I’ve encountered the following problem whilst assisting a colleague in studying for a comprehensive exam, and I am unsure of how to solve it:
Let $R$ be a principal ideal domain, let $M\in R$-$\mathrm{Mod}$ be torsion, and let $p\in R$ be a prime element. Assume that there exists $m\in M\setminus\left\{0_M\right\}$ such that $p\in\mathrm{ann}_{R}(m)$. Show that $\mathrm{ann}_{R}(M)\subseteq\langle p\rangle$.
I suspect that what I should do is multiply $\mathrm{ann}_{R}(M)$ by something not in $\langle p\rangle$ so that the product lands inside $\langle p\rangle$, but I don’t know what that something is. Or perhaps I ought to take advantage of the fact that $M$ is torsion and $R$ is a P.I.D. to conclude that $\mathrm{ann}_{R}(M)=\langle x\rangle$ for some nonzero $x\in R$, and then show that $p\mid x$. Any suggestions would be appreciated.
Your idea is a good one. Let $I := \mathrm{ann}_{R}(M) = \langle x \rangle$ for some $x \in R$, and let $p \in R$ be a prime element annihilating $0 \neq m \in M$. Then $I \subseteq J := \langle x, p \rangle$; since every prime ideal in a PID is maximal, and $J$ contains the maximal ideal $\langle p \rangle$, either $J = \langle p \rangle$ or $J = R$. If $J = R$, then $1 \in J$, i.e. there exist $q, r \in R$ such that $qx + rp = 1$. But then $0 = (qx+rp) \cdot m = 1 \cdot m = m$ since $x$ and $p$ both annihilate $m$, a contradiction.