A note on abelian stongly k-Engel π-regular rings

Abstract

Let R be an associative ring with identity and let k ≥ 1 be a fixed integer. An element (x, y) ∈ R × R is said to be left (right) k-Engel π-regular if there exists a positive integer n and an element z ∈ R such that ${[x,y]}_k^n=z{[x,y]}_k^{n+1}$ (${[x,y]}_k^n={[x,y]}_k^{n+1}z$). If every element of R × R is left (right) k-Engel π-regular, then R is said to be left (right) k-Engel π-regular. An element (x, y) ∈ R × R is strongly k-Engel π-regular if it is both left and right k-Engel π-regular. The ring R is strongly k-Engel π-regular if every element of R × R is strongly k-Engel π-regular. In this paper, we investigate properties of abelian strongly k-Engel π-regular ring.

MSC

16E50, 16D70, 16U99

Introduction

Let R be an associative ring with identity. An element x ∈ R is said to be right π-regular if there exists a positive integer n and an element y ∈ R such that xⁿ = xⁿ⁺¹y. If every element of R is right π-regular, then R is said to be right π-regular. By [1], this definition is left-right symmetric. An element of R is strongly π-regular if it is both left and right π-regular. R is strongly π-regular if every element of R is strongly π-regular. In [2], it was shown that if an element x in the ring R is strongly π-regular, then there exists a positive integer n and an element y∈R such that xⁿ = xⁿ⁺¹y and x y = y x. In the case where n = 1, the element x is said to be strongly regular.

If ${\left(x_i\right)}_{i\in \mathbb{N}}$ is a sequence of elements of R and k is a positive integer, we define $[x_1,\dots ,x_{k+1}]$ inductively as follows:

$$\begin{array}{ll}\hfill [x_1,x_2]& =x_1{x}_2-x_2{x}_1,\hfill \\ \hfill [x_1,\dots ,x_k,x_{k+1}]& =\left[\right[x_1,\dots ,x_k],x_{k+1}].\hfill \end{array}$$

If x₁= x and $x_2=\cdots =x_{k+1}=y$, the notation [x, y] _k is used to denote $[x_1,\dots ,x_{k+1}]$, and [x, y] _k is called a k-Engel element. For k = 1, [x, y] _k  =[x, y]₁ is usually just denoted by [x, y]. An element (x, y) ∈ R × R is said to be left (right) k-Engel π-regular if there exists a positive integer n and an element z ∈ R such that ${[x,y]}_k^n=z{[x,y]}_k^{n+1}$ (${[x,y]}_k^n={[x,y]}_k^{n+1}z$). If every element of R × R is left (right) k-Engel π-regular, then R is said to be left (right) k-Engel π-regular. An element (x, y) ∈ R × R is strongly k-Engel π-regular if it is both left and right k-Engel π-regular. The ring R is strongly k-Engel π-regular if every element of R × R is strongly k-Engel π-regular. Clearly, if (x, y) is strongly k-Engel π-regular, then [x, y] _k is strongly π-regular. Therefore, there exists a positive integer n and an element z ∈ R such that ${[x,y]}_k^n={[x,y]}_k^{n+1}z$ and [x, y] _k z = z[x, y] _k (by [2]).

Division rings are examples of strongly k-Engel π-regular rings. Other examples include full matrix rings over division rings and triangular matrix rings over fields. It is clear that rings which satisfy the k-Engel condition are strongly k-Engel π-regular. In [3], we studied the conditions for strongly k-Engel π-regular rings to be commutative (hence, has k-Engel condition). Now in this paper, we investigate the properties of abelian strongly k-Engel π-regular rings and obtain some characterisations of these rings. All rings in this paper are assumed to have identity. A ring is said to be abelian if all of its idempotents are central. For a ring R, we use the notation N(R) and I d(R) to denote the set of all nilpotent elements of R and the set of all idempotents of R, respectively.

Main results

Proposition 2.1. Let R be an abelian strongly k -Engel π -regular ring. Suppose that N(R) is an ideal of R.Then for each x,y ∈ R, [x, y] _k +N(R)is strongly regular (hence regular).

Proof. Let x,y∈R. Then there exist z ∈ R and a positive integer n such that ${[x,y]}_k^n={[x,y]}_k^{n+1}z$ and [x, y] _k z = z[x, y] _k . Thus, $e={[x,y]}_k^n{z}^n\in \mathit{\text{Id}}\left(R\right)$, and hence, 1−e ∈ I d(R). Then since ${[x,y]}_k^n={[x,y]}_k^{2n}{z}^n={[x,y]}_k^{n}e$, it follows that $(1-e){[x,y]}_k^n=0$, and hence, (1−e)[x,y] _k ∈N(R). Therefore,

$$\begin{array}{ll}{[x,y]}_k+N\left(R\right)& =e{[x,y]}_k+N\left(R\right)\\ ={[x,y]}_k^{n+1}{z}^n+N\left(R\right)\\ ={({[x,y]}_k+N(R\left)\right)}^2({[x,y]}_k^{n-1}{z}^n+N(R\left)\right).\end{array}$$

It follows that [x,y] _k +N(R) is strongly regular (hence regular).

