Nil g-*-Clean Rings

（School of Mathematics and Statistics， Anhui Normal University， Wuhu 241002， China）

Abstract： A ring is called nil g -*-clean ring if each of its elements is a sum of a g -projection and a nilpotent. In this paper， we study the basic properties of this ring and some relations with other rings. We prove that R is nil g -*-clean if （g （R）+J）/J=g （R/J）=and only if is nil， and is nil g -*-clean.

Key words： nil-clean ring; nil g -clean ring; *-ring; R/J nil *-clean ring; nil g -*-clean ring

Chinese Library Classification：O153.3 Document code：A

Article No.：1001-2443（2024）06-0520-07

Introduction

Clean rings were introduced by Nicholson in ［1］ in a study of exchange property of rings and modules， and have been extensively discussed in the literature. A ring is clean if each of its elements is a sum of an idempotent and a unit. As an important variant of a clean ring， a ring is called （strongly） nil-clean if each of its elements is a sum of an idempotent and a nilpotent （that commute）. This notion was introduced by Diesl in ［5］， and has been much studied. One readily notes that a ring is nil-clean iff each of its elements is a difference of a nilpotent and an idempotent. Thus， it is natural to consider the rings with the condition that every element is either a sum or a difference of a nilpotent and an idempotent. These rings are called weakly nil-clean rings and their clean analogs were called weakly clean rings earlier in ［12］. Idempotents and linear combinations of idempotents are very important in linear algebra and operator theory. So， a natural simplest generalization of idempotent is g -idempotent. A unit-picker is a map g" that associates to every ring [R] a well-defined set g [（R）] of central units in [R] which contains [1R] and is invariant under isomorphisms of rings and closed under taking inverses， and which satisfies certain set containment conditions for quotient rings， corner rings and matrix rings in ［2］. An element [a] of a ring [R] is called g -idempotent if [a2=ga] for some [g] of g [（R）]， or equivalently， [ha] is an idempotent for some [h∈] g [（R）] in ［2］. A ring is called nil g -clean if each of its elements is a sum of a g -idempotent and a nilpotent by Tang in ［2］.

An involution of a ring [R] is an operation *： [R→R] such that [（x+y）*=x*+y*]，[（xy）*=y*x*] and [（x*）*=x] for all [x，y] in [R]. Clearly， the identity mapping [idR] is an involution of [R] if and only if [R] is commutative. A ring [R] with involution * is called a *-ring. An element [p] in a * -ring [R] is called a projection if [p2=p*=p]. Obviously， 0 and 1 are projections of any *-rings.

In this paper， we adapt nil g -cleanness to *-rings. Let g" be a unit-picker. We called a [*]-ring [R] nil g -*-clean ring if for each [a∈R]，[a=vp+b] where [v∈] g [（R）]， [p2=p*=p∈R] and [b∈R] is a nilpotent. Two obvious examples of unit-pickers are g[1] and g [±1] given by g[1（R）=1R] and g[±1（R）=±1R]（for any ring [R]）. Thus， a *-ring [R] is nil *-clean iff [R] is nil g[1]-*-clean， and [R] is weakly nil-*-clean iff [R] is nil g[±1]-*-clean. In Section 2， several basic properties of nil g -*-clean rings are investigated. It is clear that every nil g - * -clean ring is NR-clean by Proposition 2.10， but the converses are not necessarily true， such as [Ζ5] and [M2（Ζ5）]. We prove that [R] is nil g - * -clean if and only if [J] is nil， （g[（R）+J）/J=]g[（R/J）]and [RJ] is nil g - * -clean. In Section 3， we research the basic properties and structures of commutative nil g - * -clean rings.

Throughout， rings are assumed to be associative with identity. The Jacobson radical， the unit group， the center and the set of nilpotents of a ring [R] is denoted by [J（R）] （or [J]）， [U（R）]， [C（R）] and nil[（R）]， respectively. Let [Uc（R）] be the set of central units of [R] and let idem[（R）] be the set of idempotents of [R]. We write [Mn（R）] for the ring of [n×n] matrices over a ring [R] whose identity is denoted by [In]. A ring is reduced if it contains no nonzero nilpotents. A ring is abelian if every idempotent is central.

