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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