The Stacks project

66.22 Birational morphisms

The following definition of a birational morphism of algebraic spaces seems to be the closest to our definition (Morphisms, Definition 29.49.1) of a birational morphism of schemes.

Definition 66.22.1. Let $S$ be a scheme. Let $X$ and $Y$ algebraic spaces over $S$. Assume $X$ and $Y$ are decent and that $|X|$ and $|Y|$ have finitely many irreducible components. We say a morphism $f : X \to Y$ is birational if

  1. $|f|$ induces a bijection between the set of generic points of irreducible components of $|X|$ and the set of generic points of the irreducible components of $|Y|$, and

  2. for every generic point $x \in |X|$ of an irreducible component the local ring map $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ is an isomorphism (see clarification below).

Clarification: Since $X$ and $Y$ are decent the topological spaces $|X|$ and $|Y|$ are sober (Proposition 66.12.4). Hence condition (1) makes sense. Moreover, because we have assumed that $|X|$ and $|Y|$ have finitely many irreducible components, we see that the generic points $x_1, \ldots , x_ n \in |X|$, resp. $y_1, \ldots , y_ n \in |Y|$ are contained in any dense open of $|X|$, resp. $|Y|$. In particular, they are contained in the schematic locus of $X$, resp. $Y$ by Theorem 66.10.2. Thus we can define $\mathcal{O}_{X, x_ i}$, resp. $\mathcal{O}_{Y, y_ i}$ to be the local ring of this scheme at $x_ i$, resp. $y_ i$.

We conclude that if the morphism $f : X \to Y$ is birational, then there exist dense open subspaces $X' \subset X$ and $Y' \subset Y$ such that

  1. $f(X') \subset Y'$,

  2. $X'$ and $Y'$ are representable, and

  3. $f|_{X'} : X' \to Y'$ is birational in the sense of Morphisms, Definition 29.49.1.

However, we do insist that $X$ and $Y$ are decent with finitely many irreducible components. Other ways to characterize decent algebraic spaces with finitely many irreducible components are given in Lemma 66.20.4. In most cases birational morphisms are isomorphisms over dense opens.

Lemma 66.22.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. If $f$ is birational then $f$ is dominant.

Proof. Follows immediately from the definitions. See Morphisms of Spaces, Definition 65.18.1. $\square$

Lemma 66.22.3. Let $S$ be a scheme. Let $f : X \to Y$ be a birational morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. If $y \in Y$ is the generic point of an irreducible component, then the base change $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is an isomorphism.

Proof. Let $X' \subset X$ and $Y' \subset Y$ be the maximal open subspaces which are representable, see Lemma 66.20.4. By Lemma 66.21.3 the fibre of $f$ over $y$ is consists of points of codimension $0$ of $X$ and is therefore contained in $X'$. Hence $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) = X' \times _{Y'} \mathop{\mathrm{Spec}}(\mathcal{O}_{Y', y})$ and the result follows from Morphisms, Lemma 29.49.3. $\square$

Lemma 66.22.4. Let $S$ be a scheme. Let $f : X \to Y$ be a birational morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. Assume one of the following conditions is satisfied

  1. $f$ is locally of finite type and $Y$ reduced (i.e., integral),

  2. $f$ is locally of finite presentation.

Then there exist dense opens $U \subset X$ and $V \subset Y$ such that $f(U) \subset V$ and $f|_ U : U \to V$ is an isomorphism.

Proof. By Lemma 66.20.4 we may assume that $X$ and $Y$ are schemes. In this case the result is Morphisms, Lemma 29.49.5. $\square$

Lemma 66.22.5. Let $S$ be a scheme. Let $f : X \to Y$ be a birational morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. Assume

  1. either $f$ is quasi-compact or $f$ is separated, and

  2. either $f$ is locally of finite type and $Y$ is reduced or $f$ is locally of finite presentation.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism.

Proof. By Lemma 66.20.4 we may assume $Y$ is a scheme. By Lemma 66.21.4 we may assume that $f$ is finite. Then $X$ is a scheme too and the result follows from Morphisms, Lemma 29.50.6. $\square$

Lemma 66.22.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. If $f$ is birational and $V \to Y$ is an ├ętale morphism with $V$ affine, then $X \times _ Y V$ is decent with finitely many irreducible components and $X \times _ Y V \to V$ is birational.

Proof. The algebraic space $U = X \times _ Y V$ is decent (Lemma 66.6.6). The generic points of $V$ and $U$ are the elements of $|V|$ and $|U|$ which lie over generic points of $|Y|$ and $|X|$ (Lemma 66.20.1). Since $Y$ is decent we conclude there are finitely many generic points on $V$. Let $\xi \in |X|$ be a generic point of an irreducible component. By the discussion following Definition 66.22.1 we have a cartesian square

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }) \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, f(\xi )}) \ar[r] & Y } \]

whose horizontal morphisms are monomorphisms identifying local rings and where the left vertical arrow is an isomorphism. It follows that in the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }) \times _ X U \ar[d] \ar[r] & U \ar[d] \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, f(\xi )}) \times _ Y V \ar[r] & V } \]

the vertical arrow on the left is an isomorphism. The horizonal arrows have image contained in the schematic locus of $U$ and $V$ and identify local rings (some details omitted). Since the image of the horizontal arrows are the points of $|U|$, resp. $|V|$ lying over $\xi $, resp. $f(\xi )$ we conclude. $\square$

Lemma 66.22.7. 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 66.20.4 we see that $X$ and $Y$ satisfy the equivalent conditions of Morphisms of Spaces, Lemma 65.49.1 and the normalizations are defined. By Morphisms of Spaces, Lemma 65.49.5 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 66.20.1) we obtain from Morphisms of Spaces, Lemma 65.49.5 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 66.6.5. 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. $\square$

Lemma 66.22.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

  1. $X$ and $Y$ are decent and have finitely many irreducible components,

  2. $f$ is integral and birational,

  3. $Y$ is normal, and

  4. $X$ is reduced.

Then $f$ is an isomorphism.

Proof. Let $V \to Y$ be an ├ętale morphism with $V$ affine. It suffices to show that $U = X \times _ Y V \to V$ is an isomorphism. By Lemma 66.22.6 and its proof we see that $U$ and $V$ are decent and have finitely many irreducible components and that $U \to V$ is birational. By Properties, Lemma 28.7.5 $V$ is a finite disjoint union of integral schemes. Thus we may assume $V$ is integral. As $f$ is birational, we see that $U$ is irreducible and reduced, i.e., integral (note that $U$ is a scheme as $f$ is integral, hence representable). Thus we may assume that $X$ and $Y$ are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma 29.53.8. $\square$

Lemma 66.22.9. Let $S$ be a scheme. Let $f : X \to Y$ be an integral birational morphism of decent algebraic spaces over $S$ which have finitely many irreducible components. Then there exists a factorization $Y^\nu \to X \to Y$ and $Y^\nu \to X$ is the normalization of $X$.

Proof. Consider the map $X^\nu \to Y^\nu $ of Lemma 66.22.7. This map is integral by Morphisms of Spaces, Lemma 65.45.12. Hence it is an isomorphism by Lemma 66.22.8. $\square$


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