Annihilators in Unit-Regular Rings

CHEN Huan-yin

(College of Science, Hangzhou Normal University, Hangzhou 310036, China)

1 Introduction

A ringRis unit-regular provided that for anyx∈R, there exists an invertibleu∈Rsuch thatx=xux(cf.[1]). The class of unit-regular rings is very large. For instance, the endomorphism ring of finitely dimensional vector space over a division ring is unit-regular. We say that a ringRhas stable range one (i.e., 1-fold stable ring, stable rank one) provided thataR+bR=Rwitha,b∈Rimplies that there exists ay∈Rsuch thata+by∈U(R)(cf.[2-5]). The concept of stable range one was initiated by Bass in his investigation of stability of the general linear group in algebraK-theory and was well known as the Bass’s first stable range condition. For example,K1(R)?U(R)/V(R) ifRis unit-regular. TheK2groups of such rings were also studied in the literatures. As is well known, a regular ringRis unit-regular if and only if it has stable range one, where a ringRis regular in case for anyx∈R, there exists ay∈Rsuch thatx=xyx. Also it is worth to note that a regular ringRis unit-regular if and only if for any finitely generated projective rightR-modulesA,BandC,A⊕B?A⊕C?B?C. That is, unit-regularity over regular rings is equivalent to cancellation problem for finitely generated projective right modules.

LetRbe an associative ring with identity, and leta∈R. We user(a) and l(a) to stand for the sets {r∈R|aa=0} and {r∈R|ra=0}, respectively. As usual,r(a)(l(a)) is called the right (left) annihilator ofa∈R. Unit-regular rings have been extensively studied by many authors (cf. [6-8]). We prove, in this note, that every unit-regular ring can be characterized by means of annihilators.

2 The Main Results

Lemma1LetRbe a regular ring, and leta,b,d∈R. Then the following are equivalent:

(1)aR+bR=dR.

(2) l(a)∩l(b)=l(d).

Proof(1)?(2) Assume thataR+bR=dR. Thend=ax+byfor somex,y∈R; hence, l(a)∩l(b)?l(d). For anyz∈l(d),zd=0, and soza=zb=0. Thus,z∈l(a)∩l(b). Hence, we conclude that l(a)∩l(b)=l(d).

(2)?(1) Assume that l(a)∩l(b)=l(d). SinceRis regular, there exists ac∈Rsuch thataR+bR=cR. Writed=dxdandc=cyc. As 1-dx∈l(d), (1-dx)a=(1-dx)b=0. Thus,a=dxaandb=dxb. Hence,aR+bR?dR. As (1-cy)c=0, we see that (1-cy)a=(1-cy)b=0. That is, 1-cy∈l(a)∩l(b). This implies that 1-cy∈l(d), and sod=cyd∈aR+bR. Hence,aR+bR=dR, as required.

LetRbe a regular ring, and leta,b,d∈R. Analogously, we deduce thatRa+Rb=Rdif and only ifr(a)∩r(b)=r(d).

Theorem1LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)∩l(b)=l(d), there existsy∈Rsuch that l(a)∩l(b)=l(a+by).

(3) Wheneverr(a)∩r(b)=r(d), there existsz∈Rsuch thatr(a)∩r(b)=r(a+zb).

Proof(1)?(2) Suppose thatRis unit-regular. Whenever l(a)∩l(b)=l(d), it follows from Lemma 1 thataR+bR=dR. Thus,ax+by′=d,a=ds,b=dtfor somex,y′,s,t∈R. Hence,dsx+dty′=d. Letw=sx+ty′-1. Thensx+ty′-w=1 withdw=0. Clearly,Ris unit-regular; whence,u:=s+(ty′-w)r∈U(R) for somer∈R. As a result, we derive thatdu=ds+dty′r=a+by′r. This infers that l(d)=l(a+by′r), i.e., l(a)∩l(b)=l(a+by), wherey=y′r.

