Jacobson radical, the right singular ideal, the left singularideal, the right Socle, and the left Socle of R, respectively.

1. GENERALIZED SEMIREGULAR RINGS

Recall that an element a ∈ R is von Neumann regular if a ∈ aRa. A ring R iscalled von Neumann regular if, for any a ∈ R, a is von Neumann regular. A ring Ris called strongly regular if, for any a ∈ R, there exists b ∈ R such that a = a2b.

Let R be a ring and M be any left R-module. A submodule K of M is saidto be small in M (Anderson and Fuller, 1974) if K + N �= M for every submoduleN �= M . The radical of M , denoted by J�M�, is defined to be the intersection of allmaximal submodules of M , and equals the set �x ∈ M � Rx is small in M�. It may

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GENERALIZATIONS OF SEMIREGULAR RINGS 3449

happen that M has no maximal submodules in which case J�M� = M . Thus, for aring R, it is easy to verify that J�R�x is small in M for each x ∈ M . In particular,J�R� is small in R and J�R� = �a ∈ R � Ra is small in R�.

A submodule N of a module M lies over a summand of M if there exists adirect decomposition M = P ⊕Q with P ⊆ N and Q ∩ N is small in M . In Nicholson(1976), the author calls a module M semiregular if Rx lies over a projectivesummand of M for any x ∈ M . Following Nicholson (1976), a ring R is said tobe semiregular if for any a ∈ R, there exists e2 = e ∈ aR such that �1− e�a ∈ J�R�.By Nicholson (1976, Lemma 2.1), we obtain the following lemma.

Lemma 1.1. The following conditions are equivalent for an element a of a ring R:

(1) There exists e2 = e ∈ aR such that �1− e�a ∈ J�R�.(2) There exists e2 = e ∈ Ra such that a�1− e� ∈ J�R�.(3) There exists a von Neumann regular element b ∈ R with a− b ∈ J�R�.(4) Ra lies over a summand of RR.

Definition. An element a of a ring R is called right generalized semiregular if thereexist two left ideals P�L of R such that lr�a� = P ⊕ L, where P ⊆ Ra and Ra ∩ L issmall in R. A ring R is called right generalized semiregular if each of its elements isright generalized semiregular. Similarly, we may define left generalized semiregularelements and left generalized semiregular rings.

Remark 1. From Nicholson and Yousif (1995), we know that a ring R is rightP-injective if and only if for any a ∈ R, lr�a� = Ra. Thus, every right P-injective ringis right generalized semiregular. In particular, every right self-injective ring is rightgeneralized semiregular.

Proposition 1.2. If R is a right AP-injective or semiregular ring, then R is rightgeneralized semiregular.

Proof. (1) Let R be a right AP-injective ring. Then lr�a� = Ra⊕ La for any a ∈ R,where La is a left ideal of R. Note that Ra ∩ La = 0 is small in R, thus R is rightgeneralized semiregular.

(2) Let R be a semiregular ring and a∈R. Then there exists e2 = e∈Ra suchthat a�1− e�∈ J�R�. Thus R=Re⊕R�1− e�, where Re⊆Ra and Ra�1− e� ⊆ J�R� issmall in R. Note that Ra ⊆ lr�a�, so by the modular law we have lr�a� = lr�a� ∩ R =lr�a� ∩ �Re⊕ R�1− e�� = Re⊕ �lr�a� ∩ R�1− e�� and Ra ∩ �lr�a� ∩ R�1− e�� =Ra ∩ R�1− e� = Ra�1− e� is small in R. Hence R is right generalized semiregular.

�

The following two examples show that right generalized semiregular rings neednot be semiregular or right AP-injective.

Example 1.3. (1) Let RVR be a bimodule over a ring R. The trivialextension of R by V is the direct sum T�R� V� = R⊕ V with multiplication�r + v��r ′ + v′� = rr ′ + �rv′ + vr ′�. It is shown in Nicholson and Yousif (2001) thatthe trivial extension R = T��� �/�� is a commutative P-injective ring, but it is not

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semiregular since R/J�R� � �. Thus R is a commutative generalized semiregularring which is not semiregular.

(2) Let R = ( �2 �20 �2

), where �2 is the ring of integers modulo 2. Then J�R� =(

0 �20 0

)and Z�RR� = Z�RR� = 0, so R is neither left nor right AP-injective. But

R/J�R� � ( �2 00 �2

)is von Neumann regular and any idempotent of R/J�R� can be

lifted to R, so R is semiregular by Nicholson (1976, Theorem 2.9). Thus R is a rightgeneralized semiregular ring which is not right AP-injective, since J�R� �= Z�RR�.

Now, we give some sufficient conditions, under which right generalizedsemiregular rings are semiregular.

Lemma 1.4. If R is a right generalized semiregular ring and for every a ∈ R thereexists e2 = e ∈ R such that r�a� = r�e�, then R is semiregular.

Proof. Let a ∈ R. Since R is right generalized semiregular, there exist two leftideals P�L of R such that lr�a� = P ⊕ L, where P ⊆ Ra and Ra ∩ L is small in R.Thus Re = P ⊕ L since r�a� = r�e�. Since r�a� = r�e�, we have lr�a� = lr�e� = Re,this gives a = ae. Take e = g + t, where g = ra ∈ P ⊆ Ra and t ∈ L. Then a = ae =ag + at = ara+ at and ra = rara+ rat, which gives ra− rara = rat ∈ P ∩ L = 0and a− ara = at ∈ Ra ∩ L ⊆ J�R�. This shows that g2 = g ∈ Ra and a�1− g� ∈J�R�, a is semiregular. Hence R is semiregular. �

Corollary 1.5. Let a be a right generalized semiregular element. If aR � eR, wheree2 = e, then a is semiregular.

Proof. Let a ∈ R and let � be the isomorphism of aR onto eR. By Page andZhou (1998, Lemma 2.12), there exists an idempotent f of R such that r�a� = r�f�.By Lemma 1.4, a is semiregular. �

If lr�a� is a direct summand of R, then there exists e2 = e ∈ R such thatlr�a� = Re. Thus r�a� = rlr�a� = r�e�, and so the following result is immediate.

Corollary 1.6. If lr�a� is a direct summand of R for any a ∈ R and R is rightgeneralized semiregular, then R is semiregular.

Kaplansky (1968) defines a ring R to be Baer if the right annihilator ofevery non-empty subset of R is generated, as a right ideal, by an idempotent. Thisdefinition is left-right symmetric. A ring R is called right (resp. left) PP (Armendariz,1974) if every principal right (resp. left) ideal of R is projective. R is a right PP ringif and only if for every a ∈ R there exists an idempotent e ∈ R such that r�a� = eR.Clearly, every Baer ring is a right and left PP ring.

