§7. Rational Maps

7.1. Rational maps and rational functions

(7.1.1) Let $X$, $Y$ be two preschemes, $U$, $V$ two dense opens in $X$, and $f$ (resp. $g$) a morphism of $U$ (resp. $V$) into $Y$; we say that $f$ and $g$ are equivalent if they coincide on a dense open in $U\cap V$. Since a finite intersection of dense opens in $X$ is a dense open in $X$, it is clear that this relation is an equivalence relation.

Definition (7.1.2). Given two preschemes $X$, $Y$, one calls a rational map of $X$ into $Y$ an equivalence class of morphisms of dense open subsets of $X$ into $Y$, for the relation defined in (7.1.1). If $X$ and $Y$ are $S$-preschemes, one says that such a class is an $S$-rational map if there exists a representative of this class which is an $S$-morphism. One calls an $S$-rational section of $X$ any $S$-rational map of $S$ into $X$. One calls a rational function on a prescheme $X$ any $X$-rational section of the $X$-prescheme $X{\otimes }_{\mathbb{Z}}\mathbb{Z}[T]$ (where $T$ is an indeterminate).

By abuse of language, when only $S$-preschemes are in question, one says “rational map” instead of “$S$-rational map” if no confusion can result.

Let $f$ be a rational map of $X$ into $Y$, and $U$ an open of $X$; if $f_1,f_2$ are morphisms belonging to the class $f$, defined respectively in dense opens $V$, $W$ of $X$, the restrictions $f_1|(U\cap V)$, $f_2|(U\cap W)$ coincide in $U\cap V\cap W$, which is dense in $U$; the class of morphisms $f$ thus defines a rational map of $U$ into $Y$, called the restriction of $f$ to $U$ and written $f|U$.

If, to each $S$-morphism $f:X\to Y$, one makes correspond the $S$-rational map to which $f$ belongs, one defines a canonical map of ${\mathrm{Hom}}_S(X,Y)$ into the set of $S$-rational maps of $X$ into $Y$. One denotes by ${\Gamma }_{\mathrm{rat}}(X/Y)$ the set of $Y$-rational sections of $X$, and one thus has a canonical map $\Gamma (X/Y)\to {\Gamma }_{\mathrm{rat}}(X/Y)$. It is moreover clear that if $X$ and $Y$ are two $S$-preschemes, the set of $S$-rational maps of $X$ into $Y$ is canonically identified with ${\Gamma }_{\mathrm{rat}}((X{\times }_{S}Y)/X)$ (3.3.14).

(7.1.3) It follows at once from (7.1.2) and (3.3.14) that the rational functions on $X$ are canonically identified with the equivalence classes of sections of the structure sheaf ${\mathcal{O}}_X$ above everywhere-dense opens of $X$, two such sections being equivalent if they coincide in an everywhere-dense open contained in the intersection of their sets of definition. It follows in particular that the rational functions on $X$ form a ring $R(X)$.

(7.1.4) When $X$ is an irreducible prescheme, every nonempty open is dense in $X$; one may also say that the nonempty opens of $X$ are the open neighborhoods of the generic point $x$ of $X$. To say that two morphisms of nonempty open subsets of $X$ into $Y$ are equivalent thus means in this case that they have the same germ at the point $x$. In other words, the rational maps (resp. $S$-rational maps) $X\to Y$ are identified with the germs of morphisms (resp. of $S$-morphisms) of nonempty open subsets of $X$ into $Y$ at the generic point $x$ of $X$. In particular:

Proposition (7.1.5). If $X$ is an irreducible prescheme, the ring $R(X)$ of rational functions on $X$ is canonically identified with the local ring ${\mathcal{O}}_x$ of the generic point $x$ of $X$. It is a local ring of dimension $0$, and consequently an artinian local ring when $X$ is noetherian; it is a field when $X$ is integral, and it is identified with the field of fractions of $A(X)$ when moreover $X$ is an affine scheme.

Proof. In view of what precedes and of the identification of rational functions with sections of ${\mathcal{O}}_X$ above an everywhere-dense open, the first assertion is none other than the definition of the stalk of a sheaf at a point. For the other assertions, one may restrict to the case where $X$ is affine with ring $A$; then ${\mathfrak{j}}_x$ is the nilradical of $A$, and ${\mathcal{O}}_x$ is therefore of dimension $0$; if $A$ is integral, ${\mathfrak{j}}_x=(0)$, and ${\mathcal{O}}_x$ is therefore the field of fractions of $A$. Finally, if $A$ is noetherian, one knows ([11], p. 127, cor. 4) that ${\mathfrak{j}}_x$ is nilpotent and ${\mathcal{O}}_x=A_{\mathfrak{p}}$ is artinian.

