Proof. Let $f : X \to S$ be an affine morphism. Let $S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times _ S X \to S'$ the base change of $f$. For every $s' \in S'$ there exists an open affine neighbourhood $s' \in V \subset S'$ which maps into some open affine $U \subset S$. By assumption $f^{-1}(U)$ is affine. By the material in Schemes, Section 26.17 we see that $f^{-1}(U)_ V = V \times _ U f^{-1}(U)$ is affine and equal to $(f')^{-1}(V)$. This proves that $S'$ has an open covering by affines whose inverse image under $f'$ is affine. We conclude by Lemma 29.11.3 above. $\square$

Comment #8490 by on

I think one could open with the following sentence (or something similar): "the result follows from Lemmas 27.4.6 and 29.11.3, but we can give a direct proof."

There are also:

• 5 comment(s) on Section 29.11: Affine morphisms

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).