The Stacks project

Lemma 66.22.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

  1. $X$ and $Y$ are decent and have finitely many irreducible components,

  2. $f$ is integral and birational,

  3. $Y$ is normal, and

  4. $X$ is reduced.

Then $f$ is an isomorphism.

Proof. Let $V \to Y$ be an ├ętale morphism with $V$ affine. It suffices to show that $U = X \times _ Y V \to V$ is an isomorphism. By Lemma 66.22.6 and its proof we see that $U$ and $V$ are decent and have finitely many irreducible components and that $U \to V$ is birational. By Properties, Lemma 28.7.5 $V$ is a finite disjoint union of integral schemes. Thus we may assume $V$ is integral. As $f$ is birational, we see that $U$ is irreducible and reduced, i.e., integral (note that $U$ is a scheme as $f$ is integral, hence representable). Thus we may assume that $X$ and $Y$ are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma 29.53.8. $\square$


Comments (0)


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Your email address will not be published.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 0B4E. Beware of the difference between the letter 'O' and the digit '0'.