1 Nil [G]-*-clean rings

We recall from ［2］ that a unit-picker is a map g that associates to every ring [R] a well-defined set g [（R）] of central units in [R] such that g [（R）] is invariant under isomorphism of rings and the following are satisfied：

（1） [1R∈] g[（R）] and [g-1∈] g[（R）] whenever [g∈] g[（R）]；

（2） （g[（R）+I）/I⊆]g[（R/I）]；

（3） eg[（R）⊆]g[（eR e）] for any [e2=e∈R]；

（4） g[（Mn（R））⊆]g[（eMn（C（R）））]for any [ngt;1].

Example 1.1 ［2， Example 2.2］ The following g[1]， g[±1]， g[1n]， g[1∞]， g[1n∞] and g[f] are examples of unit-pickers[：]

（1） g[1（R）=1]；

（2） g[±1（R）=1，-1]；

（3） g[1n（R）=g∈Uc（R）：gn=1]， where [n] is a fixed positive integer；

（4） g[1∞（R）=g∈Uc（R）：gk=1] for some [k≥1]；

（5） g[1n∞（R）=g∈Uc（R）：gnk=1] for some [k≥1]， where [n] is a fixed positive integer；

（6） g[f（R）=Uc（R）]， here the symbol [f] comes from the word full.

Definition 1.2 An element [a] in a * -ring [R] is called a g -projection if [a=gp]for some [g∈] g [（R）] and [p2=p*=p]. A * -ring [R] is called g -projection， if each of its elements is a g -projection.

Let [R] be a * -ring and [a，b∈R]. Consider the following equations：

（1）[a=aba]， （2）[b=bab]， （3）[（ab）*=ab]，（4）[（ba）*=ba].

If [a] and [b] satisfy （1） and （3）， [b] is called a [1，3]-inverse of [a]； if [aa] and [b] satisfy （1） and （4）， [b] is called a [1，4]-inverse of [a]； and if [a，b] satisfy （1）-（4）， then [b] is called an MP-inverse of [a] （see also ［15］）. Generally， [1，3]-inverses and [1，4]-inverses can not imply each other.

Due to ［11］， a ring [R] is called regular （in the sense of von Neumann） if for every element [ar∈R]， there is some element [a∈R] such that [r=rar]. The set of regular elements in [R] is denoted by reg[（R）].

Due to ［13］， a * -ring [R] is called * -regular if for every [x∈R]， there exists a projection [p] such that [xR=pR] （equivalently， [R] is regular and the involution is proper）.

Due to ［4］ and ［12］， an element [a] of a * -ring [R] is strongly * -regular if [a=pu=up] with [p2=p*=p] and [u∈U（R）]. [R] is strongly * -regular if each of its elements is strongly * -regular （equivalently， [R] is strongly regular and the involution is proper）.

Due to ［9］， a commutative ring is regular if and only if it is strongly regular. Hence a commutative ring is *-regular if and only if it is strongly *-regular.

Lemma 1.3 ［14， Proposition 2.2］" " Let [R] be a * -ring. [R] is * -regular if and only if for every [a∈R]， [a] has an MP-inverse.

Proposition 1.4 Let [R] be a * -ring and g[=]gf. Then [R] is g -projection if and only if [R] is commutative * -regular.

Proof [（⇒）] For any [a∈R]， write [a=vp] where [v∈] g[f（R）] and [p2=p*=p]. Since

[（vp）（v-1p）（vp）=vp，（v-1p）（vp）（v-1p）=v-1p，]

[（（vp）（v-1p））*=（vp）（v-1p）] and [（（v-1p）（vp））*=（v-1p）（vp），]

[a=vp] has an MP-inverse. Hence [R] is * -regular by Lemma 1.3. Then we show that [R] is commutative. Note that for any [0≠a∈R]， we have [a=vp] where [v∈]g[f（R）] and [p2=p*=p]. It follows that [an=vnp≠0] for any [n∈Z+]. Hence [R] is reduced and then every idempotent of [R] is central. Thus every element [a=vvp] is a central element.

[（⇐）] Since [R] is commutative * -regular， [R] is strongly * -regular. Then for any [a∈R]， write [a=pu=up] where [u∈U（R）] and [p2=p*=p]. Hence [R] is g -projection.

Definition 1.5 An element [a] in a * -ring [R] is called nil g - * -clean if it is a sum of a g -projection and a nilpotent. A * -ring is called a nil g - * -clean ring if each of its elements is nil g - * -clean.

