WEIXING CHEN
Received 20 March 2006; Accepted 21 March 2006
In this paper we unify the structures of various clean rings by introducing the notion ofP-clean rings. Some properties ofP-clean rings are investigated, which generalize the known results on clean rings, semiclean rings,n-clean rings, and so forth. By the way, we answer a question of Xiao and Tong onn-clean rings in the negative.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
Throughout this paperRdenotes an associative ring with identity and all modules are unitary. We use the symbolU(R) to denote the group of units ofRand Id(R) the set of idempotents ofR,Un(R) the set of elements which are the sum ofnunits ofR,UΣ(R)
the set of elements each of which is the sum of finitely many units inR,RE(R) (URE(R)) the set of regular (unit regular) elements ofR, and Peri(R) the set of periodic elements ofR. The Jacobson radical and the prime radical ofRare denoted byJ(R) and Nil∗(R),
respectively.
Following Han and Nicholson [4], an elementxof a ringRis called clean ifx=e+u wheree∈Id(R) andu∈U(R). A ringRis clean if every element ofRis clean. This notion was first introduced by Nicholson [5] as early as 1977 in his study of lifting idempotents and exchange rings. Since then, a great deal is known about clean rings and their gener-alizations (cf. [1–9]).
According to Ye [9], a ringRis called semiclean if every element ofRhas the form x= f +u, whereu∈U(R) and f is periodic, that is, fp= fqfor two different positive integerspandq. In [8], an elementxof a ringRis calledn-clean ifx=e+u1+···+un wheree∈Id(R), ui∈U(R), andnis a positive integer. The ringR is calledn-clean if every element ofRisn-clean for some fixed positive integern. WhileRis calledΣ-clean, if thenis a positive integer depending onx. Also Zhang and Tong in [10] definedRto be G-clean, if eachx∈Rhas the formx=a+uwhereais unit regular andu∈U(R).
Motivated by the results of Han and Nicholson [4] on clean rings, Ye [9] on semiclean rings, Xiao and Tong [8] onn-clean rings andΣ-clean rings, and Zhang and Tong [10] on G-clean rings, in this paper we unify the structures of various clean rings by introducing
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 75247, Pages1–7
the notion ofP-clean rings and the common properties of those rings. By the way, we answer a question of Xiao and Tong [8] in the negative and extend some known results of [8,9].
2.P-clean rings
We start this section by the following definitions.
For two subsetsAandBof a ringR, the sum ofAandB is defined as follows:A+ B= {a+b|a∈A, b∈B}. The sum of more than two subsets of anRcan be defined inductively.
LetPbe a property which is meaningful for elements of a ring. For any ringR, letP(R) be the subset{a∈R|ahas propertyP}ofR.
Definition 2.1. PropertyPwill be called admissible if the following conditions are satis-fied.
(1) For any ring homomorphismσ:R→S,σ(P(R))⊆P(S). (2) For any ringsR⊆S,P(R)⊆P(S).
(3) For anye∈Id(R),P(eRe) +P((1−e)R(1−e))⊆P(R).
For convenience, an element ofP(R) is called aP-element inR. In this paperP will always be an admissible property.
Proposition 2.2. (1) Ifσis a ring isomorphism fromRontoS, thenσ(P(R))=P(S). (2) Ife1,e2,. . .,enare orthogonal complete idempotents, that is,eiej=0 wheneveri=j
andei2=ei, ande1+e2+···+en=1, thenP(e1Re1) +P(e2Re2) +···+P(enRen)⊆ P(R).
Proof. (1) ByDefinition 2.1,σ(P(R))⊆P(S), henceσ−1(σ(P(R)))⊆σ−1(P(S))⊆P(R).
It follows thatP(R)⊆σ−1(P(S))⊆P(R), which givesσ(P(R))⊆P(S)⊆σ(P(R)) and so
σ(P(R))=P(S).
(2) We prove this by using induction onn. In fact, the casen=2 is condition (3) ofDefinition 2.1. Assume (2) holds forn−1. Lete1+e2+···+en−1= f. Then
multi-plied byeion the two sides of the above equation, we haveeif = f ei=ei, which gives ei= f eif and soei∈Id(f R f). Note that f R f is a ring with identity f. It yields that P(e1Re1) +P(e2Re2) +···+P(en−1Ren−1)⊆P(f R f) by inductive assumption. On the
other hand, f +en=1 impliesP(f R f) +P(enRen)⊆P(R) byDefinition 2.1(3). Hence P(e1Re1) +P(e2Re2) +···+P(en−1Ren−1) +P(enRen)⊆P(f R f) +P(enRen)⊆P(R). Definition 2.3. A ringRis calledP-clean if everyx∈Rhas the formx=p+u, where p∈P(R) andu∈U(R).
