Prove that if M is irreducible but not cyclic, then for every $r \in R$ and $m \in M$, $rm = 0.$
This is what i write but i don't know how to continue
Suppose there exists an $r \in R$ and $m \in M$ such that $x = rm \neq 0 \in M$. It is clear that $Mx$ is a $R$-submodule of $M$. Since $M$ is irreducible, $Mx = \{0\}$ or $Mx = M$.
I think i need to show that there exists a nonzero element in M to derive a contradiction such that M is cyclic. However, I cannot find a way to do this. Can someone give me some help please?
You are on the right track. If there are $r \in R$ and $m \in M$ such that $rm \ne 0$, then the $R$-submodule $Rm \subseteq M$ (what is $Mx$? How to multiply module elements?) is not the zero module (as $0 \ne rm \in Rm$). Hence, as $M$ is irreducible, $Rm = M$. But $Rm$ is cyclic, generated by $m$, where $M$ is not. Contradiction.
Hence, $rm = 0$ for all $r \in R$, $m \in M$.