Lemma 86.19.7. 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 66.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu$ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 86.9.1. 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 79.4.1. 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 86.9.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 86.17.3). 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.

Comment #7373 by DatPham on

I think it would be better to indicate where we use the assumption that $f$ is representable by algebraic spaces. I guess this is used to ensure that $X\times_Y T$ is a quasi-compact algebraic space, hence admits an \' etale cover by an affine scheme $U$; the composite $U\to Y$ then factors through some $Y_{\mu}$, and the same is true for $X\times_Y T$ by the sheaf property.

Comment #7397 by on

Well, already in order to say what it means that $f$ is affine using Bootstrap, Definition 79.4.1 we need to know that $f$ is representable by algebraic spaces. The factorization of $X \times_Y T \to X$ through some $X_\lambda$ follows from Lemma 86.9.4. So I think the proof is fine.

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