§7. Valuative criteria

In this section we give valuative criteria of separation and properness for a morphism — that is, criteria that bring in an auxiliary variable scheme of the form $\mathrm{Spec}(A)$, where $A$ is a valuation ring. Under suitable "Noetherian" hypotheses, one may restrict in these criteria to the case where $A$ is a discrete valuation ring. This will doubtless be the only case we shall have to apply in what follows, and we have introduced arbitrary valuation rings, in the general case, only in order to make the connection with classical developments.

7.1. Reminders on valuation rings

(7.1.1)

Among the various equivalent properties characterising valuation rings, we shall use the following: a ring $A$ is called a valuation ring if it is an integral ring that is not a field and if, in the set of local rings contained in the field of fractions $K$ of $A$ and distinct from $K$, $A$ is maximal for the relation of domination (I, 8.1.1). Recall that a valuation ring is integrally closed. If $A$ is a valuation ring, then $A_{\mathfrak{p}}$ is also a valuation ring for every prime ideal $\mathfrak{p}\ne (0)$ of $A$.

(7.1.2)

Let $K$ be a field and $A$ a local subring of $K$ that is not a field;

there then exists a valuation ring having $K$ as its field of fractions and dominating $A$ ([1], p. 1-07, lemma 2).

On the other hand, let $B$ be a valuation ring, $k$ its residue field, $K$ its field of fractions, and $L$ an extension of $k$. Then there exists a complete valuation ring $C$ that dominates $B$ and whose residue field is $L$. Indeed, $L$ is an algebraic extension of a pure transcendental extension $L^{\prime }=k(T_{\mu }{)}_{\mu \in M}$; we know that one can extend the valuation of $K$ corresponding to $B$ to a valuation of $K^{\prime }=K(T_{\mu }{)}_{\mu \in M}$ in such a way that $L^{\prime }$ is the residue field of this valuation ([24], p. 98); replacing $B$ by the completion of the ring of this extended valuation, we see that we may restrict to the case where $B$ is complete and $L$ is an algebraic closure of $k$. If $\bar{K}$ is an algebraic closure of $K$, we can then extend to $\bar{K}$ the valuation that defines $B$, and the corresponding residue field is an algebraic closure of $k$, as one sees by lifting to $\bar{K}$ the coefficients of a monic polynomial of k[T]. We are thus finally reduced to the case where $L=k$, and it then suffices to take $C$ to be the completion of $B$ in order to settle the question.

(7.1.3)

Let $K$ be a field and $A$ a subring of $K$; the integral closure $A^{\prime }$ of $A$ in $K$ is the intersection of the valuation rings containing $A$ and having $K$ as field of fractions ([11], p. 51, th. 2). Proposition (7.1.2) is expressed in the following equivalent geometric form:

Proposition.

Let $Y$ be a prescheme, $p:X\to Y$ a morphism, $x$ a point of $X$, $y=p(x)$, and $y^{\prime }\ne y$ a specialisation (0, 2.1.2) of $y$. There then exists a local scheme $Y^{\prime }$, spectrum of a valuation ring, and a separated morphism $f:Y^{\prime }\to Y$ such that, if $a$ denotes the unique closed point of $Y^{\prime }$ and $b$ the generic point of $Y^{\prime }$, one has $f(a)=y^{\prime }$ and $f(b)=y$. One may further suppose that one of the two additional properties below is satisfied:

(i) $Y^{\prime }$ is the spectrum of a complete valuation ring whose residue field is algebraically closed, and there exists a $\kappa (y)$-homomorphism $\kappa (x)\to \kappa (b)$.

(ii) There exists a $\kappa (y)$-isomorphism $\kappa (x)\stackrel{\sim }{\to }\kappa (b)$.

Proof. Let $Y_1$ be the closed reduced subprescheme of $Y$ having $\bar{y}$ as its underlying space (I, 5.2.1), and let $X_1$ be the closed subprescheme given by the inverse image $p^{-1}(Y_1)$; since $y^{\prime }\in \bar{y}$ by hypothesis and since $\kappa (x)$ is the same in $X$ and in $X_1$, we may suppose $Y$ integral, with generic point $y$; ${\mathcal{O}}_{y^{\prime }}$ is then an integral local ring that is not a field and whose field of fractions is ${\mathcal{O}}_y=\kappa (y)$, and $\kappa (x)$ is an extension of $\kappa (y)$. In order to satisfy the conditions $f(a)=y^{\prime }$ and $f(b)=y$ as well as the additional condition (i) (resp. (ii)), we take $Y^{\prime }=\mathrm{Spec}(A^{\prime })$, where $A^{\prime }$ is a valuation ring dominating ${\mathcal{O}}_{y^{\prime }}$ and which is complete with residue field algebraically closed extension of $\kappa (x)$ (resp. a valuation ring $A^{\prime }$ dominating ${\mathcal{O}}_{y^{\prime }}$ and whose field of fractions is $\kappa (x)$); the existence of $A^{\prime }$ is guaranteed by (7.1.2).

