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?
Best Answer
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. $\square$