On Semiabelian Π-regular Rings

A ring R is defined to be semiabelian if every idempotent of R is either right semicentral or left semicentral. It is proved that the set N(R) of nilpotent elements in a π-regular ring R is an ideal of R if and only if R/J(R) is abelian, where J(R) is the Jacobson radical of R. It follows that a semiabelian ring R is π-regular if and only if N(R) is an ideal of R and R/N(R) is regular, which extends the fundamental result of Badawi (1997). Moreover, several related results and examples are given. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Introduction
Throughout this paper, rings are associative with unity and modules are unitary.Given a ring R, we use the symbol Id(R) to denote the set of idempotents in R, U(R) its unit group.The Jacobson radical, the prime radical, and the set of nilpotent elements of a ring R are denoted by J(R), P(R), and N(R), respectively.The symbol Max(R) (resp., Max r (R)) stands for the set of maximal (resp., maximal right) ideals of a ring R. As usual, the symbol M n (R) denotes the ring of n × n matrices over a ring R, UTM n (R) denotes the ring of n × n upper triangular matrices over R, and E i j (1 ≤ i, j ≤ n) denotes the n × n matrix units over R. Let M be an R-R bimodule and A = (a i j ) n×n ∈ M n (R), we write MA = {(ma i j ) n×n | m ∈ M}, and write V = n−1 i=1 E i,i+1 for n ≥ 2. And we use the symbol T n (R,M) to denote the ring of n × n upper triangular matrices whose principal diagonal elements are identical and belong to R and the other elements belong to M, and write V n (R,M) = RI n + MV + ••• + MV n−1 for n ≥ 2 where I n is the n × n identity matrix over R.Moreover, we use the symbol Z p to denote the ring of integers modulo a prime p.
Following [1], an idempotent e in a ring R is called right (resp., left) semicentral if for every x ∈ R, ex = exe (resp., xe = exe).And the set of right (resp., left) semicentral idempotents of R is denoted by S r (R) (resp., S l (R)).We define a ring R to be semiabelian if Id(R) = S r (R) ∪ S l (R), this notion is a proper generalization of that of an abelian ring.
Recall that a ring R is called π-regular if for every x ∈ R, there exist an element y ∈ R and a positive integer n such that x n = x n yx n .In the case of n = 1 for all x ∈ R, then R is regular.An element a in a ring R is strongly π-regular if there exist b ∈ R and a positive integer n such that a n = a n+1 b with ab = ba.And a ring R is strongly π-regular if every element of R is strongly π-regular.Clearly, a strongly π-regular ring is a π-regular ring.A ring is called right (resp., left) quasiduo if every maximal right (resp., left) ideal is an ideal.And a ring is quasiduo if it is right and left quasiduo.A ring R is called an exchange ring if for every a ∈ R, there exists e ∈ Id(R) such that e ∈ aR and 1 − e ∈ (1 − a)R.It is known that a π-regular ring is an exchange ring (see [2,Example 2.3]).A ring is reduced if it has no nonzero nilpotent elements.And a ring is abelian if every idempotent is central.It is well known that a reduced ring is an abelian ring.For the above notions we refer the reader to [3,4].
In [5], Badawi studied abelian π-regular rings and obtained some interesting results.The fundamental result is that an abelian ring R is π-regular if and only if N(R) is an ideal of R and R/N(R) is regular.In this paper, we study semiabelian π-regular rings, extending some of the main results of [5].It is proved that for every such ring R, N(R) is an ideal of R if and only if R/J(R) is abelian.It follows that if R is a semiabelian ring, then R is π-regular if and only if N(R) is an ideal of R and R/N(R) is regular.Moreover, several related results and examples are given.

Extensions of semiabelian rings
We start this section with the following definition.
Clearly, an abelian ring is semiabelian.But the converse is not true in general as the following example shows.
Proof.Clearly, UTM 2 (R) is not abelian.And it is quite easy to check that and that 1 a 0 0 is a left semicentral idempotent and 0 b 0 1 is a right semicentral idempotent for any a,b ∈ R. Hence, UTM 2 (R) is semiabelian.
One may expect that the conclusion of Example 2.2 is true for n ≥ 3, but this is not the case.In fact, for any ring R, idempotent E 11 + E 33 is neither right nor left semicentral in UTM 3 (R).This implies that for any n ≥ 3, UTM n (R) is not semiabelian.Also the direct sum of two nonabelian semiabelian rings is not semiabelian.Now let R 1 and R 2 be semiabelian rings which are not abelian.Take e 1 ∈ R 1 to be a right semicentral idempotent which is not central and e 2 ∈ R 2 to be a left semicentral idempotent which is not central, then the idempotent (e 1 ,e 2 ) is neither right nor left semicentral in R 1 R 2 .
Weixing Chen 3 In view of this situation, it is necessary for us to study how to obtain more examples of nonabelian semiabelian rings from a given nonabelian semiabelian ring.Clearly, the direct sum of an abelian ring and a nonabelian semiabelian ring is a semiabelian ring which is not abelian.Next we consider several extensions of semiabelian rings.
Multiplying two sides of the equation ea 1 + a 1 e = a 1 by e, we have ea 1 e = 0, which gives ea 1 e = ea 1 = 0 and so a 1 = a 1 e.Assume that ea i = 0 and a i = a i e hold for all 1 ≤ i ≤ n − 1.We claim that ea n = 0 and a n e = a n .In fact, comparing the coefficients of x n in the equation f (x) 2 = f (x), we have a n e = a n .Since a i = a i e for 1 ≤ i ≤ n − 1 in the above expression, we have a i a n−i = a i ea n−i e = 0.It follows that ea n + a n e = a n .Multiplying both sides of this equation by e, then ea n e = ea n = 0, which gives a n = a n e.By induction, we have f (x) = f (x)e and e f (x) = e.Now for any g(x The only if part of the proof is trivial since the subring of a semiabelian ring is semiabelian.And the proof is complete.

