Centrally clean elements and central Drazin inverses in a ring

Li Wende Chen Jianlong

(School of Mathematics, Southeast University, Nanjing 211189, China)

AbstractElement a in ring R is called centrally clean if it is the sum of central idempotent e and unit u. Moreover, a=e+u is called a centrally clean decomposition of a and R is called a centrally clean ring if every element of R is centrally clean. First, some characterizations of centrally clean elements are given. Furthermore, some properties of centrally clean rings, as well as the necessary and sufficient conditions for R to be a centrally clean ring are investigated. Centrally clean rings are closely related to the central Drazin inverses. Then, in terms of centrally clean decomposition, the necessary and sufficient conditions for the existence of central Drazin inverses are presented. Moreover, the central cleanness of special rings, such as corner rings, the ring of formal power series over ring R, and a direct product ∏Rα of ring Rα, is analyzed. Furthermore, the central group invertibility of combinations of two central idempotents in the algebra over a field is investigated. Finally, as an application, an example that lists all invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of the group ring Z2S3 is given.

Key wordscentrally clean element; centrally clean ring; central Drazin inverse; central group inverse

In the research on ring theory, the cleanness of a ring is a basic but important topic. Clean rings originated from the study of exchange rings, which play an important role in the cancellation of modules. The interesting characterizations and properties of clean rings have motivated many scholars to conduct further investigations. The concept of clean rings was first introduced by Nicholson[1] in 1977. Subsequently, Nicholson et al.[2] proved that the linear transformation of a countable vector space over a division ring is clean. In 1999, Nicholson[3] introduced a strongly clean ring and presented some equivalent characterizations of strongly clean elements and rings. In 2001, Han et al.[4] investigated the cleanness of group rings, the ring of formal power series over a ring, and a direct product of rings. In 2011, Hiremath et al.[5] presented some characterizations of strongly clean rings. More details concerning the cleanness of the rings can be found in Refs.[6-12].

Throughout the paper,R denotes an associative ring with unity 1. The center of R is denoted by C(R)={xR:ax=xa for all aR}. The element eR is considered idempotent if e2=e. In contrast, the element eR is considered central idempotent if e2=e and eC(R). The symbols E(R), CE(R), U(R), and J(R) denote the sets of all idempotents, central idempotents, invertible elements, and Jacobson radicals of R, respectively. Recall that the element aR is considered clean if uU(R) and eE(R) exist such that a=u+e. The element aR is considered strongly clean if uU(R) and eE(R) exist such that a=u+e and ue=eu. In this case, a=e+u is considered a strongly clean decomposition.

Drazin[13] introduced the concept of pseudo-inverse (usually called Drazin inverse) in rings and semigroups. The element aR is considered a Drazin invertible if xR and the nonnegative integer k exist such that xax=x, ax=xa, xak+1=ak. Such x is unique if it exists and is considered the Drazin inverse of a. The smallest nonnegative integer k satisfying the previously presented equations is called the Drazin index of a. If k=1, then x is considered the group inverse of a.

Further research showed that there is a close connection between clean rings and Drazin inverses. For example, Zhu et al.[14] proved that aR is a Drazin invertible if and only if uU(R), eE(R) and the positive integer n exist such that an=u+e is a strongly clean decomposition and anReR=0. Moreover, many scholars investigated the Drazin invertibility of combinations of idempotents. For instance, Liu et al.[15] analyzed this topic in complex matrices, i.e., Drazin invertibility of aP+bQ+cPQ+dQP+ePQP+fQPQ+gQPQP for idempotent complex matrices P and Q under the conditions (PQ)2=(QP)2. More details concerning Drazin inverses can be found in Refs.[16-26].

In 2019, to analyze the commutative properties of Drazin inverses (see Example 2.8 in Ref.[27]), Wu et al.[28] introduced the concept of central Drazin inverses.

