The Stacks project

Lemma 87.7.1. Let $S$ be a scheme. Let $Y$ be a Noetherian integral algebraic space over $S$. Assume there exists an alteration $f : X \to Y$ with $X$ regular. Then the normalization $Y^\nu \to Y$ is finite and $Y$ has a dense open which is regular.

Proof. By ├ętale localization, it suffices to prove this when $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian domain. Let $B$ be the integral closure of $A$ in its fraction field. Set $C = \Gamma (X, \mathcal{O}_ X)$. By Cohomology of Spaces, Lemma 67.20.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties of Spaces, Lemma 64.25.4) we see that $C$ is normal domain (Spaces over Fields, Lemma 70.4.6). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian.

There exists a nonempty open $V \subset Y$ such that $f^{-1}V \to V$ is finite, see Spaces over Fields, Definition 70.8.3. After shrinking $V$ we may assume that $f^{-1}V \to V$ is flat (Morphisms of Spaces, Proposition 65.32.1). Thus $f^{-1}V \to V$ is faithfully flat. Then $V$ is regular by Algebra, Lemma 10.164.4. $\square$


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Your email address will not be published.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