The Stacks project

Lemma 37.19.3 (Normalization and smooth morphisms). Let $X \to Y$ be a smooth morphism of schemes. Assume every quasi-compact open of $Y$ has finitely many irreducible components. Then the same is true for $X$ and there is a unique isomorphism $X^\nu = X \times _ Y Y^\nu $ over $X$ where $X^\nu $, $Y^\nu $ are the normalizations of $X$, $Y$.

Proof. By Descent, Lemma 35.16.3 every quasi-compact open of $X$ has finitely many irreducible components. Note that $X_{red} = X \times _ Y Y_{red}$ as a scheme smooth over a reduced scheme is reduced, see Descent, Lemma 35.18.1. Hence we may assume that $X$ and $Y$ are reduced (as the normalization of a scheme is equal to the normalization of its reduction by definition). Next, note that $X' = X \times _ Y Y^\nu $ is a normal scheme by Descent, Lemma 35.18.2. The morphism $X' \to Y^\nu $ is smooth (hence flat) thus the generic points of irreducible components of $X'$ lie over generic points of irreducible components of $Y^\nu $. Since $Y^\nu \to Y$ is birational we conclude that $X' \to X$ is birational too (because $X' \to Y^\nu $ induces an isomorphism on fibres over generic points of $Y$). We conclude that there exists a factorization $X^\nu \to X' \to X$, see Morphisms, Lemma 29.54.5 which is an isomorphism as $X'$ is normal and integral over $X$. $\square$


Comments (2)

Comment #5347 by Hao on

It's better to mention what is in Lemma 07TD.


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Your email address will not be published.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 07TD. Beware of the difference between the letter 'O' and the digit '0'.