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.
Best Answer
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.