Corollary 1.7. Let R be a right PP ring. If R is a right generalized semiregular, thenR is semiregular.

From Page and Zhou (1998, Corollary 2.3), we see that if R is a rightAP-injective ring then J�R� = Z�RR�.

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GENERALIZATIONS OF SEMIREGULAR RINGS 3451

Proposition 1.8. If R is a right generalized semiregular ring, then Z�RR� ⊆ J�R�.

Proof. Let 0 �= a∈Z�RR�. Then for each element b of R, ba∈Z�RR�. Let u= 1− ba,then u �= 0, and r�u� = 0 since r�ba� ∩ r�u� = 0. Thus R = lr�u� = P ⊕ L, where P ⊆Ru and Ru ∩ L is small in R. We have that P = Re with e2 = e ∈ R, hence it issufficient to prove that e = 1. If not, there exists 0 �= �1− e�r ∈ �1− e�R ∩ r�ba�since r�ba� is essential in R. This gives u�1− e�r = �1− e�r. Put u = se+ t for somes ∈ R� t ∈ L. Then u�1− e�r = t�1− e�r. Hence �1− e�r = t�1− e�r, and therefore�1− t��1− e�r = 0. Note that t = u− se ∈ Ru ∩ L ⊆ J�R�, so 1− t is a unit, whichimplies �1− e�r = 0, a contradiction. So, e = 1 and R = P = Ru. Thus a ∈ J�R�.

�

Corollary 1.9. If R is a semiregular ring, then Z�RR� ⊆ J�R� and Z�RR� ⊆ J�R�.

Remark 2. Example 1.3(2) shows that there exists a semiregular ring with J�R� �=Z�RR� = Z�RR�. By Proposition 1.8, if R is a right generalized semiregular ring withJ�R� ⊆ Z�RR�, then J�R� = Z�RR�.

A ring R is called right mininjective (Nicholson and Yousif, 1997) if everyisomorphism between simple right ideals is given by multiplication by an element ofR. Equivalently, if lr�K� = K for every left ideal K = Rk for which kR is simple.

Remark 3. By Nicholson and Yousif (1997, Remark 1.2), the ring � of integersis a commutative mininjective ring but it is not right generalized semiregular as weshow in Example 4.8 below. By Nicholson and Yousif (1997, Remark 1.6), we knowthat there is a right P-injective ring which is not left mininjective. Thus there existsa right generalized semiregular ring which is not left mininjective.

Since every semiperfect ring is semiregular, and thus is right generalizedsemiregular, by Proposition 1.2 the following result extends Nicholson and Yousif(1997, Theorem 2.9).

Theorem 1.10. If R is a right generalized semiregular, right mininjective ring andSoc�RR� is an essential submodule of RR, then J�R� = Z�RR�.

Proof. Since R is right mininjective, Soc�RR� ⊆ Soc�RR� by Nicholson and Yousif(1997, Theorem 1.14). By Nicholson and Yousif (1997, Proposition 2.8 (1)), J�R� ⊆Z�RR�. Since R is right generalized semiregular, Z�RR� ⊆ J�R� by Proposition 1.8.Thus J�R� = Z�RR�. �

According to Nicholson (1977, Corollary 2.3), we know that if R is asemiregular ring then so is eRe for any idempotent e of R. Hence if the n× n matrixring Mn�R� over a ring R is semiregular for some n ≥ 1, then so is R. An idempotentelement e ∈ R is left (resp. right) semicentral in R (Birkenmeier, 1983) if Re = eRe(resp. eR = eRe).

Proposition 1.11. Let R be a right generalized semiregular ring. If an idempotent eof R is right semicentral, then eRe is right generalized semiregular.

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Proof. Let a ∈ eRe. Then there exist two left ideals P�L of R such that lr�a� =P⊕L, where P⊆Ra and Ra∩L small in R. We claim that leRereRe�a�= eP⊕ eL.In fact, eP ∩ eL ⊆ P ∩ L = 0. Take any y ∈ eP ⊆ ePe, where y = ey1� y1 ∈ P ⊆lr�a�. Then for any x ∈ reRe�a� ⊆ r�a�, y1x = 0, which gives yx = ey1x = 0. Hencey ∈ leRereRe�a�, eP ⊆ leRereRe�a�. Similarly, eL ⊆ leRereRe�a�. On the other hand,take x ∈ leRereRe�a�. Then for any y ∈ r�a�, we have aeye = aye = 0 since a∈ eRe.So xeye = 0, which gives xy = xey = xeye = 0 since x� y ∈ eRe and e is rightsemicentral. Thus leRereRe�a� ⊆ lr�a�. Take x = s + t, where s ∈ P� t ∈ L. Thenx= ex= es+ et ∈ eP+ eL. This shows that leRereRe�a� = eP ⊕ eL since the sumeP + eL is direct.

It remains to prove that eRea ∩ eL is small in eRe since eP ⊆ eRa = eRea.Since e is right semicentral, we have eRea ∩ eL ⊆ e�eRea ∩ eL�e. But eRea ∩ eL ⊆Ra ∩ L ⊆ J�R�, so eRea ∩ eL ⊆ eJ�R�e = J�eRe�. It follows from the fact thatJ�eRe� is small in eRe that eRea ∩ eL is small in eRe. Thus eRe is right generalizedsemiregular. �

Proposition 1.12. Let e� f be orthogonal central idempotents of R. If eR� fR areright generalized semiregular, then g = e+ f is central idempotent and gR = eR⊕ fRis right generalized semiregular.

Proof. Let a ∈ gR. Then ea ∈ eR and fa ∈ fR. Take x ∈ lgRrgR�a�. Then for anyy ∈ reR�ea�, we have eay = 0 and hence ay = aey = eay = 0, this implies agy =ayg = 0 and gy ∈ rgR�a�. Thus xy = xgy = 0 and exy = xye = 0, so ex ∈ leRreR�ea�.By assumption, ex ∈ leRreR�ea� = Pe ⊕ Le, where Pe ⊆ eRea = eRa and eRa ∩ Le ⊆J�eRe�. Similarly, fx ∈ lfRrfR�fa� = Pf ⊕ Lf , where Pf ⊆ fRa and fRa ∩ Lf ⊆J�fRf�. Then x = gx = ex + fx ∈ Pe ⊕ Pf ⊕ Le ⊕ Lf since e� f are orthogonal.For any x ∈ Le and any y ∈ rgR�a�. Then ay = 0, which implies aey = 0, andthus xey = 0 since Le ⊆ leRreR�ea�. Note that Le ⊆ eR ⊆ gR and x = xe, so xy = 0and Le ⊆ lgRrgR�a�. Similarly, Lf ⊆ lgRrgR�a�. On the other hand, Pe ⊕ Pf ⊆ eRa⊕fRa = gRa. This proves that lgRrgR�a� = Pe ⊕ Pf ⊕ Le ⊕ Lf . It is easily verified thatgRa ∩ �Le ⊕ Lf� ⊆ J�eR�⊕ J�fR� = J�gR�. Since gR is a ring with identity, J�gR� issmall in gR and hence gRa ∩ �Le ⊕ Lf� is small in gR. �

Corollary 1.13. Let e be a nonzero central idempotent of a ring R. Then eRe and�1− e�R�1− e� are right generalized semiregular if and only if so is R.

