# 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.

• You have to prove (or know) that the $M_p$ are also finitely generated Jun 4 '18 at 14:30

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.