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An element $a$ in a ring $R$ is called strongly regular if $a \in a^2R$ and $a \in Ra^2$, in other words $a = a^2x$ and $a = ya^2$ for some elements $x,y \in R$. Say that $R$ is a unital ring. $a \in R$ is a strongly regular element if and only if a single element $b$ can be chosen for $a$ such that $a = a^2b$, $a = ba^2$, and $ab = ba$. The proof that I've seen for this depends on the fact that $R \cong \text{End}_R(_RR)$ which only holds if $R$ is a unital ring.

Is there a similar characterization or elementwise proof for this in the case where $R$ is a non-unital ring?

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  • $\begingroup$ Do you know whether the characterisation you mention also holds in the non-unital case? $\endgroup$
    – LSpice
    Commented Apr 7, 2017 at 18:06
  • $\begingroup$ I do not know, but I don't have any reason to believe that it does not hold. $\endgroup$
    – dbossaller
    Commented Apr 7, 2017 at 18:54

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Here is a short proof in the general (non-unital) case.

We have $$ (1) \qquad a^2x=a. $$ $$ (2)\qquad ya^2=a. $$ If we reduce the monomial $ya^2x$ using (1), we get $ya$, but if we use (2) we get $ax$. Thus $$ (3) \qquad ya=ax. $$ Combining (3) and (2) we get $$ (4) \qquad axa=a. $$ I claim that $b:=yax$ satisfies the condition you seek. Indeed, we compute $$ ab=ayax=a^2x^2=ax $$ (where we used (3) for the second equality and (1) for the last equality). On the other hand $$ ba=yaxa=ya=ax $$ (where the used (4) for the second equality and (3) for the last equality). Thus $ab=ba$. Finally, $aba=axa=a$.

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    $\begingroup$ By the way, the answer to LSpice's question is yes, the given criterion is the standard definition of strongly regular elements even for non-unital rings. $\endgroup$
    – Pace Nielsen
    Commented Apr 7, 2017 at 19:36

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