Sun Xiaoqing,Liu Xia,Hua Xiujuan
(School of Science,Xi′an University of Technology,Xi′an 710054,China)
Abstract:The notions of Q-clean elements in a ring and Q-clean ring are defined.Some basic properties of Q-clean rings are obtained.In particular,it is shown that every element in a Q-clean ring with 2 invertible is a sum of a quasi-invertible element and a square root of 1.Let Rbe a ring in which every element x∈R,x=xuxwhere u∈ .Then Ris Q-clean.Furthermore,we prove that the ideals of a Q-clean ring are also Q-clean.
Keywords:quasi-invertible,Q-clean ring,ideal
Following Nicholson a ringRwith identity is called clean if every element ofRcan be written as the sum of an idempotent and a unit inR.A ring with central idempotents is clean if and only if it is exchange[1].Many authors investigated exchange rings and clean rings in references[1-4].Letdenotes the quasi-invertible elements,that is,
The definition of quasi-invertible element was used by Ara to study purely infinite ring as an infinite analogue of rings of stable rank one in references[5-6].We call a ringRis Q-clean if every element ofRcan be written as the sum of an idempotent and a quasi-invertible element inR.Clean rings are Q-clean by.Some basic properties of Q-clean rings are obtained in section 2.It is shown that every element in a Q-clean ring with 2 invertible is a sum of a quasi-invertible element and a square root of 1.And letRbe a ring in which every elementx∈R,x=xuxwhereu.ThenRis Q-clean.
In section 3,in order to study the ideal of Q-clean ring,the definition of Q-clean on rings without identity is given.According to this,it is shown that the ideal of a Q-clean ring(unitial or not)is also Q-clean.
Throughout this paper,Rdenotes an associative ring with identity.We denote byU(R),Id(R)andJ(R)the set of all units ofR,the set of all idempotents inR,and the Jacobson radical,respectively.
Let us start by recalling the concept of quasi-invertibility.We say that elementsxandyin a unital ringRare centrally orthogonal provided thatxRy=yRx=0,and we writex⊥y.An elementuin an arbitrary ringRis said to be quasi-invertible if there exist elementsa,binRsuch that
The set of quasi-invertible elements inRwill be denoted by
Ifu∈,then we have the equation(1−ua)u(1−bu)=0.Takingv=a+b−aubthis implies thatu=uvu.By computation
we have the relation(1−uv)⊥(1−vu).We say in this situation thatvis a quasi-inverse foru.
An elementxof a ringRis clean in[1]if it can be written as the sum of a unit and an idempotent inR.A ring is called a clean ring if every element ofRis clean.Many authors extend it in references[7-11].In this paper,we will extend it to a large class of rings,Q-clean rings.
Definition 2.1An elementrof a ringRis called Q-clean if there existe∈Id(R)andu∈such thatr=e+u.The ringRis called Q-clean if every element ofRis Q-clean.
The class of Q-clean ring is very large.For example,clean rings,one-sided unitregular rings,and regular QB-rings.Since,we have ifais Q-clean,then so is 1−a.IfRis commutative,then.Thus Q-clean commutative rings are clean rings.LetRbe a QB-ring and have stable range one.ThenRis Q-clean precisely that it is clean.
Now,we give some basic properties of Q-clean rings.
Lemma 2.1[5]IfIis an ideal in a unital ringRand∏:R→R/Idenotes the quotient morphism,then
Theorem 2.1The following hold:
(1)Every homomorphic image of a Q-clean ring is Q-clean;
Proof(1)It is straightforward from Lemma 2.1.
(2)Suppose that eachRiis a Q-clean ring.Letx=(xi)∈∏Ri.For eachi,writexi=ei+uiwhereei∈Id(Ri)and.Thenand.Hencex=e+uis a Q-clean expression ofx.Thereforea Q-clean ring.
The converse part is obvious from(1).This,we complete the proof.
LetS(R)be the nonempty set of all proper ideals ofRgenerated by central idempotents.An idealP∈S(R)is called a Pierce ideal ofRifPis a maximal(with respect to inclusion)element of the setS(R).IfPis a Pierce ideal ofR,then the factor ringR/Pis called a Pierce stalk ofR.The following result shows that the Q-clean property needs to be checked only for indecomposable rings or Pierce stalks.