Lemma 2.4. LetRbe a ring ande∈Id(R). Then the following hold. (1) Ifu∈U(eRe) andv∈U((1−e)R(1−e)), thenu+v∈U(R).
(2) Ife1∈Id(eRe) ande2∈Id((1−e)R(1−e)), thene1+e2∈Id(R).
(7) Ifx∈Id(eRe) +Un(eRe) and y∈Id((1−e)R(1−e)) +Un((1−e)R(1−e)), then x+y∈Id(R) +Un(R).
(8) Ifx∈Id(eRe) +UΣ(eRe) and y∈Id((1−e)R(1−e)) +UΣ((1−e)R(1−e)), then x+y∈Id(R) +UΣ(R).
Proof. We only prove (3) and (8), the others are very similar.
(3) Let f ∈Peri(eRe) andg∈Peri((1−e)R(1−e)). Then there exist positive integers m > n and p > q such that fm= fn andgp=gq. By Ye [9, Lemma 5.2], fn(m−n) and
gq(p−q) are both idempotents. Sett=2n(m−n)q(p−q). Then f2t= ft andg2t=gt. Since f g=g f =0, (f+g)2t=f2t+g2t= ft+gt=(f +g)t. Hence f +g⊆Peri(R).
(8) Assumex∈Id(eRe) +UΣ(eRe) andy∈Id((1−e)R(1−e)) +UΣ((1−e)R(1−e)). Then x= f +u1+···+un and y=g+v1+···+vm where f ∈Id(eRe), g ∈Id((1− e)R(1−e)),ui∈U(eRe), andvj∈U((1−e)R(1−e)). It is easy to show that ann-clean element ism-clean whenevern≤m, since for anye∈Id(R),e=(1−e) + (2e−1) where 1−e∈Id(R) and (2e−1)2=1. So without loss of generality, we can assume n=m.
Using (1), f +g∈Id(R). And from (6),ui+vi∈U(R), hencex+y=(f +g) + (u1+v1)
+···+ (un+vn)∈Id(R) +UΣ(R).
UsingLemma 2.4, it is easy to check that for any ringR, 0,R, Id(R), Peri(R),U(R), RE(R), URE(R), Un(R), Id(R) +Un−1(R) forn≥2, Id(R) +UΣ(R) are all subsets of R
defined by a suitable admissible propertyP.
From the above arguments, the following proposition is immediate.
Proposition 2.5. LetRbe a ring. Then the following conclusions hold. (1) Id(R)-clean rings are precisely clean rings.
(2)Pri(R)-clean rings are precisely semiclean rings. (3)U(R)-clean rings are precisely (S, 2)-rings. (4)Ure(R)-clean rings are preciselyG-clean rings.
(5) Id(R) +Un−1(R)-clean rings are preciselyn-clean rings whenn≥2.
(6) Id(R) +UΣ(R)-clean rings are preciselyΣ-clean rings.
Note that here an (S, 2)-ring is a ring in which every element can be expressed as a sum of two units ofR. While in some literature it referred to a ring in which every element can be written as a sum of no more than two units.
Proposition 2.6. Any homomorphic image of aP-clean ring isP-clean.
Proof. LetR be aP-clean ring and let f :R→S be a ring surjective homomorphism. Then for anyy∈S, there existsx∈Rsuch that f(x)=y. SinceRisP-clean,x=p+u with p∈P(R) andu∈U(R). Hence f(x)= f(p) + f(u). Obviously f(u)∈U(S) and f(p)∈f(P)⊆P(S) byDefinition 2.1, the proof is complete.
Proposition 2.7. A finite direct productR=n
i=1Ri of ringsRi isP-clean if and only if eachRiisP-clean.
for eachi. ByProposition 2.2, we can identifyRiwith (. . ., 0,Ri, 0,. . .) canonically. Letei= (. . ., 0, 1, 0,. . .). Then (pi)=(p1, 0,. . ., 0) + (0,p2,. . ., 0) +···+ (0, 0,. . .,pn)∈P(e1Re1) +
P(e2Re2) +···+P(enPen)⊆P(R). Nowx=(xi)=(pi+ui)=(pi) + (ui) with (pi)∈P(R)
and (ui)∈U(R), so we are done.
It should be noted thatProposition 2.7 is not true for an infinite direct product of ringsRi. For example, the ringZof integers is aΣ-clean ring, butR=∞i=1Zis notΣ
-clean since (1, 2,. . .,n,. . .) is obviously notΣ-clean.
