If $I\subset J(R)$, ann$(M/IM)\subset p$, then $p$ is maximal?

Question

Let $R$ be a noetherian ring, $I\subset J(R)$ where $J(R)$ is the intersection of all maximal ideals of $R$, and $M$ be a finite $R$-module. Then,

if ann$(M/IM)\subset p$, $p$ is maximal?

What I know is $\sqrt{ann(M/IM)}=\sqrt{annM+I}$, so $annM+I\subset p$. I don't know if the above question is true and I have been trying to prove $Supp_R(M/IM)=Ass_R(M/IM)$, which may not be true either.

Answer 1

Set $R=M=\mathbb{Z[x, y]}$. Let $I=0\subseteq J(R)$. Hence we have $ann(M/IM)=ann(\mathbb{Z}[x, y]/0)=ann(\mathbb{Z}[x, y])=0\subseteq \langle x\rangle$, but $\langle x\rangle$ is not a maximal ideal of $R$ since $\langle x\rangle\subset \langle x, y\rangle$.

Note that for this example $Supp(M/IM)=Supp(\mathbb{Z}[x, y])=Spec(R)$ since for any ring $R$ and prime ideal $P$ of $R$ we have $R_P\not=o$, but $Ass(M/IM)=Ass(\mathbb{Z}[x, y])=\{0\}$ since the zero ideal a prime and so $0=0$ is a primary decomposition for $0$.