Support of $\operatorname{Hom}(R/I, M)$ Let $R$ be a Noetherian ring, $I$ be an ideal of $R$ and $M$ be an $R$-module. Is the following formula true? 

$\operatorname{Supp}\operatorname{Hom}(R/I, M)=\operatorname{Supp}(M) \cap V(I)$

If not, under what conditions the above formula is true?
Thanks.
 A: Update: The previous answer was incorrect - the following is a slightly different approach.
$\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{\Ass}{\operatorname{Ass}}$
$\newcommand{\Supp}{\operatorname{Supp}}$

The relevant facts to answer this question are (assuming $M$ is a finite $R$-module):
1) $\Hom$ localizes, in the sense that $\Hom_R(R/I,M)_p \cong \Hom_{R_p}((R/I)_p, M_p)$ for any prime $p$. This shows that $\Supp(\Hom_R(R/I,M)) \subseteq \Supp(M) \cap V(I)$.
2) For any $M$, $\Hom_R(R/I,M) \cong 0 :_M I$ is the submodule of $M$ which annihilates $I$. Thus, $\Hom_R(R/I,M) \ne 0$ iff $I$ consists of zerodivisors on $M$ iff $I \subseteq p$ for some $p \in \Ass(M)$ iff $\Ass(M) \cap V(I) \ne \emptyset$. If $\Supp(M) \cap V(I) \ne \emptyset$, then this occurs iff $\text{depth}_I(M) = 0$.
3) Combining (1) and (2) gives: if $p \in \Supp(M) \cap V(I)$, then $p \in \Supp(\Hom_R(R/I,M))$ iff $\text{depth}_{I_p}(M_p) = 0$. Since closed sets are the closures of their minimal elements, we have:

$\Supp(\Hom_R(R/I,M)) = \Supp(M) \cap V(I) \iff \text{depth}_{I_p}(M_p) = 0$ for every minimal element $p$ of $\Supp(M) \cap V(I) \iff \Supp(M) \cap V(I)$ and $\Ass(M) \cap V(I)$ have the same minimal elements.

Note that $\Supp(M) \cap V(I)$ always has only finitely many minimal elements, since $R$ is Noetherian. However, these need not coincide with minimal elements of $\Ass(M) \cap V(I)$, even if the latter is nonempty. Explicitly, take a chain of prime ideals $p \subsetneq q \subsetneq p'$ in a Noetherian ring $R$, and set $M := R/p \oplus R/p'$, $I := q$. Then $\Ass(M) = \{p, p'\}$, so $\Ass(M) \cap V(I) = \{p'\}$ and $\Supp(\Hom_R(R/I,M)) = V(p')$, but $\Supp(M) \cap V(I) = V(q) \supsetneq V(p')$. In this case, $\text{depth}_I(M) = 0$, but $\text{depth}_{I_q}{M_q} \ne 0$ (and $q$ is a minimal element of $\Supp(M) \cap V(I)$).
