# The Stacks Project

## Tag 02R7

### 41.14. Preparation for flat pullback

Recall that a morphism $f : X \to Y$ which is locally of finite type is said to have relative dimension $r$ if every nonempty fibre is equidimensional of dimension $r$. See Morphisms, Definition 28.28.1.

Lemma 41.14.1. Let $(S, \delta)$ be as in Situation 41.8.1. 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 $$\dim_\delta(f^{-1}(Z)) = \dim_\delta(Z) + r.$$ 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 28.24.9 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 28.27.1 we conclude that the transcendence degree of $\kappa(\xi)$ over $\kappa(\eta)$ is $r$. In other words, $\delta(\xi) = \delta(\eta) + r$ as desired. $\square$

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 41.14.2. Let $(S, \delta)$ be as in Situation 41.8.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a morphism. Assume $\{Z_i\}_{i \in I}$ is a locally finite collection of closed subsets of $Y$. Then $\{f^{-1}(Z_i)\}_{i \in I}$ 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 $$\{i \in I \mid f^{-1}(Z_i) \cap U \not = \emptyset \} \subset \{i \in I \mid Z_i \cap V \not = \emptyset \}.$$ Since the right hand side is finite by assumption we win. $\square$

The code snippet corresponding to this tag is a part of the file chow.tex and is located in lines 3664–3741 (see updates for more information).

\section{Preparation for flat pullback}
\label{section-preparation-flat-pullback}

\noindent
Recall that a morphism $f : X \to Y$ which is locally of finite type
is said to have relative dimension $r$ if every nonempty fibre
is equidimensional of dimension $r$. See
Morphisms, Definition \ref{morphisms-definition-relative-dimension-d}.

\begin{lemma}
\label{lemma-flat-inverse-image-dimension}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
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
$$\dim_\delta(f^{-1}(Z)) = \dim_\delta(Z) + r.$$
If $Z$ is irreducible and $Z' \subset f^{-1}(Z)$ is an irreducible
component, then $Z'$ dominates $Z$ and
$\dim_\delta(Z') = \dim_\delta(Z) + r$.
\end{lemma}

\begin{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 \ref{morphisms-lemma-fppf-open}
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 \ref{morphisms-lemma-dimension-fibre-at-a-point}
we conclude that the transcendence
degree of $\kappa(\xi)$ over $\kappa(\eta)$ is $r$.
In other words, $\delta(\xi) = \delta(\eta) + r$ as desired.
\end{proof}

\noindent
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.

\begin{lemma}
\label{lemma-inverse-image-locally-finite}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$, $Y$ be locally of finite type over $S$.
Let $f : X \to Y$ be a morphism.
Assume $\{Z_i\}_{i \in I}$ is a locally
finite collection of closed subsets of $Y$.
Then $\{f^{-1}(Z_i)\}_{i \in I}$ is a locally finite
collection of closed subsets of $X$.
\end{lemma}

\begin{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
$$\{i \in I \mid f^{-1}(Z_i) \cap U \not = \emptyset \} \subset \{i \in I \mid Z_i \cap V \not = \emptyset \}.$$
Since the right hand side is finite by assumption we win.
\end{proof}

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