Proposition 2.8. The ringRisP-clean if and only if the ringR[[x]] of formal power series overRisP-clean.
Proof. IfR[[x]] isP-clean, thenRisP-clean byProposition 2.6. Now ifRisP-clean, then for anyf(x)∈R[[x]],f(x)=a0+a1x+···+anxn+···. By assumption,a0=p+uwith
p∈P(R) andu∈U(R). Hence f(x)=p+u+a1x+···+anxn+··· withp∈P(R)⊆ P(R[[x]]) andu+a1x+···+anxn+··· ∈U(R[[x]]), as desired.
The following corollary extends [8, Proposition 2.5] which states that for a commuta-tive ringR,Risn-clean if and only ifR[[x]] isn-clean.
Corollary 2.9. Risn-clean (Σ-clean) if and only ifR[[x]] isn-clean (Σ-clean).
It has been proved by Han and Nicholson in [4] that ifeis an idempotent in a ringR such thateReand (1−e)R(1−e) are both clean rings, thenRis clean. Hence the ring of n×nmatrices overRis clean. Similar results hold for semiclean rings,n-clean rings, and Σ-clean rings. We now extend these results toP-clean rings.
Lemma 2.10. Lete∈Id(R) be such thateReand (1−e)R(1−e) are bothP-clean rings. ThenRis aP-clean ring.
Proof. For convenience, write ¯r=1−rfor eachr∈R. We use the Pierce decomposition of the ringR:
R=eRe+eRe¯+ ¯eRe+ ¯eRe.¯ (2.1)
Let x=a+b+c+d wherea∈eRe,b∈eRe¯,c∈eRe¯ , and d∈eR¯ e¯. By hypothesis, writea=p+uwherep∈P(eRe) andu∈U(eRe) with inverseu1. Thend−cu1b∈eR¯ e¯,
so writed−cu1b=q+vwhereq∈P( ¯eRe¯) andv∈U( ¯eRe¯) with inversev1. Hencex=
(p+q) +u+b+c+v+cu1band it suffices to show thatu+b+c+v+cu1bis a unit inR.
To this end compute
u+b+c+v+cu1b
u1+u1bv1cu1−u1bv1−v1cu1+v1
=e+bv1cu1−bv1
+−bv1cu1+bv1
+cu1+cu1bv1cu1−cu1bv1
+−cu1+ 1−e+−cu1bv1cu1+cu1bv1=1.
(2.2)
Similarly, (u1+u1bv1cu1−u1bv1−v1cu1+v1) (u+b+c+v+cu1b)=1.
Theorem 2.11. If 1=e1+e2+···+enin a ringRwhereeiare orthogonal idempotents and eacheiReiisP-clean, thenRisP-clean.
The following two results are direct consequences ofTheorem 2.11
Corollary 2.12. IfRis aP-clean ring, so also is the matrix ringMn(R).
Corollary 2.13. If M=M1⊕M2⊕ ··· ⊕Mnare modules andend(Mi) isP-clean for eachi, then end (M) isP-clean.
Since any homomorphic image of aP-clean ring is againP-clean, withTheorem 2.11, this gives the following.
Corollary 2.14. IfAandBare rings andV=AVBis a bimodule, the split-null extension RisP-clean if and only if bothAandBareP-clean, where
R=
A V
0 B
. (2.3)
In particular, induction shows that for eachn≥1, a ringRisP-clean if and only if the ring of alln×nupper triangular matrices overRisP-clean.
LetRbe a ring and letI be an ideal ofR. We sayP-elements inR/Ilift moduloI, if for anyp∈P(R/I) there existsa∈P(R) such thatπ(a)=pwhereπis the canonical ring homomorphism fromRontoR/I.
We close this section with the following proposition whose proof is very easy.
Proposition 2.15. LetRbe a ring and letIbe an ideal contained inJ(R). IfR/Iis aP-clean ring andP-elements lift moduloI, thenRis a P-clean ring.
3. Some remarks
It is known by [4, Proposition 6] that a ringRis clean if and only ifR/Iis clean for any idealI⊆J(R) and idempotents lift moduloI. Xiao and Tong [8] naturally claimed that they do not know whether for anyn-clean ringR, idempotents ofR/Ilift moduloIwhere Iis any ideal ofRcontained inJ(R). The following counterexample shows that the answer is negative.
Example 3.1. There is a 4-clean ringRin which idempotents ofR/J(R) cannot be lifted toR.
