29.49 Rational maps
Let $X$ be a scheme. Note that if $U$, $V$ are dense open in $X$, then so is $U \cap V$.
Definition 29.49.1. Let $X$, $Y$ be schemes.
Let $f : U \to Y$, $g : V \to Y$ be morphisms of schemes defined on dense open subsets $U$, $V$ of $X$. We say that $f$ is equivalent to $g$ if $f|_ W = g|_ W$ for some $W \subset U \cap V$ dense open in $X$.
A rational map from $X$ to $Y$ is an equivalence class for the equivalence relation defined in (1).
If $X$, $Y$ are schemes over a base scheme $S$ we say that a rational map from $X$ to $Y$ is an $S$-rational map from $X$ to $Y$ if there exists a representative $f : U \to Y$ of the equivalence class which is an $S$-morphism.
We say that two morphisms $f$, $g$ as in (1) of the definition define the same rational map instead of saying that they are equivalent. In some cases rational maps are determined by maps on local rings at generic points.
Lemma 29.49.2. Let $S$ be a scheme. Let $X$ and $Y$ be schemes over $S$. Assume $X$ has finitely many irreducible components with generic points $x_1, \ldots , x_ n$. Let $s_ i \in S$ be the image of $x_ i$. Consider the map
\[ \left\{ \begin{matrix} S\text{-rational maps}
\\ \text{from }X\text{ to }Y
\end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} (y_1, \varphi _1, \ldots , y_ n, \varphi _ n)\text{ where } y_ i \in Y\text{ lies over }s_ i\text{ and}
\\ \varphi _ i : \mathcal{O}_{Y, y_ i} \to \mathcal{O}_{X, x_ i} \text{ is a local }\mathcal{O}_{S, s_ i}\text{-algebra map}
\end{matrix} \right\} \]
which sends $f : U \to Y$ to the $2n$-tuple with $y_ i = f(x_ i)$ and $\varphi _ i = f^\sharp _{x_ i}$. Then
If $Y \to S$ is locally of finite type, then the map is injective.
If $Y \to S$ is locally of finite presentation, then the map is bijective.
If $Y \to S$ is locally of finite type and $X$ reduced, then the map is bijective.
Proof.
Observe that any dense open of $X$ contains the points $x_ i$ so the construction makes sense. To prove (1) or (2) we may replace $X$ by any dense open. Thus if $Z_1, \ldots , Z_ n$ are the irreducible components of $X$, then we may replace $X$ by $X \setminus \bigcup _{i \not= j} Z_ i \cap Z_ j$. After doing this $X$ is the disjoint union of its irreducible components (viewed as open and closed subschemes). Then both the right hand side and the left hand side of the arrow are products over the irreducible components and we reduce to the case where $X$ is irreducible.
Assume $X$ is irreducible with generic point $x$ lying over $s \in S$. Part (1) follows from part (1) of Lemma 29.42.4. Parts (2) and (3) follow from part (2) of the same lemma.
$\square$
Definition 29.49.3. Let $X$ be a scheme. A rational function on $X$ is a rational map from $X$ to $\mathbf{A}^1_{\mathbf{Z}}$.
See Constructions, Definition 27.5.1 for the definition of the affine line $\mathbf{A}^1$. Let $X$ be a scheme over $S$. For any open $U \subset X$ a morphism $U \to \mathbf{A}^1_{\mathbf{Z}}$ is the same as a morphism $U \to \mathbf{A}^1_ S$ over $S$. Hence a rational function is also the same as a $S$-rational map from $X$ into $\mathbf{A}^1_ S$.
Recall that we have the canonical identification $\mathop{\mathrm{Mor}}\nolimits (T, \mathbf{A}^1_{\mathbf{Z}}) = \Gamma (T, \mathcal{O}_ T)$ for any scheme $T$, see Schemes, Example 26.15.2. Hence $\mathbf{A}^1_{\mathbf{Z}}$ is a ring-object in the category of schemes. More precisely, the morphisms
\begin{eqnarray*} + : \mathbf{A}^1_{\mathbf{Z}} \times \mathbf{A}^1_{\mathbf{Z}} & \longrightarrow & \mathbf{A}^1_{\mathbf{Z}} \\ (f, g) & \longmapsto & f + g \\ * : \mathbf{A}^1_{\mathbf{Z}} \times \mathbf{A}^1_{\mathbf{Z}} & \longrightarrow & \mathbf{A}^1_{\mathbf{Z}} \\ (f, g) & \longmapsto & fg \end{eqnarray*}
satisfy all the axioms of the addition and multiplication in a ring (commutative with $1$ as always). Hence also the set of rational maps into $\mathbf{A}^1_{\mathbf{Z}}$ has a natural ring structure.
Definition 29.49.4. Let $X$ be a scheme. The ring of rational functions on $X$ is the ring $R(X)$ whose elements are rational functions with addition and multiplication as just described.
For schemes with finitely many irreducible components we can compute this.
