A question on minimal right ideals in a simple ring

Question

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.)

Answer 1

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.

Answer 2

I think, if the right ideal $(1-g)eR$ is non-zero, then its minimality can be proved as follows: we try to prove that this ideal is generated by any of its (non-zero) element, which will imply that it is simple right $R$-module, which proves the assertion.

So, take $0\neq (1-g)ex\in (1-g)eR$.

Then $0\neq ex\in eR$, so $exR\subseteq eR$

By minimality of $eR$, we get $exR=eR$.

Hence $(1-g)exR=(1-g)eR$.

Which means, any non-zero $(1-g)ex$ in right ideal $(1-g)eR$ generates the ideal; so this ideal is simple (hence minimal).