$M$ is $\bigcap \operatorname{Ass}(M)$-primary Let $R$ be noetherian ring and $M$ an $R$-module such that $\operatorname{Ass}(M)$ is a finite set. Prove that $M$ is $\mathfrak{b}$-primary, where $\mathfrak{b}=\bigcap \operatorname{Ass}(M)=\bigcap_{\mathfrak{p}\in\operatorname{Ass}(M)} \mathfrak{p}$. > $M$ is $\mathfrak{b}$-primary, that is $M=\sum_{i=1}^\infty \operatorname{Ann}_M(\mathfrak{b^i})$. Let $x\in M$, I want to show that $x$ will be killed by some power of $\mathfrak b$. We have $$\operatorname{Ann}_R(Rx)\subseteq \bigcap \operatorname{Ass}(Rx),$$ and $$ \mathfrak{b}=\bigcap \operatorname{Ass}(M)\subseteq\bigcap \operatorname{Ass}(Rx)$$ but I can't show that $$\mathfrak{b}^k\subseteq \operatorname{Ann}(Rx)$$ for some $k$. Thank you.

> Let $R$ be a noetherian ring, $M$ an $R$-module such that the set $\operatorname{Ass}(M)$ is finite, and $\mathfrak{b}=\bigcap_{\mathfrak{p}\in\operatorname{Ass}(M)} \mathfrak{p}$. Then $M=\bigcup_{i\ge 1}(0:_M\mathfrak{b}^i)$.

This is not hard to prove: consider $(0)=\bigcap_{i=1}^nQ_i$ a primary decomposition of the zero submodule of $M$, where $Q_i$ are $\mathfrak p_i$-primary submodules. Then $r_M(0)=\mathfrak{b}$. On the other side, $r_M(0)$ is an ideal of $R$ and therefore it is finitely generated. This shows that for every $x\in M$ there exists $k\ge 1$ such that $\mathfrak{b}^kx=0$.