Let $L$ be a left ideal of a ring R such that $ RL \neq 0$. Then $L$ is simple as an R-module if and only if $L$ is a minimal left ideal?

Question

Lemma: Let $M$ be a nonzero R-module , then $ M$ is simple hf only if $ 0 \neq x \in M $, $ M = R x $.

Can we prove the following statement accordingly the lemma ?

Let $L$ be a left ideal of a ring R such that $ RL \neq 0$. Then $L$ is simple as an R-module if and only if $L$ is a minimal left ideal?

Answer 1

Specifying that $RL\ne0$ suggests you're dealing with rings not necessarily with identity.

The proof is very easy: a submodule of a left ideal is also a left ideal.

The lemma you quote is not necessary: just use the definition of a simple module $M$ as a module with $RM\ne\{0\}$ having no other submodule than $\{0\}$ and $M$.