Annihilators in discrete valuation rings

Question

Let $A$ be a discrete valuation ring and $M$ be an $A$-module. Let $a \in A$ and $m \in M$ such that $am \neq 0$. Is it true that $\operatorname{Ann}(m) = a \operatorname{Ann}(am)$?

Answer 1

I think you don't need the valuation ring to be discrete.

If $b\in\operatorname{Ann}(am)$, then $b(am)=0$ and therefore $(ab)m=0$, so $ab\in\operatorname{Ann}(m)$. Conversely, if $b\in\operatorname{Ann}(m)$ there are two cases: $a\mid b$, so $b=ac$ and from $bm=0$ we get $c(am)=0$, that is, $c\in\operatorname{Ann}(am)$; $b\mid a$, so $a=bd$ and from $bm=0$ we obtain $am=0$, a contradiction.