A * -ring is called * -clean if every element is a sum of a unit and a projection（see ［10］）. A * -ring is called （strongly） nil * -clean if every element is a sum of a nilpotent and a projection （that commute）（see ［3］ and ［16］）.

For example， nil g [1]- * -clean rings are just nil- * -clean rings.

If [R] is nil g - * -clean， [R] is * -clean. Note that [R] is nil g - * -clean， for any [a∈R]， write [a=vp+b] where [v∈]

g[（R）]， [p2=p*=p] and [b∈] nil[（R）]. Then [a=vp+b=（1-p）+vp-（1-p）+b]， where [vp-（1-p）+b∈U（R）] and [1-p] is a projection. Hence [R] is * -clean. So， we have the following implications：

strongly nil * -clean [⇒] nil * -clean [⇒] nil g - * -clean [⇒] * -clean.

Lemma 1.6 ［2， Proposition 2.21］ Let g [=]g[f] and [D] a division ring. Then [M2（D）] is nil g -clean iff [D] is a perfect field of characteristic [2] （i.e. a field of characteristic [2] and whose every element is a square）.

Example 1.7 Let [R=T2（Z2）] be the [2×2] upper triangular matrix ring over [Z2] and *： [R→R] given by" [ab0c↦cb0a] is an involution. Then the followings hold[：]

（1） [R] is a nil g -clean ring；

（2） [R] is not a nil g - * -clean ring.

Proof （1） By Lemma 1.6， [R=T2（Z2）] is nil [G]-clean.

（2） Obviously， [R] is a *-ring. Assume that [R] is nil g- * -clean. For any [a∈R]， write [a=vp+b] where [v∈]

g [f（R）]， [p2=p*=p] and [b∈] nil[（R）]. Since nil[（R）=0000，0100] and [Uc（R）=1001]， we have [b*=b]， [v*=v] for any [b∈] nil[（R）]， [v∈] g[（R）]. Then， [a*=（vp+b）*=p*v*+b*=vp+b=a]. So， * =[1R]. Hence， [R] is commutative， a contradiction.

Proposition 1.8 Let [R] be a * -ring， [u∈U（R）] and [a∈R] is a nil g -*-clean element. If [u*=u-1]， then [u-1au] is nil g -*-clean.

Proof Since [a] is nil g - * -clean， write [a=vp+b] where [v∈]g[（R）]， [p2=p*=p] and [b∈] nil[（R）]. As [b] is nilpotent， we have [bm=0] for some [m∈Z+]. Then， [（u-1bu）m=u-1bmu=0]， [（u-1pu）2=u-1pu]and [（u-1pu）*=u*p（u-1）*=u-1pu]. So， [（u-1pu）2=（u-1pu）*=u-1pu] and hence [u-1au=u-1vpu+u-1bu=vu-1pu+u-1bu] is nil g - * -clean.

Proposition 1.9 Let [R] be a * -ring， for [γ∈J（R）]， if [γ=vp+b] is a nil g - * -clean decomposition， where [v∈] g [（R）]， [p2=p*=p] and [b∈] nil[（R）]， then [p=0].

Proof Since [b∈ ]nil[（R）]， [bm=0] for some [m∈Z+]. Hence [p=pm=（v-1（γ-b））m∈J（R）]. So [p=0].

Due to ［9］， an element [a] of a ring [R] is called a quasi-idempotent if [a2=ka] for some central unit [k] of [R]. An element [a] of a ring [R] is called quasi-clean if [a] is a sum of a unit and a quasi-idempotent； [R] is called quasi-clean if each of its elements is quasi-clean.

Due to ［11］， an element [aa] of a ring [R] is called NR-clean if [a=r+b] where [r∈] reg[（R）] and [b∈] nil[（R）]； [R] is called an NR-clean ring if every element is NR-clean.

Proposition 1.10 Let [R] be a * -ring， then the followings hold[：]

（1） If [R] is a nil g - * -clean ring that is indecomposable， then for any [aa∈R]， [a∈] nil [（R）] or [ab∈U（R）]；

（2） Every nil g - * *-clean ring is quasi-clean；

（3） Every nil g - * -clean ring is NR-clean.

