Lemma 29.11.8. The base change of an affine morphism is affine.

**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.
$\square$

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## Comments (1)

Comment #8490 by ElĂas Guisado on

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