The Stacks project

Lemma 66.49.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is flat and locally of finite type and assume $Y$ satisfies the equivalent conditions of Lemma 66.49.1. Then $X$ satisfies the equivalent conditions of Lemma 66.49.1 and for $x \in |X|$ we have: $x$ has codimension $0$ in $X \Rightarrow f(x)$ has codimension $0$ in $Y$.

Proof. The last statement follows from Lemma 66.49.3. Choose a surjective ├ętale morphism $V \to Y$ where $V$ is a scheme. Choose a surjective ├ętale morphism $U \to X \times _ Y V$ where $U$ is a scheme. It suffices to show that every quasi-compact open of $U$ has finitely many irreducible components. We will use the results of Properties of Spaces, Lemma 65.11.1 without further mention. By what we've already shown, the codimension $0$ points of $U$ lie above codimension $0$ points in $U$ and these are locally finite by assumption. Hence it suffices to show that for $v \in V$ of codimension $0$ the codimension $0$ points of the scheme theoretic fibre $U_ v = U \times _ V v$ are locally finite. This is true because $U_ v$ is a scheme locally of finite type over $\kappa (v)$, hence locally Noetherian and we can apply Lemma 66.49.2 for example. $\square$

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