Lemma 67.10.8 (06MG)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.10: Monomorphisms
Lemma 67.10.8 (cite)

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Lemma 67.10.8. Let $S$ be a scheme. Let $k$ be a field and let $Z \to \mathop{\mathrm{Spec}}(k)$ be a monomorphism of algebraic spaces over $S$ . Then either $Z = \emptyset$ or $Z = \mathop{\mathrm{Spec}}(k)$ .

Proof. By Lemmas 67.10.3 and 67.4.9 we see that $Z$ is a separated algebraic space. Hence there exists an open dense subspace $Z' \subset Z$ which is a scheme, see Properties of Spaces, Proposition 66.13.3. By Schemes, Lemma 26.23.11 we see that either $Z' = \emptyset$ or $Z' \cong \mathop{\mathrm{Spec}}(k)$ . In the first case we conclude that $Z = \emptyset$ and in the second case we conclude that $Z' = Z = \mathop{\mathrm{Spec}}(k)$ as $Z \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism which is an isomorphism over $Z'$ . $\square$

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