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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