Theorem 1.14. Let 1 = f1 + · · · + fn in R, where fi’s are orthogonal centralidempotents. Then R is right generalized semiregular if and only if each fiR is rightgeneralized semiregular.

Theorem 1.15. Let e be an idempotent of R such that ReR = R. If R is a rightgeneralized semiregular ring, then eRe is right generalized semiregular.

Proof. Let a ∈ eRe. Then there exist two left ideals P�L of R such that lr�a� =P ⊕ L, where P⊆Ra and Ra∩L small in R. We claim that leRereRe�a�= ePe⊕ eLe.Since 1− e ∈ r�a�, we see that t�1− e� = 0 for any t ∈ L, which impliesL = Le. Similarly, P = Pe. Thus ePe ∩ eLe = 0. Clearly, ePe ⊆ leRereRe�a� and eLe ⊆leRereRe�a� since Pe = P and Le = L. On the other hand, Take x ∈ leRereRe�a�, and

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GENERALIZATIONS OF SEMIREGULAR RINGS 3453

write 1 = ∑ni=1 aiebi for some ai� bi in R. Then for any y ∈ r�a�, we get aeyaie =

ayaie = 0 for each i. This implies that xeyaie = 0 for each i, which gives xy = xey =xey

∑ni=1 aiebi = 0 since x ∈ eRe. Thus leRereRe�a� ⊆ lr�a�. Take x = s + t, where

s ∈ P� t ∈ L. Hence x = exe = ese+ ete ∈ ePe+ eLe. This shows that leRereRe�a� =ePe⊕ eLe, and the claim is complete. It remains to prove that eRea ∩ eLe is smallin eRe since ePe ⊆ eRea. Since L = Le, we have eRea ∩ eLe ⊆ Ra ∩ L ⊆ J�R�, soalso eRea ∩ eLe ⊆ eJ�R�e = J�eRe�. It follows then that eRea ∩ eLe is small in eRe.Thus eRe is right generalized semiregular. �

Corollary 1.16. Let Mn�R� be the n× n matrix ring over a ring R. If Mn�R� is rightgeneralized semiregular for some n ≥ 1, then so is R.

Proof. Let S = Mn�R�. Then S = Se11S is right generalized semiregular, so isR � e11Se11 by Theorem 1.15. �

2. GENERALIZED SEMIREGULAR RINGS AND SELF-INJECTIVE RINGS

In this section, we give an example to answer the question raised byCamillo and Yu (1994) in the negative, and characterize the self-injectivity of rightgeneralized semiregular rings.

An idempotent e in a ring R is called primitive if eRe has no properidempotents, and e is called local if eRe is a local ring. Let n be a positive integer.A ring R is said to be orthogonally n-finite (Tuganbaev, 2002) if every set of nonzeroorthogonal idempotents of R contains at most n elements. A ring is said to beorthogonally finite (Tuganbaev, 2002) if it is orthogonally k-finite for some positiveinteger k, that is, a ring is orthogonally finite if it contains no infinite set oforthogonal idempotents.

For a convenience, we call a ring R primitively finite if there exist finiteprimitive-orthogonal-idempotents e1� e2� � en such that 1 = e1 + e2 + · · · + en. It iseasy to see that, if a ring R is orthogonally finite, then

1 = e1 + e2 + · · · + en

with ei (i = 1� 2� � n) primitive-orthogonal-idempotents. Camillo and Yu (1994)asked whether the converse implication holds in general. Now, we give an exampleto answer the question raised by Camillo and Yu (1994) in the negative.

Example 2.1. By Shepherdson (1951, Theorem 1.0), there exists a domain S suchthat R = M2�S� with ab = e� ba �= e, where a� b ∈ R, and e is the unit of M2�S�.Let e = e11 + e22, where

e11 =(1 00 0

)� e22 =

(0 00 1

)

Then e11 and e22 are primitive idempotents of R since S is a domain. Hence Ris primitively finite. Denote by fn = an�1− ba�bn� n = 1� 2� 3� , then fn isan idempotent. Hence f1� f2� � fn� is a sequence of orthogonal nonzeroidempotents. This shows that R is a primitively finite ring which is not orthogonallyfinite.

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Utumi (1965) identified two conditions enjoyed by any right self-injectivering R:

C1 Every right ideal is essential in a direct summand of RR.C2 If a right ideal M is isomorphic to a direct summand of RR then M is a direct

summand.Except that, from Mohammed and Müller (1990), we have the third condition:

C3 If A� B are direct summands of RR with zero intersection, then A⊕ B is adirect summand.

From Page and Zhou (1998, Proposition 2.13(1)), we know that if R is a rightAP-injective ring, then RR satisfies C2. In order to prove this result is true for a rightgeneralized semiregular ring with J�R� ⊆ Z�RR�, the following well-known fact willbe needed.

Lemma 2.2. If MR is both projective and singular, then M = 0.

Proof. By Goodearl (1976, Proposition 1.20), there exists a short exact sequence0 → A → B → M → 0 with f�A� is an essential submodule of B, where f is themonomorphism of A into B. But M is projective, so B = f�A�⊕ I , where I � M .Thus M = 0 since f�A� is essential. �

Theorem 2.3. Let R be a right generalized semiregular ring with J�R� ⊆ Z�RR�, then

�1� If e2 = e ∈ R, � aR → eR is a right R-isomorphism, then there exists g2 = g ∈ Rsuch that aR = gR, i.e., RR satisfies C2.

�2� If e2 = e ∈ R, f 2 = f ∈ R and eR ∩ fR = 0, then there exists g2 = g ∈ R such thateR⊕ fR = gR, i.e., RR satisfies C3.

