Uniqueness of right-quasi-regular inverse

Asked Sep 21 '23 at 19:04

Active Sep 21 '23 at 21:05

Viewed 42 times

I've been thinking about the following exercise that my teacher left:

Let $R$ a ring. An element $a \in R$ is right-quasi-regular if there is $b \in R$ such that $$a + b+ab = 0.$$ In this case $b$ is said to be the right-quasi-inverse of $a$ (this definition follows from Herstein's Non-commutative Rings book). Then, every right-quasi-inverse of a right-quasi-regular element is unique.

I can't prove it neither find a counter example. What I can prove is that if $a\in R$ is a right and left-quasi-regular then the left and right-quasi-inverses are the same. Can it help me to solve the exercise? If not, is there any kind of counter-example or solution?

edited Sep 21 '23 at 21:05

asked Sep 21 '23 at 19:04

Guilherme Costa

Does $R$ have an identity? – Freddie Sep 21 '23 at 19:11
@Freddie, not necessarily – Guilherme Costa Sep 21 '23 at 19:17
1

The statement is actually false. As in the answer to this question, we have $(g_v-1)+(f-1)+(g_v-1)(f-1)=0$ for all $v\in V$. – Freddie Sep 21 '23 at 20:11
That is to say, even in rings with identity, rq-inverses are not necessarily unique. – rschwieb Sep 21 '23 at 20:36
@Freddie I think that this counter example works! Thank you! – Guilherme Costa Sep 21 '23 at 21:07

Uniqueness of right-quasi-regular inverse

0 Answers0