## A fact about the annihilators of torsion modules over a PID

Let $R$ be a principal ideal domain, let $B$ be a torsion (left, unital) $R$-module, and let $p \in R$ be prime. Prove that if $pb = 0$ for some nonzero $b \in B$, then $\mathsf{Ann}(B) \subseteq (p)$.

[Note: I’m not sure why D&F assume that $B$ is torsion. This proof doesn’t use that fact as far as I can tell. Read with caution.]

Recall that $\mathsf{Ann}(B) \subseteq R$ is an ideal (by this previous exercise). Since $R$ is a principal ideal domain, we have $\mathsf{Ann}(B) = (t)$ for some $t \in R$.

Consider the cyclic submodule $(b)_R$ generated by $b$. Again, $\mathsf{Ann}((b)_R) \subseteq R$ is an ideal, so $\mathsf{Ann}((b)_R) = (s)$ for some $s$. By our assumption, $p \in (s)$, so that $p = sr$ for some $r \in R$. Since $p$ is prime, either $p|r$ or $p|s$. If $p|r$, then we have $r = pq$ for some $q \in R$. Now $pqs = p$, so that $qs = 1$; in particular, $s$ is a unit, and so $\mathsf{Ann}((b)_R) = R$. But then $0 = 1 \cdot b = b$, a contradiction. So $p|s$, and thus $(s) \subseteq (p)$.

Since $\mathsf{Ann}(B) \subseteq \mathsf{Ann}((b)_R)$, we have $(t) \subseteq (s)$, and so $(t) \subseteq (p)$, as desired.