# Prove $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$

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?

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)
• 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! Feb 4 at 20:11