Definition 1[28] Element aR is considered a central Drazin invertible if xR and the nonnegative integer k exist such that xax=x, xaC(R), xak+1=ak. Such x is unique if it exists and is considered the central Drazin inverse of a, denoted by ac. The smallest nonnegative integer k satisfying the previously presented equations is still the Drazin index of a, denoted by ind(a). If k=1, then x is called the central group inverse of a, denoted by a©.

In Ref.[28], a centrally clean element and a centrally clean ring are also introduced.

Definition 2[28] Let aR. If uU(R) and e∈CE(R) exist such that a=u+e, then a is considered centrally clean. In this case, a=u+e is considered a centrally clean decomposition of a. Thus, centrally clean is clean. If every element of R is centrally clean, then R is considered a centrally clean ring.

Moreover, aR is a central Drazin invertible if and only if uU(R), e∈CE(R) and the positive integer n exist such that a=u+e is a centrally clean decomposition, and anReR=0, or equivalently, uU(R), e∈CE(R) and the positive integer n exist such that a=u+e is a centrally clean decomposition and ae is nilpotent. Subsequently, Zhao et al.[29] investigated the one-sided central Drazin inverses.

Motivated by the previous studies, we investigated centrally clean elements and central Drazin inverses in R. We first give an example and characterizations of centrally clean elements. Then, we analyze the properties of centrally clean rings and provide some equivalent characterizations. Moreover, we present the necessary and sufficient conditions for the existence of central Drazin inverses in terms of centrally clean decompositions. In addition, we investigate the central group invertibility of combinations of two central idempotents. Finally, we calculate all invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of the group ring Z2S3.

1 Characterization of Centrally Clean Elements

First, we provide an example of centrally clean elements.

Example 1

1) Units are centrally clean.

2) The elements in J(R) are centrally clean.

3) Nilpotent elements are centrally clean.

4) Central idempotents are centrally clean.

Proof 1) and 3) are obvious.

2) Let xJ(R). Notably, J(R)={xR: 1-ax is left invertible for any aR} and J(R)={xR: 1-xa is right invertible for any aR}.

Then, we take a=1, and it follows that 1-xU(R). Hence, x is centrally clean.

4) Let e∈CE(R). Given that (2e-1)2=1 and (1-e)2=1-e, it follows that 2e-1∈U(R) and 1-e∈CE(R). Then, e=(2e-1)+(1-e). Hence, e is centrally clean.

Let aR. Then, we use Ra and aR to denote the left and right ideals generated by a, respectively. We use l(a) and r(a) to denote the left and right annihilators of a, respectively. That is,

Ra={ra: rR}, aR={ar: rR}

l(a)={xR: xa=0}, r(a)={xR: ax=0}

Nicholson[3] proved that if eE(R) and aeRe is strongly clean, then aR is also strongly clean. Moreover, he provided some characterizations of strongly clean elements. Then, we investigate the relevant characterizations of centrally clean elements.

Lemma 1[3] Let aR and eE(R) with ea=ae. Then, the following conditions are equivalent:

1) aeU(eRe).

2) eRa and l(a)⊆l(e).

3) eaR and r(a)⊆r(e).

Theorem 1 Let aR. Then, the following conditions are equivalent:

1) a is centrally clean.

2) e∈CE(R) exists such that l(a)⊆R(1-e)⊆R(1-a) and l(1-a)⊆ReRa.

3) e∈CE(R) exists such that r(a)⊆(1-e)R⊆(1-a)R and l(1-a)⊆eRaR.

4) e∈CE(R) exists such that eaU(eR) and (1-e)(1-a)∈U((1-e)R).

5) e∈CE(R) exists such that ea is centrally clean in eR and (1-e)(1-a) is centrally clean in (1-e)R.

6) e∈CE(R) exists such that ea is centrally clean in eR and (1-e)a is centrally clean in (1-e)R.

7) The decomposition 1=e1+e2+…+en exists, where n is a positive integer, e is a centrally orthogonal idempotent, and eia is centrally clean in eiR for each positive integer i.

