Generalized N-Semiregular Rings

YIN Xiao-bin, WANG Rui

(College of Mathematics and Computer Science, Anhui Normal University, Wuhu 241000, China)

Generalized N-Semiregular Rings

YIN Xiao-bin, WANG Rui

(College of Mathematics and Computer Science, Anhui Normal University, Wuhu 241000, China)

This article introduces a generalization of AP-injective rings-generalized N-semiregular rings, and mainly obtains thatRis a strongly regular ring if and only ifRis a reduced and generalied N-semiregular ring. The paper also studies some properties of generalized N-semiregular rings are extends some results about AP-injective rings.

AP-injective rings; generalized N-semiregular rings; strongly regular rings

Throughout this article,Rdenotes an associative ring with identity and modules are unitary. J, Z (resp.Y), Soc(RR)(Soc(RR)) will denote respectively the Jacobson radical, the left singular ideal(resp. right singular ideal) the left socle (the right socle) ofR.l(X)(r(X)) denotes the left (right) annihilator ofXinR. IfX={a}, we will write it forl(a)(r(a)). Nil(R) denotes the biggest nil ideal. Recall thatRis said to be left (right) nonsingular ifZ=0(Y=0). A ringRis called (von Neumann) regular[1]if for anya∈R, there existsb∈Rsuch thata=aba. A ringRis called strongly regular[2]if for anya∈R, there existsb∈Rsuch thata=a2b.

Definition1A ringRis called N-semiregular, if for alla∈R, there exists an idempotente∈aRsuch that (1-e)a∈Nil(R).

Definition2a∈Ris call generalized N-semiregular if there exists two right idealsP,LofR, such thatrl(a)=P⊕LwithP⊆aRandaR∩Lis nil. A ringRis called generalized N-semiregular if each of its elements is generalized N-semiregular.

Proposition1IfRis a left AP-injective or N-semiregular ring, thenRis a generalized N-semiregular.

Proof1) LetRbe a left AP-injective ring. Thenrl(a)=aR⊕Lafor anya∈R, whereLais a right ideal ofR. Note thataR∩Lais nil, thusRis generalized N-semiregular.

2) LetRbe a N-semiregular ring anda∈R. Then there existse2=e∈aRsuch that (1-e)a∈Nil(R). ThusR=eR⊕(1-e)R, whereeR⊆aRand (1-e)aRis nil. Note thataR⊆rl(a). By modular law, we haverl(a)=rl(a)∩R=rl(a)∩(eR⊕(1-e)R)=eR⊕(rl(a)∩(1-e)R) andaR∩(rl(a)∩(1-e)R)=aR∩(1-e)R=(1-e)aRis nil. ThusRis generalized N-semiregular.

Lemma1IfRis a generalized N-semiregular ring and for everya∈Rthere existse2=e∈Rsuch thatl(a)=l(e), thenRis N-semiregular.

ProofFor anya∈R. SinceRis generalized N-semiregular, there exists two right idealsP,LofRsuch thatrl(a)=P⊕L, whereP⊆aRandaR∩Lis nil. ThuseR=P⊕Lsincel(e)=l(a). Asl(e)=l(a), we haverl(a)=rl(e)=eR, this givesa=ea. Takee=g+t, whereg=ar∈P⊆aRandt∈L.Thena=ea=ga+ta=ara+taandar=arar+tarwhich implies thatar-arar=tar∈P∩L=0 anda-ara=ta∈aR∩L⊆Nil(R). This shows thatg2=g∈aRand (1-g)a∈Nil(R), a is N-semiregular. HenceRis N-semiregular.

Corollary1LetRbe a generalized N-semiregular element. IfRa≅Re, wheree2=e, thenais N-semiregular.

ProofFor anya∈R, letφbe the isomorphism ofRaontoRe. By ([3], Lemma 2.12), there exists an idempotentfofRsuch thatl(a)=l(f). By Lemma 1,ais N-semiregular.

Ifrl(a) is a direct summand ofR, then there existse2=e∈Rsuch thatrl(a)=eR.Thusl(a)=lrl(a)=l(e) and so the following results is immediate.

Corollary2Ifrl(a) is a direct summand ofRfor anya∈RandRis generalized N-semiregular, thenRis N-semiregular.

A ringRto be Baer[4]if the right annihilator of every non-empty subset ofRis generated, as a right ideal, by an idempotent. This definition is left-right symmetric. A ringRis called right (resp. left) PP[5]if every principal right (resp.left) ideal ofRis projective.Ris a right PP ring if and only if for everya∈Rthere exists an idempotente∈Rsuch thatr(a)=eR. Clearly, every Baer ring is right and left PP ring.

Corollary3LetRbe a left PP ring. IfRis generalized N-semiregular, thenRis N-semiregular.

From ([3], Corollary 2.3), we see that ifRis a right AP-injective ring thenJ=Y.

Proposition2IfRis a generalized N-semiregular ring, thenZ⊆J.

ProofLet 0≠a∈Z. For anyb∈R, we have thatba∈Z. Putu=1-ba. Thenu≠0, andl(u)=0 sincel(ba)∩l(u)=0. ThusR=rl(u)=P⊕L, whereP⊆uRanduR∩Lis nil. We have thatP=eRwithe2=e∈R. Hence it is sufficient to prove thate=1. If not, then there exists 0≠r(1-e)∈R(1-e)∩l(ba) sincel(ba) is essential inR. This givesr(1-e)u=r(1-e). Putu=es+tfor somes∈R,t∈L. Thenr(1-e)u=r(1-e)t. Hencer(1-e)=r(1-e)tand thereforr(1-e)(1-t)=0. Note thatt=u-es∈uR∩L⊆Nil(R). Then 1-tis unit, which impliesr(1-e)=0, a contradiction. So we have thate=1 andR=P=uR. Thusa∈J.

