Lemma 10.40.9. Let $R$ be a ring and let $M$ be an $R$-module.
If $M$ is finite then the support of $M/IM$ is $\text{Supp}(M) \cap V(I)$.
If $N \subset M$, then $\text{Supp}(N) \subset \text{Supp}(M)$.
If $Q$ is a quotient module of $M$ then $\text{Supp}(Q) \subset \text{Supp}(M)$.
If $0 \to N \to M \to Q \to 0$ is a short exact sequence then $\text{Supp}(M) = \text{Supp}(Q) \cup \text{Supp}(N)$.
Proof. The functors $M \mapsto M_{\mathfrak p}$ are exact. This immediately implies all but the first assertion. For the first assertion we need to show that $M_\mathfrak p \not= 0$ and $I \subset \mathfrak p$ implies $(M/IM)_{\mathfrak p} = M_\mathfrak p/IM_\mathfrak p \not= 0$. This follows from Nakayama's Lemma 10.20.1. $\square$
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