Proof 1)⟹2). Given that a is centrally clean, we can suppose that a=(1-e)+u, where e∈CE(R) and uU(R). If ra=0, then r(1-e)+ru=0, and it follows that r=ruu-1=[-r(1-e)]u-1R(1-e), i.e., l(a)⊆R(1-e). Moreover, from ae=[(1-e)+u]e=ue, we derive e=u-1ae=u-1eaRa, i.e., ReRa.

Rewrite a=(1-e)+u as 1-a=e+(-u). Then, by a similar argument, we can obtain l(1-a)⊆Re and R(1-e)⊆R(1-a).

2)⟹4). From eReRa and l(a)⊆R(1-e)=l(e), we can obtain eaU(eR) based on Lemma 1. Similarly, we can derive (1-e)(1-a)∈U((1-e)R).

1)⟹3)⟹4) are similar to 1)⟹2)⟹4).

4)⟹5). This is obvious from 1) of Example 1.

5)⟹6). Given that (1-e)(1-a) is centrally clean in (1-e)R, it follows that (1-e)a=(1-e)-(1-e)(1-a) is also centrally clean in (1-e)R.

6)⟹7) Write e1=e and e2=1-e. Then, e1e2=0 and 1=e1+e2. Moreover, eia is centrally clean in eiR for each i.

7)⟹1). For each positive integer i, eia is centrally clean in eiR. Then, we suppose that eia=fi+ui, where fi∈CE(eiR) and uiU(eiR), and it follows that vieiR exists such that viui=uivi=ei. Given that ei is a centrally orthogonal idempotent, we derive f=∑fi∈CE(R), u=∑uiU(R), and u-1=∑vi. Hence, a=∑eia=∑(fi+ui)=∑fi+∑ui=f+u. Therefore, a is centrally clean.

Proposition 1 Let aR. Then, the following conditions are equivalent:

1) a is centrally clean.

2) vU(R) and f∈CE(R) exist such that f=fva and 1-f=-(1-f)v(1-a).

3) uU(R) and f∈CE(R) exist such that f=fua and 1-f=(1-f)u(1-a).

4) f∈CE(R) and x,yR exist such that f=fxa, 1-f=(1-f)y(1-a), and fx-(1-f)yU(R).

Proof 1)⟹2). Let a=u+e, where uU(R) and e∈CE(R). Write f=1-e and v=u-1. Then,

f=(1-e)u-1a=fva

and

1-f=-eu-1(1-a)=-(1-f)v(1-a)

2)⟹1). Write e=1-f. Then, v(a-e)=[fv+(1-f)v](a-1+f)=fva-fv+fv-(1-f)v(1-a)=f+1-f=1. Hence, a-e=v-1, i.e., a is centrally clean.

2)⟹3). Write u=(2f-1)v. Then, f=fva=fua and 1-f=-(1-f)v(1-a)=(1-f)u(1-a).

3)⟹2) is similar to 2)⟹3).

2)⟹4). Write x=v and y=-v. Then, f=fxa and 1-f=(1-f)y(1-a).

4)⟹2). Write v=fx-(1-f)yU(R). Then, fx=fv and -(1-f)y=(1-f)v, and it follows that f=fxa=fva and 1-f=(1-f)y(1-a)=-(1-f)v(1-a).

2 Characterizations of Centrally Clean Rings

Recall that if R/J(R) is a division ring, then R is considered a local ring.

Proposition 2 Every local ring is a centrally clean ring.

Proof Let aR. If aJ(R), then a is centrally clean based on 2) of Example 1. If aJ(R), then a+J(R)∈U(R/J(R)). Hence, x+J(R)∈R/J(R) exists such that

(a+J(R))(x+J(R))=1+J(R)

and it follows that ax+J(R)=1+J(R), i.e., ax-1∈J(R). Therefore, ax=1-(1-ax)∈U(R). Then, a is right invertible in R. Similarly, we can deduce that a is left invertible in R. It follows that aU(R), and hence, it is centrally clean.

