Unit-regular ring

Remarks

A unit-regular ring is von Neumann regular. Example: a division ring $D$; for $a \ne 0$ take $u = a^{-1}$, and for $a = 0$ take $u = 1$. Example: the matrix ring $M_n(F)$ over a field $F$. Write $A = PEQ$ with $P, Q$ invertible and $E = \operatorname{diag}(1, \ldots, 1, 0, \ldots, 0)$; then $u = Q^{-1} P^{-1}$ gives $AuA = PE^2Q = PEQ = A$. A unit-regular ring is Dedekind-finite: if $ab = 1$ and $a = aua$, then $1 = ab = auab = au$, so $a = u^{-1}$ and $ba = 1$. Non-example: let $V$ be a vector space with countable basis $e_1, e_2, \ldots$ over a field, and $R = \End(V)$. The ring $R$ is von Neumann regular, but the shift $s(e_i) = e_{i+1}$ and $t$ with $t(e_1) = 0$, $t(e_{i+1}) = e_i$ satisfy $ts = 1 \ne st$. So $R$ is not Dedekind-finite, hence not unit-regular.