(7.1.5)

Recall that a local ring $A$ is said to be of dimension 1 if there exists a prime ideal distinct from the maximal ideal $\mathfrak{m}$ and if every prime ideal of $A$ distinct from $\mathfrak{m}$ is a minimal prime ideal; when $A$ is integral, this amounts to saying that $\mathfrak{m}$ and (0) are the only prime ideals and $\mathfrak{m}\ne (0)$; in other words, $Y=\mathrm{Spec}(A)$ is reduced to two

points $a$ and $b$: $a$ is the unique closed point, ${\mathfrak{j}}_a=\mathfrak{m}$, and $\kappa (a)=k$ is the residue field $k=A/\mathfrak{m}$; $b$ is the generic point of $Y$, ${\mathfrak{j}}_b=(0)$, the set $b$ being the unique open subset of $Y$ distinct from both $\varnothing$ and $Y$ (an open subset which is therefore everywhere dense), and $\kappa (b)=K$ is the field of fractions of $A$.

(7.1.6)

For a local ring $A$, Noetherian and of dimension 1, we know ([1], p. 2-08 and 17-01) that the following conditions are equivalent:

(a) $A$ is normal;

(b) $A$ is regular;

(c) $A$ is a valuation ring;

furthermore, $A$ is then a discrete valuation ring. Propositions (7.1.2) and (7.1.3) then have the following analogues for discrete valuation rings:

Proposition.

Let $A$ be an integral local Noetherian ring that is not a field, $K$ its field of fractions, and $L$ an extension of finite type of $K$. There then exists a discrete valuation ring having $L$ as its field of fractions and dominating $A$.

Proof. Suppose first $L=K$. Let $\mathfrak{m}$ be the maximal ideal of $A$, $(x_1,\cdots ,x_n)$ a system of nonzero generators of $\mathfrak{m}$, and $B$ the subring $A[x_2/x_1,\cdots ,x_n/x_1]$ of $K$, which is Noetherian. It is immediate that the ideal $\mathfrak{m}B$ of $B$ is identical to the principal ideal $x_{1}B$; if $\mathfrak{p}$ is a minimal prime ideal of $x_{1}B$, we know that $\mathfrak{p}$ is of rank 1 ([13], t. I, p. 277); in other words, $B_{\mathfrak{p}}$ is a local Noetherian ring of dimension 1; it is clear that $\mathfrak{p}{B}_{\mathfrak{p}}\cap A$ is an ideal of $A$ containing $\mathfrak{m}$ and not containing 1, hence equal to $\mathfrak{m}$, so that $B_{\mathfrak{p}}$ dominates $A$ (I, 8.1.1). It follows from the Krull–Akizuki theorem ([25], p. 293) that the integral closure $C$ of $B_{\mathfrak{p}}$ is a Noetherian ring (although $C$ is not necessarily a $B_{\mathfrak{p}}$-module of finite type); if $\mathfrak{n}$ is a maximal ideal of $C$, then $C_{\mathfrak{n}}$ is a normal local Noetherian ring of dimension 1 ([25], p. 295), hence a discrete valuation ring, which dominates $B_{\mathfrak{p}}$ and a fortiori $A$.

If now $L$ is an extension of finite type of $K$, we may, by what precedes, restrict to the case where $A$ is already a discrete valuation ring. Let $w$ be a valuation of $K$ associated to $A$; there exists a discrete valuation $w^{\prime }$ of $L$ that extends $w$: indeed, by induction on the number of generators of $L$, we reduce to the case $L=K(\alpha )$, and then the proposition is classical ([24], p. 106).

Corollary.

Let $A$ be a Noetherian integral ring, $K$ its field of fractions, and $L$ an extension of finite type of $K$. Then the integral closure of $A$ in $L$ is the intersection of the discrete valuation rings having $L$ as field of fractions and containing $A$.

Proof. Indeed, such a discrete valuation ring, being normal, contains a fortiori every element of $L$ that is integral over $A$. It thus suffices to prove that if $x\in L$ is not integral over $A$, then there exists a discrete valuation ring $C$ having $L$ as field of fractions, containing $A$, and not containing $x$. The hypothesis on $x$ means that $x\notin B=A[1/x]$, in other words, that $1/x$ is not invertible in the Noetherian ring $B$. There is thus a prime ideal $\mathfrak{p}$ of $B$ containing $1/x$. The integral local ring $B_{\mathfrak{p}}$ is Noetherian and contained in $L$, which is an extension of finite type of the field of fractions of $B_{\mathfrak{p}}$ (the latter containing $K$). By virtue of (7.1.7) there is then a discrete valuation ring $C$ dominating $B_{\mathfrak{p}}$ and having $L$ as field of fractions; since $1/x\in \mathfrak{p}{B}_{\mathfrak{p}}$ belongs to the maximal ideal of $C$, we have $x\notin C$, which concludes the proof.

The geometric form of (7.1.7) is the following:

Proposition.