In 2004, Nicholson et al.[12] proved that if R≠0, then R[x] is not clean. Then, it is obvious that R[x] is not centrally clean when R≠0.

Proposition 3 The following conditions are equivalent:

1) 2∈U(R), and R is centrally clean.

2) For any aR, uU(R) and xC(R) exist, with x2=1, such that a=u+x.

Proof 1)⟹2). Let aR. Given that R is centrally clean, uU(R) and e∈CE(R) exist such that and it follows that a=(2e-1)+2u. From 2∈U(R), we derive 2uU(R). Moreover, (2e-1)2=1 and 2e-1∈C(R).

2)⟹1). Hypothetically, we have 1=x+v, where vU(R), x2=1 and xC(R). Then, (1-v)2=x2=1, and it follows that v2=2v. Hence, from vU(R), we derive 2=vU(R). Let aR. Then, 2a-1=y+w, where wU(R), y2=1 and yC(R). Therefore, is centrally clean.

Then, we provide some characterizations of centrally clean rings.

Theorem 2 The following conditions are equivalent:

1) R is centrally clean.

2) Every element xR can be written as x=u-e, where uU(R) and e∈CE(R).

3) Every element xR can be written as x=u+e, where uU(R)∪0 and e∈CE(R).

4) Every element xR can be written as x=u-e, where uU(R)∪0 and e∈CE(R).

Proof 1)⟹2). Let xR. Then, -x=e+v, and it follows that x=-v-e, u=-vU(R) and e∈CE(R).

2)⟹3) and 3)⟹4) are similar to 1)⟹2).

4)⟹1). Let xR. Then, we derive -x=u-e based on the assumption, where uU(R)∪0 and e∈CE(R). Hence, x=(-u)+e. The case when u=0 follows from 4) of Example 1.

Recall that if eE(R) exists such that eaR and 1-e∈(1-a)R for any aR, then R is an exchange ring. Moreover, if every idempotent of R is central, then R is called abelian.

Theorem 3 The following conditions are equivalent:

1) R is centrally clean.

2) R is an exchange and abelian.

3) R is clean and abelian.

4) For any aR, e∈CE(R) exists such that eaR and 1-e∈(1-a)R.

Proof 1)⟹3). It suffices to prove that every idempotent of R is central. Let eE(R). Then, we derive e=f+u, where f∈CE(R) and uU(R), and it follows that if f+u=(f+u)2=f+2fu+u2, then 1=u+2f. Hence, we obtain e=f+u=f+1-2f=1-f∈CE(R).

3)⟹4). Given that clean rings are exchange rings, it follows that, for any aR, eE(R) exists such that eaR and 1-e∈(1-a)R. Given that R is abelian, we derive e∈CE(R).

4)⟹2). This is enough to show that R is abelian. Let fE(R). Then, according to the assumption, e∈CE(R) exists such that efR and 1-e∈(1-f)R. Hence, we obtain fe=e and (1-f)(1-e)=1-e. Then, f=e∈CE(R). Therefore, R is abelian.

2)⟹1). Let xR. Given that R is an exchange ring, eE(R) exists such that exR and 1-e∈(1-x)R. Let e=xa′, where a′∈R. Then, e=e2=xaxa′. Write a=axa′, and it follows that e=xa and ae=axaxa′=axa′=a. Then, axa=a. Given that R is abelian, we derive ax=axax=xa(ax)=xaxa=xa. By a similar argument, we can obtain (1-e)=(1-x)b, b(1-e)=b, and (1-x)b=b(1-x). Furthermore, we can obtain [x-(1-e)](a-b)=xa-xb-(1-e)a+(1-e)b=e+(1-x)b=1 and (a-b)[x-(1-e)]=1. That is, [x-(1-e)]-1=a-b. Then, x=x-(1-e)+(1-e). Hence, R is centrally clean.

Proposition 4 Let p∈CE(R). Then, apR is centrally clean in R if and only if a is centrally clean in pR.

