As the title says, I am trying to prove the following:
Let $\mathcal{F} = \widetilde{M}$ on $\text{Spec}{A}$ where $M$ is a finitely generated $A$-module. Show that the support of $\mathcal{F}$ is equal to $V(\text{Ann}(M))$.
Here is my proof. I am very inexperienced in commutative algebra, so please let me know if there are any errors/improvements I can make.
Suppose $\mathfrak{p} \in \text{supp}(\mathcal{F})$. This means that $M_{\mathfrak{p}} \neq 0$, so we pick some $m \in M$ whose germ in $M_{\mathfrak{p}}$ is nonzero. What this means is that $sm \neq 0$ for all $s \in A \setminus \mathfrak{p}$, so the elements of $\text{Ann}(m)$ must be contained in $\mathfrak{p}$. Therefore we have $\mathfrak{p} \in V(\text{Ann}(M))$. Conversely, suppose $\mathfrak{p} \notin \text{supp}(\mathcal{F})$. Then $M_{\mathfrak{p}} = 0$, so for each generator $x_i$ of $M$, we have $f_ix_i = 0$ for some $f_i \in A\setminus \mathfrak{p}$. Since $\mathfrak{p}$ is a prime ideal, and the set of generators is finite, we have $f:=\prod_{i}f_i \in A\setminus \mathfrak{p}$. However $f \in \text{Ann}(M)$, so $\mathfrak{p} \notin V(\text{Ann}(M))$.