On clean ideals

PII. S0161171203211339 http://ijmms.hindawi.com © Hindawi Publishing Corp.

ON CLEAN IDEALS

HUANYIN CHEN and MIAOSEN CHEN

Received 21 November 2002

We introduce the notion of clean ideal, which is a natural generalization of clean rings. It is shown that every matrix ideal over a clean ideal of a ring is clean. Also we prove that every ideal having stable range one of a regular ring is clean. These generalize the corresponding results for clean rings.

2000 Mathematics Subject Classification: 16E50, 16U99.

1. Introduction. LetRbe a unital ring. We say thatRis a clean ring in case every element ofR is a sum of an idempotent and a unit. It is well known that every endomorphism ring of a countably generated vector space over a division ring is a clean ring (cf. [7, Theorem]). A ringRis said to be unit reg-ular in case for everyx ∈R, there exists a unit u∈R such that x=xux. Answering a question of Nicholson, Camillo and Yu [3, Theorem 5] claimed that every unit regular ring is clean. But there was a gap in their proof. In [2, Theorem 1], Camillo and Khurana proved this result by a new route and gave a characterization of unit regular rings.

In this paper, a natural problem asked whether there is a nonclean ringR while some element ofRis a sum of an idempotent and a unit. So as to deal with such rings, we will introduce a notion of clean ideals. We show that clean ideals have similar properties to clean rings.

Throughout, all rings are associative with identity and all modules are right modules. We always useU(R)to denote the set of units ofR.

Definition_1.1. _{An ideal}I_{of a unital ring}R_{is clean in case every element} inIis a sum of an idempotent and a unit ofR.

Clearly, every ideal of clean rings is clean. But there exist nonclean rings which contain some clean ideals. LetR1be a clean ring andR2not a clean ring.

SetR=R1⊕R2. ThenRis not a clean ring. SetI=R1⊕0. Given any(x,0)∈I,

we have an idempotente∈R1and a unitu∈R1such thatx=e+ubecause R1 is clean. Hence, (x,0)=(e,1)+(u,−1). Clearly, (e,1)=(e,1)2∈R, and (u,−1)∈R is invertible. Therefore, we conclude thatI is a clean ideal ofR. Hence, the notion of clean ideals is a nontrivial generalization of clean rings.

Recall that an idealIof a unital ringRis an exchange ideal provided that for anyx∈I, there exists an idempotente∈Isuch thate−x∈R(x−x2_).

Proof. _{Given any}x∈I_{, we have an idempotent}e_{and a unit}u∈U(R)_such thatx=e+u. Clearly,u(x−u−1_(1−e)u)=x2_{−x. Setf=u}−1_{(1−e)u. Then}

we havef=f2_{∈R. Furthermore,f=x−u}−1_(x2_−x)=(1−u−1_(1−x))x∈ Rx; hence,f =f2_{∈RxRx⊆Ix. In addition, 1−f=1−x+u}−1_(x2_−x)∈ R(1−x). Therefore,Iis an exchange ideal ofRby [1, Lemma 2.1].

Theorem_1.3. _LetR_{be a unital ring and}I_{an ideal in which every idempotent} is central. Then the following are equivalent:

(1) Iis a clean ideal, (2) Iis an exchange ideal.

Proof. ₍₁₎⇒_{(2) is clear byLemma 1.2.}

(2)⇒(1). Given anyx∈I, we have an idempotente∈Rsuch thate∈Rxand 1−e∈R(1−x). Assume thate=axand 1−e=b(1−x). In addition, we may assume thatea=aand(1−e)b=b. Hence,axa=ea=a, soax,xa∈Iare central idempotents. Thus, we haveax=axax=(ax)(xa)=x(ax)a=xa. Likewise,b(1−x)b=b, so thatb(1−x)and(1−x)b, and hence 1−b(1−x) and 1−(1−x)b, are idempotents. But 1−b(1−x)=e∈I and 1−(1−x)b= e−bx+xb∈I; hence, 1−b(1−x)and 1−(1−x)bare central. Thus,b(1−x) and(1−x)bare central; and arguing as foraxandxaabove, we getb(1−x)= (1−x)b. One easily checks that(a−b)(x−(1−e))=(x−(1−e))(a−b)=1. Setu=x−(1−e). Thenx=(1−e)+uwith 1−e=(1−e)2_{∈Randu∈U(R),}

as required.