Let $Y$ be a locally Noetherian prescheme, $p:X\to Y$ a morphism locally of finite type, $x$ a point of $X$, $y=p(x)$, and $y^{\prime }\ne y$ a specialisation of $y$. There then exists a local scheme $Y^{\prime }$, spectrum of a discrete valuation ring, a separated morphism $f:Y^{\prime }\to Y$, and a $Y$-rational map $g$ from $Y^{\prime }$ to $X$ such that, denoting the closed point of $Y^{\prime }$ by $a$ and its generic point by $b$, one has $f(a)=y^{\prime }$, $f(b)=y$, $g(b)=x$, and such that in the commutative diagram

                       κ(x)
                      ↗     ↘ γ
                   π /        ↘
                    /           ↘
              κ(y) ────φ────→ κ(b)                                       (7.1.9.1)

(where $\pi$, $\varphi$, $\gamma$ are the homomorphisms corresponding to $p$, $f$, $g$ respectively), $\gamma$ is a bijection.

Proof. As in (7.1.4), we may reduce to the case where $Y$ is integral with generic point $y$ (taking (I, 6.3.4, (iv)) into account); since the question is local on $X$ and $Y$, we may assume $p$ of finite type; we are then in the situation of (7.1.4), with the additional property that $\kappa (x)$ is an extension of finite type of $\kappa (y)$ (I, 6.4.11) and ${\mathcal{O}}_{y^{\prime }}$ is Noetherian; this permits us to apply (7.1.7) and take $Y^{\prime }=\mathrm{Spec}(A^{\prime })$, where $A^{\prime }$ is a discrete valuation ring dominating ${\mathcal{O}}_{y^{\prime }}$ and whose field of fractions is $\kappa (x)$. We have thus defined a commutative diagram (7.1.9.1) where $\gamma$ is a bijection, and $\pi$ and $\varphi$ correspond to the morphisms $p$ and $f$. Furthermore, since $X$ and $Y$ are locally Noetherian (I, 6.6.2) and $Y^{\prime }$ is integral, there is exactly one $Y$-rational map $g$ from $Y^{\prime }$ to $X$ to which corresponds the isomorphism $\gamma$ (I, 7.1.15), which finishes the proof.

7.2. Valuative criterion of separation

Proposition.

Let $X$, $Y$ be two preschemes, and $f:X\to Y$ a quasi-compact morphism. The following two conditions are equivalent:

(a) The morphism $f$ is closed.

(b) For every $x\in X$ and every specialisation $y^{\prime }$ of $y=f(x)$ distinct from $y$, there exists a specialisation $x^{\prime }$ of $x$ such that $f(x^{\prime })=y^{\prime }$.

Proof. Condition (b) expresses that $f(\bar{x})=\bar{y}$ and is thus a consequence of (a). To show that (b) implies (a), consider a closed subset $X^{\prime }$ of the underlying space $X$; let $Y^{\prime }=f\overline{X^{\prime }}$ and let us show that $Y^{\prime }=f(X^{\prime })$. Consider the closed reduced subpreschemes of $X$ and $Y$ having $X^{\prime }$ and $Y^{\prime }$ respectively as their underlying spaces (I, 5.2.1); there is then a morphism $f^{\prime }:X^{\prime }\to Y^{\prime }$ such that the diagram

   X' ──f'──→ Y'
   │          │
   ↓          ↓
   X  ──f──→  Y

commutes (I, 5.2.2), and since $f$ is quasi-compact, so too is $f^{\prime }$. We are thus reduced to proving that if $f$ is a quasi-compact and dominant morphism, then

condition (b) implies that $f(X)=Y$. Indeed, let $y^{\prime }$ be a point of $Y$, and let $y$ be the generic point of an irreducible component of $Y$ containing $y^{\prime }$; by (b), it suffices to show that $f^{-1}(y)$ is not empty. But this property is a consequence of the fact that $f$ is dominant and quasi-compact (I, 6.6.5).

Corollary.

Let $f:X\to Y$ be a quasi-compact immersion morphism. For the underlying space $X$ to be closed in $Y$, it is necessary and sufficient that it contain every specialisation (in $Y$) of each of its points.

Proposition.

Let $Y$ be a prescheme (resp. a locally Noetherian prescheme), $f:X\to Y$ a morphism (resp. a morphism locally of finite type). The following conditions are equivalent:

(a) $f$ is separated.

(b) The diagonal morphism $X\to X{\times }_{Y}X$ is quasi-compact, and for every $Y$-prescheme of the form $Y^{\prime }=\mathrm{Spec}(A)$, where $A$ is a valuation ring (resp. a discrete valuation ring), any two $Y$-morphisms from $Y^{\prime }$ to $X$ that coincide at the generic point of $Y^{\prime }$ are equal.

(c) The diagonal morphism $X\to X{\times }_{Y}X$ is quasi-compact, and for every $Y$-prescheme of the form $Y^{\prime }=\mathrm{Spec}(A)$, where $A$ is a valuation ring (resp. a discrete valuation ring), any two $Y^{\prime }$-sections of $X^{\prime }=X_{(Y^{\prime })}$ that coincide at the generic point of $Y^{\prime }$ are equal.

