Correction to “Left uniquely generated elements in rings” [Commun. Algebra, DOI: 10.1080/00927872.2021.1908549]

Abstract

In this corrigendum, statement (5) of Proposition 4.4 is reformulated and an example is given to show that this correction is necessary.

Keywords:

• Abelian ring
• idempotent

2000 Mathematics Subject Classification:

• Primary 16U99

This article refers to:

Left uniquely generated elements in rings

There is an incorrect statement in Proposition 4.4 of the paper.

Correction 1.

(1) Replace statement (5) of Proposition 4.4 with: If ${(ab)}^2=ab$ and ${(ba)}^2=ba$, then ab = ba.

• (2) in line 1 of the proof of “$(5)\Rightarrow (1)$”, replace “…, we have $1+ex(1-e))e=e$” by “…, we have $(1+ex(1-e))e=e$ and $e(1+ex(1-e))=e+ex(1-e)$”.

The correction is necessary because of the following example.

Example 2.

Let $R=\left\{(r_{ij})\in {\mathbb{T}}_3({\mathbb{Z}}_2):r_{11}=r_{22}=r_{33}\right\}.$ Then R has the trivial idempotents only, so R is an abelian ring. But R does not satisfy Proposition 4.4(5). Indeed, if $a=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$ and $b=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),$ then ${(ab)}^2=ab=0,$ but $ba=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\cancel{=}0.$ So $ab\cancel{=}ba.$

Acknowledgment

The author is indebted to Professor T.Y. Lam for his careful reading of the paper and for drawing to the author’s attention the fact that the proof of “$(2)\Rightarrow (5)$” in Proposition 4.4 has a flaw.