Proof. (1) Let a ∈ R such that aR � eR, where e2 = e ∈ R. By Page and Zhou(1998, Lemma 2.12), there exists an idempotent f of R such that a = afand r�a� = r�f�. By Lemmas 1.4 and 1.1, there exists an idempotent g of Rwith g ∈ aR and �1− g�a ∈ J�R�. Thus aR = gR⊕ S, where S = �1− g�aR ⊆ J�R�.By assumption, S ⊆ Z�RR� is a singular right R-module. Let � be the rightR-epimorphism of fR to afR given by ��fr� = afr for any r ∈ R. If afr = 0, thenfr ∈ r�a� ∩ fR = r�f� ∩ fR = 0. This shows that aR = afR � fR is a projective rightR-module. Thus S is a projective and singular right R-module, and so S = 0 byLemma 2.2. Hence aR = gR.

(2) By the assumption, there exists a right ideal L1 of R such that RR = eR⊕L1.Thus eR⊕ fR = �eR⊕ fR� ∩ R = �eR⊕ fR� ∩ �eR⊕ L1� = eR⊕ ��eR⊕ fR� ∩ L1�.Then fR � �eR⊕ fR� ∩ L1. By (1), �eR⊕ fR� ∩ L1 is a summand of RR. LetRR = ��eR⊕ fR� ∩ L1�⊕ L2, then L1 = L1 ∩ R = ��eR⊕ fR� ∩ L1�⊕ �L1 ∩ L2�.Hence RR = eR⊕ ��eR⊕ fR� ∩ L1�⊕ �L1 ∩ L2� = eR⊕ fR⊕ �L1 ∩ L2�. This givesthat eR⊕ fR is a summand of RR, thus there exists g2 = g ∈ R such thateR⊕ fR = gR. �

Proposition 2.4. If R is a primitively finite ring that satisfies C2, then R � R1 × R2,where R1 is semisimple and every simple right ideal of R2 is nilpotent.

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Proof. This follows from the proof in Nicholson and Yousif (1995, Theorem 1.4).�

The following result extends Nicholson and Yousif (1995, Theorem 1.4) andPage and Zhou (1998, Theorem 2.16), by Theorem 2.3.

Corollary 2.5. If R is a primitively finite and right generalized semiregular ring withJ�R� ⊆ Z�RR�, then R � R1 × R2, where R1 is semisimple and every simple right idealof R2 is nilpotent.

Corollary 2.6. If R is a semiperfect ring with J�R� ⊆ Z�RR�, then R � R1 × R2,where R1 is semisimple and every simple right ideal of R2 is nilpotent.

A ring R is called right MI (Yue Chi Ming, 1995) if R contains an injectivemaximal right ideal.

Theorem 2.7. If R is a right MI and right generalized semiregular ring with J�R� ⊆Z�RR�, then R is right self-injective.

Proof. By assumption, there exists e2 = e ∈ R such that RR = M ⊕ L, whereM = eR is an injective maximal right ideal of R and L = �1− e�R is a minimalright ideal of R. If LM �= 0, we claim that LR is injective, and then RR is injectivetoo. To check this, take u ∈ L such that uM �= 0, which implies that L = uM . Thusthere is a right R-epimorphism � M → L with ��x� = ux. Since L = �1− e�R isprojective, there exists T ⊆ M such that M = Ker���⊕ T with T � L. Hence L � Tis injective by the injectivity of MR, so R is injective. If LM = 0, then M = r�L�is a two-sided ideal of R. Put f = 1− e. For any 0 �= u ∈ Rf , u = uf and r�u� =M = r�f� by the maximality of M . Since R is right generalized semiregular, we havef ∈ lr�f� = lr�u� = P ⊕ L, where 0 �= P ⊆ Ru and Ru ∩ L is small in R. As in theproof of Theorem 2.3 (1), there exists g2 = g ∈ R such that uR = gR. Thereforeu = uru for some r ∈ R, which implies uf − uru = u− uru = 0 and �f − ru� ∈r�u� = r�f�, so f = fru. This shows that Rf = Ru is a minimal left ideal of R.Consequently, Re is a maximal left ideal of R. Since eR = M is a two-sided ideal of R,we get M = eR = Re by the maximality of Re. But R�R/M� � R�1− e� is projective,and so it is flat. Hence L � �R/M�R is injective, so RR = M ⊕ L is injective. �

By Theorem 2.7 and Crawley and Jónsson (1964, Corollary 24.22), we obtaineasily the following result.

Corollary 2.8. If R is a right MI ring, then R is a right generalized semiregular ringwith J�R� ⊆ Z�RR� and maximum condition on right annihilators if and only if R isquasi-Frobeniusean.

For a module MR we let E�M� be the injective hull of M . The module Mis said to be weakly injective (Jain and López-Permouth, 1990) if, for any finitelygenerated submodule N ⊆ E�M�, we have N ⊆ X � M for some X ⊆ E�M�. Clearly,injective rings are weakly injective, but the converse need not true (see Jain andLópez-Permouth, 1990). For a right generalized semiregular ring R, we prove that R

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is right weakly injective if and only if it is right self-injective. The following lemmawill be needed to prove the result above.

Lemma 2.9. Let R be a right generalized semiregular ring. If RR is an essentialsubmodule of X and X � RR, where X is a right R-module, then X = R.

Proof. Denote by � the isomorphism from RR onto X and write ��1� = b ∈ X,then Im� = X = bR. Note that 1 ∈ R ⊆ X, so there exists u ∈ R such that 1 =bu. Thus RR = buR, and r�u� = 0. By Corollary 1.5, u is semiregular. So thereexists h2 = h ∈ uR such that �1− h�u ∈ J�R�. Since u = hu+ �1− h�u, we haveuR = hR⊕ �1− h�uR. This implies that R = buR = bhR+ b�1− h�uR. It followsfrom �1− h�u ∈ J�R� that bhR = R. For any x = bhr1 = b�1− h�r2 with r1� r2 ∈ R,�−1�x� = hr1 = �1− h�r2, this implies x = bhr1 = bhhr1 = bh�1− h�r2 = 0. ThusbhR ∩ b�1− h�R = 0 and X = bR = bhR⊕ b�1− h�R. Since bhR = R is essential inX, we have X = bhR = R. �

Theorem 2.10. A ring R is right self-injective if and only if R is right generalizedsemiregular and RR is weakly injective.

Proof. One direction is obvious. Suppose that R is right generalized semiregularand RR is weakly injective. It is sufficient to prove that E�RR� ⊆ R. For anya ∈ E�RR�, R+ aR ⊆ E�RR�. Then there exists X ⊆ E�RR� such that R+ aR ⊆ Xand X � R by the assumption. By Lemma 2.9, X = R, which implies a ∈ R andR = E�RR�. �

We conclude this section with the following proposition.

Proposition 2.11. Let R be a semiprimitive ring. Then R is right generalizedsemiregular if and only if R is right AP-injective.