(2)?(1) Assume thataR+bR=R. Then l(a)∩l(b)=l(1) from Lemma 1. By hypothesis, there exists ay∈Rsuch that l(a+by)=l(1)=0. SinceRis regular, we have ac∈Rsuch thata+by=(a+by)c(a+by); hence, 1-(a+by)c∈l(a+by)=0. Thus,a+by∈Ris right invertible. Givenx′y′=1, theny′R+(1-y′x′)R=R. By the proceeding discussion, we can findz′,v′∈Rsuch thaty′+(1-y′x′)z′=u′ andu′v′=1. Thus, 1=x′y′=x′(y′+(1-y′x′)z′)=x′u′, and sox′=(x′u′)v′=v′. Hence,v′u′=1. This infers thatx′∈U(R), and theny′∈U(R). So,y′x′=1 fromy′x′y′x′=y′x′. That is,Ris directly finite. Consequently,a+by∈U(R), i.e.,Ris unit-regular.

(1)?(3) Analogously, we deduce thatRis unit-regular if and only ifr(a)∩r(b)=r(d) implies that there existsz∈Rsuch thatr(a)∩r(b)=r(a+zb).

Corollary1LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)∩l(b)=0, there existsy∈Rsuch that l(a+by)=0.

(3) Wheneverr(a)∩r(b)=0, there existsz∈Rsuch thatr(a+zb)=0.

Proof(1)?(2) Whenever l(a)∩l(b)=0, then l(a)∩l(b)=l(1). By virtue of Theorem 1, there exists ay∈Rsuch that l(a)∩l(b)=l(a+by), and so l(a+by)=0, as required.

(2)?(1) GivenaR+bR=R, then l(a)∩l(b)=l(1) from Lemma 1. By hypothesis, we can find ay∈Rsuch that l(a+by)∩l(0)=0=l(1). By using Lemma 1 again, (a+by)R+0×R=R. This implies thata+by∈Ris right invertible. Similarly to Theorem 1,Ris directly finite, and thereforeRis unit-regular.

(1)?(3) is proved in the same manner.

Leta,b∈EndZ(Z) be defined by left multiplication by 2 and 5, respectively. We note thatr(a)∩r(b)=0, whiler(a+by)≠0 for anyy∈R. In this case,EndZ(Z) is not unit-regular.

Theorem2LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)∩l(b)=l(d), there existu∈U(R),y∈Rsuch thatau+by=d.

(3) Wheneverr(a)∩r(b)=r(d), there existu∈U(R),y∈Rsuch thatua+yb=d.

Proof(1)?(2) Whenever l(a)∩l(b)=l(d), by virtue of Theorem 1, l(a+by)=l(d) for ay∈R. In light of Lemma 1, (a+by)R=dR. Writea+by=dsand (a+by)t=d. Clearly,Rhas stable range one. It follows fromst+(1-st)=1 thatu:=s+(1-st)z∈U(R). Hence,du=ds+d(1-st)z=a+by. Thereforeau-1+byu-1=d, as needed.

(2)?(1) Whenever l(a)∩l(b)=0, we get l(a)∩l(b)=l(1). Hence, there existsy∈Rsuch thata+by∈U(R). This implies thatr(a+by)=0. Thus,Ris unit-regular by Corollary 1.

(1)?(3) is symmetric.

Corollary2LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)∩l(b)=0, there existsy∈Rsuch thata+by∈U(R).

(3) Wheneverr(a)∩r(b)=0, there existsz∈Rsuch thata+zb∈U(R).

Proof(1)?(2) l(a)∩l(b)=0 if and only if l(a)∩l(b)=l(1). In light of Theorem 2, we are through.

(1)?(3) is similar to the preceding proof.

Lemma2LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Wheneverx=xyx, there exists somea∈Rsuch that 1+xa,y+a∈U(R).

ProofAs in the proof of [9, Proposition 2.5].

Recall that a ringRsatisfies unit 1-stable range provided thataR+bR=Rwitha,b∈Rimplies that there exists au∈U(R) such thata+bu∈U(R). As is well known, a ringRsatisfies unit 1-stable range if and only if for anyx,y∈R, there exists au∈U(R) such that 1+x(y-u)∈U(R) (cf. [10]). This fact should be contract to the following: a regular ringRis unit-regular if and only if wheneverx=xyx, there exists au∈U(R) such that 1+x(y-u)∈U(R).

Theorem3LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Wheneverx=xyx, there exist ae∈l(x) and au∈U(R) such thaty=e+u.

