Lemma 42.13.1. Let (S, \delta ) be as in Situation 42.7.1. Let X, Y be locally of finite type over S. Let f : X \to Y be a morphism. Assume f is flat of relative dimension r. For any closed subset Z \subset Y we have
\dim _\delta (f^{-1}(Z)) = \dim _\delta (Z) + r.
provided f^{-1}(Z) is nonempty. If Z is irreducible and Z' \subset f^{-1}(Z) is an irreducible component, then Z' dominates Z and \dim _\delta (Z') = \dim _\delta (Z) + r.
Proof.
It suffices to prove the final statement. We may replace Y by the integral closed subscheme Z and X by the scheme theoretic inverse image f^{-1}(Z) = Z \times _ Y X. Hence we may assume Z = Y is integral and f is a flat morphism of relative dimension r. Since Y is locally Noetherian the morphism f which is locally of finite type, is actually locally of finite presentation. Hence Morphisms, Lemma 29.25.10 applies and we see that f is open. Let \xi \in X be a generic point of an irreducible component of X. By the openness of f we see that f(\xi ) is the generic point \eta of Z = Y. Note that \dim _\xi (X_\eta ) = r by assumption that f has relative dimension r. On the other hand, since \xi is a generic point of X we see that \mathcal{O}_{X, \xi } = \mathcal{O}_{X_\eta , \xi } has only one prime ideal and hence has dimension 0. Thus by Morphisms, Lemma 29.28.1 we conclude that the transcendence degree of \kappa (\xi ) over \kappa (\eta ) is r. In other words, \delta (\xi ) = \delta (\eta ) + r as desired.
\square
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