Lemma 81.9.1. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a morphism over $B$. Assume $f$ is flat of relative dimension $r$. For any closed subset $T \subset |Y|$ we have

$\dim _\delta (|f|^{-1}(T)) = \dim _\delta (T) + r.$

provided $|f|^{-1}(T)$ is nonempty. If $Z \subset Y$ is an integral closed subscheme and $Z' \subset f^{-1}(Z)$ is an irreducible component, then $Z'$ dominates $Z$ and $\dim _\delta (Z') = \dim _\delta (Z) + r$.

Proof. Since the $\delta$-dimension of a closed subset is the supremum of the $\delta$-dimensions of the irreducible components, it suffices to prove the final statement. We may replace $Y$ by the integral closed subscheme $Z$ and $X$ by $f^{-1}(Z) = Z \times _ Y X$. 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 of Spaces, Lemma 66.30.6 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 $Z = Y$. Thus $Z'$ dominates $Z = Y$. Finally, we see that $\xi$ and $\eta$ are in the schematic locus of $X$ and $Y$ by Properties of Spaces, Proposition 65.13.3. 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 $0$ (we may use usual local rings as $\xi$ and $\eta$ are in the schematic loci of $X$ and $Y$). Thus by Morphisms of Spaces, Lemma 66.34.1 (and the definition of morphisms of given relative dimension) 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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