$M$ is an irreducible $R$ module $\iff$ $M$ is a cyclic module and every nonzero element is a generator.
($\rightarrow$) If $M$ is an irreducible $R$-module then it's obvious that $M$ is a cylclic module where every nonzero element is a generator, otherwise $mR <_R M$ would be a submodule of M.
($\leftarrow$) So, this would be obvious if we knew that every $M$ submodule is somehow related to the cyclic submodules of $M$, like a direct product or something. If $M$ was free then this would be easy, but as it is this suprises me, I feel like there should be some weird counterexample where there is some module $M$ with some submodule not related to its cyclic submodules. Why is this not possible?
Best Answer
Note that the statement is not really correct: a module $M$ is irreducible if and only if it is nonzero and every nonzero element of $M$ is a generator.
You're looking too far, in my opinion.
Suppose every nonzero element of $M$ (a nonzero module) is a generator. Let $L$ be a nonzero submodule of $M$; then take $x\in L$, $x\ne0$; then $M=xR\subseteq L$, forcing $L=M$.
For the converse: let $x\in M$, $x\ne0$; then $xR$ is a nonzero submodule of $M$, so $xR=M$ and $x$ is a generator.