Lemma 10.126.4. Let R be a ring. Let S \subset R be a multiplicative subset. Let M be an R-module.
If S^{-1}M is a finite S^{-1}R-module then there exists a finite R-module M' and a map M' \to M which induces an isomorphism S^{-1}M' \to S^{-1}M.
If S^{-1}M is a finitely presented S^{-1}R-module then there exists an R-module M' of finite presentation and a map M' \to M which induces an isomorphism S^{-1}M' \to S^{-1}M.
Proof. Proof of (1). Let x_1, \ldots , x_ n \in M be elements which generate S^{-1}M as an S^{-1}R-module. Let M' be the R-submodule of M generated by x_1, \ldots , x_ n.
Proof of (2). Let x_1, \ldots , x_ n \in M be elements which generate S^{-1}M as an S^{-1}R-module. Let K = \mathop{\mathrm{Ker}}(R^{\oplus n} \to M) where the map is given by the rule (a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i. By Lemma 10.5.3 we see that S^{-1}K is a finite S^{-1}R-module. By (1) we can find a finite submodule K' \subset K with S^{-1}K' = S^{-1}K. Take M' = \mathop{\mathrm{Coker}}(K' \to R^{\oplus n}). \square
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