(3) Wheneverx=xyx, there exist ae∈r(x) and au∈U(R) such thaty=e+u.

Proof(1)?(3) Suppose thatx=xyx. Obviously,Rhas stable range one. Sinceyx+(1-yx)=1, we can find az∈Rsuch thatu:=y+(1-yx)z∈U(R). Thus,y=(yx-1)z+u. Lete=(yx-1)z. Theny=e+u, wheree∈r(x) andu∈U(R).

(3)?(1) Given any regularx∈R, there exists ay∈Rsuch thatx=xyx. Hence, there exist ae∈r(x) and au∈U(R) such thaty=e+u. Leta=-e. Theny+a∈U(R) and 1+xa=1∈U(R). In light of Lemma 2,Ris unit-regular.

(1)?(2) is symmetric.

Lemma3LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)=l(b), there existsu∈U(R) such thata=bu.

(3) Wheneverr(a)=r(b), there existsu∈U(R) such thata=ub.

Proof(1)?(2) Given l(a)=l(b), then l(a)∩l(0)=l(b). By virtue of Theorem 2, there existu∈U(R),y∈Rsuch thatau+by=b. Hence,a+0×yu-1=bu-1, and soa=bu-1, as desired.

(2)?(1) For anya∈R, there exists ac∈Rsuch thata=aca. It is easy to verify that l(a)=l(ac). By hypothesis, there exists au∈U(R) such thata=acu. Hence,au-1a=aca=a. ThereforeRis unit-regular.

(1)?(3) is obtained by symmetry.

Furthermore, we can derive the following.

Theorem4LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Whenever l(a)=l(b), there existsu∈U(R) such thata=bua.

(3) Wheneverr(a)=r(b), there existsu∈U(R) such thata=aub.

Proof(1)?(2) Suppose that l(a)=l(b). AsRis regular, there existsx∈Rsuch thata=axa. Obviously, l(a)=l(ax). Hence, l(ax)=l(b). By virtue of Lemma 3, we have au∈U(R) such thatb=axu; hence,ax=bu-1. Thereforea=axa=bu-1a, as desired.

(2)?(1) For anya∈R, there exists ax∈Rsuch thata=axa. It is easy to verify that l(a)=l(ax). By hypothesis, there exists au∈U(R) such thatax=auax. Hence,a=axa=auaxa=aua. ThereforeRis unit-regular.

(1)?(3) follows by symmetry.

Corollary3LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Wheneverφ:Ra?Rb, there existsu∈U(R) such thata=φ(a)ua.

(3) For any idempotentse,f∈R,φ:eR?fRimplies that there existsu∈U(R) such thate=φ(e)ue.

Proof(1)?(2) Suppose thatφ:Ra?Rb. AsRis regular, there exists ax∈Rsuch thata=axa. Thus,φ(a)=φ(axa)=aφ(xa)?aR. On the other hand, we have ay∈Rsuch thatφ(a)=φ(a)yφ(a); hence,φ(a)=φ(φ(a)ya). Sinceφis aR-isomorphism, we geta=φ(a)y. Thus,aR?φ(a)R. ThereforeaR=φ(a)R. In light of Lemma 1, we have l(a)=l(φ(a)). According to Theorem 4, there exists au∈U(R) such thata=φ(a)ua, as required.

(2)?(3) is trivial.

(3)?(1) For anya∈R, there exists ax∈Rsuch thata=axaandx=xax. It is easy to verify thatφ:Rxa?Raxgiven byφ(rxa)=(rxa)x. Obviously,xa,ax∈Rare idempotents. By hypothesis, there exists au∈U(R) such thatxa=φ(xa)ua=xaxua. Thereforea=axa=a(xaxua)=aua, i.e.,a∈Ris unit-regular. As a result,Ris unit-regular.

Corollary4LetRbe a regular ring. Then the following are equivalent:

(1)Ris unit-regular.

(2) Wheneverφ:aR?bR, there existsu∈U(R) such thata=auφ(a).

(3) For any idempotentse,f∈R,φ:Re?Rfimplies that there existsu∈U(R) such thate=euφ(e).

ProofThis is proved in the same manner.

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