When $M$ is a finitely generated module over a commutative ring $R$, it is easy to see that the support of $\tilde{M}$ on $\mathrm{Spec}\,R$ is given by $V(\mathrm{ann}_R(M))$. This is not true for general $M$, as $\mathrm{supp}(\tilde{M})$ might not even be closed. But if we take the Zariski closure of $\mathrm{supp}(\tilde{M})$, do we retrieve $V(\mathrm{ann}_R(M))$ again? In other words, is the following identity true: $V(\mathrm{ann}_R(M))=\mathrm{Cl}_{\mathrm{Spec}\,R}\mathrm{supp}(\tilde{M})$ ?
(By support I mean the set of prime ideals $\mathfrak{p}\subset \mathrm{Spec}\,R$ such that $M_\mathfrak{p} \neq 0$.)
We always have $\overline{\mathrm{supp}(M)} \subseteq V(\mathrm{Ann}(M))$, but the converse doesn't always hold:
Consider the $\mathbb{Z}$-module $M=\bigoplus_{k \geq 0} \mathbb{Z}/p^k$ for some prime number $p$ (for example $p=57$). We have $\mathrm{Ann}(M)=\{0\}$, hence $V(\mathrm{Ann}(M))=\mathrm{Spec}(\mathbb{Z})$. Now let's compute the support.
We have $M \otimes \mathbb{Q} = 0$ (since $\mathbb{Z}/p^k \otimes \mathbb{Q}=0$). If $q$ is a prime number $\neq p$, then $M \otimes \mathbb{Z}_{(q)}=0$ (since $\mathbb{Z}/p^k \otimes \mathbb{Z}_{(q)}=0$). Therefore $\mathrm{supp}(M) = \{(p)\}$. This is already closed.