If $X$ is integral, the ring ${\mathcal{O}}_z$ is integral for every $z\in X$; every affine open $U$ containing $z$ also contains $x$, and $R(U)$, equal to the field of fractions of $A(U)$, is thus identified with $R(X)$; one concludes that $R(X)$ is also identified with the field of fractions of ${\mathcal{O}}_z$: the canonical identification of ${\mathcal{O}}_z$ with a subring of $R(X)$ consists in making correspond to each germ of section $s\in {\mathcal{O}}_z$ the unique rational function on $X$, the class of a section of ${\mathcal{O}}_X$ (necessarily defined on an everywhere-dense open) having germ $s$ at the point $z$.

(7.1.6) Suppose now that $X$ has a finite number of irreducible components $X_i$ ($1\le i\le n$) (which is the case when the space underlying $X$ is noetherian); let $X_i^{\prime }$ be the open of $X_i$ complementary, relative to $X_i$, to the union of the $X_j\cap X_i$ for $j\ne i$; $X_i^{\prime }$ is irreducible, its generic point $x_i$ is the generic point of $X_i$, and the $X_i^{\prime }$ are pairwise disjoint, their union being dense in $X$ (0, 2.1.6). For every everywhere-dense open $U$ of $X$, $U_i^{\prime }=U\cap X_i^{\prime }$ is a nonempty open dense in $X_i^{\prime }$, the $U_i^{\prime }$ being pairwise without common point, so that $U^{\prime }={\bigcup }_i{U}_i^{\prime }$ is dense in $X$. To give a morphism of $U^{\prime }$ into $Y$ amounts to giving (arbitrarily) a morphism of each of the $U_i^{\prime }$ into $Y$. Therefore:

Proposition (7.1.7). Let $X$, $Y$ be two preschemes (resp. $S$-preschemes) such that $X$ has a finite number of irreducible components $X_i$, with generic points $x_i$ ($1\le i\le n$). If $R_i$ is the set of germs of morphisms (resp. $S$-morphisms) of open subsets of $X$ into $Y$ at the point $x_i$, the set of rational maps (resp. $S$-rational maps) of $X$ into $Y$ is identified with the product of the $R_i$ ($1\le i\le n$).

Corollary (7.1.8). Let $X$ be a noetherian prescheme. The ring of rational functions on $X$ is an artinian ring, whose local components are the rings ${\mathcal{O}}_{x_i}$ of the generic points $x_i$ of the irreducible components of $X$.

Corollary (7.1.9). Let $A$ be a noetherian ring, and let $X=\mathrm{Spec}(A)$. If $Q$ is the complement of the union of the minimal prime ideals of $A$, the ring of rational functions on $X$ is canonically identified with the ring of fractions $Q^{-1}A$.

This will follow from the following lemma:

Lemma (7.1.9.1). For an element $f\in A$ to be such that $D(f)$ is everywhere dense in $X$, it is necessary and sufficient that $f\in Q$; every dense open in $X$ contains an open of the form $D(f)$, where $f\in Q$.

In fact, suppose this lemma proved; since the ring of sections $\Gamma (D(f),{\mathcal{O}}_X)$ is identified with $A_f$ (1.3.6 and 1.3.7), it follows from the fact that the $D(f)$ with $f\in Q$ form a cofinal set in the ordered set (for $\supset$) of dense opens in $X$, and from def. (7.1.1), that the ring of rational functions on $X$ is identified with the inductive limit of the $A_f$ for $f\in Q$ (for the preorder relation “$g$ is a multiple of $f$”), that is, with $Q^{-1}A$ (0, 1.4.5).

To prove (7.1.9.1), let us again denote by $X_i$ the irreducible components of $X$ ($1\le i\le n$); if $D(f)$ is dense in $X$, then $D(f)\cap X_i\ne \varnothing$ for $1\le i\le n$, and conversely; but this means that $f\notin {\mathfrak{p}}_i$ for $1\le i\le n$, on setting ${\mathfrak{p}}_i=\mathfrak{j}(X_i)$, and since the ${\mathfrak{p}}_i$ are the minimal prime ideals of $A$ (1.1.14), the relations $f\notin {\mathfrak{p}}_i$ ($1\le i\le n$) are equivalent to $f\in Q$, whence the first assertion of the lemma. On the other hand, if $U$ is a dense open in $X$, the complement of $U$ is a set of the form $V(\mathfrak{a})$, where $\mathfrak{a}$ is an ideal contained in none of the ${\mathfrak{p}}_i$; it is therefore not contained in their union ([10], p. 13), and there thus exists an $f\in \mathfrak{a}$ belonging to $Q$; whence $D(f)\subset U$, which completes the proof.

(7.1.10) Suppose again $X$ irreducible, with generic point $x$. Since every nonempty open $U$ of $X$ contains $x$, and consequently also contains every $z\in X$ such that $x\in \overline{\{z\}}$, every morphism $U\to Y$ may be composed with the canonical morphism $\mathrm{Spec}({\mathcal{O}}_x)\to X$ (2.4.1); and two morphisms into $Y$ of two nonempty open subsets of $X$, which coincide in a nonempty open subset of $X$, give by composition the same morphism $\mathrm{Spec}({\mathcal{O}}_x)\to Y$. In other words, to each rational map of $X$ into $Y$ there thus corresponds a well-determined morphism $\mathrm{Spec}({\mathcal{O}}_x)\to Y$.

