# Prove ${\sqrt {\operatorname {ann}_{R}(M)}}=\bigcap _{{{\mathfrak {p}}\in \operatorname {supp}M}}{\mathfrak {p}}$ for a finitely generated module $M$

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.

• math.stackexchange.com/questions/3363387/… It seems like $R$ doesn't have to be Noetherian.
– user682705
Sep 29 '19 at 0:00
• 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).
– user682705
Sep 29 '19 at 1:03
• @Tim Grosskreutz Note that $V({\rm ann_R(M)})={\rm Supp}_RM$ and the result follows from this fact. Sep 29 '19 at 11:00