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Let $M$ be a finitely generated $A$-module, and let $I$ be an ideal of $A$. Claim: $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$

One implication is trivial: if $p \in \operatorname{Supp}(M / IM)$, then the definition of support tells me that the localized in $p$ module $(M / IM)_p \neq 0$. So $M_p \neq 0$ as well and $I \subset p$ (otherwise $I$ would countain $a \in I$ which is invertible in $M_p$ and therefore $M_p = (aM)_p \subset IM_p$ and we obtain $(M / IM)_p = 0$, a contradiction).

The other implication I not manage to show. Can any body help? I tried to do it by contraposition. Assume $p \notin \operatorname{Supp}(M / IM)$, then $(M / IM)_p = 0$. If $M_p=0$ ($= p \notin \operatorname{Supp} M$ we are done), so we can assume $M_p \neq 0$. And I have to show that $p \notin V(I)$. Assume that $I \subset p$. And now?

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Hint: If $(M/IM)_p=0$ then $M_p=I_pM_p$. Now assume $I\subset p$ and apply Nakayama's lemma (this is where we need the finiteness assumption)

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    $\begingroup$ Then since $p =J(A_p)$ (Jacobson ideal) the conclusion $M_p= IM_p$ implies $M_p=0$ and we are in first case, yes, thank you! $\endgroup$ Feb 4 at 20:11

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