The Stacks project

67.47 Rational maps

This section is the analogue of Morphisms, Section 29.49. We will use without further mention that the intersection of dense opens of a topological space is a dense open.

Definition 67.47.1. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$.

  1. Let $f : U \to Y$, $g : V \to Y$ be morphisms of algebraic spaces over $S$ defined on dense open subspaces $U$, $V$ of $X$. We say that $f$ is equivalent to $g$ if $f|_ W = g|_ W$ for some dense open subspace $W \subset U \cap V$.

  2. A rational map from $X$ to $Y$ is an equivalence class for the equivalence relation defined in (1).

  3. Given morphisms $X \to B$ and $Y \to B$ of algebraic spaces over $S$ we say that a rational map from $X$ to $Y$ is a $B$-rational map from $X$ to $Y$ if there exists a representative $f : U \to Y$ of the equivalence class which is a morphism over $B$.

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 many cases we will consider in the future, the algebraic spaces $X$ and $Y$ will contain a dense open subspaces $X'$ and $Y'$ which are schemes. In that case a rational map from $X$ to $Y$ is the same as an $S$-rational map from $X'$ to $Y'$ in the sense of Morphisms, Definition 67.47.1. Then all of the theory developed for schemes can be brought to bear.

Definition 67.47.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. A rational function on $X$ is a rational map from $X$ to $\mathbf{A}^1_ S$.

Looking at the discussion following Morphisms, Definition 29.49.3 we find that this is the same as the notion defined there in case $X$ happens to be a scheme.

Recall that we have the canonical identification

\[ \mathop{\mathrm{Mor}}\nolimits _ S(T, \mathbf{A}^1_ S) = \mathop{\mathrm{Mor}}\nolimits (T, \mathbf{A}^1_\mathbf {Z}) = \Gamma (T, \mathcal{O}_ T) \]

for any scheme $T$ over $S$, see Schemes, Example 26.15.2. Hence $\mathbf{A}^1_ S$ is a ring-object in the category of schemes over $S$. In other words, addition and multiplication define morphisms

\[ + : \mathbf{A}^1_ S \times _ S \mathbf{A}^1_ S \to \mathbf{A}^1_ S \quad \text{and}\quad * : \mathbf{A}^1_ S \times _ S \mathbf{A}^1_ S \to \mathbf{A}^1_ S \]

satisfying the axioms of the addition and multiplication in a ring (commutative with $1$ as always). Hence also the set of rational maps into $\mathbf{A}^1_ S$ has a natural ring structure.

Definition 67.47.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The ring of rational functions on $X$ is the ring $R(X)$ whose elements are rational functions with addition and multiplication as just described.

We will define function fields for integral algebraic spaces later, see Spaces over Fields, Section 72.4.

Definition 67.47.4. Let $S$ be a scheme. Let $\varphi $ be a rational map between two algebraic spaces $X$ and $Y$ over $S$. 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.

The domain of definition is viewed as an open subspace of $X$ via Properties of Spaces, Lemma 66.4.8. 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 67.47.5. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic spaces over $S$. Assume $X$ is reduced and $Y$ is separated over $S$. Let $\varphi $ be a rational map from $X$ to $Y$ with domain of definition $U \subset X$. Then there exists a unique morphism $f : U \to Y$ of algebraic spaces representing $\varphi $.

Proof. Let $(V, g)$ and $(V', g')$ be representatives of $\varphi $. Then $g, g'$ agree on a dense open subspace $W \subset V \cap V'$. On the other hand, the equalizer $E$ of $g|_{V \cap V'}$ and $g'|_{V \cap V'}$ is a closed subspace of $V \cap V'$ because it is the base change of $\Delta : Y \to Y \times _ S Y$ by the morphism $V \cap V' \to Y \times _ S Y$ given by $g|_{V \cap V'}$ and $g'|_{V \cap V'}$. Now $W \subset E$ implies that $|E| = |V \cap V'|$. As $V \cap V'$ is reduced we conclude $E = V \cap V'$ scheme theoretically, i.e., $g|_{V \cap V'} = g'|_{V \cap V'}$, see Properties of Spaces, Lemma 66.12.4. It follows that we can glue the representatives $g : V \to Y$ of $\varphi $ to a morphism $f : U \to Y$ because $\coprod V \to U$ is a surjection of fppf sheaves and $\coprod _{V, V'} V \cap V' = (\coprod V) \times _ U (\coprod V)$. $\square$

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 spaces are irreducible and we look at dominant rational maps, then we can compose rational maps.

Definition 67.47.6. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic spaces over $S$. Assume $|X|$ and $|Y|$ are irreducible. A rational map from $X$ to $Y$ is called dominant if any representative $f : U \to Y$ is a dominant morphism in the sense of Definition 67.18.1.

We can compose a dominant rational map $\varphi $ between irreducible algebraic spaces $X$ 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 the image of $|f|$ is dense and hence must meet the nonempty open $V$) and hence dense as $|X|$ is irreducible. We define $\psi \circ \varphi $ as 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 algebraic spaces over $S$ and whose morphisms are dominant rational maps.

Definition 67.47.7. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic spaces over $S$ with $|X|$ and $|Y|$ irreducible. We say $X$ and $Y$ are birational if $X$ and $Y$ are isomorphic in the category of irreducible algebraic spaces over $S$ and dominant rational maps.

If $X$ and $Y$ are birational irreducible algebraic spaces, then the set of rational maps from $X$ to $Z$ is bijective with the set of rational map from $Y$ to $Z$ for all algebraic spaces $Z$ (functorially in $Z$). For “general” irreducible algebraic spaces this is just one possible definition. Another would be to require $X$ and $Y$ have isomorphic rings of rational functions; sometimes these two notions are equivalent (insert future reference here).

Lemma 67.47.8. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic space over $S$ with $|X|$ and $|Y|$ irreducible. Then $X$ and $Y$ are birational if and only if there are nonempty open subspaces $U \subset X$ and $V \subset Y$ which are isomorphic as algebraic spaces over $S$.

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 $X$. After shrinking $U$ we may assume $f : U \to Y$ factors through $V$. 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 open we may assume that $U \to V \to X$ is the inclusion of $U$ into $X$. By symmetry we find there exists an open subspace $V' \subset V$ such that $g|_{V'} : V' \to X$ factors through $U \subset X$ and such that $V' \to U \to Y$ is the identity. The inverse image of $|V'|$ by $|U| \to |V|$ is an open of $|U|$ and hence equal to $|U'|$ for some open subspace $U' \subset U$, see Properties of Spaces, Lemma 66.4.8. Then $U' \subset U \to V$ factors as $U' \to V'$. Similarly $V' \to U$ factors as $V' \to U'$. The reader finds that $U' \to V'$ and $V' \to U'$ are mutually inverse morphisms of algebraic spaces over $S$ and the proof is complete. $\square$

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