Lemma 89.7.1. 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 étale 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 69.20.2 we see that C is a finite A-module. As X is normal (Properties of Spaces, Lemma 66.25.4) we see that C is normal domain (Spaces over Fields, Lemma 72.4.6). 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 72.8.3. After shrinking V we may assume that f^{-1}V \to V is flat (Morphisms of Spaces, Proposition 67.32.1). Thus f^{-1}V \to V is faithfully flat. Then V is regular by Algebra, Lemma 10.164.4. \square
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