Element in a finitely generated torsion module on a PID with smallest non-zero annihilator This question was motivated by this answer in order to get a simple proof of the structure theorem of finitely generated torsion modules over PID.
Let $A$ be a PID and $M$ be a finitely generated torsion module.
It's well known that $\operatorname{Ann}_A(M)$ is a non-zero ideal of $A$.

I'm looking for a simple proof of the existence of an element $e\in M$ such
  that $\operatorname{Ann}_A(e)=\operatorname{Ann}_A(M)$.

Note that a such element generates a direct summand $Ae$ of $M$ (because $Ae$ is a injective $A/\operatorname{Ann}(M)$-module) hence the structure theorem follows by induction on the number of generators of $M$.

My try.
The existence of a such element $e$ is simple for finitely generated $p$-torsion module for a prime $p\in A$ because the set
$$\{n\in\Bbb N:\exists x\in M(\operatorname{Ann}_A(x)=Ap^n)\}$$
has a greatest element.
Thus if for each prime $p$ let $M_p$ denote the set of elements of $M$ annihilated by some power of $p$, there exists $e_p\in M_p$ such that $\operatorname{Ann}(e_p)=\operatorname{Ann}(M_p)$.
From the direct sum decomposition $M\cong\bigoplus_p M_p$, follows that $e=(e_p:p)$ is the required element.
 A: First, we may assume that $A$ is local. Namely, let $(p)$ be any maximal ideal containing $\mathrm{Ann}_A(M)$ and consider $A_{(p)}$  and $\mathrm{Ann}_{A_{(p)}}(M_{(p)})= \mathrm{Ann}_A(M)_{(p)}$. If we show $\mathrm{Ann}_{A_{(p)}}(M_{(p)}) = \mathrm{Ann}(m'_p)$ for some $m'_p \in M_{(p)}$, then after multiplying with some $a \in A \backslash (p)$ we get $am'_p= \tilde{m}_p \in \mathrm{im}(M) \subseteq M_{(p)}$ and for a preimage $m_p$ of $\tilde{m}_p $ in $A$ we get $\mathrm{Ann}_{A_{(p)}}(m'_p) \cap A = \mathrm{Ann}_A(m_p)$.
But we have $ \mathrm{Ann}_A(M) = (\bigcap_{\mathrm{Ann}_A(M) \subseteq (p)} \mathrm{Ann}_A(M)_{(p)}) \cap A = \bigcap \mathrm{Ann}_A(m_p)$. So it suffices to show $\bigcap \mathrm{Ann}_A(m_p) = \mathrm{Ann}_A(\sum m_p)$. The inclusion $LHS \subseteq RHS$ is clear. For $RHS \subseteq LHS$, observe that $\sum m_p \mapsto \tilde{m}_p \in A_{(p)}$ for any $(p)$ containing $\mathrm{Ann}_A(M)$. Thus, any $a \in \mathrm{Ann}_A(\sum m_p)$ is also contained in $\mathrm{Ann}_{A_{(p)}}(\tilde{m}_p) = \mathrm{Ann}_{A_{(p)}}(M_{(p)})$.
Next, clearly we have $0 \neq \mathrm{Ann}_A(M) = \bigcap_{m \in M} \mathrm{Ann}_A(m)$. But in a local PID with maximal ideal $(p)$ any non-zero ideal is of the form $(p^n)$ for some $n \geq 0$.Thus, one of the ideals in the intersection on the right hand side already equals $\mathrm{Ann}_A(M)$ and we are done.