Theorem 2.2LetRbe a ring.Then the following are equivalent:
(1)Ris Q-clean;(2)Every factor ring ofRis Q-clean;(3)Every indecomposable factor ring ofRis Q-clean;(4)Every Pierce stalk ofRis Q-clean.
Proof(1)⇒(2),(2)⇒(3)and(2)⇒(4)are directly verified.
(4)⇒(1)LetMbe the set of all proper idealsIofRsuch thatIis generated by central idempotents and the ringR/Iis not Q-clean.Suppose thatRis not Q-clean.Then 0∈Mand the setMis not empty.It is directly verified as above that the union of every ascending chain of ideals fromMbelongs toM.By Zorn′s Lemma,the setMcontains a maximal elementP.By(4),it is sufficient to prove thatPis a Pierce ideal.Assume the contrary.By the definition of the Pierce ideal,there is a central idempotenteofRsuch thatP+eRandP+(1−e)Rare proper ideals ofRwhich properly contain the idealP.Since idealsP+eRandP+(1−e)Rdo not belong toMand are generated by central idempotents,R/(P+eR)andR/(P+(1−e)R)are Q-clean.Note that(R/(P+eR))×(R/(P+(1−e)R)).It follows from Theorem 2.1(2)thatR/Pis Q-clean,a contradiction.HencePis a Pierce ideal ofR,a contradiction.This,we complete the proof.
Theorem 2.3Ifxis Q-clean in R,so thatx=e+uwheree2=e∈Randu∈with quasi-inversev∈R,then each element of the formx+a(1−vu)+(1−uv)bwitha,b∈Rwill be Q-clean inR.
ProofLetu′=u+a(1−vu)+(1−uv)b.Evidently eachu′inRof the prescribed form equation(2.1)will be a partial inverse forv.Moreover,
whence(1−u′v)⊥(1−vu′),so thatu′∈andu′is a quasi-inverse foru.Sincex+a(1−vu)+(1−uv)b=e+u+a(1−vu)+(1−uv)b,it is Q-clean inR.This,we complete the proof.
LetRbe a ring in which 2 is invertible.Camillo and Yu in reference[11]showed thatRis clean if and only if every element ofRis a sum of a unit and a square root of 1.We now take a look at the case for Q-clean rings.IfGis a group,we denote the group ring overRbyRG.It seems to be difficult to characterizeRandGfor whichRGis Q-clean in general.We here also focus onRwith 2∈U(R).This,we complete the proof.
Theorem 2.4LetRbe a ring in which 2 is invertible andG={1,g}be a group.
(1)Ris Q-clean if and only if every element ofRis a sum of a quasi-invertible element and a square root of 1;(2)RGis Q-clean if and only ifRis Q-clean.
Proof(1)SupposeRis Q-clean andx∈R,thenwheree2=eandu∈.Thusx=(2e−1)+2u,with(2e−1)2=1.It is easy to check that.Then 2u∈.
Conversely,ifx∈R,then 2x−1=f+uwheref2=1 andu∈.Thus.We have.Hencexis Q-clean,which shows thatRis a Q-clean ring.
(2)As an image ofRG,Ris Q-clean whenRGis Q-clean.
Conversely,Since 2∈U(R),we havevia the mapping
HenceRGis Q-clean by Theorem 2.1.This,we complete the proof.
Reference[12]proved that every element in a unit-regular ring is the sum of an idempotent and a unit.Reference[2]proved that every element in a one-sided unitregular ring is the sum of an idempotent and a one-sided unit.We get the similar result on Q-clean rings.
Lemma 2.2[5]IfRis a untial exchange ring the following conditions are equivalent:(1)Ris aQB-ring;(2)Every regular element inRextends to an element in;(3)For every regular elementxinRthere is avinsuch thatx=xvx.
Theorem 2.5LetRbe a ring.If for every elementxinRthere is auinsuch thatx=xux,thenRis a Q-clean ring.
ProofIt follows from Lemma 2.2 thatRis a regularQB-ring.By Theorem 5 in reference[3],for anya∈R,there exist an idempotente∈Rand au∈such thata=e+u.
LetIbe a not necessarily unital ring.There is a canonical unitization=I⊕Z,with the obvious multiplication(a,n)(b,m)=(ab+ma+nb,nm)fora,binIandn,min Z.EvidentlycontainsIas an ideal with/I=Z.We write
Heres⊥tinImeans thatst=0 andts=0.We say in this case thatsandtare centrally orthogonal.
