Are idempotents always central in a von Neumann regular ring?

Question

A ring $R$ is called von Neumann regular if for every $a \in R$; $a=axa$ for some $x \in R$.

A ring $R$ called Abelian if every idempotent in $R$ is central.

Are von Neumann regular rings abelian?

Answer 1

$M_2(F)$ is von Neumann regular (for any field $F$), but has the noncentral idempotent $\begin{bmatrix}1&0\\0&0\end{bmatrix}$.

In fact it is known that a von Neumann regular ring is abelian if and only if it is strongly regular, meaning that for every $a\in R$ there exists a $b\in R$ such that $a^2b=a$. Strongly regular rings do not have nonzero nilpotent elements, so you can see why the matrix ring example suggests itself as a counterexample.