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.