## Characterization of irreducible modules

Let $R$ be a ring with 1. A (left, unital) $R$-module $M$ is called irreducible if $M \neq 0$ and 0 and $M$ are the only $R$-submodules of $M$. Prove that $M$ is irreducible if and only if $M \neq 0$ and if $x \in M$ is nonzero, then $M = (x)$. Describe the irreducible $\mathbb{Z}$-modules.

$(\Rightarrow)$ Suppose $M$ is irreducible. Now $M \neq 0$ by definition. Now let $x \in M$ be nonzero. Note that $Rx \subseteq M$ is a nonzero submodule since $1 \cdot x = x \in Rx$; since $M$ is irreducible, $Rx = M$. Thus any nonzero element of $M$ generates $M$.

$(\Leftarrow)$ Now suppose $M \neq 0$ and that if $x \neq 0$, then $Rx = M$. Let $N \subseteq M$ be a nonzero submodule. Now there exists some nonzero $x \in N$, and $N \supseteq Rx = M$. Thus $N = M$. Since the only submodules of $M$ are 0 and $M$, $M$ is irreducible.

Recall that $\mathbb{Z}$-modules are precisely the abelian groups, and that $\mathbb{Z}$-submodules are precisely the subgroups of abelian groups. Now let $M$ be an irreducible $\mathbb{Z}$-module. In particular, $M$ is cyclic as an abelian group. So $M$ must be finite of prime order, as otherwise it has nontrivial proper subgroups. Conversely, every cyclic group of prime order is irreducible as a $\mathbb{Z}$-module.