Proof The necessity is clearly stated in Theorem 1. Conversely, assume apR is centrally clean in R. Then, e∈CE(R) and uU(R) exist such that a=e+u, and it follows that pa=pe+pu, pe∈CE(pR), and puU(pR). From pa=a, we derive a=pe+pu. Hence, a is centrally clean in pR.

Corollary 1 Let p∈CE(R). If R is a centrally clean ring, then so is pR.

Han et al.[4] proved that when eE(R), if eRe and (1-e)R(1-e) are clean rings, then so is R. Here, we consider the case of e∈CE(R).

Corollary 2 Let e∈CE(R). If eR and (1-e)R are centrally clean, then so is R.

Proof This is clearly stated in Theorem 1.

Han et al.[4] also investigated the cleanness of group rings, the ring of formal power series over a ring, and a direct product of rings. Then, we analyze the relevant results of R[[x]] and ∏Rα.

Proposition 5 The ring R[[x]] is centrally clean if and only if R is centrally clean.

Proof Let f=a+bx+cx2+…∈R[[x]]. Given that R is centrally clean, we can suppose that a=u+e, where e∈CE(R) and uU(R). Then, f=e+(u+bx+cx2+…), e∈CE(R[[x]]), and u+bx+cx2+…∈U(R[[x]]). Therefore, R[[x]] is centrally clean.

Conversely, we know that R[[x]]/(x) is centrally clean because R[[x]] is centrally clean. Hence, RR[[x]]/(x) is centrally clean.

Lemma 2 Let R, S be two rings and φ: RS be a surjective ring homomorphism. If R is centrally clean, then so is S.

Proof It is obvious.

Proposition 6 A direct product R=∏Rα is centrally clean if and only if Rα is centrally clean.

Proof Given that πα: ∏RαRα is a surjective ring homomorphism, it follows that Rα is centrally clean based on Lemma 2.

Conversely, suppose that Rα is centrally clean. Let x=(xα)∈∏Rα. Then, for each α, we derive xα=uα+eα, where uαU(Rα) and eα∈CE(Rα). Hence, we obtain x=e+u, u=(uα)∈U(∏Rα) and e=(eα)∈CE(∏Rα), and it follows that R=∏Rα is centrally clean.

Let L be a two-sided ideal of R. We suppose that the idempotents can be lifted modulo L if, given that xE(R/L), eE(R) exists such that e-xL. Similarly, we can define the concept that the central idempotents can be lifted modulo L if e∈CE(R) exists such that e-xL for x∈CE(R/L).

Proposition 7 R is centrally clean if and only if R/J(R) is centrally clean, and the central idempotent can be lifted modulo J(R).

Proof Based on Lemma 2, we confirm that the factor ring of a centrally clean ring is centrally clean. Then, R/J(R) is centrally clean. Given that a centrally clean ring is exchange, it follows that the idempotents can be lifted modulo J(R). Based on Theorem 3, we determine that the idempotents of R are central. Then, the sufficiency is proven.

Conversely, let xR and Then, r1,r2J(R) exist such that uv=1+r1 and vu=1+r2. Hence, uU(R). Given that the central idempotents can be lifted modulo J(R), we suppose that p∈CE(R) and p-eJ(R), and it follows that rJ(R) exists such that x=p+u+r=p+u(1+u-1r). Given that u-1rJ(R), we derive 1+u-1rU(R). Hence, R is centrally clean.

3 Characterizations of Central Drazin Inverses

In this section, we mainly provide some characterizations for the existence of central Drazin inverses.

Theorem 4 Let aR. Then, the following conditions are equivalent:

1) a is central Drazin invertible.

2) uU(R), e∈CE(R), and the positive integer m exist such that am=eu and au=ua.

3) vU(R) and f∈CE(R) exist such that a=f+v and afRnil.

4) p∈CE(R) exists such that apU(pR) and a(1-p)∈Rnil.