Proposition (7.1.11). Let $X$, $Y$ be two $S$-preschemes; suppose $X$ irreducible, with generic point $x$, and $Y$ of finite type over $S$. Two $S$-rational maps of $X$ into $Y$, to which corresponds the same $S$-morphism $\mathrm{Spec}({\mathcal{O}}_x)\to Y$, are identical. If one supposes in addition $S$ locally noetherian, every $S$-morphism of $\mathrm{Spec}({\mathcal{O}}_x)$ into $Y$ corresponds to an $S$-rational map (and only one) of $X$ into $Y$.

Proof. Taking account of the fact that every nonempty open of $X$ is everywhere dense, this follows at once from (6.5.1).

Corollary (7.1.12). Suppose $S$ locally noetherian, and the other hypotheses of (7.1.11) satisfied. The $S$-rational maps of $X$ into $Y$ are then identified with the points of the $S$-prescheme $Y$, with values in the $S$-prescheme $\mathrm{Spec}({\mathcal{O}}_x)$.

This is none other than (7.1.11), with the terminology introduced in (3.4.1).

Corollary (7.1.13). Suppose the conditions of (7.1.12) fulfilled. Let $s$ be the image of $x$ in $S$. To give an $S$-rational map of $X$ into $Y$ is equivalent to giving a point $y$ of $Y$ above $s$, and a local ${\mathcal{O}}_s$-homomorphism ${\mathcal{O}}_y\to {\mathcal{O}}_x=R(X)$.

This follows from (7.1.11) and (2.4.4).

In particular:

Corollary (7.1.14). Under the conditions of (7.1.12), the $S$-rational maps of $X$ into $Y$ depend (for $Y$ given) only on the $S$-prescheme $\mathrm{Spec}({\mathcal{O}}_x)$, and in particular remain the same when one replaces $X$ by $\mathrm{Spec}({\mathcal{O}}_z)$, for any $z\in X$.

Proof. In fact, since $z\in \overline{\{x\}}$, $x$ is the generic point of $Z=\mathrm{Spec}({\mathcal{O}}_z)$, and ${\mathcal{O}}_{X,x}={\mathcal{O}}_{Z,x}$.

When $X$ is integral, $R(X)={\mathcal{O}}_x=\kappa (x)$ is a field (7.1.5); the preceding corollaries then specialize to:

Corollary (7.1.15). Suppose the conditions of (7.1.12) verified and moreover that $X$ is integral. Let $s$ be the image of $x$ in $S$. Then the $S$-rational maps of $X$ into $Y$ are identified with the geometric points of $Y{\otimes }_S\kappa (s)$ with values in the extension $R(X)$ of $\kappa (s)$; in other words, each of them is equivalent to giving a point $y\in Y$ above $s$ and a $\kappa (s)$-monomorphism of $\kappa (y)$ into $\kappa (x)=R(X)$.

Proof. The points of $Y$ above $s$ are in fact identified with those of $Y{\otimes }_S\kappa (s)$ (3.6.3), and the local ${\mathcal{O}}_s$-homomorphisms ${\mathcal{O}}_y\to R(X)$ with the $\kappa (s)$-monomorphisms $\kappa (y)\to R(X)$.

More particularly:

Corollary (7.1.16). Let $k$ be a field, and $X$, $Y$ two algebraic preschemes (6.4.1) over $k$; suppose in addition $X$ integral. Then the $k$-rational maps of $X$ into $Y$ are identified with the geometric points of $Y$ with values in the extension $R(X)$ of $k$ (3.4.4).

7.2. Domain of definition of a rational map

(7.2.1) Let $X$, $Y$ be two preschemes, $f$ a rational map of $X$ into $Y$. One says that $f$ is defined at a point $x\in X$ if there exists an everywhere-dense open set $U$ containing $x$ and a morphism $U\to Y$ belonging to the equivalence class $f$. The set of points $x\in X$ where $f$ is defined is called the domain of definition of $f$; it is clear that it is an everywhere-dense open in $X$.

Proposition (7.2.2). Let $X$, $Y$ be two $S$-preschemes, such that $X$ is reduced and $Y$ separated over $S$. Let $f$ be an $S$-rational map of $X$ into $Y$, and $U_0$ its domain of definition. There then exists one and only one $S$-morphism $U_0\to Y$ belonging to the class $f$.

