If $M$ is a finitely generated module then $\sqrt{\text{ann}(M)}=\bigcap\text{supp}(M)$ 
Let $R$ be a unital commutative ring and let $M$ be a finitely generated $R$-module. Prove that
  $$
\sqrt{\text{ann}(M)}=\bigcap \text {supp}(M)
$$

Recall that:


*

*$\text{ann}(M)=\{r\in R:\forall m\in m, rm=0\}$

*$\text{supp}(M)=\{p\in\text{spec}(R):M_p\ne 0\}$

*$\forall p\in\text{spec}(R), V(p)=\{q\in\text{spec}(R):p\subseteq q\}$

*$\forall A\subseteq R,\sqrt A=\{r\in R:\exists n, r^n\in A\}$

*$\operatorname{spec}(R)=\{I\subseteq R:I\text{ is a prime ideal}\}$
According to a theorem, $$\text{supp}(M)=\bigcup_{p\in\text{ass}(M)}V(p)$$ But I don't really success to proceed.
 A: First, we show that if $p \in supp(M)$, then $ann(M) \subset p$. 
Indeed, let $x \in ann(M)$ not in $p$: then multiplication by $x$ from $M_p$ to itself is an isomorphism, since $x$ is invertible in $A_p$. On the other hand, this function is zero since $x$ is in the annihilator of $M$, a contradiction. 
Next, we show that if $ann(M) \subset p$, then $M_p \neq 0$. 
For this, notice that if $M_p=0$, this means that for all $m \in M$, there is $t \notin p$ such that $tm=0$. Since $M$ is finitely generated, this implies the existence of $t \notin p$ such that $tM=0$, so $t \in ann(M)$ but $t \notin p$, a contradiction. 
The rest is standard commutative algebra. 
A: $\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in V(\mathrm{ann}(M))}^{}p$ (Proposition 1.14 in Introduction to Commutative Algebra by M.F. Atiyah, I.G. MacDonald). But $V(\mathrm{ann}(M))=\mathrm{supp}(M)$ for a finitely generated $R$-module $M$(Theorem 1.5.5, https://faculty.math.illinois.edu/~r-ash/ComAlg/ComAlg1.pdf).
So, $\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in\mathrm{Supp}(M)}^{}p$.