Proof 1)⟹2). Write e=aac. Then, e∈CE(R). Given that a is central Drazin invertible, and it follows that the positive integer m exists such that am=amaac=ame. Write u=am+(1-e). Then, [am+(1-e)][(ac)me+(1-e)]=am(ac)me+am(1-e)+(1-e)(ac)me+(1-e)2=e+1-e=1. Hence, we have uU(R) and u-1=(ac)me+(1-e), and it follows that am=ame=[u-(1-e)]e=eu and au=a[am+(1-e)]=am+1+a(1-e)=am+1+(1-e)a=[am+(1-e)]a=ua.

2)⟹3). Write f=1-e. Then, f∈CE(R). Given that (am-f)(u-1e-f)=1, we derive am-fU(R). Then, (a-f)(am-1+am-2f+…+af+f)=am-fU(R). Hence, we obtain v=a-fU(R) and (af)m=amf=eu(1-e)=0, i.e., afRnil.

3)⟹4). Write p=1-f. Then, p∈CE(R), ap=pa=pvU(pR) and a(1-p)=afRnil.

4)⟹1). Based on this assumption, it follows that wU(pR) exists such that apw=pwa=p. From pw=w, we derive aw=wa=pC(R), waw=pw=w, and a-a2w=a(1-aw)=a(1-p)∈Rnil. Hence, a is central Drazin invertible.

Zhu et al.[14] showed that a is Drazin invertible if and only if uU(R), eE(R), and the positive integer m exist such that am=u+e is strongly clean decomposition and amReR=0. Here, we investigate the relevant results of central Drazin inverses.

Theorem 5 Let aR. Then, the following conditions are equivalent:

1) a is central Drazin invertible.

2) uU(R), e∈CE(R), and the positive integer n exist such that an=u+e and anReR=0.

3) uU(R), e∈CE(R), and the positive integer n exist such that an=u-e and anReR=0.

Proof 1)⟹2). Given that a is central Drazin invertible, we derive u=an-1+aacU(R) for any positive integer n>ind(a). Write e=1-aac. Then, an=u+e is the centrally clean decomposition. Let xanReR. Then, y,zR exist such that x=any=ez=eany=0. Hence, anReR=0.

2)⟹1). From e∈CE(R), it follows that the positive integer m exists such that:

(ane)m=(an)me=e(an)manReR=0

i.e., aneRnil. Let m be the nilpotent index of ane. Then, (an)m=um(1-e). In fact,

(an)m=(u+e)m=

um(1-e)+(an)me=

um(1-e)+(ane)m=um(1-e)

Hence, (an)m is central group invertible derived by Theorem 3.6 in Ref.[28]. Then, a is central Drazin invertible derived by Theorem 3.3 in Ref.[28].

1)⟺3). This is similar to the proof of 1)⟺2).

From Theorem 5, we derive the following corollary.

Proposition 8 Let aR. Then, the following conditions are equivalent:

1) a is central Drazin invertible.

2) e∈CE(R) and the positive integer n exist such that ane=0 and an-eU(R).

3) e∈CE(R) and the positive integer n exist such that ane=0 and an+eU(R).

Proof 1)⟹2). Given that a is central Drazin invertible, we derive u=an-1+aacU(R) for any positive integer n>ind(a). Write e=1-aac. Then, ane=0 and an-eU(R).

2)⟹1). Let xanReR. Then, y,zR exist such that x=any=ez=eany=aney=0. Hence, anReR=0. According to Theorem 5, the proof is completed.

1)⟺3). This is similar to the proof of 1)⟺2).

For the central group inverses, we also obtain the following relevant results.

Proposition 9 Let aR. Then, the following conditions are equivalent:

1) a is central group invertible.

2) uU(R) and e∈CE(R) exist such that a=u+e and aReR=0.

3) vU(R) and f∈CE(R) exist such that f=fva, 1-f=(1-f)v(1-a), and af=a.

Proof 1)⟺2). This is given in Corollary 4.6 in Ref.[28].

