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 that Ris a strongly regular ring if and only if Ris 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）of R.l（X）（r（X））denotes the left （right）annihilator of Xin R.If X＝｛a｝，we will write it for l（a）（r（a））.Nil（R）denotes the biggest nil ideal.Recall that Ris said to be left（right）nonsingular if Z＝0（Y＝0）.A ring Ris called（von Neumann）regular[1]if for any a∈R，there exists b∈Rsuch that a＝aba.A ring Ris called strongly regular[2]if for any a∈R，there exists b∈Rsuch that a＝a2b.

Definition 1 A ring Ris called N-semiregular，if for all a∈R，there exists an idempotent e∈aR such that（1－e）a∈Nil（R）.

Definition 2 a∈Ris call generalized N-semiregular if there exists two right ideals P，Lof R，such that rl（a）＝P⊕L with P?aRand aR∩Lis nil.A ring Ris called generalized N-semiregular if each of its elements is generalized N-semiregular.

Proposition 1 If Ris a left AP-injective or N-semiregular ring，then Ris a generalized N-semiregular.

Proof 1）Let Rbe a left AP-injective ring.Then rl（a）＝aR⊕Lafor any a∈R，where Lais a right ideal of R.Note that aR∩Lais nil，thus Ris generalized N-semiregular.

2）Let Rbe a N-semiregular ring and a∈R.Then there exists e2＝e∈aRsuch that（1－e）a∈Nil（R）.Thus R＝eR⊕（1－e）R，where eR?aRand（1－e）aRis nil.Note that aR?rl（a）.By modular law，we have rl（a）＝rl（a）∩R＝rl（a）∩（eR ⊕（1－e）R）＝eR ⊕（rl（a）∩（1－e）R）and aR∩（rl（a）∩（1－e）R）＝aR∩（1－e）R＝（1－e）aRis nil.Thus Ris generalized N-semiregular.

Lemma 1 If Ris a generalized N-semiregular ring and for every a∈Rthere exists e2＝e∈Rsuch that l（a）＝l（e），then Ris N-semiregular.

Proof For any a∈R.Since Ris generalized N-semiregular，there exists two right ideals P，Lof R such that rl（a）＝P⊕L，where P?aRand aR∩Lis nil.Thus eR＝P⊕Lsince l（e）＝l（a）.As l（e）＝l（a），we have rl（a）＝rl（e）＝eR，this gives a＝ea.Take e＝g＋t，where g＝ar∈P?aRand t∈L.Then a＝ea＝ga＋ta＝ara＋taand ar＝arar＋tar which implies that ar－arar＝tar∈P∩L＝0and a－ara＝ta∈aR∩L?Nil（R）.This shows that g2＝g∈aRand（1－g）a∈Nil（R），a is N-semiregular.Hence R is N-semiregular.

Corollary 1 Let Rbe a generalized N-semiregular element.If Ra?Re，where e2＝e，then ais N-semiregular.

Proof For any a∈R，letφbe the isomorphism of Ra onto Re.By（[3]，Lemma 2.12），there exists an idempotent fof Rsuch that l（a）＝l（f）.By Lemma 1，ais N-semiregular.

If rl（a）is a direct summand of R，then there exists e2＝e∈Rsuch that rl（a）＝eR.Thus l（a）＝lrl（a）＝l（e）and so the following results is immediate.

Corollary 2 If rl（a）is a direct summand of Rfor any a∈Rand Ris generalized N-semiregular，then Ris N-semiregular.

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

Corollary 3 Let Rbe a left PP ring.If Ris generalized N-semiregular，then Ris N-semiregular.

From （[3]，Corollary 2.3），we see that if Ris a right AP-injective ring then J＝Y(jié).

Proposition 2 If Ris a generalized N-semiregular ring，then Z?J.

Proof Let 0≠a∈Z.For any b∈R，we have that ba∈Z.Put u＝1－ba.Then u≠0，and l（u）＝0 since l（ba）∩l（u）＝0.Thus R＝rl（u）＝P⊕L，where P?uRand uR∩Lis nil.We have that P＝eR with e2＝e∈R.Hence it is sufficient to prove that e＝1.If not，then there exists 0≠r（1－e）∈R（1－e）∩l（ba）since l（ba）is essential in R.This gives r（1－e）u＝r（1－e）.Put u＝es＋t for some s∈R，t∈L.Then r（1－e）u＝r（1－e）t.Hence r（1－e）＝r（1－e）t and therefor r（1－e）（1－t）＝0.Note that t＝ues∈uR∩L?Nil（R）.Then 1－t is unit，which implies r（1－e）＝0，a contradiction.So we have that e＝1and R＝P＝uR.Thus a∈J.

Corollary 4 If Ris a N-semiregular ring，then Z?J.

A ring Ris called left mininjective，if every isomorphism between simple left ideals is given by multiplication by an element of R[6].Equvivalently，if rl（K）＝Kfor every right ideals K＝kRfor which Rkis simple.

Theorem 1 If Ris a generalized N-semiregular，left miniinjective ring and Soc（RR）is an essential submodule ofRR，then J＝Z.

Proof Since Ris left mininjective，Soc（RR）?Soc（RR）by[6].By（[6]，Proposition 2.8（1）），J?Z.Since Ris generalized N-semiregular，By Proposition 2，Z?J.Thus J＝Z.

An idempotent element e∈Ris left（resp.right）semicentral[7]in Rif Re＝eRe（resp.eR＝eRe）.

Proposition 3 Let R be a generalized N-semiregular ring.If an idempotent e of Ris left semicentral，then eReis generalized N-semiregular.

