Lemma 10.63.2. Let $R$ be a ring. Let $M$ be an $R$-module. Then $\text{Ass}(M) \subset \text{Supp}(M)$.

**Proof.**
If $m \in M$ has annihilator $\mathfrak p$, then in particular no element of $R \setminus \mathfrak p$ annihilates $m$. Hence $m$ is a nonzero element of $M_{\mathfrak p}$, i.e., $\mathfrak p \in \text{Supp}(M)$.
$\square$

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