Proof. Since for every morphism $U\to Y$ belonging to the class $f$ one necessarily has $U\subset U_0$, it is clear that the proposition will be a consequence of the

Lemma (7.2.2.1). Under the hypotheses of (7.2.2), let $U_1,U_2$ be two everywhere-dense opens of $X$, $f_i:U_i\to Y$ ($i=1,2$) two $S$-morphisms such that there exists an open $V\subset U_1\cap U_2$ dense in $X$ and in which $f_1$ and $f_2$ coincide. Then $f_1$ and $f_2$ coincide in $U_1\cap U_2$.

Proof. One may evidently restrict to the case where $X=U_1=U_2$. Since $X$ (hence $V$) is reduced, $X$ is the smallest closed subprescheme of $X$ majorizing $V$ (5.2.2). Let $g=(f_1,f_2{)}_S:X\to Y{\times }_{S}Y$; since by hypothesis the diagonal $T={\Delta }_Y(Y)$ is a closed subprescheme of $Y{\times }_{S}Y$, $Z=g^{-1}(T)$ is a closed subprescheme of $X$ (4.4.1). If $h:V\to Y$ is the common restriction of $f_1$ and $f_2$ to $V$, the restriction of $g$ to $V$ is $g^{\prime }=(h,h{)}_S$, which factors as $g^{\prime }={\Delta }_Y\circ h$; since ${\Delta }_Y^{-1}(T)=Y$, one has $g^{\prime -1}(T)=V$, and consequently $Z$ is a closed subprescheme of $X$ inducing $V$, hence majorizing $V$, which entails $Z=X$. From the relation $g^{-1}(T)=X$ one deduces (4.4.1) that $g$ factors as ${\Delta }_Y\circ f$, where $f$ is a morphism $X\to Y$, which entails by definition of the diagonal morphism that $f_1=f_2=f$.

It is clear that the morphism $U_0\to Y$ defined in (7.2.2) is the unique morphism of the class $f$ which cannot be prolonged to a morphism of an open subset of $X$ strictly containing $U_0$. Under the hypotheses of (7.2.2), one may thus identify the rational maps of $X$ into $Y$ with the non-prolongeable morphisms (to strictly larger opens) of everywhere-dense opens of $X$ into $Y$. With this identification, prop. (7.2.2) entails:

Corollary (7.2.3). The hypotheses on $X$ and $Y$ being those of (7.2.2), let $U$ be an everywhere-dense open of $X$. There exists a canonical one-to-one correspondence between the $S$-morphisms of $U$ into $Y$ and the $S$-rational maps of $X$ into $Y$ defined at all points of $U$.

Proof. By virtue of (7.2.2), for every $S$-morphism $f$ of $U$ into $Y$, there exists in fact one and only one $S$-rational map $\bar{f}$ of $X$ into $Y$ which prolongs $f$.

Corollary (7.2.4). Let $S$ be a scheme, $X$ a reduced $S$-prescheme, $Y$ an $S$-scheme, $f:U\to Y$ an $S$-morphism of a dense open $U$ of $X$ into $Y$. If $\bar{f}$ is the $\mathbb{Z}$-rational map of $X$ into $Y$ which prolongs $f$, then $\bar{f}$ is an $S$-morphism (and is consequently the $S$-rational map of $X$ into $Y$ prolonging $f$).

Proof. In fact, if $\phi :X\to S$, $\psi :Y\to S$ are the structure morphisms, $U_0$ the domain of definition of $\bar{f}$, $j$ the injection $U_0\to X$, it suffices to prove that $\psi \circ \bar{f}=\phi \circ j$, which follows at once from (7.2.2.1), since $f$ is an $S$-morphism.

Corollary (7.2.5). Let $X$, $Y$ be two $S$-preschemes; suppose $X$ reduced, $X$ and $Y$ separated over $S$. Let $p:Y\to X$ be an $S$-morphism (making $Y$ an $X$-prescheme), $U$ an everywhere-dense open of $X$, $f$ a $U$-section of $Y$; then the rational map $\bar{f}$ of $X$ into $Y$ prolonging $f$ is an $X$-rational section of $Y$.

Proof. One must prove that $p\circ \bar{f}$ is the identity in the domain of definition of $\bar{f}$; since $X$ is separated over $S$, this again follows from (7.2.2.1).

Corollary (7.2.6). Let $X$ be a reduced prescheme, $U$ an everywhere-dense open of $X$. There is a canonical one-to-one correspondence between the sections of ${\mathcal{O}}_X$ above $U$ and the rational functions $f$ on $X$ defined at every point of $U$.

Proof. Taking account of (7.2.3), (7.1.2), and (7.1.3), it suffices to remark that the $X$-prescheme $X{\otimes }_{\mathbb{Z}}\mathbb{Z}[T]$ is separated above $X$ (5.5.15 (iv)).

