Lemma 35.25.3 (09NP)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 35: Descent
Section 35.25: Application of fpqc descent of properties of morphisms
Lemma 35.25.3 (cite)

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Lemma 35.25.3. Let $f : X \to Y$ be a morphism of schemes. Let $X^0$ denote the set of generic points of irreducible components of $X$ . If

$f$ is flat and separated,
for $\xi \in X^0$ we have $\kappa (f(\xi )) = \kappa (\xi )$ , and
if $\xi , \xi ' \in X^0$ , $\xi \not= \xi '$ , then $f(\xi ) \not= f(\xi ')$ ,

then $f$ is universally injective.

Proof. We have to show that $\Delta : X \to X \times _ Y X$ is surjective, see Morphisms, Lemma 29.10.2. As $X \to Y$ is separated, the image of $\Delta$ is closed. Thus if $\Delta$ is not surjective, we can find a generic point $\eta \in X \times _ S X$ of an irreducible component of $X \times _ S X$ which is not in the image of $\Delta$ . The projection $\text{pr}_1 : X \times _ Y X \to X$ is flat as a base change of the flat morphism $X \to Y$ , see Morphisms, Lemma 29.25.8. Hence generalizations lift along $\text{pr}_1$ , see Morphisms, Lemma 29.25.9. We conclude that $\xi = \text{pr}_1(\eta ) \in X^0$ . However, assumptions (2) and (3) guarantee that the scheme $(X \times _ Y X)_{f(\xi )}$ has at most one point for every $\xi \in X^0$ . In other words, we have $\Delta (\xi ) = \eta$ a contradiction. $\square$

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