Minimal right ideals of a simple ring are generated by idempotents

Question

Let $M$ be a minimal right ideal of a simple ring $R$. Show that $M$ contains an idempotent $e$ such that $M = eR$.

Answer 1

A slightly more general statement:

If $M$ is a minimal right ideal of a ring $R$, then either $M^2=\{0\}$ or else $M^2=M$ and $M$ is generated by an idempotent.

If you ask for $R$ to be simple, the annihilator of $M$ is zero, so you are automatically in the second case.

Anyhow, the argument is simple (har har). Since $M^2=M\neq \{0\}$, $aM=M$ for some $a\in M$. In particular, $ae=a$ for some $e\in M$. But look: that means $ae^2=ae=a$ also, so $a(e^2-e)=0$. This means $e^2-e$ annihilates $a$ on the right. We want to show $e^2-e=0$.

If $e^2-e$ weren't zero, then the intersection $r.ann(a)\cap M\neq \{0\}$, and therefore $M\subseteq r.ann(a)$ by simplicity of $M$. But if that were the case, $aM=\{0\}$, a contradiction.

Thus $e^2=e$ and $eR=M$