Recently I have been studying the structure theorem of finitely generated modules over a principal ideal domain $R$. I am stuck with the following statement and I am looking for an elegant and nice proof:
Let $M$ be a finitely generated module over a principal ideal domain $R$, and $p$ be a prime element in $R$. If $N_{p} = \{m \in M \mid p^{i}m = 0, \text{ for some integer $i \geq 0$}\}$, then $N_{p}$ (the $p$-Primary component of $M$) is a direct sum of cyclic modules.
I can see the following: $\mathrm{Ann}_{R}(M) \neq 0$, so, suppose, $\mathrm{Ann}_{R}(M) = \langle p_{1}^{h_{1}} \cdots p_{k}^{h_{k}} \rangle$. Now, $N_{p_{k}}$ is finitely generated since $M$ is finitely generated over a PID, so, every submodule of $N_{p_{k}}$ is also finitely generated. Now, if we consider $M_{p_{k}^{j}} = \{m \in M \mid p_{k}^{j}m = 0\}$, then we have the following: $M_{p_{k}} \subseteq M_{p_{k}^{2}} \subseteq \cdots \subseteq M_{p_{k}^{h_{k}}}$, where $M_{p_{k}^{n}} = M_{p_{k}^{h_{k}}}$ whenever $n > h_{k}$. After observing these, I am stuck.
Please help me.