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 $X$. $\square$

Comment #5347 by Hao on

It's better to mention what $X^{\nu}$ is in Lemma 07TD.

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