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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