Proof （1） Since [R] is a nil g - * -clean ring that is indecomposable， for any [a∈R]， write [a=vp+b] where [v∈] g[（R）]， [p=0] or [pp=1] and [b∈] nil[（R）]. If [p=0]， then [a=b∈] nil[（R）]. If [p=1]， then [a=v+b∈U（R）].

（2） Let [R] be a nil g - * -clean ring， then for any [a∈R]， write [a-1=vp+b] where [v∈] g [（R）]， [p2=p*=p] and [b∈] nil[（R）]. Then [a=vp+（b+1）]， [b+1∈U（R）]. So， [R] is quasi-clean.

（3） Let [R] be a nil g - * -clean ring， then for any [a∈R]， write [a=vp+b] where [v∈] g [（R）]， [p2=p*=p] and [b∈] nil[（R）]. Since [vp] is regular， [R] is NR-clean.

Remark （1） The converse of Proposition 1.10（2） need not be true. For example， the ring [Z5] is quasi-clean. However， the element [2] can not be written as a sum of a g -projection and a nilpotent because g and g[f] are not necessarily equal. Hence， the ring [Z5] is not nil g - * -clean.

（2） NR-clean rings may not be nil g - * -clean rings. For example， let [R=M2（Z2）] be the [2×2] matrix ring over [Z2] and *： [R→R] given by [abcd↦dbca] is an involution. Projections of [R] are [0000]， [1001] and nil[（R）=0000，0100，0010，1111]. Clearly， [R] is NR-clean but not nil g-*-clean because [1010] is not a nil g-*-clean element.

An ideal [I] of a *-ring [R] is called a * -ideal if [I*⊆I]. For a *-ideal [I] of a *-ring [R]， [RI] is a *-ring， where *： [R/I→R/I] is given by [a*=a*]. In particular， if [R] is a *-ring， then [R/J（R）] is a *-ring.

Proposition 1.11 If [R] is nil g - * -clean， so is [RI] for any * -ideal [I⊲R].

Proof Let [a∈R]， and write [a=vvp+b] where [v∈] g [（R）]， [pp2=p*=p] and [b∈] nil[（R）]. So， in [RI]， [a=vp+b] with [v∈] g [（R/I）]， [p2=p*=p] and [b∈] nil[（R/I）].

Corollary 1.12 Let [R] be a *-ring. If [R] is nil g - * -clean， the followings hold[：]

（1） [R/J（R）] is nil g - * -clean[;]

（2） If nil[（R）] is an ideal， then [R/]nil[（R）] is nil g - * -clean.

Proof （1） In view of Proposition 1.11， it suffices to show that [J（R）] is a * -ideal. For any [a*∈（J（R））*]， we show that [a*∈J（R）]. Note that [a∈J（R）]. Take any [x∈R]， then [1-x*a∈U（R）]. Hence，[1-a*x=（1-x*a）*] is a unit of [R]， as desired.

（2） By assumption， it suffices to show that nil[（R）] is a * -ideal. For any [a*∈（]nil[（R））*]，we show that [a*∈] nil[（R）]. Since [a∈] nil[（R）]， [am=0] for some [m∈Z+]. Then[（a*）m=（am）*=0]， as desired. The rest follows from Proposition 1.11.

Lemma 1.13 ［3， Lemma 2.4］" " Let [R] be a * -ring. If idempotents lift modulo [J（R）]， then every projection of [R/J（R）] is lifted to a projection of [R].

Theorem" " 1.14" " Let [R] be a * -ring. Then [R] is nil g- * -clean if and only if [J] is nil， （g [（R）+J）/J=] g [（R/J）]and [RJ] is nil g - * -clean.

Proof" " [（⇒）] If [R] is nil g- * -clean， then [RJ] is clearly nil g - * -clean by Proposition 1.11. Let [j∈J]. Then [j=vp+b] where [v∈]g[（R）]， [p2=p*=p] and [b∈] nil[（R）]. So， in [RJ]， [vp=-b] is nilpotent， showing that [p=0]. So [p∈J] and hence [p=0]. Thus， [j=b] is a nilpotent. Since g is a unit-picker， （g[（R）+J）/J⊆]g[（R/J）]. Let [u∈]g[（R/J）]， and write [u=vp+b] where [v∈]g[（R）]， [p2=p*=p] and [b∈]nil[（R）]. Then [vp=u-b] is a unit in [RJ]， so [p=1]. Hence， [b=u-v] is central. So， [b∈J（R/J）=0]. That is，[b∈J]. Hence，g [（R/J）⊆]（g[（R）+J）/J].

