Radical of the annihilator of an element of a Noetherian module

Question

Assume $M$ is a commutative Noetherian $R$-module and $m\in M$ is such that $P=\sqrt{\operatorname{Ann}(m)}$ is a prime ideal in $R$. Is it true that $P$ is an associated prime of $M$, i.e. there is an element $m'\in M$ such that $P=\operatorname{Ann}(m')$?

Answer 1

This holds whenever $R$ is noetherian; see Bourbaki, Commutative Algebra, Chapter 4, Exercise 17(g). (In fact, it's enough to know that $P$ is minimal over $\operatorname{Ann}(m)$ for some $m\in M$, $m\ne 0$.) But this isn't the case in your question. However, it also holds under the assumption $M$ noetherian. In this case the ring $R/\operatorname{Ann}(M)$ is noetherian (why?). Since $P=\sqrt{\operatorname{Ann}(m)}$ it follows that $P$ is minimal over $\operatorname{Ann}(m)$, so $P/\operatorname{Ann}(M)$ is minimal over $\operatorname{Ann}(m)/\operatorname{Ann}(M)$. From the mentioned result we deduce that there is $m'\in M$ such that $P/\operatorname{Ann}(M)=\operatorname{Ann}_{R/\operatorname{Ann}(M)}(m')$. But $\operatorname{Ann}_{R/\operatorname{Ann}(M)}(m')=\operatorname{Ann}(m')/\operatorname{Ann}(M)$ and therefore $P=\operatorname{Ann}(m')$.