Lemma 10.126.4. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module.

1. 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 $S^{-1}M' \to S^{-1}M$.

2. 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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