Lemma 87.19.7 (0AKN)—The Stacks project

Lemma 87.19.7. 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 67.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. 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 80.4.1. 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 87.9.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 87.17.3). 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. $\square$

Comments (4)

Comment #4054 by Matthew Emerton on March 11, 2019 at 16:51

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 Johan on April 06, 2019 at 19:18

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

Comment #7373 by DatPham on May 22, 2022 at 07:44

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 Johan on May 27, 2022 at 10:09

Well, already in order to say what it means that $f$ is affine using Bootstrap, Definition 80.4.1 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 87.9.4. So I think the proof is fine.

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