Proof. One direction is obvious. Conversely, for any a ∈ R, lr�a� = P ⊕ L, whereP ⊆ Ra and Ra ∩ L is small in R. By assumption, J�R� = 0. Hence Ra ∩ L ⊆ J�R� =0. Clearly, lr�a� = Ra+ L, so lr�a� = Ra⊕ L, and this implies that R is rightAP-injective. �

3. GENERALIZED SEMIREGULAR RINGS AND VON NEUMANNREGULAR RINGS

Recall that a ring R is called right nonsingular if Z�RR� = 0; R is reduced if ithas no nonzero nilpotent elements.

In this section, we characterize the regularity of right generalized semiregularrings. Some known results about right P-injective rings are extended.

Lemma 3.1. If R is a reduced and right generalized semiregular ring with J�R� ⊆Z�RR�, then R is strongly regular.

Proof. Since R is reduced, Z�RR� = 0. By Proposition 1.8, J�R� = 0, and hence Ris right AP-injective by Proposition 2.11. Let any 0 �= a ∈ R, then a2 �= 0 and there

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GENERALIZATIONS OF SEMIREGULAR RINGS 3457

exists a left ideal La2 of R such that lr�a2� = Ra2 ⊕ La2 . But r�a� = r�a2� since Ris reduced, thus a = ra2 + x with r ∈ R and x ∈ L, which implies a2 − ara2 = ax ∈Ra2 ∩ La2 = 0, so a2 = ara2. Then 1− ar ∈ l�a2� = l�a�, so a = ara and R is vonNeumann regular. This proves that R is strongly regular since R is reduced. �

Corollary 3.2. If R is a reduced and right AP-injective ring, then R is stronglyregular.

Lemma 3.3. Let c ∈ C�R�, where C�R� is the center of R. If c is von Neumannregular in R, then so is c in C�R�.

Proof. Let c = cdc with d ∈ R. Put u = dcd, then c = cuc. We claim thatu ∈ C�R�. In fact, for any x ∈ R, ux − xu ∈ r�c2� = r�c�, so c2�xd2 − d2x� =c�xu− ux� = 0, which implies xd2 − d2x ∈ r�c2� = r�c�. Thus xu− ux = xcd2 −cd2x = c�xd2 − d2x� = 0. This completes the proof of the lemma. �

Theorem 3.4. Let R be a semiprimitive and right generalized semiregular ring.Then the center C�R� of R is von Neumann regular.

Proof. By Proposition 1.8, R is right nonsingular, thus R has a von Neumannregular maximal right quotient ring S (see, e.g. Goodearl, 1976, Corollary 2.31).Consequently, the center C�S� of S is von Neumman regular by Lemma 3.3. Forany a ∈ C�R� ⊆ C�S�, there exists s ∈ C�S� such that a = asa = a2s = sa2. ThenC�R� is reduced. By Proposition 2.11, R is right AP-injective. Thus there existsa left ideal La2 of R such that a ∈ lr�a� = lr�a2� = Ra2 ⊕ La2 . As in the proof ofProposition 3.1, a is a von Neumann regular element in R. Using Lemma 3.3 again,we conclude that a is von Neumann regular in C�R�. �

Corollary 3.5. If R is a right nonsingular and right AP-injective ring, then the centerC�R� of is strongly regular.

A right R-module M is uniform if any two nonzero submodules of M havenonzero intersection. Equivalently, M �= 0 and every nonzero submodule of M isessential in M . A ring R is called right uniform if RR is uniform.

Proposition 3.6. Let R be a right nonsingular and right uniform ring. If R is a rightgeneralized semiregular ring with J�R� ⊆ Z�RR�, then R is a division ring.

Proof. For any 0 �= a ∈ R, r�a� = 0. If not, r�a� is essential in R since R is a rightuniform ring. Then a ∈ Z�RR� = 0, which is a contradiction. By Propositions 1.8and 2.11, R is right AP-injective. Thus there exists a left ideal La of R such thatR = lr�a� = Ra⊕ La. This gives that Ra is a direct summand of R, so a = ara forsome r ∈ R. Then 1− ra ∈ r�a� = 0, hence a is left invertible. Since r �= 0, similarly,we have 1 = br for some b ∈ R, which implies that R is a division ring. �

Proposition 3.7. Let R be a right generalized semiregular ring with J�R� ⊆ Z�RR�.If for any 0 �= a ∈ R, aR = Ra and r�a� = 0, then every left ideal I of R has exactlyone maximal essential extension.

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Proof. Let K1, K2 be two essential extensions of I . By Zorn’s Lemma andGoodearl (1976, Theorem 1.10), we must prove that K1 + K2 is essential extensionof I . It is sufficient to prove that R�k1 + k2� ∩ I �= 0 for any 0 �= k1 ∈ K1, 0 �= k2 ∈ K2

with �k1 + k2� �= 0.By assumption, r�k1� = r�k1 + k2� = 0. Thus �k1 + k2� ∈ lr�k1 + k2� = lr�k1� =

P ⊕ L, where P ⊆ Rk1 = k1R and Rk1 ∩ L is small in R. Write k1 + k2 = p1 +t, where p1 = rk1 = k1r

′ ∈ P. If k1r′ = 0, then k1 + k2 ∈ L, and thus p�k1 + k2� ∈

L ∩ pR = L ∩ Rp ⊆ L ∩ P = 0 for any p ∈ P. This shows that P�k1 + k2� = 0, andthus P = 0. Hence Rk1 ∩ L = Rk1 = k1R ⊆ J�R� = Z�RR� by Proposition 1.8. It is acontradiction with k1 �= 0 since r�k1� = 0. This implies 0 �= k1r

′. Note that p1�k1 +k2�− p2

1 = p1t ∈ p1R ∩ L = Rp1 ∩ L ⊆ P ∩ L = 0. We then have that p1�k1 + k2 −p1� = 0, which implies k1 + k2 = p1 ∈ Rk1. A similar argument shows that k1 ∈R�k1 + k2�. Thus Rk1 = R�k1 + k2�, and R�k1 + k2� ∩ I = I ∩ Rk1 �= 0 since I isessential in K1. �

Corollary 3.8. Let R be a right generalized semiregular ring with J�R� ⊆ Z�RR�.If for any 0 �= a ∈ R, aR = Ra and r�a� = 0, then every uniform left ideal of R iscontained in exactly one maximal uniform left ideal of R.

Proof. Let I be a uniform left ideal of R. Then there exists exactly one maximalessential extension K by Proposition 3.7. Then K is uniform. If K′ is a maximaluniform left ideal containing I , then I is essential in K′. Thus there exists a maximalessential extension K′′ of I such that K′ is essential in K′′, and hence K′′ = K.This shows that K′ also is essential in K, and so K′ = K since K is uniform. �

As usual, a ring Q is called a classical right quotient ring of R if (a)R ⊆ Q; (b)every non-zero-divisor of R is invertible in Q; (c) for any q ∈ Q� q = ab−1� a� b ∈ R� bbeing a non-zero-divisor.

