Lemma 10.66.6. Let $R$ be a ring. Let $M$ be an $R$-module. Then

$\text{Ass}(M) \subset \text{WeakAss}(M) \subset \text{Supp}(M).$

Proof. The first inclusion is immediate from the definitions. If $\mathfrak p \in \text{WeakAss}(M)$, then by Lemma 10.66.2 we have $M_{\mathfrak p} \not= 0$, hence $\mathfrak p \in \text{Supp}(M)$. $\square$

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