The Stacks project

Lemma 10.63.16. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Via the canonical injection $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ we have

  1. $\text{Ass}_ R(S^{-1}M) = \text{Ass}_{S^{-1}R}(S^{-1}M)$,

  2. $\text{Ass}_ R(M) \cap \mathop{\mathrm{Spec}}(S^{-1}R) \subset \text{Ass}_ R(S^{-1}M)$, and

  3. if $R$ is Noetherian this inclusion is an equality.

Proof. The first equality follows, since if $m \in S^{-1}M$, then the annihilator of $m$ in $R$ is the intersection of the annihilator of $m$ in $S^{-1}R$ with $R$. The displayed inclusion and equality in the Noetherian case follows from Lemma 10.63.15 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset $ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$. $\square$

Comments (2)

Comment #8249 by Et on

I don't think the first statement of the proof about the annhilator of m is true. It does hold however if the annhilator is a prime ideal which doesn't intersect S, which is what you need anyway.

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

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