Lemma 54.13.1. Let $Y$ be a Noetherian integral scheme. 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. 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 Schemes, Lemma 30.19.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties, Lemma 28.9.4) we see that $C$ is normal domain (Properties, Lemma 28.7.9). 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 Morphisms, Definition 29.51.12. After shrinking $V$ we may assume that $f^{-1}V \to V$ is flat (Morphisms, Proposition 29.27.1). Thus $f^{-1}V \to V$ is faithfully flat. Then $V$ is regular by Algebra, Lemma 10.164.4. $\square$
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