Proof. The equivalence of (b) and (c) results from the bijective correspondence between $Y$-morphisms from $Y^{\prime }$ to $X$ and $Y^{\prime }$-sections of $X^{\prime }$ (I, 3.3.14). If $X$ is separated over $Y$, condition (b) is satisfied by virtue of (I, 7.2.2.1), since $Y^{\prime }$ is integral. It remains to prove that (b) implies that the diagonal morphism $\Delta :X\to X{\times }_{Y}X$ is closed, and it comes to the same to show that it satisfies the criterion of (7.2.2). So let $z$ be a point of the diagonal $\Delta (X)$, and $z^{\prime }\ne z$ a specialisation of $z$ in $X{\times }_{Y}X$. There exists then (7.1.4) a valuation ring $A$ and a morphism $f$ from $Y^{\prime }=\mathrm{Spec}(A)$ to $X{\times }_{Y}X$ which sends the closed point $a$ of $Y^{\prime }$ to $z^{\prime }$ and the generic point $b$ of $Y^{\prime }$ to $z$; this morphism makes $Y^{\prime }$ into an $(X{\times }_{Y}X)$-prescheme, and a fortiori a $Y$-prescheme. If we compose $f$ with the two projections of $X{\times }_{Y}X$, we obtain two $Y$-morphisms $g_1$, $g_2$ from $Y^{\prime }$ to $X$, which by hypothesis coincide at the point $b$; they are therefore equal to a single morphism $g$, which means (I, 5.3.1) that $f$ factors as $f=\Delta \circ g$, whence $z^{\prime }\in \Delta (X)$. When one supposes $Y$ locally Noetherian and $f$ locally of finite type, $X{\times }_{Y}X$ is locally Noetherian (I, 6.6.7); one may then take up the preceding argument by supposing that $A$ is a discrete valuation ring, by virtue of (7.1.9).

Remarks.

(i) The hypothesis that the morphism $\Delta$ is quasi-compact is always satisfied when $Y$ is locally Noetherian and $f$ locally of finite type, since $X{\times }_{Y}X$ is then locally Noetherian (I, 6.6.4, (i)). In the general case, it also means that for every covering $(U_{\alpha })$ of $X$ by affine opens, the sets $U_{\alpha }\cap U_{\beta }$ are quasi-compact.

(ii) For $f$ to be separated, it suffices that condition (b) or condition (c) be satisfied for a valuation ring $A$ which is complete and whose residue field is algebraically closed; this follows in fact from the proof of (7.2.3) and from (7.1.4).

7.3. Valuative criterion of properness

Proposition.

Let $A$ be a valuation ring, $Y=\mathrm{Spec}(A)$, $b$ the generic point of $Y$, $X$ an integral scheme, and $f:X\to Y$ a closed morphism such that $f^{-1}(b)$ reduces to a single point $x$ and the corresponding homomorphism $\kappa (b)\to \kappa (x)$ is bijective. Then $f$ is an isomorphism.

Proof. Since $f$ is closed and dominant, $f(X)=Y$; it suffices (I, 4.2.2) to prove that for every $y^{\prime }\ne b$ in $Y$ there exists exactly one point $x^{\prime }$ such that $f(x^{\prime })=y^{\prime }$ and the corresponding homomorphism ${\mathcal{O}}_{y^{\prime }}\to {\mathcal{O}}_{x^{\prime }}$ is bijective, for then $f$ will be a homeomorphism. Now if $f(x^{\prime })=y^{\prime }$, ${\mathcal{O}}_{x^{\prime }}$ is a local ring contained in $K=\kappa (x)=\kappa (b)$ and dominating ${\mathcal{O}}_{y^{\prime }}$; the latter is the local ring $A_{y^{\prime }}$, hence a valuation ring (7.1.1) having $K$ as field of fractions. On the other hand, ${\mathcal{O}}_{x^{\prime }}\ne K$, since $x^{\prime }$ is not the generic point of $X$ (0, 2.1.3); we conclude that ${\mathcal{O}}_{x^{\prime }}={\mathcal{O}}_{y^{\prime }}$. Since $X$ is an integral scheme, the relation ${\mathcal{O}}_{x^{\prime }}={\mathcal{O}}_{x^{\prime \prime }}$ entails $x^{\prime }=x^{\prime \prime }$ (I, 8.2.2), which concludes the proof.

(7.3.2)

Let $A$ be a valuation ring, $Y=\mathrm{Spec}(A)$, $b$ the generic point of $Y$, so that ${\mathcal{O}}_b=\kappa (b)$ is equal to $K$, the field of fractions of $A$; let $f:X\to Y$ be a morphism. We know (I, 7.1.4) that the rational $Y$-sections of $X$ are in bijective correspondence with the germs of $Y$-sections (defined in neighbourhoods of $b$) at the point $b$, whence a canonical map

$${\Gamma }_{rat}(X/Y)\to \Gamma (f^{-1}(b)/\mathrm{Spec}(K))(\mathrm{7.3.2.1})$$

the elements of $\Gamma (f^{-1}(b)/\mathrm{Spec}(K))$ being identified, by definition (I, 3.4.5), with the points of $f^{-1}(b)=X{\otimes }_{A}K$ that are rational over $K$. When $f$ is separated, it follows from (I, 5.4.7) that the map (7.3.2.1) is injective, since $Y$ is an integral scheme.

