Lemma 10.66.15. 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 $\text{WeakAss}_ R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$ and

$\text{WeakAss}(M) \cap \mathop{\mathrm{Spec}}(S^{-1}R) = \text{WeakAss}(S^{-1}M).$

Proof. Suppose that $m \in S^{-1}M$. Let $I = \{ x \in R \mid xm = 0\}$ and $I' = \{ x' \in S^{-1}R \mid x'm = 0\}$. Then $I' = S^{-1}I$ and $I \cap S = \emptyset$ unless $I = R$ (verifications omitted). Thus primes in $S^{-1}R$ minimal over $I'$ correspond bijectively to primes in $R$ minimal over $I$ and avoiding $S$. This proves the equality $\text{WeakAss}_ R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$. The second equality follows from Lemma 10.66.2 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 #5513 by Ben church on

I think you may mean Lemma 0566 not Lemma 0310 in the last line.

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