2)⟹3). Write f=1-e and v=u-1(1-2e). Then, vU(R) and f∈CE(R), and it follows that fva=(1-e)u-1(1-2e)(u+e)=(1-e)(1-2e)=1-e=f and a(1-f)=(u+e)eaReR=0. Hence, af=a and

(1-f)v(1-a)=eu-1(1-2e)(1-e-u)=

eu-1(1-2e)(-u)=e=1-f

3)⟹2). Write u=v-1(2f-1) and e=1-f. Given that 1-f=(1-f)v(1-a)=(1-f)v-(1-f)va=(1-f)v-va+fva=(1-f)v-va+f, it follows that a=v-1(1-f)v+v-1(2f-1)=1-f+v-1(2f-1)=e+u. Let xaReR. Then, r,tR exist such that x=ar=et, and it follows that fr=fvar=fv(et)=fv(1-f)t=0, i.e., r=(1-f)r. Then, x=ar=a(1-f)r=(a-af)r=0. Therefore, aReR=0.

4 Central Group Invertibility of Combinations of Two Central Idempotents

Motivated by the study conducted by Liu et al.[15], we investigate the central group invertibility of combinations of two central idempotents in this section.

In this section, F denotes a field and A denotes the algebra over F.

Theorem 6 Let p,qA be the central idempotent and a=d1p+d2q+d3pq, where diF, i=1,2,3. Then, a is central group invertible, and

where

Proof Let which suffices to prove that x is the central group inverse of a.

Given that p,q∈CE(R), we derive xaC(R). By computation, it follows that

Then, we obtain

and

[d1dd-d1+d2dd-d2+d3dd]pq=

Hence, x is the central group inverse of a, and where

Let d1=1, d2=1, d3=0. Then, we obtain the following results according to Theorem 6.

Corollary 3 Let 2∈U(R) and p,qA be the central idempotents. Then, p+q is central group invertible, and

If pq=p, then we obtain the following results according to Theorem 6. That is, we take d3=0 in Theorem 6.

Corollary 4 Let p,qA be the central idempotent and pq=p. Then, d1p+d2q is central group invertible, and

If pq=q, then we obtain the following results according to Theorem 6.

Corollary 5 Let p,qA be the central idempotent and pq=q. Then, d1p+d2q is central group invertible, and

5 An Example

In this section, we present all invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of Z2S3. For convenience, we write g1=(1), g2=(12), g3=(13), g4=(23), g5=(123), g6=(132), and

By computation, we obtain the following results:

E(Z2S3)={0,g1,g5+g6,g1+g5+g6,g2+g3+g5,g2+

g3+g6,g2+g4+g5,g2+g4+g6,g3+g4+g5,g3+

g4+g6,g1+g2+g3+g5,g1+g2+g3+g6,g1+g2+

g4+g5,g1+g2+g4+g6,g1+g3+g4+g5,g1+g3+

g4+g6}

C(Z2S3)={0,g1,g5+g6,g1+g5+g6,g2+g3+g4,

g1+g2+g3+g4,e+g1,e}

CE(Z2S3)={0,g1,g5+g6,g1+g5+g6}

Example 2 All invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of Z2S3 are listed as follows.

For convenience, we use CG(Z2S3), G(Z2S3), CD(Z2S3), and CC(Z2S3) to denote the sets of all central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of Z2S3, respectively.