Composing (7.3.2.1) with the canonical map $\Gamma (X/Y)\to {\Gamma }_{rat}(X/Y)$ (I, 7.1.2), we obtain a canonical map

$$\Gamma (X/Y)\to \Gamma (f^{-1}(b)/\mathrm{Spec}(K)).(\mathrm{7.3.2.2})$$

When $f$ is separated, this map is again injective (I, 5.4.7).

Proposition.

Let $A$ be a valuation ring with field of fractions $K$, $Y=\mathrm{Spec}(A)$, $b$ the generic point of $Y$, and $f:X\to Y$ a separated and closed morphism. Then the canonical map (7.3.2.2) is bijective (which amounts to saying that it is surjective, and implies that the rational $Y$-sections of $X$ are everywhere defined).

Proof. Let $x$ be a point of $f^{-1}(b)$ rational over $K$. Since $f$ is separated, so too is the morphism $f^{-1}(b)\to \mathrm{Spec}(K)$ corresponding to $f$ (I, 5.5.1, (iv)), and every section of $f^{-1}(b)$ being a closed immersion (I, 5.4.6), $x$ is closed in $f^{-1}(b)$. Consider the closed reduced subprescheme $X^{\prime }$ of $X$ having for underlying space the closure $\bar{x}$ of $x$ in $X$. It is clear that the restriction of $f$ to $X^{\prime }$ verifies the hypotheses of (7.3.1), and is therefore an isomorphism of $X^{\prime }$ onto $Y$, whose inverse isomorphism is the sought-for $Y$-section of $X$.

(7.3.4)

To state the two following results, we make use of a terminology that will be justified and discussed in chapter IV: if $F$ is a subset

of a prescheme $Y$, we call codimension of $F$ in $Y$, denoted $codi{m}_{Y}F$, the lower bound of the integers $\dim ({\mathcal{O}}_z)$ as $z$ runs through $F$.

Corollary.

Let $Y$ be a locally Noetherian reduced prescheme, $N$ the set of points $y\in Y$ where $Y$ is not regular (0, 4.1.4); suppose that $codi{m}_{Y}N\ge 2$. Let $f:X\to Y$ be a morphism of finite type, both separated and closed, and let $g$ be a rational $Y$-section of $X$; if $Y^{\prime }$ is the set of points of $Y$ where $g$ is not defined (a set which is closed (I, 7.2.1)), then $codi{m}_Y{Y}^{\prime }\ge 2$.

Proof. It suffices to prove that $g$ is defined at every point $z\in Y$ such that $\dim {\mathcal{O}}_z\le 1$. If $\dim {\mathcal{O}}_z=0$, $z$ is the generic point of an irreducible component of $Y$ (I, 1.1.14), so belongs to every everywhere dense open subset of $Y$, and in particular to the domain of definition of $g$. Suppose then that $\dim {\mathcal{O}}_z=1$; ${\mathcal{O}}_z$ is then by hypothesis a regular Noetherian local ring, hence (7.1.6) a discrete valuation ring. Let $Z=\mathrm{Spec}({\mathcal{O}}_z)$; since $U=Y\setminus Y^{\prime }$ is everywhere dense, $U\cap Z$ is not empty (I, 2.4.2); let $g^{\prime }$ be the rational map from $Z$ to $X$ induced by $g$ (I, 7.2.8); it suffices to show that $g^{\prime }$ is a morphism (I, 7.2.9). Now, $g^{\prime }$ may be regarded as a rational $Z$-section of the $Z$-prescheme $f^{-1}(Z)=X{\times }_{Y}Z$; it is clear that the morphism $f^{-1}(Z)\to Z$ corresponding to $f$ is closed, and it follows from (I, 5.5.1, (i)) that it is separated; we conclude from (7.3.3) that $g^{\prime }$ is everywhere defined; since $Z$ is reduced and $X$ is separated over $Y$, $g^{\prime }$ is a morphism (I, 7.2.2).

Corollary.

Let $S$ be a locally Noetherian prescheme, and $X$, $Y$ two $S$-preschemes; suppose $Y$ reduced and, furthermore, that the set $N$ of points $y\in Y$ where $Y$ is not regular is such that $codi{m}_{Y}N\ge 2$; suppose finally that the structure morphism $X\to S$ is proper. Let $f$ be an $S$-rational map from $Y$ to $X$, and let $Y^{\prime }$ be the set of points of $Y$ where $f$ is not defined; then $codi{m}_Y{Y}^{\prime }\ge 2$.

Proof. We know (I, 7.1.2) that the $S$-rational maps from $Y$ to $X$ may be identified with the $Y$-rational sections of $X{\times }_{S}Y$; since the structure morphism $X{\times }_{S}Y\to Y$ is closed (5.4.1), we may apply (7.3.5), whence the corollary.