Corollary1.4. Every exchange ideal without nonzero nilpotent elements is a clean ideal.

Proof. _LetI_{be an exchange ideal of a ring}R_{. Assume that every nilpotent} element inIis zero. Lete∈Ibe an idempotent andx∈R. We see thatex(1− e)∈Iis a nilpotent element; hence,ex=exe. Likewise,xe=exe. Therefore, every idempotent inIis central and we complete the proof byTheorem 1.3.

Recall that an ideal I of a ringR is left quasi-duo provided that for any maximal left idealJofR,JI⊆J. FromTheorem 1.3, it follows that every left quasi-duo exchange ideal of a ring is clean. An idealIof a unital ringRisπ -regular provided that for anyx∈I, there exists a positive integern(x)such thatxn(x)_=xn(x)_yxn(x)_{for ay∈I. It follows byCorollary 1.4that everyπ} -regular ideal without nonzero nilpotent elements is a clean ideal. Recall that a ringRis regular in case for anyx∈R, there exists ay∈Rsuch thatx=xyx. Let I be an ideal of a regular ring R. We say that I has stable range one if aR+bR=R witha∈1+I,b∈R implies thata+by∈U(R)for ay∈R. It is well known that an idealI of a regular ring R has stable range one if and only if eReis unit regular for any idempotent e∈I. Let V1and V2 be

is its ideal having stable range one. We know that every unit regular ring is clean (cf. [2, Theorem 1]). Now, we extend this result to ideals of regular rings having stable range one.

Lemma1.5. LetRbe a regular ring andIan ideal ofR. Then the following are equivalent:

(1) Ihas stable range one,

(2) eReis unit regular for all idempotentse∈I,

(3) I+C(R)is unit regular, whereC(R)is the center ofR.

Proof. See [6, Lemma 1.4].

Theorem1.6. LetR be a regular ring andI an ideal ofR. IfI has stable range one, thenIis a clean ideal ofR.

Proof. _{Let C}(R)_{be the center of}R_{and let}R=I+_C(R)_{. In view ofLemma} 1.5,Ris unit regular. By [2, Theorem 1], we know thatRis a clean ring. Given anyx∈I, we havex∈R; hence, there exist an idempotente∈R and a unit u∈R such that x=e+u. We easily check thate=e2_{∈R and u∈U(R).}

Therefore,Iis a clean ideal ofR.

Corollary_1.7. _LetR_{be a regular, right self-injective ring and let}I= {x∈

R|xRis directly finite}. ThenIis a clean ideal ofR.

Proof. _{By [4, Corollary 9.21],}I_{is an ideal of}R_{. Given any idempotent}e∈I_, we have thateRis directly finite; hence,eReEndR(eR)is a directly finite ring. Using [4, Theorem 9.17],eReis unit regular. It follows byLemma 1.5thatIhas stable range one. Therefore, we complete the proof byTheorem 1.6.

LetRbe a regular, right self-injective ring andxRdirectly finite. Thenxis a sum of an idempotent and a unit byCorollary 1.7. Recall that an idealIof a ringRis of bounded index if there is a positive integernsuch thatxn_{=0 for} any nilpotentxofI.

Corollary_1.8. _LetI_{be an ideal of a regular ring}R_{. If}I_{is of bounded index,} thenIis a clean ideal ofR.

Proof. _{Given any idempotent}e∈I_{, we have}eRe⊆I_{. Assume now that} the nilpotent bounded index of I is n. If(er e)m =0 ineRe, then we have (er e)n_{=0. Hence,eRe is a regular ring of bounded index. According to [4,} Corollary 7.11],eReis unit regular. It follows byLemma 1.5thatI has stable range one. UsingTheorem 1.6, we get thatIis a clean ideal ofR.

Theorem_1.9. _LetI_{be a clean ideal of a unital ring}R_{. ThenM}n(I)_{is a clean} ideal ofMn(R).

Proof. _{Clearly, the result holds for}n=_{1. Assume now that the result holds} forn=k−1(k≥2). Suppose thatA∈Mk(I), writeA=a qp B, wherea∈I, B∈Mk−1(I). SinceI is a clean ideal of R, we havee=e2∈R andu∈U(R)

such thata=e+u. SinceB−pu−1_q∈M

k−1(I), there existF =F2∈Mk−1(R)

andV∈GL_k−1(R)such thatB−pu−1q=F+V. Set

E=

e 0

0 F

, U=

u q

p V+pu−1_q

. (1.1)

It is easy to verify thatE=E2_and

U



u−1+u−1qV−1pu−1 −u−1qV−1

−V−1_pu−1 _V−1

 

=



u−1+u−1qV−1pu−1 −u−1qV−1

−V−1_pu−1 _V−1

 _U

=

 1 0

0 Ik−1



 in Mk(R).