U(Z2S3)={g1,g2,g3,g4,g5,g6,e+g1,e+g2,e+g3,

e+g4,e+g5,e+g6}

CG(Z2S3)={0,U(Z2S3),g2+g3,g2+g4,g3+g4,

g5+g6,g1+g5,g1+g6}

G(Z2S3)={CG(Z2S3),g1+g2+g5,g1+g2+g6,

g1+g3+g5,g1+g3+g6,g1+g4+g5,g1+g4+g6,

g2+g3+g5,g2+g3+g6,g2+g4+g5,g2+g4+g6,

g3+g4+g5,g3+g4+g6,e+g4+g6,e+g4+g5,

e+g3+g6,e+g3+g5,e+g2+g6,e+g2+g5,e}

CD(Z2S3)={CG(Z2S3),g1+g2,g1+g3,g1+g4,

g1+g2+g3,g1+g2+g4,g1+g3+g4,g2+g5+g6,

g3+g5+g6,g4+g5+g6,e+g3+g4,e+g2+g4,

e+g2+g3,e+g1+g4,e+g1+g3,e+g1+g2,e}

CC(Z2S3)={0,U(Z2S3),g1+g2,g1+g3,g1+g4,

g1+g5,g1+g6,g5+g6,g2+g3,g2+g4,g3+g4,

g1+g2+g3,g1+g2+g4,g1+g3+g4,g2+g3+g4,

g1+g5+g6,g2+g5+g6,g3+g5+g6,g4+g5+g6,

g2+g3+g4,e+g1+g6,e+g1+g5,e+g1+g4,

e+g1+g3,e+g1+g2,e+g3+g4,e+g2+g4,

e+g2+g3,e+g6,e+g5,e}

Proof Given that Z2S3 is finite, it follows that Z2S3 is strongly π-regular. Hence, the elements in Z2S3 are Drazin invertible.

Then, we calculate the units of Z2S3.

Let α=x1g1+x2g2+x3g3+x4g4+x5g5+x6g6 and β=y1g1+y2g2+y3g3+y4g4+y5g5+y6g6. From αβ=g1, we can obtain

which has a unique solution.

Denote Notably, x1+x2+x3+x4+x5+x6=1. Hence, we can obtain

(1+x4+x6)(1+x4+x5)+(1+x1+x3)(1+x1+x2)

Then, from |A+B|≠0, it follows that xi=0 for certain i∈{1,2,…,6} and the others are 1, or xi=1 for certain i∈{1,2,…,6} and the others are 0. Hence, we have

U(Z2S3)={g1,g2,g3,g4,g5,g6,e+g1,e+g2,e+g3,

e+g4,e+g5,e+g6}

Therefore,based on Theorem 3.6 in Ref.[28], Proposition 8.24 in Ref.[30], Theorem 4.5 in Ref.[28], and the definition of centrally clean elements, we can present the sets of CG(Z2S3), G(Z2S3), CD(Z2S3), and CC(Z2S3), respectively.

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环中的中心clean元和中心Drazin逆

李文德 陈建龙

(东南大学数学学院, 南京 211189)

摘要:如果环R中元素a是一个中心幂等元e和一个可逆元u的和,那么称a是中心clean元, 并且称a=u+ea的一个中心clean分解.如果环R中所有元素都是中心clean元,则称环R是中心clean环.首先,给出了中心clean元的等价刻画,并进一步研究了中心clean环的一些性质以及环R是中心clean环的充分必要条件.中心clean环与中心Drazin逆有着紧密的联系.接着,从中心clean分解的角度给出了中心Drazin逆存在的充分必要条件,并研究了角环、幂级数环和环Rα的笛卡尔积∏Rα等特殊环的中心clean性.此外,还研究了一般域上代数中的2个中心幂等元组合的中心群可逆性.最后,作为一个应用,分别计算出了群环Z2S3中所有的可逆元、中心群可逆元、群可逆元、中心Drazin可逆元以及中心clean元.

关键词:中心clean元;中心clean环;中心Drazin逆;中心群逆

DOI:10.3969/j.issn.1003-7985.2022.03.014

Received 2022-04-11,

Revised 2022-07-22.

Biographies:Li Wende (1993—), male, Ph.D. candidate; Chen Jianlong (corresponding author), male, doctor, professor, jlchen@seu.edu.cn.

Foundation items:The National Natural Science Foundation of China (No. 12171083, 11871145, 12071070), the Qing Lan Project of Jiangsu Province.

CitationLi Wende, Chen Jianlong. Centrally clean elements and central Drazin inverses in a ring[J].Journal of Southeast University (English Edition),2022,38(3):315-322.DOI:10.3969/j.issn.1003-7985.2022.03.014.

中图分类号:O153.3