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?



Commutative inner inverse for non-unital strongly regular ring
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, $aba=axa=a$.


