The Stacks project

Lemma 68.22.7. Let $S$ be a scheme. Let $f : X \to Y$ be a birational morphism between algebraic spaces over $S$ which are decent and have finitely many irreducible components. Then the normalizations $X^\nu \to X$ and $Y^\nu \to Y$ exist and there is a commutative diagram

\[ \xymatrix{ X^\nu \ar[r] \ar[d] & Y^\nu \ar[d] \\ X \ar[r] & Y } \]

of algebraic spaces over $S$. The morphism $X^\nu \to Y^\nu $ is birational.

Proof. By Lemma 68.20.4 we see that $X$ and $Y$ satisfy the equivalent conditions of Morphisms of Spaces, Lemma 67.49.1 and the normalizations are defined. By Morphisms of Spaces, Lemma 67.49.8 the algebraic space $X^\nu $ is normal and maps codimension $0$ points to codimension $0$ points. Since $f$ maps codimension $0$ points to codimension $0$ points (this is the same as generic points on decent spaces by Lemma 68.20.1) we obtain from Morphisms of Spaces, Lemma 67.49.8 a factorization of the composition $X^\nu \to X \to Y$ through $Y^\nu $.

Observe that $X^\nu $ and $Y^\nu $ are decent for example by Lemma 68.6.5. Moreover the maps $X^\nu \to X$ and $Y^\nu \to Y$ induce bijections on irreducible components (see references above) hence $X^\nu $ and $Y^\nu $ both have a finite number of irreducible components and the map $X^\nu \to Y^\nu $ induces a bijection between their generic points. To prove that $X^\nu \to Y^\nu $ is birational, it therefore suffices to show it induces an isomorphism on local rings at these points. To do this we may replace $X$ and $Y$ by open neighbourhoods of their generic points, hence we may assume $X$ and $Y$ are affine irreducible schemes with generic points $x$ and $y$. Since $f$ is birational the map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is an isomorphism. Let $x^\nu \in X^\nu $ and $y^\nu \in Y^\nu $ be the points lying over $x$ and $y$. By construction of the normalization we see that $\mathcal{O}_{X^\nu , x^\nu } = \mathcal{O}_{X, x}/\mathfrak m_ x$ and similarly on $Y$. Thus the map $\mathcal{O}_{X^\nu , x^\nu } \to \mathcal{O}_{Y^\nu , y^\nu }$ is an isomorphism as well. $\square$

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