Intersection of associated prime ideals is radical of annulator

abstract-algebramaximal-and-prime-ideals

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:
$$\bigcap_{\mathfrak{p} \in \text{Ass}(M)} \mathfrak{p} = \sqrt{\text{Ann}(M)}$$

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 $$\operatorname{Supp}(M)$$, 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 $$\sqrt{I} = \bigcap_{\mathfrak{p} \in \text{Spec}R, I \subseteq \mathfrak{p}} \mathfrak{p}$$ for an ideal $$I$$. That didn't really seem to work. What can I try here?

Lemma. Let $$M$$ be a finite module over a Noetherian ring $$R$$. Then $$\sqrt{\operatorname{Ann}M}=\bigcap\operatorname{Supp}M=\bigcap\operatorname{Ass}M.$$
Proof. Since $$M$$ is finite, then $$V(\operatorname{Ann}M)=\operatorname{Supp}M$$, 00L2. Thus, by 00E0.7, we have $$\sqrt{\operatorname{Ann}M}=\bigcap V(\operatorname{Ann}M)=\bigcap\operatorname{Supp} M$$. Suppose we have shown that for $$S\in\{\operatorname{Ass}M,\operatorname{Supp}M\}$$, every prime $$\mathfrak{p}\in S$$ contains a minimal prime of $$S$$. Then it will follow $$\bigcap\operatorname{Supp} M=\bigcap\operatorname{Ass}M$$ from 02CE. On the one hand, the assertion is true for $$S=\operatorname{Ass}M$$ by 00LC. On the other hand, each point in $$S=\operatorname{Supp}M$$ is contained in some irreducible component of $$S$$ by 0052 ($$S$$ is Noetherian since $$\operatorname{Spec}R$$ is). Since $$\operatorname{Supp}M=V(\operatorname{Ann}M)$$ is sober (use 0B31 and that $$\operatorname{Spec}R$$ is sober), this means that every point in $$\operatorname{Supp}M$$ generalizes (in the sense of 0061) to a minimal prime of $$\operatorname{Supp}M$$, as we wanted to show. $$\square$$