Subring generated by idempotents Assume that $R$ is a simple ring possessing an idempotent $e\not =0,1$. If $R'$ is the subring of $R$ generated by the idempotents, one could prove that $eRe⊆R'$ (and of course, that $(1-e)R(1-e)⊆R'$). How could one deduce that $R=R'$?
Thanks!
 A: There's a cute but slightly fiddly proof in "Characterizing homomorphisms, derivations and multipliers in rings with idempotents" by Matej Brešar, Proc. Royal Soc. Edinburgh 137A, 9-21 (2007):
Suppose $e\in R$ is an idempotent, and $x\in R$. Then $e+ex-exe$ and $e+xe-exe$ are also idempotents, and their difference is $[e,x]=ex-xe$. So $[e,x]$ is in the subring generated by idempotents.
For any $r,s\in R$,
$$r\left[e,x\right]s=\left[e,r\left[e,\left[e,x\right]\right]s\right] 
-\left[e,r\right]\left[e,\left[e,x\right]s\right]
-\left[e,r\left[e,x\right]\right]\left[e,s\right]+2\left[e,r\right]\left[e,x\right]\left[e,s\right]$$
(as I'm sure you know well!) which shows that the ideal generated by $[e,x]$ is contained in the subring generated by $\left\{\left[e,y\right]\vert y\in R\right\}$ and hence is contained in the subring generated by idempotents.
But if $e\not\in\{0,1\}$ and $e$ is central, then $R=Re\times R(1-e)$ and so $R$ can't be simple. So if $R$ is simple, we can choose $x$ so that $[e,x]\neq0$ and so the ideal generated by $[e,x]$ is $R$.
