As an exercise I have to prove:
Let $M$ be a finitely generated $R$-module, where $R$ is a noetherian ring (commutative and with $1$). Then: $$\cap_{\mathfrak{p} \in \text{Ass}(M)} \mathfrak{p} = \sqrt{\text{Ann}(M)}$$
Where $\text{Ass}(M) = \{\mathfrak{p}\in \text{Spec}R: \mathfrak{p} = \text{Ann}(m) \text{ for some } m \in M\}$.
The "$\supseteq$" inclusion is trivial, however I am stuck on the other inclusion. I have seen a similar result formulated using supp$(M)$, however we have not encountered this yet. My idea was to show that $\text{Ass}(M) = \{\mathfrak{p} \in \text{Spec}R : \text{Ann}(M) \subseteq \mathfrak{p} \}$, because we know that $\sqrt{I} = \cap_{\mathfrak{p} \in \text{Spec}R, I \subseteq \mathfrak{p}} \mathfrak{p}$ for an ideal $I$. That didn't really seem to work. What can I try here?