Corollary (7.2.7). Let $Y$ be a reduced prescheme, $f:X\to Y$ a separated morphism, $U$ an everywhere-dense open of $Y$, $g:U\to f^{-1}(U)$ a $U$-section of $f^{-1}(U)$, $Z$ the reduced subprescheme of $X$ having $g(U)$ as underlying space (5.2.1). For $g$ to be the restriction of a $Y$-section of $X$ (in other words (7.2.5), for the rational map of $Y$ into $X$ prolonging $g$ to be defined everywhere), it is necessary and sufficient that the restriction of $f$ to $Z$ be an isomorphism of $Z$ onto $Y$.

Proof. The restriction of $f$ to $f^{-1}(U)$ is a separated morphism (5.5.1, (i)), so $g$ is a closed immersion (5.4.6), and consequently $g(U)=Z\cap f^{-1}(U)$ and the subprescheme induced by $Z$ on the open $g(U)$ of $Z$ is identical with the closed subprescheme of $f^{-1}(U)$ associated with $g$ (5.2.1). It is then clear that the condition of the statement is sufficient, for if it is fulfilled and if $h:Z\to Y$ is the restriction of $f$ to $Z$ and $\bar{g}:Y\to Z$ the inverse isomorphism, $\bar{g}$ prolongs $g$. Conversely, if $g$ is the restriction to $U$ of a $Y$-section $h$ of $X$, $h$ is a closed immersion (5.4.6), so $h(Y)$ is closed, and since it is contained in $Z$, it is equal to $Z$, and it follows from (5.2.1) that $h$ is necessarily an isomorphism of $Y$ onto the closed subprescheme $Z$ of $X$.

(7.2.8) Let $X$, $Y$ be two $S$-preschemes, $X$ being supposed reduced and $Y$ separated over $S$. Let $f$ be an $S$-rational map of $X$ into $Y$, and let $x$ be a point of $X$; one may compose $f$ with the canonical $S$-morphism $\mathrm{Spec}({\mathcal{O}}_x)\to X$ (2.4.1) provided that the trace on $\mathrm{Spec}({\mathcal{O}}_x)$ of the domain of definition of $f$ is dense in $\mathrm{Spec}({\mathcal{O}}_x)$ (identified with the set of $z\in X$ such that $x\in \overline{\{z\}}$ (2.4.2)). This will be the case in the following situations:

1° $X$ is irreducible (hence integral), for then the generic point $\xi$ of $X$ is the generic point of $\mathrm{Spec}({\mathcal{O}}_x)$; since the domain of definition $U$ of $f$ contains $\xi$, $U\cap \mathrm{Spec}({\mathcal{O}}_x)$ contains $\xi$, hence is dense in $\mathrm{Spec}({\mathcal{O}}_x)$.

2° $X$ is locally noetherian; our assertion in fact follows then from the

Lemma (7.2.8.1). Let $X$ be a prescheme whose underlying space is locally noetherian, $x$ a point of $X$. The irreducible components of $\mathrm{Spec}({\mathcal{O}}_x)$ are the traces on $\mathrm{Spec}({\mathcal{O}}_x)$ of the irreducible components of $X$ containing $x$. For an open $U\subset X$ to be such that $U\cap \mathrm{Spec}({\mathcal{O}}_x)$ is dense in $\mathrm{Spec}({\mathcal{O}}_x)$, it is necessary and sufficient that it meet the irreducible components of $X$ containing $x$ (which is the case in particular if $U$ is dense in $X$).

Proof. The second assertion evidently follows from the first, and it thus suffices to prove the latter. Since $\mathrm{Spec}({\mathcal{O}}_x)$ is contained in every affine open $U$ containing $x$, and since the irreducible components of $U$ containing $x$ are the traces on $U$ of the irreducible components of $X$ containing $x$ (0, 2.1.6), one may suppose $X$ affine with ring $A$. Since the prime ideals of $A_x$ correspond bijectively to the prime ideals of $A$ contained in ${\mathfrak{j}}_x$ (0, 1.2.6), the minimal prime ideals of $A_x$ correspond to the minimal prime ideals of $A$ contained in ${\mathfrak{j}}_x$, whence the lemma (1.1.14).

This being so, suppose that one is in one of the two cases cited above. If $U$ is the domain of definition of the $S$-rational map $f$, let us denote by $f^{\prime }$ the rational map of $\mathrm{Spec}({\mathcal{O}}_x)$ into $Y$ which coincides (taking account of (2.4.2)) with $f$ in $U\cap \mathrm{Spec}({\mathcal{O}}_x)$; we shall say that this rational map is induced by $f$.

Proposition (7.2.9). Let $S$ be a locally noetherian prescheme, $X$ a reduced $S$-prescheme, $Y$ an $S$-scheme of finite type. Suppose in addition $X$ irreducible or locally noetherian. Let then $f$ be an $S$-rational map of $X$ into $Y$, and $x$ a point of $X$. For $f$ to be defined at the point $x$, it is necessary and sufficient that the rational map $f^{\prime }$ of $\mathrm{Spec}({\mathcal{O}}_x)$ into $Y$, induced by $f$ (7.2.8), be a morphism.

