## 81.9 Preparation for flat pullback

This section is the analogue of Chow Homology, Section 42.13.

Recall that a morphism of algebraic spaces is said to have relative dimension $r$ if étale locally on the source and the target we get a morphism of schemes which has relative dimension $r$. The precise definition is equivalent, but in fact slightly different, see Morphisms of Spaces, Definition 66.33.2.

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. $\square$

Here is the lemma that we will use to prove that the flat pullback of a locally finite collection of closed subschemes is locally finite.

Lemma 81.9.2. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a morphism over $B$. Assume $\{ T_ i\} _{i \in I}$ is a locally finite collection of closed subsets of $|Y|$. Then $\{ |f|^{-1}(T_ i)\} _{i \in I}$ is a locally finite collection of closed subsets of $X$.

Proof. Let $U \subset |X|$ be a quasi-compact open subset. Since the image $|f|(U) \subset |Y|$ is a quasi-compact subset there exists a quasi-compact open $V \subset |Y|$ such that $|f|(U) \subset V$. Note that

$\{ i \in I : |f|^{-1}(T_ i) \cap U \not= \emptyset \} \subset \{ i \in I : T_ i \cap V \not= \emptyset \} .$

Since the right hand side is finite by assumption we win. $\square$

Comment #7277 by Anonymous on

Typo: "Recall that a morphism of algebraic spaces is...relative dimension $r$ if...has relative dimension $d$."

The $r$ and $d$ should be the same.

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