Lemma 29.49.5. Let $X$ be a scheme with finitely many irreducible components $X_1, \ldots , X_ n$. If $\eta _ i \in X_ i$ is the generic point, then
\[ R(X) = \mathcal{O}_{X, \eta _1} \times \ldots \times \mathcal{O}_{X, \eta _ n} \]
If $X$ is reduced this is equal to $\prod \kappa (\eta _ i)$. If $X$ is integral then $R(X) = \mathcal{O}_{X, \eta } = \kappa (\eta )$ is a field.
Proof.
Let $U \subset X$ be an open dense subset. Then $U_ i = (U \cap X_ i) \setminus (\bigcup _{j \not= i} X_ j)$ is nonempty open as it contained $\eta _ i$, contained in $X_ i$, and $\bigcup U_ i \subset U \subset X$ is dense. Thus the identification in the lemma comes from the string of equalities
\begin{align*} R(X) & = \mathop{\mathrm{colim}}\nolimits _{U \subset X\text{ open dense}} \mathop{\mathrm{Mor}}\nolimits (U, \mathbf{A}^1_\mathbf {Z}) \\ & = \mathop{\mathrm{colim}}\nolimits _{U \subset X\text{ open dense}} \mathcal{O}_ X(U) \\ & = \mathop{\mathrm{colim}}\nolimits _{\eta _ i \in U_ i \subset X\text{ open}} \prod \mathcal{O}_ X(U_ i) \\ & = \prod \mathop{\mathrm{colim}}\nolimits _{\eta _ i \in U_ i \subset X\text{ open}} \mathcal{O}_ X(U_ i) \\ & = \prod \mathcal{O}_{X, \eta _ i} \end{align*}
where the second equality is Schemes, Example 26.15.2. The final statement follows from Algebra, Lemma 10.25.1.
$\square$
Definition 29.49.6. Let $X$ be an integral scheme. The function field, or the field of rational functions of $X$ is the field $R(X)$.
We may occasionally indicate this field $k(X)$ instead of $R(X)$. We can use the notion of the function field to elucidate the separation condition on an integral scheme. Note that by Lemma 29.49.5 on an integral scheme every local ring $\mathcal{O}_{X, x}$ may be viewed as a local subring of $R(X)$.
Lemma 29.49.7. Let $X$ be an integral separated scheme. Let $Z_1$, $Z_2$ be distinct irreducible closed subsets of $X$. Let $\eta _ i$ be the generic point of $Z_ i$. If $Z_1 \not\subset Z_2$, then $\mathcal{O}_{X, \eta _1} \not\subset \mathcal{O}_{X, \eta _2}$ as subrings of $R(X)$. In particular, if $Z_1 = \{ x\} $ consists of one closed point $x$, there exists a function regular in a neighborhood of $x$ which is not in $\mathcal{O}_{X, \eta _{2}}$.
Proof.
First observe that under the assumption of $X$ being separated, there is a unique map of schemes $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2}) \to X$ over $X$ such that the composition
\[ \mathop{\mathrm{Spec}}(R(X)) \longrightarrow \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2}) \longrightarrow X \]
is the canonical map $\mathop{\mathrm{Spec}}(R(X)) \to X$. Namely, there is the canonical map $can : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2}) \to X$, see Schemes, Equation (26.13.1.1). Given a second morphism $a$ to $X$, we have that $a$ agrees with $can$ on the generic point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$ by assumption. Now $X$ being separated guarantees that the subset in $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$ where these two maps agree is closed, see Schemes, Lemma 26.21.5. Hence $a = can$ on all of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$.
Assume $Z_1 \not\subset Z_2$ and assume on the contrary that $\mathcal{O}_{X, \eta _{1}} \subset \mathcal{O}_{X, \eta _{2}}$ as subrings of $R(X)$. Then we would obtain a second morphism
\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _{2}}) \longrightarrow \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _{1}}) \longrightarrow X. \]
By the above this composition would have to be equal to $can$. This implies that $\eta _2$ specializes to $\eta _1$ (see Schemes, Lemma 26.13.2). But this contradicts our assumption $Z_1 \not\subset Z_2$.
$\square$
Definition 29.49.8. Let $\varphi $ be a rational map between two schemes $X$ and $Y$. We say $\varphi $ is defined in a point $x \in X$ if there exists a representative $(U, f)$ of $\varphi $ with $x \in U$. The domain of definition of $\varphi $ is the set of all points where $\varphi $ is defined.
With this definition it isn't true in general that $\varphi $ has a representative which is defined on all of the domain of definition.
Lemma 29.49.9. Let $X$ and $Y$ be schemes. Assume $X$ reduced and $Y$ separated. Let $\varphi $ be a rational map from $X$ to $Y$ with domain of definition $U \subset X$. Then there exists a unique morphism $f : U \to Y$ representing $\varphi $. If $X$ and $Y$ are schemes over a separated scheme $S$ and if $\varphi $ is an $S$-rational map, then $f$ is a morphism over $S$.
