Lemma 53.2.4 (0CCL)—The Stacks project

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Table of contents
Part 3: Topics in Scheme Theory
Chapter 53: Algebraic Curves
Section 53.2: Curves and function fields
Lemma 53.2.4 (cite)

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Lemma 53.2.4. Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes over $k$ . Assume

$Y$ is separated over $k$ ,
$X$ is proper of dimension $\leq 1$ over $k$ ,
$f(Z)$ has at least two points for every irreducible component $Z \subset X$ of dimension $1$ .

Then $f$ is finite.

Proof. The morphism $f$ is proper by Morphisms, Lemma 29.41.7. Thus $f(X)$ is closed and images of closed points are closed. Let $y \in Y$ be the image of a closed point in $X$ . Then $f^{-1}(\{ y\} )$ is a closed subset of $X$ not containing any of the generic points of irreducible components of dimension $1$ by condition (3). It follows that $f^{-1}(\{ y\} )$ is finite. Hence $f$ is finite over an open neighbourhood of $y$ by More on Morphisms, Lemma 37.44.2 (if $Y$ is Noetherian, then you can use the easier Cohomology of Schemes, Lemma 30.21.2). Since we've seen above that there are enough of these points $y$ , the proof is complete. $\square$

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