Lemma 35.20.10. The property $\mathcal{P}(f) =$“$f$ is locally of finite type” is fpqc local on the base.

Proof. Being locally of finite type is preserved under base change, see Morphisms, Lemma 29.15.4. Being locally of finite type is Zariski local on the base, see Morphisms, Lemma 29.15.2. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is locally of finite type. Let $U \subset X$ be an affine open. Then $U' = S' \times _ S U$ is affine and of finite type over $S'$. Write $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$, $U = \mathop{\mathrm{Spec}}(A)$, and $U' = \mathop{\mathrm{Spec}}(A')$. We know that $R \to R'$ is faithfully flat, $A' = R' \otimes _ R A$ and $R' \to A'$ is of finite type. We have to show that $R \to A$ is of finite type. This is the result of Algebra, Lemma 10.125.1. It follows that $f$ is locally of finite type. Therefore Lemma 35.19.4 applies and we win. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02KX. Beware of the difference between the letter 'O' and the digit '0'.