Proof. The condition being evidently necessary (since $\mathrm{Spec}({\mathcal{O}}_x)$ is contained in every open containing $x$), let us prove that it is sufficient. By virtue of (6.5.1), there exists an open neighborhood $U$ of $x$ in $X$ and an $S$-morphism $g$ of $U$ into $Y$, inducing $f^{\prime }$ on $\mathrm{Spec}({\mathcal{O}}_x)$. If $X$ is irreducible, $U$ is dense in $X$, and by virtue of (7.2.3) one may suppose that $g$ is an $S$-rational map. Moreover, the generic point of $X$ belongs to $\mathrm{Spec}({\mathcal{O}}_x)$ and to the domain of definition of $f$, so $f$ and $g$ coincide at this point, and consequently in a nonempty open set of $X$ (6.5.1). But since $f$ and $g$ are $S$-rational maps, they are identical (7.2.3), so $f$ is defined at $x$.

If now one supposes $X$ locally noetherian, one may suppose $U$ noetherian; there is then only a finite number of irreducible components $X_i$ of $X$ containing $x$ (7.2.8.1), and one may suppose that these are the only ones meeting $U$, by replacing $U$ if necessary by a smaller open (since there are only a finite number of irreducible components of $X$ meeting $U$, $U$ being noetherian). One sees then, as above, that $f$ and $g$ coincide in a nonempty open of each of the $X_i$. Taking account of the fact that each of the $X_i$ is contained in $U$, let us then consider the morphism $f_1$, defined in a dense open of $U\cup (X-U)$, equal to $g$ in $U$ and to $f$ in the intersection of $X-U$ and the domain of definition of $f$. Since $U\cup (X-U)$ is dense in $X$, $f_1$ and $f$ coincide in a dense open of $X$, and since $f$ is a rational map, $f$ is an extension of $f_1$ (7.2.3), hence is defined at the point $x$.

7.3. Sheaf of rational functions

(7.3.1) Let $X$ be a prescheme. For every open $U\subset X$, let us denote by $R(U)$ the ring of rational functions on $U$ (7.1.3); it is a $\Gamma (U,{\mathcal{O}}_X)$-algebra. Moreover, if $V\subset U$ is a second open of $X$, every section of ${\mathcal{O}}_X$ above an everywhere-dense open subset of $U$ gives, by restriction to $V$, a section above an everywhere-dense open subset of $V$, and if two sections coincide above an everywhere-dense open subset of $U$, their restrictions to $V$ coincide above an everywhere-dense open subset of $V$. One thus defines a $\mathrm{di}$-homomorphism of algebras ${\rho }_{V,U}:R(U)\to R(V)$, and it is clear that if $U\supset V\supset W$ are three opens of $X$, one has ${\rho }_{W,U}={\rho }_{W,V}\circ {\rho }_{V,U}$; the $R(U)$ thus define a presheaf of algebras on $X$.

Definition (7.3.2). One calls the sheaf of rational functions on a prescheme $X$, and one denotes by $\mathcal{R}(X)$, the ${\mathcal{O}}_X$-Algebra associated with the presheaf formed by the $R(U)$.

For every prescheme $X$ and every open $U\subset X$, it is clear that the induced sheaf $\mathcal{R}(X)|U$ is none other than $\mathcal{R}(U)$.

Proposition (7.3.3). Let $X$ be a prescheme such that the family $(X_{\lambda })$ of its irreducible components is locally finite (which is in particular the case when the space underlying $X$ is locally noetherian). Then the ${\mathcal{O}}_X$-Module $\mathcal{R}(X)$ is quasi-coherent, and for every open $U$ of $X$ meeting only a finite number of components $X_{\lambda }$, $R(U)$ is equal to $\Gamma (U,\mathcal{R}(X))$ and is canonically identified with the direct composite of the local rings of the generic points of the $X_{\lambda }$ such that $U\cap X_{\lambda }\ne \varnothing$.

Proof. One may evidently limit oneself to the case where $X$ has only a finite number of irreducible components $X_i$, with generic points $x_i$ ($1\le i\le n$). The fact that $R(U)$ is canonically identified with the direct composite of the ${\mathcal{O}}_{x_i}=R(X_i)$ such that $U\cap X_i\ne \varnothing$ then follows from (7.1.7). Let us show moreover that the presheaf $U\to R(U)$ verifies the sheaf axioms, which will prove that $R(U)=\Gamma (U,\mathcal{R}(X))$. In fact, it verifies (F 1) according to what precedes. To see that it satisfies (F 2), consider a cover of an open $U$ of $X$ by opens $V_{\alpha }\subset U$; if the $s_{\alpha }\in R(V_{\alpha })$ are such that the restrictions of $s_{\alpha }$ and $s_{\beta }$ to $V_{\alpha }\cap V_{\beta }$ coincide for every pair of indices, one concludes that for every index $i$ such that $U\cap X_i\ne \varnothing$, the components in $R(X_i)$ of all the $s_{\alpha }$ such that $V_{\alpha }\cap X_i\ne \varnothing$ are the same; denoting by $t_i$ this component, it is clear that the element of $R(U)$ having the $t_i$ as components has restriction $s_{\alpha }$ on each $V_{\alpha }$. Finally, to see that $\mathcal{R}(X)$ is quasi-coherent, one may limit oneself to the case where $X=\mathrm{Spec}(A)$ is affine; taking for $U$ the affine opens of the form $D(f)$, where $f\in A$, it follows from what precedes and from the definition (1.3.4) that one has $\mathcal{R}(X)=\tilde{M}$, where $M$ is the direct sum of the $A$-modules $A_{x_i}$.