The following lemma is well known and can be found for example in p. 72 of [4]. □

Lemma 2.2. Let R be a ring and I a nil ideal of R.Then idempotents of R/I can be lifted to R.

Proposition 2.3. Let R be an abelian ring. If N(R) is an ideal of R and for each x, y ∈ R, [x, y] _k +N(R) is regular, then R is strongly k-Engel π-regular.

Proof. Let x,y ∈ R. Since [x,y] _k +N(R) is regular, there exist some z ∈ R such that [x,y] _k z[x,y] _k +N(R) = [x,y] _k +N(R). Clearly, (z[x,y] _k )²+N(R) = z[x,y] _k +N(R). By Lemma 2.2, there is an idempotent e ∈ R such that e+N(R) = z[x,y] _k +N(R), that is, e−z[x,y] _k ∈N(R). Thus, there exists an integer m≥1 such that (e−z[x,y] _k )^m = 0. Since e is central, e = t[x,y] _k for some t∈R.

Now [x,y] _k +N(R) = ([x,y] _k z[x,y] _k )+N(R) = [x,y] _k e+N(R) gives us [x,y] _k −[x,y] _k e∈N(R). Hence, there exist some integer n≥1 with $0={({[x,y]}_k-{[x,y]}_{k}e)}^n={[x,y]}_k^n-{[x,y]}_k^{n}e$. Therefore, ${[x,y]}_k^n={[x,y]}_k^{n}e=e{[x,y]}_k^n=t{[x,y]}_k^{n+1}$. Thus, R is strongly k-Engel π-regular.

By Propositions 2.1 and 2.3, we readily have the following: □

Theorem 2.4. Let R be an abelian ring such that N(R) is an ideal of R. Then R is strongly k-Engel π-regular if and only if for each x,y∈R, [x,y] _k +N(R) is regular.

Proposition 2.5. Let R be an abelian strongly k-Engel π-regular ring and let P be a prime ideal of R. Then for each x,y ∈ R, [x,y] _k +P is nilpotent or a unit.

Proof. Let x, y ∈ R. Since R is strongly k-Engel π-regular, by the proof of Theorem 2.1 in [3], we may write [x,y] _k  = f u = u f for some near idempotent f and some unit u ∈ R. By near idempotent we mean that there exists a positive integer n such that e = fⁿis an idempotent. Then ${[x,y]}_k^n=e{u}^n=u^{n}e$. Since $(1-e)\mathit{\text{Re}}=\left\{0\right\}\subseteq P$ and P is a prime ideal, it follows that e ∈ P or 1 − e ∈ P. If e ∈ P, then ${[x,y]}_k^n=e{u}^n\in P$; hence, [x,y] _k +P is nilpotent. If 1 − e ∈ P, then ${[x,y]}_k^n+P=e{u}^n+P=(e+P)(u^n+P)=u^n+P$ is a unit in R/P. It follows that [x,y] _k +P is a unit in R/P. □

Proposition 2.6. Let R be a strongly k-Engel π-regular ring and I an ideal of R. Then I is strongly k-Engel π-regular as a ring.

Proof. Let x,y ∈ I. Since R is strongly k-Engel π-regular, then there exist z ∈ R and a positive integer n such that ${[x,y]}_k^n={[x,y]}_k^{n+1}z$ and [x,y] _k z = z[x,y] _k . If n = 1, let t = [x,y] _k z². Then t ∈ I, [x,y] _k t = t[x,y] _k and ${[x,y]}_k^{2}t={[x,y]}_k^2{[x,y]}_k{z}^2={[x,y]}_k\left({[x,y]}_k^{2}z\right)z={[x,y]}_k^{2}z={[x,y]}_k$. If n ≥ 2, let $t={[x,y]}_k^{n-1}{z}^n\in I$. Then [x,y] _k t = t[x,y] _k . Therefore, ${[x,y]}_k^n={[x,y]}_k^{n+1}z=\mathrm{...}={[x,y]}_k^{n+1}$$\left({[x,y]}_k^{n-1}{z}^n\right)={[x,y]}_k^{n+1}t$. Thus, I is strongly k-Engel π-regular. □

Authors’ information

AYMC and SS are lecturers in their respective institutions.