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
Comment #5588 by Johan on