Corollary 2.4. A ring R is semiabelian if and only if the ring R[x] of polynomials over R is semiabelian.
It is known by [6, Propositions 2.4 and 2.5] that if ), then e ∈ S l (R), e f (x) = f (x), and f (x)e = e.This is true, in particular, for a polynomial , then e ∈ S r (R), f (x)e= f (x), and e f (x)=e.And this is true especially when f (x) ∈ S r (R[x]).
From the proof of Theorem 2.3 and the above argument, we obtain a characterization of left (resp., right) semicentral idempotents in R ) with the constant term e.Then one has the following conclusions: (1) Proof.Assume that R is a semiabelian ring and E n ∈ Id(T n (R,M)).Then where e ∈ Id(R) and a i j ∈ M. We claim that if e ∈ S r (R), then E n ∈ S r (T n (R,M)).First we prove that E n = E n e is true by induction on n.This is trivial in the case of n = 1.Assume that E n−1 = E n−1 e holds for any n ≥ 2. In the case of n, then E n = En−1 α 0 e where α = (a 1n ,a 2n ,...,a n−1,n ) T .Since E n is an idempotent, we have E n−1 α + αe = α, which gives E n−1 αe = 0. On the other hand, since R,M).In the case of n, we write B n = Bn−1 β 0 bnn where B n−1 ∈ T n−1 (R,M) and b nn ∈ R. Hence we have the following equations: Hence, E n B n E n = E n B n and so E n ∈ S r (T n (R,M)).Similarly, it can be proved that if e ∈ S l (R), then E n ∈ S l (T n (R,M)).Therefore T n (R,M) is semiabelian.The only if part of the proof is trivial.

Corollary 2.8. A ring R is semiabelian if and only if the trivial extension
Proof.It is trivial in the case of n = 1.If n ≥ 2, then there exists a ring isomorphism θ:

