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Let $M$ be a finitely generated module over a noetherian ring $R$. I want to find out why the formula

$${\sqrt {\operatorname {ann}_{R}(M)}}=\bigcap _{{{\mathfrak {p}}\in \operatorname {supp}M}}{\mathfrak {p}}$$

is true. recall the definitions $\operatorname{Supp}(M)= \{\mathfrak{p} \in \operatorname{Spec}(R): M_{\mathfrak{p}} \neq 0 \}$ and $\operatorname{Ann}_R(M) = \{r \in R \mid \forall m \in M \text{ holds } r \cdot m =0 \}$

the "$\subset$" is trivial: $a \in {\sqrt {\operatorname {ann}_{R}(M)}}$ then $a^n \in {\operatorname {ann}_{R}(M)}$ for certain $n$ and consequently $a^n \cdot m=0$ for all $m \in M$. let $\mathfrak {p}\in {\operatorname {supp}M}$ then $M_{\mathfrak{p}} \neq 0$ and therefore $a^n \not \in R \backslash \mathfrak{p}$. consequently $a^n \in \mathfrak{p}$, therefore $a \in \mathfrak{p}$.

The "$⊃$" I don't know.

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  • $\begingroup$ math.stackexchange.com/questions/3363387/… It seems like $R$ doesn't have to be Noetherian. $\endgroup$
    – user682705
    Sep 29 '19 at 0:00
  • $\begingroup$ But when $M$ is a finitely generated module over a Noetherian ring $R$, since $\mathrm{Ass}(M)$ is finite nonempty and the set of minimal elements of $\mathrm{Ass}(M)$ and of $\mathrm{Supp}(M)$ coincide, the additional equality in the formula in your link holds(p39, Theorem 6.5., Commutative Ring Thoery by Hideyuki Matsumura). $\endgroup$
    – user682705
    Sep 29 '19 at 1:03
  • $\begingroup$ @Tim Grosskreutz Note that $V({\rm ann_R(M)})={\rm Supp}_RM$ and the result follows from this fact. $\endgroup$ Sep 29 '19 at 11:00

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