Lemma 66.4.11 (0H2X)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 66: Properties of Algebraic Spaces
Section 66.4: Points of algebraic spaces
Lemma 66.4.11 (cite)

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Lemma 66.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ . Let $p : \mathop{\mathrm{Spec}}(K) \to X$ and $q : \mathop{\mathrm{Spec}}(L) \to X$ be morphisms where $K$ and $L$ are fields. Assume $p$ and $q$ determine the same point of $|X|$ and $p$ is a monomorphism. Then $q$ factors uniquely through $p$ .

Proof. Since $p$ and $q$ define the same point of $|X|$ , we see that the scheme

$Y = \mathop{\mathrm{Spec}}(K) \times _{p, X, q} \mathop{\mathrm{Spec}}(L)$

is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphism $Y \to \mathop{\mathrm{Spec}}(L)$ is a monomorphism. Hence $Y = \mathop{\mathrm{Spec}}(L)$ , see Schemes, Lemma 26.23.11. We conclude that $q$ factors through $p$ . Uniqueness comes from the fact that $p$ is a monomorphism. $\square$

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