The Stacks project

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., $\mathfrak p \in \text{Supp}(M)$. $\square$

Comments (2)

Comment #8381 by Fawzy N. Hegab on

I think this does not need a proof, since it is a corollary of this lemma https://stacks.math.columbia.edu/tag/07Z5 proved earlier.

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  • 14 comment(s) on Section 10.63: Associated primes

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