Remark.

The hypotheses made on $Y$ in (7.3.5) and (7.3.6) will be satisfied in particular when $Y$ is normal (0, 4.1.4), by virtue of (7.1.6).

We can characterise the universally closed (resp. proper) morphisms by a converse of (7.3.3):

Theorem.

Let $Y$ be a prescheme (resp. a locally Noetherian prescheme), $f:X\to Y$ a quasi-compact separated morphism (resp. a morphism of finite type). The following conditions are equivalent:

(a) $f$ is universally closed (resp. proper).

(b) For every $Y$-scheme of the form $Y^{\prime }=\mathrm{Spec}(A)$, where $A$ is a valuation ring (resp. a discrete valuation ring) with field of fractions $K$, the canonical map

  Hom_Y(Y', X) → Hom_Y(Spec(K), X)

corresponding to the canonical injection $A\to K$ is surjective (resp. bijective).

(c) For every $Y$-scheme of the form $Y^{\prime }=\mathrm{Spec}(A)$, where $A$ is a valuation ring (resp. a discrete valuation ring), the canonical map (7.3.2.2) relative to the $Y^{\prime }$-prescheme $X_{(Y^{\prime })}$ is surjective (resp. bijective).

Proof. The equivalence of (b) and (c) follows immediately from (I, 3.3.14). (a) implies (c), since (a) entails, in either hypothesis, that $f_{(Y^{\prime })}$ is separated (I, 5.5.1, (iv)) and closed, and it suffices to apply (7.3.3). It remains to prove that (b) implies (a). Place ourselves first in the case where $Y$ is arbitrary, $f$ separated and quasi-compact. If condition (b) is satisfied by $f$, it is also satisfied by $f_{(Y^{\prime \prime })}:X_{(Y^{\prime \prime })}\to Y^{\prime \prime }$, where Y'' is an arbitrary $Y$-prescheme, by virtue of the equivalence of (b) and (c) and of the fact that $X_{(Y^{\prime \prime })}{\times }_{Y^{\prime \prime }}{Y}^{\prime }=X{\times }_Y{Y}^{\prime }$ for every morphism $Y^{\prime }\to Y^{\prime \prime }$ (I, 3.3.9.1); since moreover $f_{(Y^{\prime \prime })}$ is separated and quasi-compact when $f$ is (I, 5.5.1, (iv) and 6.6.4, (iii)), we are reduced to proving that (b) implies that $f$ is closed. For this, it suffices to verify condition (b) of (7.2.1). Let then $x\in X$, $y^{\prime }$ a specialisation of $y=f(x)$, distinct from $y$; by virtue of (7.1.4), there is a scheme $Y^{\prime }$, spectrum of a valuation ring, and a separated morphism $g:Y^{\prime }\to Y$ such that, if $a$ denotes the closed point and $b$ the generic point of $Y^{\prime }$, one has $g(a)=y^{\prime }$, $g(b)=y$, and there exists a $\kappa (y)$-homomorphism $\kappa (x)\to \kappa (b)$. The latter corresponds canonically to a $Y$-morphism $\mathrm{Spec}(\kappa (b))\to X$ (I, 2.4.6), and it follows from (b) that there exists a $Y$-morphism $h:Y^{\prime }\to X$ to which the previous morphism corresponds. We then have $h(b)=x$; if we set $h(a)=x^{\prime }$, then $x^{\prime }$ is a specialisation of $x$, and we have $f(x^{\prime })=f(h(a))=g(a)=y^{\prime }$.

If now $Y$ is locally Noetherian and $f$ of finite type, hypothesis (b) implies first that $f$ is separated, by virtue of (7.2.3), the diagonal morphism $X\to X{\times }_{Y}X$ being quasi-compact (7.2.4). Moreover, to verify that $f$ is proper, it suffices to show that $f_{(Y^{\prime \prime })}:X_{(Y^{\prime \prime })}\to Y^{\prime \prime }$ is closed for every $Y$-prescheme Y'' of finite type, taking (5.6.3) into account. Since Y'' is then locally Noetherian, we may resume the reasoning made in the first case, taking for $Y^{\prime }$ a spectrum of a discrete valuation ring and applying (7.1.9) instead of (7.1.4).

Remarks.

(i) When $Y$ is an arbitrary prescheme and $f$ a separated morphism, for $f$ to be universally closed, it suffices that condition (b) or condition (c) be satisfied for the valuation rings $A$ that are complete and whose residue field is algebraically closed; this follows from the proof above and from (7.1.4).