(1.2)

Hence,U ∈GLk(R). Clearly, A=E+U. Therefore, Mk(I)is a clean ideal of Mn(R). By induction, we complete the proof.

Corollary_1.10. _LetR_{be a unital ring and}I _{an exchange ideal in which} every idempotent is central. ThenMn(I)is a clean ideal ofMn(R).

Proof. _{The proof is clear by Theorems1.3and1.9.}

A Morita context denoted by(A,B,M,N,ψ,φ)consists of two ringsA,B, two bimodulesANB,BMA, and a pair of bimodule homomorphisms (called pairings) ψ:N⊗BM→Aandφ:M⊗AN→B which satisfy the following associativity: ψ(n⊗m)n =nφ(m⊗n),φ(m⊗n)m =mψ(n⊗m), for anym,m ∈M, n,n ∈N. These conditions insure that the setTof generalized matrices_{m b}a n, a∈A,b∈B,m∈M, andn∈N, forms a ring, called the ring of the context. Haghany studied hopficity and co-hopficity for Morita contexts with zero pair-ings. Now, we investigate clean Morita contexts with zero pairpair-ings.

Lemma _1.11. _LetT _{be the ring of a Morita context}(A,B,M,N,ψ,φ) _with zero pairings. IfIandJare clean ideals ofAandB, respectively, then_{M J}I Nis a clean ideal ofT.

Proof. SinceT is the ring of a Morita context(A,B,M,N,ψ,φ)with zero pairings, we check that_{M J}I Nis an ideal ofT. LetA=m ba n

∈I N M J

a∈I,b∈J,m∈M, andn∈N. AsIis a clean ideal ofA, we havee=e2_∈A

andu∈U(A)such thata=e+u. Inasmuch asb∈J, there existf=f2_∈B

andv∈U(B)such thatb=f+v. Set

E=

 e 0

0 f



_{, U=}  u n

m v



_. (1.3)

It is easy to verify thatE=E2_∈Tand

U



 u−1 −u−1nv−1

−v−1_mu−1 _v−1

 ₌



 u−1 −u−1nv−1

−v−1_mu−1 _v−1

 _U

=

 1 0

0 1

  inT .

(1.4)

Hence,U∈U(T ). Obviously, we haveA=E+U. Therefore, we get the result.

LetA1,A2, andA3be associative rings with identities andM21,M31, andM32

be(A2,A1)-,(A3,A1)-, and(A3,A2)-bimodules, respectively. Letφ:M32⊗_A2M21

→M31be an(A3,A1)-homomorphism and letT=

A1 0 0 M21 A2 0 M31M32A3

with usual

ma-trix operations.

Theorem_1.12. _{The following are equivalent:}

(1) I,J, andKare clean ideals ofA1,A2, andA3, respectively,

(2) the formal triangular matrix ideal MI21 0J 00 M31M32K

is a clean ideal of

A1 0 0 M21 A2 0 M31M32A3

.

Proof. (1)⇒(2). Clearly,

I 0 0 M21 J 0 M31M32K

is an ideal of MA211 A02 00 M31M32A3

. Let B =

_A 2 0 M32A3

andM=M21 M31

. SinceJandKare clean ideals ofA2andA3,

respec-tively, byLemma 1.11, we see that_MJ_32K0is a clean ideal ofB. ByLemma 1.11

again, MI21 0J 00 M31M32K

is a clean ideal ofA10 M B

, as required.

(2)⇒(1). Inasmuch as MI21 0J 00 M31M32K

is an ideal of A1

0 0

M21 A2 0 M31M32A3

, we show thatI, J, andKare ideals ofA1,A2, andA3, respectively. For anyx∈J, we have

    

0 0 0

0 x 0

0 0 0

    ∈     

I 0 0

M21 J 0

M31 M32 K

   

Thus, we have idempotent e∗1e0 02 0 ∗ ∗e3

∈T and a unit u∗1u02 00 ∗ ∗ u3

∈T such that

   

0 0 0

0 x 0

0 0 0

   =    

e1 0 0

∗ e2 0 ∗ ∗ e3

   +    

u1 0 0

∗ u2 0

∗ ∗ u3

  

. (1.6)

Clearly,e2=e22andu2∈U(A2). Furthermore, we havex=e2+u2. Therefore, Jis a clean ideal ofA2. Likewise, we claim thatIandKare clean ideals ofA1

andA3, respectively.