Theorem 3.9. If R is a right generalized semiregular ring with J�R� ⊆ Z�RR�, andhas a classical right quotient ring Q, then Q is strongly regular if and only if R isreduced.

Proof. One direction is obvious. Suppose that R is reduced. By Lemma 3.1, R isstrongly regular. We claim that Q is reduced. Take any q = ab−1 ∈ Q with q2 = 0,where a� b ∈ R. Then ab−1ab−1 = 0 and so ab−1a = 0. Since Q is a classical rightquotient ring, there exist c� d ∈ R such that b−1a = dc−1, so ac = bd and adc−1 = 0.Thus dbd = dac and ad = 0, so da = 0 since R is reduced. Hence dbd = 0, bd =0 because R is reduced. Thus ac = 0� a = 0� q = ab−1 = 0. Therefore Q is reduced,and our claim is complete. For any q = ab−1 ∈ Q� a� b ∈ R, there exists r ∈ R suchthat a = ara = ab−1bra = qbra. This implies that q = ab−1 = qbrab−1 = qbrq, andbr ∈ R ⊆ Q. Hence Q is strongly regular. �

Recall that R is called a left (resp. right) GPP (i.e., generalized PP-ring)(Hirano, 1983) if for any a ∈ R, there exists a positive integer m such that Ram (resp.amR) is a projective R-module. A ring R is called �-regular if for any a ∈ R, thereexists a positive integer m such that am = amabam for some b ∈ R.

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GENERALIZATIONS OF SEMIREGULAR RINGS 3459

Definition. A ring R is said to satisfy �∗� if for any 0 �= a ∈ R, there exists apositive integer m with am �= 0 such that amR is projective.

Lemma 3.10. Let R be a right PP ring. If R is a right generalized semiregular ringwith J�R� ⊆ Z�RR�, then R is von Neumann regular.

Proof. For any a ∈ R, r�a� = eR for some idempotent e. Denote f = 1− e, thenr�f� = r�a� and a = af . As in the proof of Theorem 2.3(1), there exists g2 = g ∈ R

such that aR = gR. Thus a = ara for some r ∈ R, and hence R is von Neumannregular. �

The following lemma is clear.

Lemma 3.11. If a� b ∈ R and d = a− aba is von Neumann regular, then so is a.

Theorem 3.12. A ring R is von Neumann regular if and only if R is a rightgeneralized semiregular ring with J�R� ⊆ Z�RR� and that satisfies condition �∗�.

Proof. If R is von Neumann regular, then R is semiregular and hence rightgeneralized semiregular by Proposition 1.2. Conversely, suppose that R is a rightgeneralized semiregular ring with J�R� ⊆ Z�RR� and that satisfies condition �∗�. Leta be a nonzero element with a2 = 0. Then aR is projective. As in the proof ofLemma 3.10, a = ara for some r ∈ R. For any 0 �= a ∈ R, there exists a positiveinteger m with am �= 0 such that r�am� = eR, where e2 = e ∈ R. Let b = am, and asin the proof of Lemma 3.10 again, we have that b = brb for some r ∈ R. Thusam = amram. If m = 1, we are done. If m > 1, we denote c = ar, then am = am−1cam.This implies that �am−1 − am−1cam−1�2 = 0. If am−1 − am−1cam−1 �= 0, then am−1 −am−1cam−1 is von Neumann regular, which implies am−1 is von Neumann regular byLemma 3.11. Thus, in all cases, am−1 = am−1cam−1 for some c ∈ R. By induction, ais von Neumann regular. �

Corollary 3.13. A ring R is von Neumann regular if and only if R is a rightgeneralized semiregular ring with J�R� ⊆ Z�RR� and R is a right PP ring.

Example 3.14. There exists a commutative right GPP and generalized semiregularring with J�R� = Z�RR� which is not von Neumann regular. Let R = �q2 be thering of integers modulo q2, where q is a prime number. Then R is a commutativeAP-injective ring. Thus R is a commutative generalized semiregular ring with J�R� =Z�RR�. Note that R is �-regular, so R is a GPP-ring. But R is not von Neumannregular.

From the proof in Theorem 3.12, we easily obtain the following result.

Corollary 3.15. If R is a right GPP and right AP-injective ring, then R is �-regular.

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4. SEMI-�-REGULAR RINGS

Definition. An element a is semi-�-regular in a ring R if there exists a positiveinteger n such that an is semiregular. A ring R is called semi-�-regular if all itselements are semi-�-regular.

Clearly, a ring R is semi-�-regular if and only if, for any a ∈ R, there existse2 = e ∈ anR for some positive integer n such that �1− e�an ∈ J�R�. By Lemma 1.1,the following result is clear.

Proposition 4.1. The following are equivalent for an element a of a ring R and anypositive integer n:

(1) There exists e2 = e ∈ anR such that �1− e�an ∈ J�R�.(2) There exists e2 = e ∈ Ran such that an�1− e� ∈ J�R�.(3) There exists a von Neumann regular element b ∈ R with an − b ∈ J�R�.(4) There exists b ∈ R with b = banb and an − anban ∈ J�R�.(5) Ran lies over a summand of RR.

The following two results are easy consequences of Proposition 4.1.

Corollary 4.2. If R is a semi-�-regular, so also is every homomorphic image of R andso is every subring of the form eRe, where e2 = e.

Corollary 4.3. If a− b ∈ J�R� and b is semi-�-regular, then so is a.

Proof. Since a− b ∈ J�R�, we have a2 − b2 = a�a− b�+ �a− b�b ∈ J�R�. Assumethat an − bn ∈ J�R�, then an+1 − bn+1 = a�an − bn�+ �a− b�bn ∈ J�R�. Thus an −bn ∈ J�R� for any positive integer n. By condition (1) in Proposition 4.1, there exista positive integer m and a von Neumann regular element c such that bm − c ∈ J�R�.Then am − c = am − bm + bm − c ∈ J�R�. This proves that a is semi-�-regular. �

Theorem 4.4. The following are equivalent for a ring R:

(1) R is semi-�-regular.(2) R/J�R� is �-regular and idempotents can be lifted modulo J�R�.