Semiabelian π-regular rings
For convenience of the reader, we list some known facts which are necessary for the study of π-regularity of rings.
Proof.(⇒) Since N(R) is a nil ideal of R, N(R) ⊆ J(R) holds.On the other hand, we have J(R) ⊆ N(R) by the assumption.It follows that J(R) = N(R) and so R/J(R) is reduced and hence it is abelian.(⇐) Because R is an exchange ring and R/J(R) is abelian, R/J(R) is an abelian exchange ring and so it is reduced by Lemma 3.6.Hence N(R) ⊆ J(R).On the other hand, J(R) ⊆ N(R) by the assumption.So N(R) = J(R) is an ideal of R.
It is known and easy to prove that the Jacobson radical of a π-regular ring is nil.Hence Theorem 3.8 implies that for a π-regular ring R, N(R) is an ideal of R if and only if R/J(R) is abelian.And [2,Example 4.16] shows that the class of exchange rings with J(R) nil properly contains the class of π-regular rings.
Badawi [5,Theorem 2] proved that if R is an abelian π-regular ring, then N(R) is an ideal of R. In fact, the similar result is true for a right (resp., left) quasiduo π-regular ring.
Proof.Since R is a right (resp., left) quasiduo ring, R/J(R) is reduced by [12, Corollary 2] and hence it is abelian.And since R is π-regular, it is an exchange ring with J(R) nil.Hence, N(R) is an ideal of R by Theorem 3.8.
Proof.Clearly, R is a semiabelian exchange ring with J(R) nil, and R/J(R) is abelian by Lemma 3.7.By Theorem 3.8, N(R) is an ideal of R. Theorem 3.11.Let R be a semiabelian ring.Then R is π-regular if and only if N(R) is an ideal of R and R/N(R) is regular.
Proof.(⇒) Suppose that R is π-regular.By Corollary 3.10, N(R) is an ideal of R and so R/N(R) is reduced and π-regular.Let x ∈ R/N(R).Then there exist ȳ ∈ R/N(R) and a positive integer Then R is abelian regular (and hence unit regular) since it is reduced.To prove R is π-regular, it is sufficient to prove that R/P is strongly π-regular for every prime ideal P of R by Lemma 3.2.For any x ∈ R, then x = x + N(R) ∈ R is unit regular.So we have x = e u = u e with e ∈ Id(R) and u ∈ U(R) since idempotents and units of R can be lifted modulo N(R).Hence, x = eu + w 1 = ue + w 2 where w 1 ,w 2 ∈ N(R), which implies ex = e(u + w 1 ) and xe = (u + w 2 )e, and (1 x n = 0, and hence x n = ex n .If e ∈ P, then x n ∈ P and x = x + P ∈ N(R/P), so x is strongly π-regular in R/P.If e / ∈ P, then 0 = eR(1 − e) ⊆ P by Lemma 3.1, which gives 1 − e ∈ P and so ē = 1 in R/P.This implies x = ē x = e(u + w 1 ) = u + w 1 in R/P.Hence, x is a unit and so it is a strongly π-regular element in R/P.If e ∈ Sr(R), then 1 − e ∈ S l (R).Equation [x(1 − e)] n = 0 implies x n (1 − e) = 0, and hence x n = x n e.Note that xe = (u + w 2 )e.Similar to the above Weixing Chen 7 proof, it can be shown that x is a nilpotent element or a unit in R/P.And the proof is completed.
It is known by Example 2.2 that UTM 2 (R) is a nonabelian semiabelian π-regular ring for any π-regular local ring (e.g., an artinian local ring by Lemma 3.2) R. From this we can construct more nonabelian semiabelian π-regular rings by using Theorems 2.7 and 3.11.
Corollary 3.12 (see [5,Theorem 3]).Let R be an abelian ring.Then R is π-regular if and only if N(R) is an ideal of R and R/(R) is regular.
The following corollary is an immediate result of Theorem 3.11.
Corollary 3.13.Let R be a semiabelian π-regular ring.Then for any prime ideal P of R, every element in R/P is either a nilpotent element or a unit, and hence R is strongly π-regular with J(R) = N(R).
In light of Theorem 3.11, we naturally ask the following question.Question 3.14.Let R be any ring.If N(R) is an ideal of R and R/N(R) is regular, then is R π-regular?
There are many partially positive solutions to this question (see [13][14][15] for the details).For a ring R with bounded index (i.e., there exists a positive integer n such that a n = 0 for all a ∈ N(R)), the answer is also positive.Proposition 3.15.Let R be a ring with bounded index.If N(R) is an ideal of R and R/N(R) is regular, then R is strongly π-regular.
Proof.It is proved in [16,Lemma 11] that if I is a right ideal of a ring R and n is a positive integer such that a n = 0 for all a ∈ I, then a n−1 Ra n−1 = 0. Now since R/N(R) is reduced and regular, it is strongly π-regular.By Lemma 3.2, it is sufficient to prove that N(R) = P(R).Let m be the bounded index (the least positive integer m such that a m = 0 for all a ∈ N(R)) of R. If m = 1, then P(R) = N(R) = 0.If m ≥ 2, then N(R) = 0. We claim that P(R) = N(R) is also true.If not, then N(R)/P(R) is a nonzero nil ideal of R = R/P(R) with the bounded index n ≥ 2. Thus there exists a nonzero element ā ∈ N(R)/P(R) such that ān = 0 and ān−1 = 0, so R ān−1 R = 0.By [16,Lemma 11], ān−1 R ān−1 = 0 and so (R ān−1 R) 2 = 0, which is impossible since R/P(R) is a semiprime ring.So P(R) = N(R), and the proof is completed.

Theorem 3 . 16 .
Let R be a semiabelian ring.Then R is π-regular if and only if there exists a nil idealI of R such that R/I is π-regular.Proof.(⇒) If R is π-regular, then I = N(R) is an ideal of Rand R/I is regular by Theorem 3.11, and so we are done.(⇐) If R/I is π-regular for some nil ideal I of R, then R/I is semiabelian π-regular by Lemma 3.7.According to Theorem 3.11, N(R/I) = N(R)/I is an ideal of R/I.So N(R) is a nil ideal of R. Since R/I is π-regular, R/N(R) is π-regular.And since R/N(R) is reduced and π-regular, R/N(R) is regular by [4, Proposition 23.5].Therefore R is π-regular by Theorem 3.11.
and only if e ∈ S r (R), f (x)e= f (x) and e f (x)=e.Similar to the proof of Theorem 2.3, it is easy to prove the next theorem.
Theorem 2.6.A ring R is semiabelian if and only if the group ring RC ∞ is semiabelian, where C ∞ is the infinite cyclic group.Theorem 2.7.A ring R is semiabelian if and only if T n (R,M) is semiabelian, where M is an R-R bimodule.