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The Stacks project

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


Comments (2)

Comment #5513 by Ben church on

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

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  • 3 comment(s) on Section 10.66: Weakly associated primes

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