A question on minimal right ideals in a simple ring Let $R$ be a simple ring with $1$. Let $e,g$ be idempotents in $R$ with $eR$ a minimal right ideal of $R$. In a paper [ _A short proof of the Wedderburn-Artin theorem - Tsiu-Kwen Lee_ ], the author asserts: > If $(1-g)eR\neq 0$, then it is a minimal right ideal of $R$. I didn't get proof of this. What we know is following: $e=(1-g)e + ge$ and so $$eR\subseteq (1-g)eR+geR.$$ Further the sum on the right side is direct (easy to see). But, how minimality of $(1-g)eR$ is asserted, when it is non-zero, is not clear to me. (This could be easy, but not sparking arguments to me.)

Phrased more simply, left multiplication by an element of $R$ on a right ideal $T$ of $R$ is a homomorphism of $T\to R$ as right $R$-modules.

Clearly then $(1-g)eR$ is a homomorphic image of the simple module $eR$, and therefore it can only be the zero module or a simple module.