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The Stacks project

Lemma 10.83.2. Let R \to S be a faithfully flat ring map. Let M be an R-module. Then

  1. if the S-module M \otimes _ R S is of finite type, then M is of finite type,

  2. if the S-module M \otimes _ R S is of finite presentation, then M is of finite presentation,

  3. if the S-module M \otimes _ R S is flat, then M is flat, and

  4. add more here as needed.

Proof. Assume M \otimes _ R S is of finite type. Let y_1, \ldots , y_ m be generators of M \otimes _ R S over S. Write y_ j = \sum x_ i \otimes f_ i for some x_1, \ldots , x_ n \in M. Then we see that the map \varphi : R^{\oplus n} \to M has the property that \varphi \otimes \text{id}_ S : S^{\oplus n} \to M \otimes _ R S is surjective. Since R \to S is faithfully flat we see that \varphi is surjective, and M is finitely generated.

Assume M \otimes _ R S is of finite presentation. By (1) we see that M is of finite type. Choose a surjection R^{\oplus n} \to M and denote K the kernel. As R \to S is flat we see that K \otimes _ R S is the kernel of the base change S^{\oplus n} \to M \otimes _ R S. As M \otimes _ R S is of finite presentation we conclude that K \otimes _ R S is of finite type. Hence by (1) we see that K is of finite type and hence M is of finite presentation.

Part (3) is Lemma 10.39.8. \square


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