Hot answers tagged noncommutative-algebra
The key here is that $M(R)$ is generated by the left and right multiplication maps of $R$. Let $L_r$ and $R_r$ denote left and right multiplication by $r \in R$. $R$ is a left $R$-module under the natural action $r \cdot x = rx$ (action = left multiplication). Likewise, $R$ is a right $R$-module under the natural action $x \cdot r = xr$ (action = right ...
There are maps $\alpha:I\mapsto eIe$ from the set of ideals of $R$ to the set of ideals of $eRe$, and $\beta:J\mapsto RJR$ from the set of ideals of $eRe$ to the set of ideals of $R$. They aren't bijections, but $\alpha\circ\beta$ is the identity, so $\beta$ is injective, which is enough to deduce that $R$ simple implies $eRe$ simple.
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