Corollary (7.3.4). Let $X$ be a reduced prescheme having only a finite number of irreducible components, and let $X_i$ ($1\le i\le n$) be the reduced closed subpreschemes of $X$ having for underlying spaces the irreducible components of $X$ (5.2.1). If $h_i$ is the canonical injection $X_i\to X$, then $\mathcal{R}(X)$ is the direct composite of the ${\mathcal{O}}_X$-Algebras $(h_i{)}_{\ast }(\mathcal{R}(X_i))$.

Corollary (7.3.5). If $X$ is irreducible, every quasi-coherent $\mathcal{R}(X)$-Module $\mathcal{F}$ is a simple sheaf.

Proof. It suffices to show that every $x\in X$ admits a neighborhood $U$ such that $\mathcal{F}|U$ is a simple sheaf (0, 3.6.2), in other words one is reduced to the case where $X$ is affine; one may moreover suppose that $\mathcal{F}$ is the cokernel of a homomorphism $(\mathcal{R}(X){)}^{(I)}\to (\mathcal{R}(X){)}^{(J)}$ (0, 5.1.3), and everything comes down to seeing that $\mathcal{R}(X)$ is a simple sheaf; but this is evident since $\Gamma (U,\mathcal{R}(X))=R(X)$ for every nonempty open $U$, $U$ containing the generic point of $X$.

Corollary (7.3.6). If $X$ is irreducible, for every quasi-coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$, $\mathcal{F}{\otimes }_{{\mathcal{O}}_X}\mathcal{R}(X)$ is a simple sheaf; if in addition $X$ is reduced (hence integral), $\mathcal{F}{\otimes }_{{\mathcal{O}}_X}\mathcal{R}(X)$ is isomorphic to a sheaf of the form $(\mathcal{R}(X){)}^{(I)}$.

Proof. The second assertion follows from the fact that $R(X)$ is then a field.

Proposition (7.3.7). Suppose that the prescheme $X$ is locally integral or locally noetherian. Then $\mathcal{R}(X)$ is a quasi-coherent ${\mathcal{O}}_X$-Algebra; if in addition $X$ is reduced (which is the case when $X$ is locally integral), the canonical homomorphism ${\mathcal{O}}_X\to \mathcal{R}(X)$ is injective.

Proof. The question being local, the first assertion follows from (7.3.3); the second follows at once from (7.2.3).

(7.3.8) Let $X$, $Y$ be two preschemes each having a finite number of irreducible components, and let $f:X\to Y$ be a morphism whose restriction to the set of generic points of the irreducible components of $X$ is a surjection onto the set of generic points of the irreducible components of $Y$. Then one has

$$f^{\ast }(\mathcal{R}(Y))=\mathcal{R}(X).\text{\hspace{1em}(7.3.8.1)}$$

In fact, one is reduced (by virtue of (7.3.3)) to the case where $X$ and $Y$ are irreducible, with generic points $x$, $y$, with $f(x)=y$; hence $(f^{\ast }(\mathcal{R}(Y)){)}_x={\mathcal{O}}_y{\otimes }_{{\mathcal{O}}_y}{\mathcal{O}}_x={\mathcal{O}}_x$ (0, 4.3.1), which proves (7.3.8.1) by virtue of (7.3.5).

7.4. Torsion sheaves and torsion-free sheaves

(7.4.1) Let $X$ be an integral prescheme. For every ${\mathcal{O}}_X$-Module $\mathcal{F}$, the canonical homomorphism ${\mathcal{O}}_X\to \mathcal{R}(X)$ defines by tensorization a homomorphism (again called canonical) $\mathcal{F}\to \mathcal{F}{\otimes }_{{\mathcal{O}}_X}\mathcal{R}(X)$ which, on each fiber, is none other than the homomorphism $z\to z\otimes 1$ of ${\mathcal{F}}_x$ into ${\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_x}R(X)$. The kernel $\mathcal{T}$ of this homomorphism is a sub-${\mathcal{O}}_X$-Module of $\mathcal{F}$, called the torsion sheaf of $\mathcal{F}$; it is quasi-coherent if $\mathcal{F}$ is quasi-coherent (4.1.1 and 7.3.6). One says that $\mathcal{F}$ is torsion-free if $\mathcal{T}=0$ and that $\mathcal{F}$ is a torsion sheaf if $\mathcal{T}=\mathcal{F}$. For every ${\mathcal{O}}_X$-Module $\mathcal{F}$, $\mathcal{F}/\mathcal{T}$ is torsion-free. One deduces from (7.3.5) that:

