## Characterize the irreducible torsion modules over a PID

Let $R$ be a principal ideal domain and let $M$ be a torsion $R$-module. Prove that $M$ is irreducible if and only if $M = Rm$ for some nonzero element $m \in M$ where the annihilator of $m$ in $R$ is a nonzero prime ideal $P$.

Suppose $M$ is an irreducible, torsion $R$-module. By this previous exercise, we have $M \cong_R R/I$ where $I \subseteq R$ is a maximal ideal. Since $R$ is a principal ideal domain, in fact $I = (p)$ for some prime element $p \in R$. Let $\psi : R/(p) \rightarrow M$ be an isomorphism, and let $m = \psi(1 + (p))$. Now $M = Rm$, and moreover $\mathsf{Ann}_R(m) = \mathsf{Ann}_R(1 + (p)) = (p)$.

Conversely, consider the module $M = Rm$, where $\mathsf{Ann}_R(m) = (p)$ is a prime ideal. Define $\psi : R \rightarrow (m)_R$ by $r \mapsto r \cdot m$. Certainly we have $\mathsf{ker}\ \psi = \mathsf{Ann}_R(m) = (p)$. By the First Isomorphism Theorem for modules, we have $\overline{\psi} : R/(p) \rightarrow M$ an isomorphism. Again by this previous exercise, $R/(p)$ is irreducible.