Lemma 87.20.1. Let S be a scheme. Let X \to Y be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, and flat. Then X is countably indexed if and only if Y is countably indexed.
Proof. Assume X is countably indexed. We write X = \mathop{\mathrm{colim}}\nolimits X_ n as in Lemma 87.10.1. Write Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda as in Definition 87.9.1. For every n we can pick a \lambda _ n such that X_ n \to Y factors through Y_{\lambda _ n}, see Lemma 87.9.4. On the other hand, for every \lambda the scheme Y_\lambda \times _ Y X is affine (Lemma 87.19.7) and hence Y_\lambda \times _ Y X \to X factors through X_ n for some n (Lemma 87.9.4). Picture
If we can show the dotted arrow exists, then we conclude that Y = \mathop{\mathrm{colim}}\nolimits Y_{\lambda _ n} and Y is countably indexed. To do this we pick a \mu with \mu \geq \lambda and \mu \geq \lambda _ n. Thus both Y_\lambda \to Y and Y_{\lambda _ n} \to Y factor through Y_\mu \to Y. Say Y_\mu = \mathop{\mathrm{Spec}}(B_\mu ), the closed subscheme Y_\lambda corresponds to J \subset B_\mu , and the closed subscheme Y_{\lambda _ n} corresponds to J' \subset B_\mu . We are trying to show that J' \subset J. By the diagram above we know J'A_\mu \subset JA_\mu where Y_\mu \times _ Y X = \mathop{\mathrm{Spec}}(A_\mu ). Since X \to Y is surjective and flat the morphism Y_\lambda \times _ Y X \to Y_\lambda is a faithfully flat morphism of affine schemes, hence B_\mu \to A_\mu is faithfully flat. Thus J' \subset J as desired.
Assume Y is countably indexed. Then X is countably indexed by Lemma 87.19.10. \square
Comments (3)
Comment #1955 by Brian Conrad on
Comment #1958 by jojo on
Comment #2009 by Johan on