Proposition (7.4.2). If $X$ is an integral prescheme, every quasi-coherent torsion-free ${\mathcal{O}}_X$-Module $\mathcal{F}$ is isomorphic to a subsheaf $\mathcal{G}$ of a simple sheaf of the form $(\mathcal{R}(X){)}^{(I)}$, generated (as $\mathcal{R}(X)$-Module) by $\mathcal{G}$.

The cardinal of $I$ is called the rank of $\mathcal{F}$; for every nonempty affine open $U$ of $X$, the rank of $\mathcal{F}$ is equal to the rank of $\Gamma (U,\mathcal{F})$ as a $\Gamma (U,{\mathcal{O}}_X)$-module, as one sees at once by considering the generic point of $X$, contained in $U$. In particular:

Corollary (7.4.3). On an integral prescheme $X$, every quasi-coherent torsion-free ${\mathcal{O}}_X$-Module of rank $1$ (in particular every invertible ${\mathcal{O}}_X$-Module) is isomorphic to a sub-${\mathcal{O}}_X$-Module of $\mathcal{R}(X)$, and conversely.

Corollary (7.4.4). Let $X$ be an integral prescheme, $\mathcal{L}$, ${\mathcal{L}}^{\prime }$ two torsion-free ${\mathcal{O}}_X$-Modules, $f$ (resp. $f^{\prime }$) a section of $\mathcal{L}$ (resp. ${\mathcal{L}}^{\prime }$) above $X$. For $f\otimes f^{\prime }=0$, it is necessary and sufficient that one of the sections $f,f^{\prime }$ be zero.

Proof. Let $x$ be the generic point of $X$; one has by hypothesis $(f\otimes f^{\prime }{)}_x=f_x\otimes f_x^{\prime }=0$. Since ${\mathcal{L}}_x$ and ${\mathcal{L}}_x^{\prime }$ are identified with sub-${\mathcal{O}}_x$-Modules of the field ${\mathcal{O}}_x$, the preceding relation entails $f_x=0$ or $f_x^{\prime }=0$, and consequently $f=0$ or $f^{\prime }=0$ since $\mathcal{L}$ and ${\mathcal{L}}^{\prime }$ are torsion-free (7.3.5).

Proposition (7.4.5). Let $X$, $Y$ be two integral preschemes, $f:X\to Y$ a dominant morphism. For every quasi-coherent torsion-free ${\mathcal{O}}_X$-Module $\mathcal{F}$, $f_{\ast }(\mathcal{F})$ is a torsion-free ${\mathcal{O}}_Y$-Module.

Proof. Since $f_{\ast }$ is left exact (0, 4.2.1), it suffices, by virtue of (7.4.2), to prove the proposition when $\mathcal{F}=(\mathcal{R}(X){)}^{(I)}$. Now, every nonempty open $U$ of $Y$ contains the generic point of $Y$, so $f^{-1}(U)$ contains the generic point of $X$ (0, 2.1.5), so one then has $\Gamma (U,f_{\ast }(\mathcal{F}))=\Gamma (f^{-1}(U),\mathcal{F})=(R(X){)}^{(I)}$; in other words, $f_{\ast }(\mathcal{F})$ is the simple sheaf with fiber $(R(X){)}^{(I)}$, considered as $\mathcal{R}(Y)$-Module, and it is evidently torsion-free.

Proposition (7.4.6). Let $X$ be an integral prescheme, $x$ its generic point. For every quasi-coherent ${\mathcal{O}}_X$-Module of finite type $\mathcal{F}$, the following conditions are equivalent: a) $\mathcal{F}$ is a torsion sheaf; b) ${\mathcal{F}}_x=0$; c) $\mathrm{Supp}(\mathcal{F})\ne X$.

Proof. By virtue of (7.3.5) and (7.4.1), the relations ${\mathcal{F}}_x=0$ and $\mathcal{F}{\otimes }_{{\mathcal{O}}_X}\mathcal{R}(X)=0$ are equivalent, so a) and b) are equivalent; on the other hand, $\mathrm{Supp}(\mathcal{F})$ is closed in $X$ (0, 5.2.2), and since every nonempty open of $X$ contains $x$, b) and c) are equivalent.

(7.4.7) One extends (by abuse of language) the definitions of (7.4.1) to the case where $X$ is a reduced prescheme having only a finite number of irreducible components; it then follows from (7.3.4) that the equivalence of a) and c) in (7.4.6) is still valid for such a prescheme.