Corollary4IfRis a N-semiregular ring, thenZ⊆J.

A ringRis called left mininjective, if every isomorphism between simple left ideals is given by multiplication by an element ofR[6]. Equvivalently, ifrl(K)=Kfor every right idealsK=kRfor whichRkis simple.

Theorem1IfRis a generalized N-semiregular, left miniinjective ring and Soc(RR) is an essential submodule ofRR, thenJ=Z.

ProofSinceRis left mininjective , Soc(RR)⊆Soc(RR) by [6]. By ([6], Proposition 2.8(1)),J⊆Z. SinceRis generalized N-semiregular, By Proposition 2,Z⊆J. ThusJ=Z.

An idempotent elemente∈Ris left (resp.right) semicentral[7]inRifRe=eRe(resp.eR=eRe).

Proposition3LetRbe a generalized N-semiregular ring. If an idempotenteofRis left semicentral, theneReis generalized N-semiregular.

ProofLeta∈eRe, then there exists two right idealsPandLofR, such thatrl(a)=P⊕L, whereP⊆aRandaR∩Lis nil. We claim thatreReleRe(a)=Pe⊕Le. In fact,Pe∩Le⊆P∩L=0. Take anyy∈Pe⊆ePe, wherey=y1e,y1∈P⊆rl(a). Then for anyx∈leRe(a)⊆l(a),xy1=0, which givesxy=xy1e=0. Hencey∈reReleRe(a),Pe⊆reReleRe(a). Similarly, we haveLe⊆reReleRe(a), on the other hand, takex∈reReleRe(a), then for anyy∈l(a), we haveeyea=0 sincea∈eRe. Soeyex=0, which givesyx=eyex=0 sincex∈eReandeis left semicentral. ThusreReleRe(a)⊆rl(a). Takex=s+t, wheres∈P,t∈L. Thenx=xe=se+te∈Pe+Le. This shows thatreReleRe(a)=Pe⊕Le.

It remains to prove thataeRe∩Leis nil.SincePe⊆aRe=aeRe. Since e is left semicentral, we haveaeRe∩Le⊆e(aeRe∩Le)e⊆eRe. ButaeRe∩Le⊆aR∩Lis nil. ThuseReis generalized N-semiregular.

Theorem2Letebe an idempotent ofRsuch thatReR=R. IfRis a generalized N-semiregular ring,eReis generalized N-semiregular.

Proposition4If Nil(R)=0, thenRis a generalized N-semiregular ring if and only ifRis a left AP-injective ring.

ProofOne direction is obvious. Conversely, for anya∈R,rl(a)=P⊕L, whereP⊆aRandaR∩Lis nil. By assumption, Nil(R)=0, henceaR∩L⊆Nil(R)=0. It is clearly thatrl(a)=aR+L. Sorl(a)=aR⊕L, and this implies thatRis a left AP-injective ring.

Proposition5Ris a reduced and generalized N-semiregular ring if and only ifRis a strongly regular.

ProofOne direction is obvious. Conversely, sinceRis reduced,Z=0. By Proposition 4,Ris left AP-injective andZ=J=0. Let any 0≠a∈R, thena2≠0 and there exists a right idealLa2ofRsuch thatrl(a2)=a2R⊕La2. Butl(a)=l(a2) sinceRis reduced, thusa=a2r+xwithr∈Randx∈La2, which impliesa2-a2ra=xa∈a2R⊕La2=0, soa2=a2ra. Then 1-ra∈r(a2)=r(a), soa=ara. This proved thatRis a strongly regular ring.

Lemma2[8]Letc∈C(R), whereC(R) is center ofR. Ifcis von Neumann regular inR, then so iscinC(R).

Theorem3LetRbe a semiprimitive and generalized N-semiregular ring, then the centerC(R) ofRis von Neumann regular.

ProofBy Proposition 2,Ris left nonsingular, thusRhas von Neumann regular maximal left quotient rings (see, e.g.Corollary 2.31[1]). Consequently, the centerC(S) ofSis von Neumman regular by Lemma 2. For anya∈C(R)⊆C(S), there existss∈C(S) such thata=asa=a2s=sa2. ThenC(R) is reduced. By Proposition 4,Ris left AP-injective. Thus there exists a right idealLa2ofRsuch thata∈rl(a)=rl(a2)=a2R⊕La2. As in the proof of Proposition 5,ais a von Neumann regular element inR. Using Lemma 2 again, we conclude thatais von Neumann regular inC(R).

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广义N-半正则环

殷晓斌,王 瑞

(安徽师范大学数学与计算机科学学院,安徽 芜湖 241000)

介绍了AP-内射环的推广-广义N-半正则环,主要得到了R是强正则环当且仅当R是约化的广义N-半正则环.文章研究了广义N-半正则环的性质且对AP-内射环的某些结果进行了推广.

AP-内射环;广义N-半正则环;强正则环

10.3969/j.issn.1674-232X.2011.02.001

date： 2010-09-10

Supported by National Natural Science Foundation of China (10871106,10901002) and NSF of Anhui Province Education Committee (KJ2008A026).

Biography： YIN Xiao-bin(1972—), male, born in Zongyang, Anhui province, associate professor, engaged in algebra. E-mail: xbyinzh@gmail.com

O153.3MSC201016E50ArticlecharacterA

1674-232X(2011)02-0097-04