Lemma 3.1LetSbe any unital ring containingIas a subring,and letRdenote the unital subring ofSgenerated byIand 1.Then we have the equalities.
ProofStraightforward computations based on the equation
This,we complete the proof.
Corollary 3.1Ifx∈we can find a single elementyinRsuch that−(x+y+xy)and−(x+y+yx)are centrally orthogonal idempotents inR.
ProofCombine Proposition 2.2 in reference[6]and Lemma 3.1.
Lemma 3.2IfIis an ideal in a unital ringR,andt∈I,thent∈if and only if 1+t∈.
ProofIft∈,then by Corollary 3.1 we can find centrally orthogonal idempotents inIof the formp=−(t+s+ts)andq=−(t+s+st)for somesinI.SinceIis an ideal inRthis implies thatpRq=p2Rq2⊂pIq=0.SimilarlyqRp=0.It follows that 1+t∈with quasi-inverse 1+s.
Conversely,if 1+t∈we can find a quasi-inverse inR,written in the form 1+s,such that(1−(1+t)(1+s))⊥(1−(1+s)(1+t)).The equation 1+t=(1+t)(1+s)(1+t)shows thats=−t−t2−ts−st−tst∈I,whencet∈.
Note thatJ(I)⊆.IfRhas a unity,then;in factviax 7→1+x.Recall thatRis Q-clean if every elementa∈Rcan be written asa=e+uwheree2=e∈Randu∈.This,we complete the proof.
Theorem 3.1The following are equivalent for a ringR:
(1)Ris Q-clean;(2)For eacha∈R,a=e+xwheree2=eandx∈.
ProofLeta∈R.IfRis Q-clean,writea+1=e+uwheree2=eandu∈.Thena=e+xwherex=u−1∈by Lemma 3.2.
Conversely,ifa−1=e+xwheree2=eandx∈,thena=e+uwhereu=x+1∈by Lemma 3.2.
With this in mind,an elementainI(unitial or not)is called a Q-clean element ifa=e+xwheree∈Id(I)andx∈,andIis called a Q-clean ring if every element is Q-clean.Clearly,every homomorphic image of a Q-clean ring(unitial or not)is Q-clean,and the direct productof rings(unitial or not)is Q-clean if and only if eachIjis Q-clean.This,we complete the proof.
Lemma 3.3The following are equivalent forI:
(1)Iis Q-clean;
(2)WheneverI▹RwhereRis a ring,eacha∈Iis a Q-clean element ofR;
(3)(a,0)is Q-clean infor everya∈I.
Proof(1)⇒(2).Givena∈I▹R,let−a=e+xwheree2=eandx∈Thena=(1−e)+(−1−x)where(1−e)2=1−e.It follows from Lemma 3.2 that 1+x∈.Then−1−x∈.
(2)⇒(3).It is clear that takeR=.
(3)⇒(1).It is easy to check that the idempotent elements ofare the form of(f,0)or(−f,1)wheref2=f.Since(a,0)is Q-clean infor everya∈I,we have either(a,0)=(f,0)+(u,0)or(a,0)=(−f,1)+(v,−1)wheref2=fand(u,0),(v,−1)∈.If(u,0)∈,then there exist elements(a,n),(b,m)∈such that((0,1)−(u,0)(a,n))⊥((0,1)−(b,m)(u,0)).Thus,
Since−a∈I,we have(−a,0)=(−f,1)+(v,−1)wheref2=fand(v,−1)∈.Then(−a,0)=(−f,1)+(v,−1)=(0,1)+(−f,0)+(v,0)−(0,1)=(−f,0)+(v,0).Since(v,−1)∈,we have(−v,1)∈.It follows from Lemma 3.2 that
Thena=f+(−v)wheref2=f∈Iand−v∈.Therefore,Iis Q-clean.This,we complete the proof.
Theorem 3.2LetIbe a Q-clean ring(unitial or not)andA▹I.ThenAis Q-clean.
ProofWriteR=.Then(A,0)▹R.It suffices by(3)of Lemma 3.3 to show that(a,0)is Q-clean inRfor eacha∈A.But if−a=f+xwheref∈Id(I)andx∈,then(a,0)=(−f,1)+(−x,−1).This proves(1)of Lemma 3.3 because(−f,1)is an idempotent ofRand(−x,−1)=−(x,1)=−[(x,0)+(0,1)]∈by Lemma 3.2.This,we complete the proof.