Proof. LetR be the subring of rational numbers Qgiven by R= {m/n∈Q|(m,n)=
(n, 6)=1}. ThenRhas only two maximal ideals: 2Rand 3R, soJ(R)=6R. Denote the ring of integers modulonbyZn, thenR/J(R)∼=Z2×Z3, which has four idempotents. But
Rhas only two idempotents. This shows that idempotents ofR/J(R) cannot be lifted toR moduloJ(R). But it can be shown thatRis a 4-clean ring.
Clearly,x=m/n∈U(R) if and only if (m, 6)=(n, 6)=1. Now for anyx∈R,xhas the formx=3p2qm/nwhere (m, 6)=(n, 6)=1. Ifp,q≥1, then 3p2qm=(3p2q−1 + 1)m= (3p2q−1)m+m∈U
2(R), sox∈U2(R). Ifx=3pm/nwithp≥1 and (m, 6)=(n, 6)=1,
then 3pm=(3p−2 + 2)m=(3p−2)m+m+m, which impliesx∈U
the case ofx=2qm/nwhere (m, 6)=(n, 6)=1 andq≥1, then 2q=2q−3 + 3=2q−3 + 1 + 1 + 1. It follows thatx∈U4(R). Since anyn-clean element must bem-clean for any
n≤m(cf. the proof ofLemma 2.4(8)),Ris a 4-clean ring from the above arguments. This ringRis clearly ak-clean for anyk≥4 as the proof shows. The following two results are obtained by Xiao and Tong [8] for commutative rings, now we extend them to 2-primal rings (rings whose prime radical coincides with the set of nilpotent elements).
Proposition 3.2. For any 2-primal ringR, the polynomial ringR[x] is notΣ-clean. Proof. Assume the contrary, then x=e(x) +u1(x) +u2(x) +···+un(x) where e(x)∈
Id(R[x]),ui(x)∈U(R[x]) for eachi, andnis a positive integer. Let
e(x)=e0+a1x+···+amxm, u1(x)=u10+u11x+···+u1mxm,
.. .
un(x)=un0+un1x+···+unmxm.
(3.1)
SinceRis 2-primal, a polynomial overRis invertible if and only if its constant term is inU(R) and the other coefficients are in Nil∗(R) by [3, Theorem 2.4], soui j∈Nil∗(R)
for eachj≥1. Hencex=e(x) +u1(x) +···+un(x) givesa1+u11+···+un1=1, soa1is
a unit inR, anda2+u12+···+un2=0 impliesa2∈Nil∗(R). On the other hand,e(x)2=
e(x) impliese2
0=e0, ande(x)2=e0+ (e0a1+a1e0)x+ (e0a2+a21+a2e0)x2+···+a2mx2m. Soa2=e0a2+a21+a2e0by comparing the coefficient ofx2ine(x)2=e(x). Note that the
sum of a unit and a nilpotent element must be a unit ande0a2+a2e0∈Nil∗(R). It follows
thata2∈U(R). This is a contradiction, and the proof is complete.
FromProposition 3.2, the following corollary is immediate.
Corollary 3.3. For any 2-primal ringR, the polynomial ringR[x] is notn-clean. We conclude this paper with the following proposition.
Proposition 3.4. For any 2-primal ringR, the polynomial ringR[x] is not semiclean. Proof. Assume the contrary, thenx=p(x) +u(x) wherep(x) is a periodic element and u(x) is a unit. Letp(x)=p0+p1x+···+pnxnandu(x)=u0+u1x+···+unxn. SinceR is 2-primal,ui∈Nil∗(R) for eachi≥1 by [3, Theorem 2.4]. By comparing the coefficient
ofx=p(x) +u(x), we havep1+u1=1, which implies p1is a unit inR, and pi+ui=0 givespi∈Nil∗(R) for eachi≥2. Clearly we can assume thatp(x)s=p(x)t for positive
integerss > t≥2. Then a routine calculation shows that the coefficient ofxsin p(x)sis
i1+i2+···+is=s
pi1pi2. . . pis=p s
While the coefficient ofxsinp(x)tis
j1+j2+···+jt=s
pj1pj2. . . pjt =b for someb∈Nil∗(R). (3.3)
Comparing the coefficients ofxson two sides of p(x)s=p(x)t, we havep
1∈Nil∗(R),
which is a contradiction.
The above result is obtained by Ye [9] only for a commutative ring.
Acknowledgments
This research was partially supported by the Doctorate Foundation of China Education Ministry (Grant no. 20020284009). The author wishes to thank his advisor Professor Wenting Tong for his helpful suggestions. Also he wishes to thank the referees for valuable comments.
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Weixing Chen: Mathematics and Information Science School, Shandong Institute of Business and Technology, Yantai 264005, China
