You may be used to the notion of a birational map of varieties having the property that it is an isomorphism over an open subset of the target. However, in general a birational morphism may not be an isomorphism over any nonempty open, see Example 29.50.4. Here is the formal definition.
Definition 29.50.1. Let $X$, $Y$ be schemes. Assume $X$ and $Y$ have finitely many irreducible components. We say a morphism $f : X \to Y$ is birational if
$f$ 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
for every generic point $\eta \in X$ of an irreducible component of $X$ the local ring map $\mathcal{O}_{Y, f(\eta )} \to \mathcal{O}_{X, \eta }$ is an isomorphism.
We will see below that the fibres of a birational morphism over generic points are singletons. Moreover, we will see that in most cases one encounters in practice the existence a birational morphism between irreducible schemes $X$ and $Y$ implies $X$ and $Y$ are birational schemes.
Lemma 29.50.2. Let $f : X \to Y$ be a morphism of schemes having finitely many irreducible components. If $f$ is birational then $f$ is dominant.
Proof. Follows from Lemma 29.8.2 and the definition. $\square$
Lemma 29.50.3. Let $f : X \to Y$ be a birational morphism of schemes having 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. We may assume $Y = \mathop{\mathrm{Spec}}(B)$ is affine and irreducible. Then $X$ is irreducible too. If we prove the result for any nonempty affine open $U \subset X$, then the result holds for $X$ (small argument omitted). Hence we may assume $X$ is affine too, say $X = \mathop{\mathrm{Spec}}(A)$. Let $y \in Y$ correspond to the minimal prime $\mathfrak q \subset B$. By assumption $A$ has a unique minimal prime $\mathfrak p$ lying over $\mathfrak q$ and $B_\mathfrak q \to A_\mathfrak p$ is an isomorphism. It follows that $A_\mathfrak q \to \kappa (\mathfrak p)$ is surjective, hence $\mathfrak p A_\mathfrak q$ is a maximal ideal. On the other hand $\mathfrak p A_\mathfrak q$ is the unique minimal prime of $A_\mathfrak q$. We conclude that $\mathfrak p A_\mathfrak q$ is the unique prime of $A_\mathfrak q$ and that $A_\mathfrak q = A_\mathfrak p$. Since $A_\mathfrak q = A \otimes _ B B_\mathfrak q$ the lemma follows. $\square$
Example 29.50.4. Here are two examples of birational morphisms which are not isomorphisms over any open of the target. First example. Let $k$ be an infinite field. Let $A = k[x]$. Let $B = k[x, \{ y_{\alpha }\} _{\alpha \in k}]/ ((x-\alpha )y_\alpha , y_\alpha y_\beta )$. There is an inclusion $A \subset B$ and a retraction $B \to A$ setting all $y_\alpha $ equal to zero. Both the morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$ and the morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ are birational but not an isomorphism over any open. Second example. Let $A$ be a domain. Let $S \subset A$ be a multiplicative subset not containing $0$. With $B = S^{-1}A$ the morphism $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is birational. If there exists an open $U$ of $\mathop{\mathrm{Spec}}(A)$ such that $f^{-1}(U) \to U$ is an isomorphism, then there exists an $a \in A$ such that each every element of $S$ becomes invertible in the principal localization $A_ a$. Taking $A = \mathbf{Z}$ and $S$ the set of odd integers give a counter example.
Lemma 29.50.5. Let $f : X \to Y$ be a birational morphism of schemes having finitely many irreducible components over a base scheme $S$. Assume one of the following conditions is satisfied
$f$ is locally of finite type and $Y$ reduced,
$f$ 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. In particular if $X$ and $Y$ are irreducible, then $X$ and $Y$ are $S$-birational.
