Lemma 35.20.15. The property $\mathcal{P}(f) =$“$f$ is flat” is fpqc local on the base.

Proof. Being flat is preserved under arbitrary base change, see Morphisms, Lemma 29.25.8. Being flat is Zariski local on the base by definition. 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 flat. Let $U \subset X$ be an affine open. Then $U' = S' \times _ S U$ is affine. 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 flat. Goal: Show that $R \to A$ is flat. This follows immediately from Algebra, Lemma 10.39.8. Hence $f$ is flat. Therefore Lemma 35.19.4 applies and we win. $\square$

Comment #92 by Keenan Kidwell on

In the proof of this result, it says that $U^\prime=S^\prime\times_SU$ is affine and of finite type over $S^\prime$. I think the and of finite type" is not supposed to be there since there are no finiteness hypotheses involved here.

There are also:

• 2 comment(s) on Section 35.20: Properties of morphisms local in the fpqc topology on the target

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).