(ii) From criterion (c) of (7.3.8) we deduce a new proof of the fact that a projective morphism $X\to Y$ is closed (5.5.3), closer to the classical methods. One may indeed suppose $Y$ affine, and consequently $X$ identified with a closed subprescheme of a projective bundle ${\mathbb{P}}_Y^n$ (5.3.3); to prove that $X\to Y$ is closed, it suffices to verify that the structure morphism ${\mathbb{P}}_Y^n\to Y$ is closed, and criterion (c) of (7.3.8), combined with (4.1.3.1), shows that we are reduced to proving the following fact: if $Y$ is the spectrum of a valuation ring $A$ with field of fractions $K$, every point of ${\mathbb{P}}_Y^n$ with values in $K$ comes (by restriction to the generic point of $Y$) from a point of ${\mathbb{P}}_Y^n$ with values in $A$. Indeed, every invertible ${\mathcal{O}}_Y$-module is trivial (I, 2.4.8); therefore, by (4.2.6), a point of ${\mathbb{P}}_Y^n$ with values in $K$ is identified with a class of elements $(\zeta c_0,\zeta c_1,\cdots ,\zeta c_n)$ of $K$, where $\zeta \ne 0$ and the $c_i$ are elements of $K$ not all zero. Now, by multiplying the $c_i$ by an element of $A$ of

suitable valuation, one may suppose that the $c_i$ all belong to $A$ and that at least one of them is invertible. But then (4.2.6) the system $(c_0,\cdots ,c_n)$ also defines a point of ${\mathbb{P}}_Y^n$ with values in $A$, which proves our assertion.

(iii) The criteria (7.2.3) and (7.3.8) are especially convenient when one considers the data of a $Y$-prescheme $X$ as equivalent to the data of the functor

$$X(Y^{\prime })={\mathrm{Hom}}_Y(Y^{\prime },X)$$

in a $Y$-prescheme $Y^{\prime }$; these criteria will allow us, for example, to prove that under certain conditions the "Picard schemes" are proper.

Corollary.

Let $Y$ be an integral scheme (resp. an integral locally Noetherian scheme), $X$ an integral scheme, and $f:X\to Y$ a dominant morphism.

(i) If $f$ is quasi-compact and universally closed, every valuation ring whose field of fractions is the field $R(X)$ of rational functions on $X$ and which dominates a local ring of $Y$ also dominates a local ring of $X$.

(ii) Conversely, suppose $f$ of finite type, and that the property stated in (i) is satisfied by every valuation ring (resp. every discrete valuation ring) having $R(X)$ as field of fractions. Then $f$ is proper.

Proof. Note first that the hypotheses imply, in any case, that $f$ is separated (I, 5.5.9).

(i) Let $K=R(Y)$, $L=R(X)$, $y$ a point of $Y$, $A$ a valuation ring having $L$ as field of fractions and dominating ${\mathcal{O}}_y$; the injection ${\mathcal{O}}_y\to A$ defines a morphism $h$ of $Y^{\prime }=\mathrm{Spec}(A)$ into $Y$ (I, 2.4.4) such that $h(a)=y$, in which we denote by $a$ the closed point of $Y^{\prime }$; moreover, if $\eta$ is the generic point of $Y$, which is also that of $\mathrm{Spec}({\mathcal{O}}_y)$, one has $h(b)=\eta$, denoting by $b$ the generic point of $Y^{\prime }$ (since $K\subset L$ by hypothesis). If $\xi$ is the generic point of $X$, one has $\kappa (\xi )=\kappa (b)=L$ by hypothesis, whence a $Y$-morphism $g:\mathrm{Spec}(L)\to X$ such that $g(b)=\xi$; by virtue of (7.3.8), $g$ comes from a $Y$-morphism $g^{\prime }:Y^{\prime }\to X$. If $x=g^{\prime }(a)$, it is clear that $A$ dominates ${\mathcal{O}}_x$.

(ii) The question being local on $Y$, we may always suppose $Y$ affine (resp. affine and Noetherian). Since $f$ is of finite type, we may apply Chow's lemma (5.6.1) in both cases. There is then a projective morphism $p:P\to Y$, an immersion morphism $j:X^{\prime }\to P$, and a projective, surjective, and birational morphism $g:X^{\prime }\to X$ (with $X^{\prime }$ integral) such that the diagram

   P ←──j── X'
   │        │
  p│       g│
   ↓        ↓
   Y ←──f── X

commutes. It suffices to prove that $j$ is a closed immersion, for then $f\circ g=p\circ j$ will be a projective morphism, hence proper, and since $g$ is surjective, $f$ will also be proper (5.4.3). Let $Z$ be the closed reduced subprescheme of $P$ having $j\overline{X^{\prime }}$ as underlying space (I, 5.2.1); since $X^{\prime }$ is integral, $j$ factors as $i\circ h$, where $i:Z\to P$ is the canonical injection, $h:X^{\prime }\to Z$ a dominant open immersion (I, 5.2.3), and $Z$ is integral;

moreover, $Z$ is projective over $Y$, and we see that we may restrict to the case where $P$ is integral and $j$ dominant and birational, and everything reduces to seeing that $j$ is surjective. So let $z\in P$; ${\mathcal{O}}_z$ is an integral (resp. integral and Noetherian) local ring whose field of fractions is

  L = R(P) = R(X') = R(X).

