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The Stacks project

Lemma 10.126.3. Let R be a ring. Let I \subset R be an ideal. Let S \subset R be a multiplicative subset. Set R' = S^{-1}(R/I) = S^{-1}R/S^{-1}I.

  1. For any finite R'-module M' there exists a finite R-module M such that S^{-1}(M/IM) \cong M'.

  2. For any finitely presented R'-module M' there exists a finitely presented R-module M such that S^{-1}(M/IM) \cong M'.

Proof. Proof of (1). Choose a short exact sequence 0 \to K' \to (R')^{\oplus n} \to M' \to 0. Let K \subset R^{\oplus n} be the inverse image of K' under the map R^{\oplus n} \to (R')^{\oplus n}. Then M = R^{\oplus n}/K works.

Proof of (2). Choose a presentation (R')^{\oplus m} \to (R')^{\oplus n} \to M' \to 0. Suppose that the first map is given by the matrix A' = (a'_{ij}) and the second map is determined by generators x'_ i \in M', i = 1, \ldots , n. As R' = S^{-1}(R/I) we can choose s \in S and a matrix A = (a_{ij}) with coefficients in R such that a'_{ij} = a_{ij} / s \bmod S^{-1}I. Let M be the finitely presented R-module with presentation R^{\oplus m} \to R^{\oplus n} \to M \to 0 where the first map is given by the matrix A and the second map is determined by generators x_ i \in M, i = 1, \ldots , n. Then the map M \to M', x_ i \mapsto x'_ i induces an isomorphism S^{-1}(M/IM) \cong M'. \square


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