Proof.
Let $(V, g)$ and $(V', g')$ be representatives of $\varphi $. Then $g, g'$ agree on a dense open subscheme $W \subset V \cap V'$. On the other hand, the equalizer $E$ of $g|_{V \cap V'}$ and $g'|_{V \cap V'}$ is a closed subscheme of $V \cap V'$ (Schemes, Lemma 26.21.5). Now $W \subset E$ implies that $E = V \cap V'$ set theoretically. As $V \cap V'$ is reduced we conclude $E = V \cap V'$ scheme theoretically, i.e., $g|_{V \cap V'} = g'|_{V \cap V'}$. It follows that we can glue the representatives $g : V \to Y$ of $\varphi $ to a morphism $f : U \to Y$, see Schemes, Lemma 26.14.1. We omit the proof of the final statement.
$\square$
In general it does not make sense to compose rational maps. The reason is that the image of a representative of the first rational map may have empty intersection with the domain of definition of the second. However, if we assume that our schemes are irreducible and we look at dominant rational maps, then we can compose rational maps.
Definition 29.49.10. Let $X$ and $Y$ be irreducible schemes. A rational map from $X$ to $Y$ is called dominant if any representative $f : U \to Y$ is a dominant morphism of schemes.
By Lemma 29.8.6 it is equivalent to require that the generic point $\eta \in X$ maps to the generic point $\xi $ of $Y$, i.e., $f(\eta ) = \xi $ for any representative $f : U \to Y$. We can compose a dominant rational map $\varphi $ between irreducible schemes $X$ and $Y$ with an arbitrary rational map $\psi $ from $Y$ to $Z$. Namely, choose representatives $f : U \to Y$ with $U \subset X$ open dense and $g : V \to Z$ with $V \subset Y$ open dense. Then $W = f^{-1}(V) \subset X$ is open nonempty (because it contains the generic point of $X$) and we let $\psi \circ \varphi $ be the equivalence class of $g \circ f|_ W : W \to Z$. We omit the verification that this is well defined.
In this way we obtain a category whose objects are irreducible schemes and whose morphisms are dominant rational maps. Given a base scheme $S$ we can similarly define a category whose objects are irreducible schemes over $S$ and whose morphisms are dominant $S$-rational maps.
Definition 29.49.11. Let $X$ and $Y$ be irreducible schemes.
We say $X$ and $Y$ are birational if $X$ and $Y$ are isomorphic in the category of irreducible schemes and dominant rational maps.
Assume $X$ and $Y$ are schemes over a base scheme $S$. We say $X$ and $Y$ are $S$-birational if $X$ and $Y$ are isomorphic in the category of irreducible schemes over $S$ and dominant $S$-rational maps.
If $X$ and $Y$ are birational irreducible schemes, then the set of rational maps from $X$ to $Z$ is bijective with the set of rational map from $Y$ to $Z$ for all schemes $Z$ (functorially in $Z$). For “general” irreducible schemes this is just one possible definition. Another would be to require $X$ and $Y$ have isomorphic rings of rational functions. For varieties these conditions are equivalent, see Lemma 29.50.6.
Lemma 29.49.12. Let $X$ and $Y$ be irreducible schemes.
The schemes $X$ and $Y$ are birational if and only if they have isomorphic nonempty opens.
Assume $X$ and $Y$ are schemes over a base scheme $S$. Then $X$ and $Y$ are $S$-birational if and only if there are nonempty opens $U \subset X$ and $V \subset Y$ which are $S$-isomorphic.
Proof.
Assume $X$ and $Y$ are birational. Let $f : U \to Y$ and $g : V \to X$ define inverse dominant rational maps from $X$ to $Y$ and from $Y$ to $X$. We may assume $V$ affine. We may replace $U$ by an affine open of $f^{-1}(V)$. As $g \circ f$ is the identity as a dominant rational map, we see that the composition $U \to V \to X$ is the identity on a dense open of $U$. Thus after replacing $U$ by a smaller affine open we may assume that $U \to V \to X$ is the inclusion of $U$ into $X$. It follows that $U \to V$ is an immersion (apply Schemes, Lemma 26.21.11 to $U \to g^{-1}(U) \to U$). However, switching the roles of $U$ and $V$ and redoing the argument above, we see that there exists a nonempty affine open $V' \subset V$ such that the inclusion factors as $V' \to U \to V$. Then $V' \to U$ is necessarily an open immersion. Namely, $V' \to f^{-1}(V') \to V'$ are monomorphisms (Schemes, Lemma 26.23.8) composing to the identity, hence isomorphisms. Thus $V'$ is isomorphic to an open of both $X$ and $Y$. In the $S$-rational maps case, the exact same argument works.
$\square$
If $U \subset X$ is a dense open of a scheme, then $U^0 \subset X^0$ need not be an equality, but if $X$ is an object of our category, then this is the case. Thus given two morphisms in our category, the composition is well defined and a morphism in our category.
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