[（⇐）] Let [a∈R]. In [RJ]， write [a=vvp+b] where [v∈]g[（R/J）]， [p2=p*=p] and [b∈]nil[（R/J）]. Since [J] is nil， idempotents lift modulo [J]. Projection lift modulo [J] by Lemma 2.13. Thus， we may assume that [b∈] nil[（R）] and [p2=p*=p]. As （g[（R）+J）/J=]g[（R/J）]， we may further assume that [v∈] g[（R）]. Now， [a=vp+（b+j）] for some [j∈J]. As [J] is nil， [b+j∈]nil[（R）]. So， [a] is nil g - * -clean.

Let [R=Πni=1Ri] be the direct product of [Ri（i=1，2，…，n）]and [*i：Ri→Ri] is the involution of [Ri]. Define * ： [R→R] by [（ai）*↦（a*i）]. Then [R] is a * -ring.

Proposition 1.15 Let [R] be a * -ring and [i=1，2，…，n]. Then [R=Πni=1Ri] is nil g - * -clean if and only if every ring [Ri] is nil g -*-clean.

Proof [（⇒）] For any [i=1，2，…，n]， write [I=R1×…×Ri-1×0×Ri+1×…×Rn]， then [I] is an ideal of [R] and [R/I≅Ri]. Hence， [Ri] is nil g- * -clean by Proposition 1.11.

[（⇐）] [a=（r1，r2，…，rn）∈R] for [ri∈Ri]， [i=1，2，…，n]. Since [Ri] is nil g- * -clean， write [ri=vipi+bi] where [vi∈G（Ri）]， [p2=p*=p] and [b∈] nil[（Ri）]. Let [v=（v1，v2，…，vn）]， [p=（p1，p2，…，pn）] and [b=（b1，b2，…，bn）]， then [a=vp+b] where [v∈] g [（R）]， [p2=p*=p] and [b∈] nil[（R）]. Hence， [R] is nil g - * -clean.

For a * -ring [R]， there is an induced involution of [Mn（R）] given by [（aij）*=（a*ij）T]， where [AT] denotes the transpose of a matrix [A].

Theorem 1.16 Let [R] be a * -ring， g=g[f] and [e] is a central projection of [R]， then [R] is nil g - * -clean if and only if [eR e] and [（1-e）R （1-e）] are nil g - * -clean.

Proof [（⇒）] Let [eae∈eR e]. Then [a=vp+b] where [v∈G（R）]， [p2=p*=p] and [b∈] nil[（R）]. Hence， [eae=e（vp+b）e=（eve）（epe）+ebe]. Since [ab] is nilpotent， we have [bm=0] for some [m∈Z+]. Then， [（ebe）m=ebme=0] and [（epe）2=（epe）*=epe]. Since g is a unit-picker， [e] g[（R）⊆G（eR e）]for any [e2=e*=e∈R]. Hence， [eve=ev∈=]g [（eR e）]. So， [eae] is nil g - * -clean. Similarly， [（1-e）a （1-e）] is nil g - * -clean.

[（⇐）] Since [e∈R] is an idempotent， we have a pierce decomposition of [R]：

[R=eR e⊕eR （1-e）⊕（1-e）R e⊕（1-e）R（1-e）].

It follows from [e∈c（R）] that [R≅eR e⊕（1-e）R（1-e）≅eR e00（1-e）R（1-e）]. For any [A∈R]， write [A=a00b] where [a∈eR e] and [b∈（1-e）R（1-e）]. Because [eR e] and [（1-e）R（1-e）] are nil g -*-clean， [a=v1p1+b1] and [b=v2p2+b2] where [v1∈] g [（eR e）]， [v2∈] g[（（1-e）R （1-e））]， [p12=p1*=p1]， [p22=p2*=p2] and [b1∈] nil[（eR e）]， [b2∈] nil[（（1-e）R （1-e））]. Then we have that

[A=a00b=v1p1+b100v2p2+b2=v100v2p100p2+b100b2].

Obviously， [b100b2∈]nil[（R）]， [v100v2∈Uc（R）=] g [（R）]and [p100p22=p100p2=p100p2T]. So [A] is nil g - * -clean and [R] is nil g - * -clean.