Corollary1.13. LetRbe a unital ring andI an ideal ofR. Then the fol-lowing are equivalent:

(1) Iis a clean ideal ofR, (2) triangular matrix ideal

      

I 0 ··· 0

R I ··· 0

..

. ... . .. ...

R R ··· I

       n×n (1.7)

is a clean ideal of the ring of alln×nlower triangular matrices overR.

Proof. The proof is an immediate consequence ofTheorem 1.12and the induction.

Analogously, we can derive the following proposition.

Proposition_1.14. _{An ideal}I_{of a ring}R_{is clean if and only if the ideal of} alln×nlower triangular matrices overIis a clean ideal of the ring of alln×n lower triangular matrices overR.

Similarly, we deduce that an idealI of a ringR is clean if and only if the ideal of alln×nupper triangular matrices overI is a clean ideal of the ring of alln×nupper triangular matrices overR. We say that an idealIof a ring R is stronglyπ-regular if for any x∈I, there exists a positive integern(x) such thatxn(x)_=xn(x)+1_{yfor ay∈I. It is proved that an idealIof a ringRis}

stronglyπ-regular if and only if there exists a positive integern(x)such that xn(x)_=xn(x)+1_{yandxy=yx, for ay∈I. So every stronglyπ-regular ideal}

of a ring is left-right symmetric. Hence all stronglyπ-regular ideals of a ring are clean ideals. Thus, we see that the ideal of alln×nupper (lower) triangular matrices over a stronglyπ-regular ideal of a ring is a clean ideal of the ring of alln×nlower triangular matrices overR.

A finite orthogonal set of idempotents e1,...,en in a ring R is said to be

complete in casee1+ ··· +en=1∈R. Using the method inTheorem 1.9, we

Proposition _1.15. _LetR _{be a unital ring and}I _{an ideal of}R_{. Then the} following are equivalent:

(1) Iis a clean ideal ofR,

(2) there exists a complete set{e1,...,en}of idempotents such thateiIeiis a

clean ideal ofeiReifor alli.

Proof. ₍₁₎⇒_{(2) is clear by choosing}n=_1.

(2)⇒(1). Suppose that{e1,...,en}is a complete set of idempotents such that eiIeiis a clean ideal ofeiReifor alli. It suffices to show that the result holds forn=2. Clearly, Ie1Ie1e1Ie2

e2Ie1e2Ie2

and R e1Re1e1Re2 e2Re1e2Re2

. Let A=a11a12 a21a22

∈

e1Ie1e1Ie2 e2Ie1e2Ie2

. As e1Ie1 is a clean ideal ofe1Re1, we havee=e2∈e1Re1 and u∈U(e1Re1) such thata11=e+u. Inasmuch as a22−a21u−1a12 ∈e2Re2,

there existf=f2_∈e

2Re2andv∈U(e2Re2)such thata22−a21u−1a12=f+v.

Set

E=

e 0

0 f

, U=

u a12

a21 v+a21u−1a12

. (1.8)

It is easy to verify thatE=E2_and

U



u−1+u−1a12v−1a21u−1 −u−1a12v−1 −v−1_a

21u−1 v−1

 

=



u−1+u−1a21v−1a12u−1 −u−1a21v−1 −v−1_a

12u−1 v−1

 _U

=

e1 0

0 e2

.

(1.9)

Hence,Uis invertible. Clearly,A=E+U, so we get the result.

Acknowledgments. _{The authors are grateful to the referee for useful} sug-gestions which led to the new version ofTheorem 1.3. These helped us to im-prove the presentation considerably. This work was supported by the Natural Science Foundation (NSF) of Zhejiang Province.

References

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Huanyin Chen: Department of Mathematics, Zhejiang Normal University, Jinhua, Zhe-jiang 321004, China

E-mail address:[email protected]

Miaosen Chen: Department of Mathematics, Zhejiang Normal University, Jinhua, Zhe-jiang 321004, China