Lemma 10.82.2. 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 10.38.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).