Proof. �1� ⇒ �2� By Corollary 4.2, R/J�R� is semi-�-regular. But J�R/J�R�� = 0,so for any a ∈ R/J�R�, there exists a positive integer n such that an is regular.Thus R/J�R� is �-regular. Let a ∈ R, a2 − a ∈ J�R�. Then a3 − a2 = a�a2 − a� ∈J�R�, which implies a3 − a ∈ J�R�. By induction, an − a ∈ J�R� for any positiveinteger n ≥ 2. By the definition of semi-�-regular, we may choose a positive integern and e2 = e ∈ anR such that an − ean ∈ J�R�. Then ean − a ∈ J�R�. We claim thate− ane ∈ J�R�. If n = 1, we are done since e ∈ aR and a2 − a ∈ J�R�. If n > 1, lete = anr for some r ∈ R, then e− ane = anr − a2nr = �a− an+1�an−1r ∈ J�R�. Thisproves e− ane ∈ J�R�. Let f = e+ ean�1− e�. Then f 2 = f and f − a = e− eane+ean − a ∈ J�R�.

�2� ⇒ �1� Let a ∈ R/J�R� is �-regular, where a ∈ R. Then there exist apositive integer n and b ∈ R/J�R� such that an = anban. Since anb = �anb�2,

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GENERALIZATIONS OF SEMIREGULAR RINGS 3461

by assumption, there exists e2=e∈R such that anb−e∈J�R�. Then an−anban ∈ J�R�and an − anbean ∈ J�R�. Put u= 1− e+ any. Then u is a unit and u−1�anb�e= e. Letf = anbeu−1, so f 2 = f ∈ anR. Hence an − fan = an − anbeu−1an = an − anbean +anbean − anbeu−1an = an − anbean + anbeu−1�anbean − an� ∈ J�R�. This proves that(1) holds. �

A left R-module M is said to have the exchange property (Crawley andJónsson, 1964) if for any module X and decompositions

X = M ′ ⊕ Y = ⊕i∈I

Ni�

where M ′ � M , there exist submodules N ′i ⊆ Ni for each i such that

X = M ′ ⊕(⊕

i∈IN ′

i

)

If this condition holds for finite sets I (equivalently for �I� = 2), the module M is saidto have the finite exchange property. In Warfield (1972), R is called an exchange ringif RR has the (finite) exchange property. A module has the finite exchange propertyif and only if its endomorphism ring is an exchange ring. By Nicholson (1977), R isan exchange ring if and only if R/J�R� is an exchange ring and idempotents can belifted modulo J�R�.

The following example shows that there exists a semi-�-regular ring which isneither semiregular nor �-regular.

Example 4.5. (1) Let M = M2��2�, where �2 is the ring of integers modulo 2,and I = ( �2 �2

0 �2

). Then M is von Neumann regular and I is �-regular, but

not von Neumann regular, since J�I� = (0 �20 0

). Put R = ��x1� x2� � xn� x� x� � �

x1� x2� � xn ∈ M�x ∈ I�. Then J�R� = 0 since each nonzero ideal of R contains anonzero idempotent. Since R has I as a homomorphic image, we see that R is notvon Neumann regular. However, R is �-regular with component-wise operations.Hence R is semi-�-regular but not semiregular.

(2) Let C be a commutative von Neumann regular ring with no minimal idealand M a maximal ideal. Let R be the matrix ring of the form

(C CM C

). The ring R is not

von Neumann regular with J�R� = 0. By Snider (2001, Example 2), R is �-regular.Thus R is a semi-�-regular ring but not semiregular.

(3) Let R1 be the ring R in (2), R2 = �2��x be the formal power series ringover the ring �2, and R = R1 ⊕ R2. Then R2/J�R2� = �2, and hence R2 is semiregular.However, R2 is not �-regular since x is not �-regular. By (2), R1 is �-regular. Notethat R/J�R� = �2 ⊕ R1 since J�R1� = 0, it is easy to verify that R/J�R� is �-regular.By Corollary 4.6, R1 and R2 are exchange rings. Thus R = R1 ⊕ R2 is an exchangering, and hence idempotents can be lifted modulo J�R�. By Theorem 4.4, R is semi-�-regular. But R1 is not semiregular by (2), thus R is not semiregular by Nicholson(1976, Corollary 2.3) since R1 is a homomorphic image of R. Likewise, we see thatR2 is not �-regular. Therefore R is a semi-�-regular ring which is neither semiregularnor �-regular.

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Since �-regular rings are exchange rings (see Tuganbaev, 2002), the followingresult is clear.

Corollary 4.6. If R is a semi-�-regular ring, then R is an exchange ring.

Example 4.7. Let R�Q�L� be the ring in Nicholson (1977, Example 1.7), where Qdenotes the rational numbers, L denotes the ring of all rational fractions with odddenominators. Then R�Q�L� is a commutative exchange ring with J�R�Q�L�� = 0which is not von Neumann regular. Since R�Q�L� is reduced, we have that R�Q�L�is not �-regular. This shows that R�Q�L� is an exchange ring but not semi-�-regular.By Corollary 3.2, R�Q�L� is not AP-injective. Since R�Q�L� is semiprimitive,R�Q�L� is not generalized semiregular by Proposition 2.11. Therefore, there existsa commutative exchange ring which is neither semi-�-regular nor generalizedsemiregular.

The following example shows that a right generalized semiregular ring neednot be an exchange ring (and hence not a semi-�-regular ring). It also shows that theproperty of being right generalized semiregular is not preserved under the passageto quotients.

Example 4.8. Let R be the ring in Example 1.3(1). Then R is right generalizedsemiregular, and R/J�R� � � is not an exchange ring. Thus R is not an exchangering. If � is right generalized semiregular, then, by Proposition 2.11, � is right AP-injective since J��� = 0. Since � is reduced, this implies that � is strongly regularby Corollary 3.2, which is a contradiction. This shows that R/J�R� is not rightgeneralized semiregular.

Recall that a ring R is semiperfect if and only if R/J�R� is semisimple andidempotents can be lifted modulo J�R�. The following result is well-known for vonNeumann regular, semiperfect, semiregular, and exchange rings.

Corollary 4.9. In a semi-�-regular ring every primitive idempotent is local.

By Corollary 4.6 and Camillo and Yu (1994, Proposition 3), we obtain thefollowing result.

Corollary 4.10. A ring R is semiperfect if and only if R is a primitively finite andsemi-�-regular ring.

If M is a module, denote the endomorphism of M by End�M�. By Nicholson(1977, Theorem 2.1) and Corollary 4.6, the following result is immediate.

Corollary 4.11. Let M be a left R-module such that End�M� is semi-�-regular.Then M is a module with the finite exchange property.

Theorem 4.12. Let M be a projective left R-module, and let E = End�M�. ThenJ�E� = �� ∈ E � �M is small in M�. Moreover, E is semi-�-regular if and only if forevery � ∈ E there exists a positive integer n such that �nM lies over a summand of M .

