The Stacks project

Lemma 35.23.18. The property $\mathcal{P}(f) =$“$f$ is affine” is fpqc local on the base.

Proof. A base change of an affine morphism is affine, see Morphisms, Lemma 29.11.8. Being affine is Zariski local on the base, see Morphisms, Lemma 29.11.3. Finally, let $g : S' \to S$ 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 affine. In other words, $X'$ is affine, say $X' = \mathop{\mathrm{Spec}}(A')$. Also write $S = \mathop{\mathrm{Spec}}(R)$ and $S' = \mathop{\mathrm{Spec}}(R')$. We have to show that $X$ is affine.

By Lemmas 35.23.1 and 35.23.6 we see that $X \to S$ is separated and quasi-compact. Thus $f_*\mathcal{O}_ X$ is a quasi-coherent sheaf of $\mathcal{O}_ S$-algebras, see Schemes, Lemma 26.24.1. Hence $f_*\mathcal{O}_ X = \widetilde{A}$ for some $R$-algebra $A$. In fact $A = \Gamma (X, \mathcal{O}_ X)$ of course. Also, by flat base change (see for example Cohomology of Schemes, Lemma 30.5.2) we have $g^*f_*\mathcal{O}_ X = f'_*\mathcal{O}_{X'}$. In other words, we have $A' = R' \otimes _ R A$. Consider the canonical morphism

\[ X \longrightarrow \mathop{\mathrm{Spec}}(A) \]

over $S$ from Schemes, Lemma 26.6.4. By the above the base change of this morphism to $S'$ is an isomorphism. Hence it is an isomorphism by Lemma 35.23.17. Therefore Lemma 35.22.4 applies and we win. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 35.23: Properties of morphisms local in the fpqc topology on the target

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Your email address will not be published.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 02L5. Beware of the difference between the letter 'O' and the digit '0'.