The Stacks project

Proof. Let $f : X \to Y$ be a flat monomorphism of algebraic spaces. To prove $f$ is representable, we have to show $X \times _ Y V$ is a scheme for every scheme $V$ mapping to $Y$. Since being a scheme is local (Properties of Spaces, Lemma 66.13.1), we may assume $V$ is affine. Thus we may assume $Y = \mathop{\mathrm{Spec}}(B)$ is an affine scheme. Next, we can assume that $X$ is quasi-compact by replacing $X$ by a quasi-compact open. The space $X$ is separated as $X \to X \times _{\mathop{\mathrm{Spec}}(B)} X$ is an isomorphism. Applying Limits of Spaces, Lemma 70.17.3 we reduce to the case where $B$ is local, $X \to \mathop{\mathrm{Spec}}(B)$ is a flat monomorphism, and there exists a point $x \in X$ mapping to the closed point of $\mathop{\mathrm{Spec}}(B)$. Then $X \to \mathop{\mathrm{Spec}}(B)$ is surjective as generalizations lift along flat morphisms of separated algebraic spaces, see Decent Spaces, Lemma 68.7.4. Hence we see that $\{ X \to \mathop{\mathrm{Spec}}(B)\} $ is an fpqc cover. Then $X \to \mathop{\mathrm{Spec}}(B)$ is a morphism which becomes an isomorphism after base change by $X \to \mathop{\mathrm{Spec}}(B)$. Hence it is an isomorphism by fpqc descent, see Descent on Spaces, Lemma 74.11.15. $\square$


Comments (4)

Comment #1563 by Matthew Emerton on

Unless I'm misunderstanding, the ring morphs into the ring partway through the argument.

Comment #2457 by Matthieu Romagny on

First sentence of proof: flat morphism --> flat monomorphism.


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