Lemma 35.20.18. 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 35.20.1 and 35.20.6 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 35.20.17. Therefore Lemma 35.19.4 applies and we win. $\square$

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