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GENERALIZATIONS OF SEMIREGULAR RINGS 3463

Proof. Since M is projective, it is well-known that, J�E� = �� ∈ E � �M is smallin M�.

Assume that �nM lies over a direct summand of M for each � ∈ E anda positive integer n. Then M = P ⊕Q, where P ⊆ �nM and �nM ∩Q is smallin M . Take �2 = � ∈ E with �M = P and ker� = Q. Then ��n M → P is anepimorphism, so there exists � ∈ E such that ��n� = � since P is projective.Put � = �n��. Then �2 = � ∈ �nE ⊆ �M , �M ⊆ �nM and ker� = ker� = Q. Thus��n − ��n�M ⊆ �nM ∩Q, and hence �n − ��n ∈ J�E�. So E is semi-�-regular.

Conversely, assume that E is semi-�-regular. Given � ∈ E, there exist a positiveinteger n and �2 = � ∈ �nE such that �n − ��n ∈ J�E�. Then ��n − ��n�M is small inM . This proves that M = �M ⊕ �1− ��M , where �M ⊆ �nM and �nM ∩ �1− ��M =�1− ���nM is small in M . �

Proposition 4.13. Let M be a projective left R-module such that End�M� issemi-�-regular. Then J�M� is small in M .

Proof. If J�M�+ K = M and � M → M/K is the natural map, then thereexists � M → J�M� with �� = �. Then � = �� = · · · = ��n for any positiveinteger n. By assumption, there exists �2 = � ∈ �n�End�M�� such that �n − ��n ∈J�End�M��. Then �M ⊆ �nM ⊆ �M ⊆ J�M�. Since �M is projective, it follows fromProposition 2.7 in Bass (1960) that �M = 0. Hence �n ∈ J�End�M�� and so � = 0.This shows that J�M� is small in M . �

Mares (1963) calls a projective module semiperfect if every homomorphic imagehas a projective cover. By Nicholson (1976, Corollary 3.7) and Proposition 4.13, thefollowing result is immediate.

Corollary 4.14. If R is a semiperfect ring, then the following are equivalent for aprojective left R-module M:

(1) M is semiperfect.(2) End�M� is semiregular.(3) End�M� is semi-�-regular.(4) J�M� is small in M .

An ideal L in a ring R is left T -nilpotent if, given elements a1� a2� from L,there exists a positive integer n such that a1a2 · · · an = 0.

Corollary 4.15. Let R be a ring and let M be a countably generated free leftR-module. If End�M� is semi-�-regular, then J�R� is left T -nilpotent.

Proof. Let F be a countably generated free left R-module. Then J�F� is small inF by Proposition 4.13, thus J�R�F is small in F by Anderson and Fuller (1974,Corollary 15.18). Therefore by Anderson and Fuller (1974, Lemma 28.3), J�R� is leftT -nilpotent. �

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ACKNOWLEDGMENTS

The first author would like to thank the referee for his/her valuablesuggestions. He also would like to thank Professor Nanqing Ding and DoctorWeixing Chen for their useful advice. This work is supported by the DoctorateFoundation of China Education Ministry (Grant No. 20020284009).

REFERENCES

Anderson, F. W., Fuller, K. R. (1974). Rings and Categories of Modules. New York: Springer-Verlag.

Armendariz, E. P. (1974). A note on extensions of Baer and pp-rings. J. Austral. Math. Soc.18:470–473.

Bass, H. (1960). Finitistic dimension and a homological generalization of semiprimary rings.Trans. Amer. Math. Soc. 95:466–488. MR 28 �1212.

Birkenmeier, G. F. (1983). Idempotents and completely semiprime ideals. Comm. Algebra11:567–580.

Camillo, V. P., Yu, H. P. (1994). Exchange rings, units and idempotents. Comm. Algebra22(12):4737–4749.

Chen, J. L., Ding, N. Q. (2001). On regularity of rings. Algebra Colloquium 4:267–274.Crawley, P., Jónsson, B. (1964). Refinements for infinite direct decompositions of algebraic

systems. Pacific J. Math. 14:797–855. MR 30 � 49.Faith, C. (1976). Algebra (II). Ring Theory. New York, Berlin: Springer-Verlag.Goodearl, K. R. (1976). Ring Theory: Nonsingular Rings and Modules. Pure and Appl. Maths.,

Vol. 33. New York: Marcel Dekker.Hirano, Y. (1983). On generalized PP-rings. Math. J. Okayama Univ. 25:7–11.Jain, S. K., López-Permouth, S. R. (1990). Rings whose cyclics are essentially embeddable

in projectives. J. Algebra 128:257–269.Kaplansky, I. (1968). Rings of Operators. New York: W. A. Benjamin.Mares, E. A. (1963). Semiperfect modules. Math. Zeitschr 82:347–360.Mohammed, S. H., Müller, B. J. (1990). Continuous and Discrete Modules. Cambridge Univ.

Press.Nicholson, W. K. (1976). Semiregular modules and rings. Can. J. Math. XXXIII(5):

1105–1120.Nicholson, W. K. (1977). Lifting idempotents and exchange rings. Trans. Amer. Math. Soc.

229:269–278.Nicholson, W. K., Yousif, M. F. (1995). Principally injective rings. J. Algebra 174:77–93.Nicholson, W. K., Yousif, M. F. (1997). Mininjective rings. J. Algebra 187:548–578.Nicholson, W. K., Yousif, M. F. (2001). Weakly continuous and C2-rings. Comm. Algebra

29(6):2429–2446.Page, S. S., Zhou, Y. Q. (1998). Generalizations of principally injective rings. J. Algebra

206:706–721.Shepherdson, J. C. (1951). Inverses and zero divisors in matrix rings. Proc. London. Math.

Soc. 1(3):71–85.Snider, R. L. (2001). Semiprimitive �-regular rings of bounded index are left max rings.

Comm. Algebra 29(12):5433–5437.Tuganbaev, A. (2002). Rings Close to Regular. Dordrecht: Kluwer Academic.Utumi, Y. (1965). On continuous and self-injective rings. Trans. Amer. Math. Soc. 118:

158–173.

Dow

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ded

by [

Flor

ida

Stat

e U

nive

rsity

] at

09:

18 1

1 M

ay 2

013

GENERALIZATIONS OF SEMIREGULAR RINGS 3465

Warfield, R. B. Jr. (1972). Exchange rings and decompositions of modules. Math. Ann.199:31–36. MR 48 � 11218.

Xiao, G. S., Yin, X. B., Tong, W. T. (2003). A note on AP-injective rings. J. Math. Researchand Exposition 23(2):211–216.

Yue Chi Ming, R. (1995). On injectivity and P-injectivity(II). Soochow J. Math. 21(4):401–412.

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