Let $M$ be a finitely generated module over a commutative Noetherian ring $R$ such that that $M_P=0$ for every associated prime $P$ of $R$. Then, is it true that for every $m\in M$, there exists a non-zero-divisor $s\in R$ (i.e. $s\notin \bigcup_{P\in\mathrm{Ass}(R)} P$) such that $sm=0$?
My thoughts: Let $P_1,…,P_n$ be the associated primes of $R$.
Let $m\in M$. Then, $m/1=0$ in each $M_{P_i}$, so for every $P_i$, there exists $s_i\notin P_i$ such that $s_im=0$. Now from here I am not sure if it is possible to get $s\notin \bigcup_{i} P_i$ such that $sm=0$.
Please help.
Best Answer
$s_im=0$ is equivalent to $\mathrm{Ann}(m)\nsubseteq P_i$. Since this holds for all $i=1,\dots,n$, by Prime Avoidance Lemma we get $\mathrm{Ann}(m)\nsubseteq\bigcup_{i=1}^nP_i$.