Proof Let a∈eRe，then there exists two right ideals Pand Lof R，such that rl（a）＝P ⊕L，where P?aRand aR∩Lis nil.We claim that reReleRe（a）＝Pe⊕Le.In fact，Pe∩Le?P∩L＝0.Take any y∈Pe?ePe，where y＝y(tǒng)1e，y1∈P?rl（a）.Then for any x∈leRe（a）?l（a），xy1＝0，which gives xy＝xy1e＝0.Hence y∈reReleRe（a），Pe?reReleRe（a）.Similarly，we have Le?reReleRe（a），on the other hand，take x∈reReleRe（a），then for any y∈l（a），we have eyea＝0since a∈eRe.So eyex＝0，which gives yx＝eyex＝0since x∈eRe and eis left semicentral.Thus reReleRe（a）?rl（a）.Take x＝s＋t，where s∈P，t∈L.Then x＝xe＝se＋te∈Pe＋Le.This shows that reReleRe（a）＝Pe⊕Le.

It remains to prove that aeRe∩Leis nil.Since Pe?aRe＝aeRe.Since e is left semicentral，we have aeRe∩Le?e（aeRe∩Le）e?eRe.But aeRe∩Le?aR∩Lis nil.Thus eReis generalized N-semiregular.

Theorem 2 Let e be an idempotent of Rsuch that ReR＝R.If Ris a generalized N-semiregular ring，eReis generalized N-semiregular.

Proof Let a∈eRe，then there exists two right ideals Pand Lof Rsuch that rl（a）＝P⊕L，where P?aRand aR∩Lis nil.We claim that reReleRe（a）＝ePe⊕eLe.Since 1－e∈l（a），we see that（1－e）t＝0for any t∈L，which implies L＝eL.Similarly，P＝eP，Thus ePe∩eLe＝0.Clearly，ePe?reReleRe（a）and eLe?reReleRe（a）since eP＝Pand eL＝L.On the other hand，take x∈reReleRe（a），and write 1＝∑n

i＝1aiebifor some ai，biin R.Then for any y∈l（a），we get ebiyea＝ebiya＝0for each i.This implies that ebiyex＝0for each i，which give yx＝y(tǒng)ex＝∑ni＝1aiebiyex＝0since x∈eRe.Thus reReleRe（a）?rl（a）.Take x＝s＋t，where s∈P，t∈L.Hence x＝exe＝ese＋ete∈ePe＋eLe.This shows that reReleRe（a）＝ePe⊕eLe，and the claim is complete.It remains to prove that aeRe∩eLeis nil since ePe?aeRe.Since L＝eL，we have aeRe∩eLe?aR∩Lis nil.Thus eReis generalized N-semiregular.

Proposition 4 If Nil（R）＝0，then Ris a generalized N-semiregular ring if and only if Ris a left AP-injective ring.

Proof One direction is obvious.Conversely，for any a∈R，rl（a）＝P⊕L，where P?aRand aR∩Lis nil.By assumption，Nil（R）＝0，hence aR∩L?Nil（R）＝0.It is clearly that rl（a）＝aR＋L.So rl（a）＝aR ⊕L，and this implies that Ris a left AP-injective ring.

Proposition 5 Ris a reduced and generalized N-semiregular ring if and only if Ris a strongly regular.

Proof One direction is obvious.Conversely，since Ris reduced，Z＝0.By Proposition 4，Ris left AP-injective and Z＝J＝0.Let any 0≠a∈R，then a2≠0and there exists a right ideal La2of Rsuch that rl（a2）＝a2R⊕La2.But l（a）＝l（a2）since Ris reduced，thus a＝a2r＋x with r∈Rand x∈La2，which implies a2－a2ra＝xa∈a2R⊕La2＝0，so a2＝a2ra.Then 1－ra∈r（a2）＝r（a），so a＝ara.This proved that Ris a strongly regular ring.

Lemma 2[8]Let c∈C（R），where C（R）is center of R.If cis von Neumann regular in R，then so is cin C（R）.

Theorem 3 Let Rbe a semiprimitive and generalized N-semiregular ring，then the center C（R）of R is von Neumann regular.

Proof By Proposition 2，Ris left nonsingular，thus Rhas von Neumann regular maximal left quotient rings（see，e.g.Corollary 2.31[1]）.Consequently，the center C（S）of Sis von Neumman regular by Lemma 2.For any a∈C（R）?C（S），there exists s∈C（S）such that a＝asa＝a2s＝sa2.Then C（R）is reduced.By Proposition 4，Ris left AP-injective.Thus there exists a right ideal La2of Rsuch that a∈rl（a）＝rl（a2）＝a2R ⊕La2.As in the proof of Proposition 5，ais a von Neumann regular element in R.Using Lemma 2again，we conclude that ais von Neumann regular in C（R）.

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廣義N-半正則環(huán)

殷曉斌，王 瑞
（安徽師范大學數(shù)學與計算機科學學院，安徽 蕪湖 241000）

介紹了AP－內(nèi)射環(huán)的推廣－廣義N－半正則環(huán)，主要得到了R是強正則環(huán)當且僅當R是約化的廣義N－半正則環(huán).文章研究了廣義N－半正則環(huán)的性質(zhì)且對AP－內(nèi)射環(huán)的某些結(jié)果進行了推廣.

AP-內(nèi)射環(huán)；廣義N－半正則環(huán)；強正則環(huán)

O153.3 MSC2010：16E50Article character：A

1674-232X（2011）02-0097-04

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

10.3969／j.issn.1674-232X.2011.02.001