Let [R] be a *-ring. Define * ： [R[x]/（xn+1）→R[x]/（xn+1）] by [a0+a1x+…+anxn+（xn+1）↦a0*+a1*x+…+an*xn+（xn+1）]. Then [R[x]/（xn+1）] is a *-ring.

Proposition 1.17 Let [R] be a * -ring. Then [R] is nil g - * -clean if and only if [R[x]/（xn+1）] is nil g - * -clean for every [n≥0].

Proof [（⇒）] One can check that

nil[（R[x]/（xn+1））=a0+a1x+…+anxn+（xn+1）a0∈] nil[（R），a1，…an∈R].

Let [f（x）=a0+a1x+…+anxn+（xn+1）∈R[x]/（xn+1）]. Write" [a0=vp+b] where [v∈] g [f（R）]， [pp2=p*=p] and [b∈] nil[（R）]. Then [f（x）=vp+（b+a1x+…+anxn+（xn+1））∈R[x]/（xn+1）] where [v∈] g [f（R）]， [p2=p*=p] and [b+a1x+…+anxn+（xn+1）∈] nil[（R[x]/（xn+1））]. Hence [f（x）] is nil g - * -clean.

[（⇐）] Since a homomorphic image of [R[x]/（xn+1）] is [R]， then [R] is nil g - * -clean.

Let [R] be a * -ring and let [T（R，R）] be the trivial extension of [R] by [R]， i.e.，[T（R，R）=ab0a a，b∈R] . Define * ： [T（R，R）→T（R，R）] by [（x，y）↦（x*，y*）]. Then [T（R，R）] is a * -ring.

Corollary" " 1.18" " Let [R] be a *-ring. Then [R] is nil g -*-clean if and only if so is [T（R，R）].

Proof" " It is obvious from [T（R，R）≅R[x]/（x2）] and Proposition 1.17.

2 Commutative" nil [G]-*-clean rings

In this section， we focus on the study of commutative nil g - * -clean rings.

Let [R] be a commutative ring and [S] be a multiple closed subset of [R]. Define an equivalence relation on the set [R×S] by

[（a，s）～（b，t）⇔（at-bs）u=0] for some [u∈S].

The equivalence class of [（a，s）] in [R×S] is denoted by [as]. We define a ring structure on [S-1R] by

[（a/s）+（b/t）=（at+bs）/st]；

[（a/s）（b/t）=ab/st].

Then [S-1R] is called the fraction ring， also called localization.

By ［9， Proposition 3.1］， a localization of a commutative clean ring need not to be a clean ring as every indecomposable commutative clean ring is local and every localization of a commutative quasi-clean ring is quasi-clean.

Proposition 3.1 Let [R] be a commutative *-ring. Then every localization of a commutative nil g - * -clean ring is nil g - * -clean for g=gf.

Proof" nbsp; Let [S] be a multiple closed subset of [R]. Set [T=S-1R]. For any [x∈T]， [x=s-1a]， where [a∈R] and [s∈S]. Since [R] is nil g -*-clean， we know that [a] has a nil g - * -clean decomposition [a=vp+b] in [R]. Then we have [s-1vp+s-1b] is a nil g - * -clean decomposition in [T].

We use [N（R）] to denote the prime radical of [R]， i.e. it is the intersection of all prime ideals of [R]. For a commutative ring [R]， we have [N（R）=] nil[（R）].

Proposition 2.2 Let [R] be a commutative *-ring. Then [R] is nil g - * -clean if and only if （g[（R）+N（R））/N（R）=]g[（R/N（R））] and [R/N（R）] is nil g - * -clean.

Proof It follows from Theorem 1.14.

Due to ［12］， An element [a] of a * -ring [R] is called strongly [π]- * -regular if it satisfies the conditions in [[12]， Theorem [3.2]]； [R] is called strongly [π]-*-regular if every element is strongly [π]- * -regular.

Lemma" " 2.3" " ［12， Theorem 3.6］" " Let [R] be a *-ring. Then [R] is strongly [π]-*-regular if and only if for any [a∈R]， there exist [p2=p*=p] and [u∈U（R）] such that [a=p+u]， [ap∈] nil[（R）]； and [v-1qv] is a projection for all [v∈U（R）] and all [q2=q*=q].

