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. For $m \in S^{-1}M$, let $I \subset R$ and $J \subset S^{-1}R$ be the annihilators of $m$. Then $I$ is the inverse image of $J$ by the map $R \to S^{-1}R$ and $J = S^{-1}I$. The equality in (1) follows by the description of the map $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ in Lemma 10.17.5. For $m \in M$, let $I \subset R$ be the annihilator of $m$ in $R$ and let $J \subset S^{-1}R$ be the annihilator of $m/1 \in S^{-1}M$. We have $J = S^{-1}I$ which implies (2). The equality in the Noetherian case follows from Lemma 10.63.15 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset$ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$. $\square$

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.

Comment #8250 by on

Hint: intersection means inverse image by the ring map $R \to S^{-1}R$.

Comment #9034 by on

Regarding "the displayed inclusion and equality in the Noetherian case follows from Lemma 10.63.15": I think the displayed inclusion is independent of the invoked lemma (one observes that the $R$-annihilator $\mathfrak{p}$ of $m$ in $M$ is the $R$-annihilator of $m$ in $S^{-1}M$ if $\mathfrak{p}\cap S=\varnothing$), and it is only the converse inclusion $\text{Ass}_R(M) \cap \operatorname{Spec}(S^{-1}R) \supset \text{Ass}_R(S^{-1}M)$ what requires the mentioned lemma.

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

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