The Stacks project

A scheme admitting a surjective integral map from an affine scheme is affine.

Proposition 31.11.2. Let $f : X \to S$ be a morphism of schemes. Assume that $f$ is surjective and integral, and assume that $X$ is affine. Then $S$ is affine.

Proof. Since $f$ is surjective and $X$ is quasi-compact we see that $S$ is quasi-compact. Since $X$ is separated and $f$ is surjective and universally closed (Morphisms, Lemma 28.42.7), we see that $S$ is separated (Morphisms, Lemma 28.39.11).

By Lemma 31.7.2 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with $X_ i \to S$ finite. By Lemma 31.4.13 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to S$ factors through each $X_ i$ we see that $X_ i \to S$ is surjective. Hence we conclude that $S$ is affine by Lemma 31.11.1. $\square$


Comments (1)

Comment #3025 by Brian Lawrence on

Suggested slogan: A scheme admitting a surjective integral map from an affine scheme is affine.


Post a comment

Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05YU. Beware of the difference between the letter 'O' and the digit '0'.