Proposition 2.4 Let [R] be a commutative *-ring， g=gf" and [r∈R]. If [r=a+b]， where [a] is a strongly [π]-*-regular element of [R] and [b∈] nil[（R）]， then [r] is nil g -*-clean.

Proof Since [a] is strongly [π]-*-regular， we can write [a=p+u] by Lemma 2.3， where [p2=p*=p]， [u∈U（R）] and [ap∈] nil [（R）]. Then we can write

[r=a+b=pa+（1-p）a+b=pa+（1-p）u+b=u（1-p）+（b+pa）].

Since [R] is a commutative *-ring， [b+pa∈] nil[（R）]. Hence， this is a nil g - * -clean decomposition.

Let [R] be a *-ring and [I] be a *-ideal. [RI] is called a *-decomposable ring if [R/I≅R/I1⊕R/I2]" where [I1] and [I2] are *-ideals.

Theorem 2.5 Let [R] be a commutative *-ring and g=gf . Then [R] is nil g - * -clean iff for each *-ideal [I⊲R] with [RI] *-indecomposable， every element of [RI] is either a unit or a nilpotent.

Proof [（⇒）] Suppose that [RI] is an *-indecomposable image of [R]. Then [RI] has only the trivial projections. Thus， the g -projection in [RI] are either zero or units. So， for [x∈R/I]， either [x] is zero plus a nilpotent， which is a nilpotent， or [x] is a unit plus a nilpotent， which is a unit.

[（⇐）] Assume that [a∈R] is not nil g - * -clean. Then

[ℱ]=[{I⊲R：a+I∈R/I]is not nil g - * -clean[}]

is not empty. For a chain [Iλ] of elements of [ℱ]. Let [I=UλIλ]， then [I] is an ideal of [R]. Assume that [a+I∈R/I] is nil g - * -clean. Thus there exist [p，v，v′，b∈R] such that

[a+I=（v+I）（p+I）+（b+I）]， where

[（p+I）2=（p+I）*=p+I]， [（v+I）（v+I）′=1+I]，[（b+I）m=0] for some [m∈Z+].

Thus， all the following elements are in [I]：

[a-vp-b，p2-p，p*-p，vv′-1，bm].

Because [Iλ] is a chain， there exists some [Iλ] such that all these elements are in [Iλ]. So， [a+Iλ=（v+Iλ）（p+Iλ）+（b+Iλ）∈R/Iλ] is nil g - * -clean. This contradiction shows that [I] is in [ℱ]. By Zorn’s Lemma， [ℱ] has a maximal element [I]. Next， we show that [RI] is * -indecomposable.

Assume that [RI] is * -decomposable. Then there exist *-ideals [Ij⊉I（j=1，2）] such that

[R/I≅R/I1⊕R/I2;r+I↦（r+I1，r+I2）].

By the maximality of [I] in [ℱ]， for [j=1，2，a+Ij∈R/Ij] is nil g - * -clean， so

[a+Ij=（vj+Ij）（pj+Ij）+（bj+Ij）]，

where [vj+Ij] is a unit in [RIj]， [pj+Ij] is an projection in [RIj] and [bj+Ij] is a nilpotent in [RIj]. Hence

[（a+I1，a+I2）=（v1+I1，v2+I2）（p1+I1，p2+I2）+（b1+I1，b2+I2）]

is a sum of a g -projection and a unit in [R/I1⊕R/I2]. This shows that [a+I∈R/I] is nil g - * -clean， a contradiction. Hence [RI] is *-indecomposable， with [a+I∈R/I] not nil g - * -clean. So [a+I∈R/I] is neither a unit nor a nilpotent， a contradiction. Hence， [R] is nil g -*-clean.

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诣零g-*-clean环

李筱璇，殷晓斌

（安徽师范大学 数学与统计学院， 安徽 芜湖 241002）

摘要：如果一个环的每一个元素都可以表示成一个g -投射元和一个幂零元之和，那么这个环被称为诣零 g - * -clean环。本文研究了该环的基本性质以及与其它环的一些关系，证明了[R]是诣零 g - * -clean环当且仅当[J]诣零， g [（R）+J）/J=] g [（R/J）]且[RJ]是诣零g - * -clean环。

关键词：诣零clean环； 诣零g - clean环； * -环； 诣零 * -clean环； 诣零g - * -clean环

（责任编辑：马乃玉）