Processing math: 100%

The Stacks project

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

\xymatrix{ Y_\lambda \times _ Y X \ar[r] \ar[d] & X_ n \ar[r] \ar[d] & X \ar[d] \\ Y_\lambda \ar@{..>}[r] \ar@/_1pc/[rr] & Y_{\lambda _ n} \ar[r] & Y }

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

There are a lot of arrows in the diagram pointing the wrong way: both left horizontal arrows and the lower curved arrow, while the upper curves arrow plays no role and so can be removed. Maybe reorganize the diagram?

Likewise, the contained being proved is that , not that , and the diagram gives that (rather than the other way around).

Comment #1958 by jojo on

In the statement doesn't seem to be used explicitly.

Comment #2009 by on

@#1955 Thanks very much. Indeed it was very confusing. I've simplified the diagram and now it is clearer. See changes.

@#1958 You are in good company with your complaint. But please read the post on the stacks project blog with title "Base scheme for spaces" for the explanation.


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.