Proof. There is an immediate reduction to the case where $X$ and $Y$ are irreducible which we omit. Moreover, after shrinking further and we may assume $X$ and $Y$ are affine, say $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. By assumption $A$, resp. $B$ has a unique minimal prime $\mathfrak p$, resp. $\mathfrak q$, the prime $\mathfrak p$ lies over $\mathfrak q$, and $B_\mathfrak q = A_\mathfrak p$. By Lemma 29.50.3 we have $B_\mathfrak q = A_\mathfrak q = A_\mathfrak p$.
Suppose $B \to A$ is of finite type, say $A = B[x_1, \ldots , x_ n]$. There exist a $b_ i \in B$ and $g_ i \in B \setminus \mathfrak q$ such that $b_ i/g_ i$ maps to the image of $x_ i$ in $A_\mathfrak q$. Hence $b_ i - g_ ix_ i$ maps to zero in $A_{g_ i'}$ for some $g_ i' \in B \setminus \mathfrak q$. Setting $g = \prod g_ i g'_ i$ we see that $B_ g \to A_ g$ is surjective. If moreover $Y$ is reduced, then the map $B_ g \to B_\mathfrak q$ is injective and hence $B_ g \to A_ g$ is injective as well. This proves case (1).
Proof of (2). By the argument given in the previous paragraph we may assume that $B \to A$ is surjective. As $f$ is locally of finite presentation the kernel $J \subset B$ is a finitely generated ideal. Say $J = (b_1, \ldots , b_ r)$. Since $B_\mathfrak q = A_\mathfrak q$ there exist $g_ i \in B \setminus \mathfrak q$ such that $g_ i b_ i = 0$. Setting $g = \prod g_ i$ we see that $B_ g \to A_ g$ is an isomorphism. $\square$
Lemma 29.50.6. Let $S$ be a scheme. Let $X$ and $Y$ be irreducible schemes locally of finite presentation over $S$. Let $x \in X$ and $y \in Y$ be the generic points. The following are equivalent
$X$ and $Y$ are $S$-birational,
there exist nonempty opens of $X$ and $Y$ which are $S$-isomorphic, and
$x$ and $y$ map to the same point $s$ of $S$ and $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ are isomorphic as $\mathcal{O}_{S, s}$-algebras.
Proof. We have seen the equivalence of (1) and (2) in Lemma 29.49.12. It is immediate that (2) implies (3). To finish we assume (3) holds and we prove (1). By Lemma 29.49.2 there is a rational map $f : U \to Y$ which sends $x \in U$ to $y$ and induces the given isomorphism $\mathcal{O}_{Y, y} \cong \mathcal{O}_{X, x}$. Thus $f$ is a birational morphism and hence induces an isomorphism on nonempty opens by Lemma 29.50.5. This finishes the proof. $\square$
Lemma 29.50.7. Let $S$ be a scheme. Let $X$ and $Y$ be integral schemes locally of finite type over $S$. Let $x \in X$ and $y \in Y$ be the generic points. The following are equivalent
$X$ and $Y$ are $S$-birational,
there exist nonempty opens of $X$ and $Y$ which are $S$-isomorphic, and
$x$ and $y$ map to the same point $s \in S$ and $\kappa (x) \cong \kappa (y)$ as $\kappa (s)$-extensions.
Proof. We have seen the equivalence of (1) and (2) in Lemma 29.49.12. It is immediate that (2) implies (3). To finish we assume (3) holds and we prove (1). Observe that $\mathcal{O}_{X, x} = \kappa (x)$ and $\mathcal{O}_{Y, y} = \kappa (y)$ by Algebra, Lemma 10.25.1. By Lemma 29.49.2 there is a rational map $f : U \to Y$ which sends $x \in U$ to $y$ and induces the given isomorphism $\mathcal{O}_{Y, y} \cong \mathcal{O}_{X, x}$. Thus $f$ is a birational morphism and hence induces an isomorphism on nonempty opens by Lemma 29.50.5. This finishes the proof. $\square$
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Comment #9757 by Fanjun Meng on
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