How to prove that every simple left $R$-module is isomorphic to a minimal left ideal of $R$

Question

We know that:

$T$ is a simple left $R$-module $\Longleftrightarrow T\cong R/M$, where $M$ is a maximal left ideal of $R$.

So please tell me how to prove that every simple left $R$-module is isomorphic to a minimal left ideal of $R$?

Thanks!

Answer 1

Rings don't have minimal ideals, in general. For instance, $\mathbb{Z}$ has none, because $n^2\mathbb{Z}\subset n\mathbb{Z}$ properly, when $n>1$.

If $R$ is a semisimple ring, then the assertion is true: any submodule of $R$ splits; if $M$ is a maximal left ideal of $R$, then $R=M\oplus T$ for some left ideal $T$; then $$ R/M=(M\oplus T)/M\cong T $$ and so $T$ is a minimal left ideal which is isomorphic to $R/M$.