Lemma 85.15.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces. Then $f$ is representable (by schemes) and affine.

Proof. We will show that $f$ is affine; it will then follow that $f$ is representable and affine by Morphisms of Spaces, Lemma 65.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu$ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 85.5.1. Let $T \to Y$ be a morphism where $T$ is a scheme over $S$. We have to show that $X \times _ Y T \to T$ is affine, see Bootstrap, Definition 78.4.1. To do this we may assume that $T$ is affine and we have to prove that $X \times _ Y T$ is affine. In this case $T \to Y$ factors through $Y_\mu \to Y$ for some $\mu$, see Lemma 85.5.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 85.13.3). Hence $X \times _ Y T \to X$ factors through $X_\lambda$ for some $\lambda$. Similarly $X_\lambda \to Y$ factors through $Y_\mu$ after increasing $\mu$. Then $X \times _ Y T = X_\lambda \times _{Y_\mu } T$. We conclude as fibre products of affine schemes are affine. $\square$

Comment #4054 by Matthew Emerton on

Lemma 0AIG has a locally quasi-finite hypothesis, which doesn't seem to hold in the generality of the present lemma (unless I'm missing something). But it seems that the second argument just works anyway to show that if $T$ is a scheme, then the fibre product $X\times_Y T$ is an affine scheme.

Comment #4137 by on

Yes, you are totally right! I have fixed this more or less as you suggested. See changes here.

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