We may restrict to the case where $z$ is not the generic point of $P$. There is consequently (7.1.2 and 7.1.7) a valuation ring (resp. a discrete valuation ring) $A$ having $L$ as field of fractions and dominating ${\mathcal{O}}_z$. A fortiori, $A$ dominates ${\mathcal{O}}_y$, putting $y=p(z)$, and by hypothesis there is then an $x\in X$ such that $A$ dominates ${\mathcal{O}}_x$. Since $g$ is proper, the first part of the proof shows that $A$ also dominates some ${\mathcal{O}}_{x^{\prime }}$ for an $x^{\prime }\in X^{\prime }$; it follows that ${\mathcal{O}}_z$ and ${\mathcal{O}}_{j(x^{\prime })}={\mathcal{O}}_{x^{\prime }}$ are allied (I, 8.1.4), and since $P$ is a scheme, this entails $z=j(x^{\prime })$ (I, 8.2.2), which concludes the proof.

Corollary.

Let $X$, $Y$ be two integral schemes, $f:X\to Y$ a dominant, quasi-compact and universally closed morphism. Suppose moreover $Y$ affine of (integral) ring $B$. Then $\Gamma (X,{\mathcal{O}}_X)$ is canonically isomorphic to a subring of the integral closure of $B$ in $R(X)$.

Proof. Indeed (I, 8.2.1.1), $\Gamma (X,{\mathcal{O}}_X)$ is identified with the intersection of the ${\mathcal{O}}_x$ for $x\in X$; by virtue of (7.3.10), (7.1.2) and (7.1.3), $\Gamma (X,{\mathcal{O}}_X)$ is consequently contained in the intersection of the valuation rings containing $B$ and having $R(X)$ as field of fractions; the conclusion then follows from (7.1.3).

Remarks.

Under the hypotheses of (7.3.11), and when one supposes that $R(X)$ is an extension of finite type of $R(Y)$, one will be able in many cases to conclude that $\Gamma (X,{\mathcal{O}}_X)$ is a module of finite type over the ring $B=\Gamma (Y,{\mathcal{O}}_Y)$. This will be the case for example when $B$ is an algebra of finite type over a field, for one knows then that the integral closure of $B$ in an extension of finite type of its field of fractions is a $B$-module of finite type ([13], t. I, p. 267, th. 9); the conclusion then follows from (7.3.11) and from the fact that $B$ is Noetherian.

In particular, a scheme $X$ proper and affine over a field $K$ is finite. Indeed, by virtue of (1.6.4), (5.4.6) and (I, 6.4.4, c)), one may restrict to the case where $X$ is reduced. Furthermore, it will suffice to prove that each of the closed subpreschemes of $X$ having for underlying space an irreducible component of $X$ (of which there are finitely many) is finite over $K$, so that (taking (5.4.5) into account) one is finally reduced to the case where $X$ is integral. But then the result follows from the remarks made above.

In chapter III we shall recover this last proposition by other methods and as a consequence of more general results, showing that if $f:X\to Y$ is proper and $Y$ locally Noetherian, $f_{\ast }(\mathcal{F})$ is coherent for every coherent ${\mathcal{O}}_X$-module $\mathcal{F}$ (III, 4.4.2).

Let us note finally that the criterion (7.3.10) is taken as the definition of proper morphisms in classical algebraic geometry. We have mentioned it only for that reason, criterion (7.3.8) seeming more manageable in all the applications known to us.

7.4. Algebraic curves and function fields of dimension 1

The aim of this number is to show how the classical notion of algebraic curve (as it is presented, for example, in the book of C. Chevalley [23]) is formulated in the language of schemes. Throughout this number, $k$ denotes a field, all schemes considered are $k$-schemes of finite type, and all morphisms are $k$-morphisms.

Proposition.

Let $X$ be a prescheme of finite type over $k$ (hence Noetherian); let $x_i$ ($1\le i\le n$) be the generic points of the irreducible components $X_i$ of $X$, and let $K_i=\kappa (x_i)$ ($1\le i\le n$). The following conditions are equivalent:

(a) Each of the $K_i$ is an extension of $k$ whose transcendence degree is equal to 1.

(b) For every closed point $x$ of $X$, the local ring ${\mathcal{O}}_x$ is of dimension 1 (7.1.5).

(c) The irreducible closed subsets of $X$ distinct from the $X_i$ are the closed points of $X$.

Proof. Since $X$ is quasi-compact, every irreducible closed subset $F$ of $X$ contains a closed point (0, 2.1.3). By virtue of (I, 2.4.2), there is a bijective correspondence between the prime ideals of ${\mathcal{O}}_x$ and the irreducible closed subsets of $X$ containing $x$ (I, 1.1.14); the equivalence of (b) and (c) follows immediately. On the other hand, if ${\mathfrak{p}}_{\alpha }$ ($1\le \alpha \le r$) are the minimal prime ideals of the Noetherian local ring ${\mathcal{O}}_x$, the local rings ${\mathcal{O}}_x/{\mathfrak{p}}_{\alpha }$ are integral and have for fields of fractions the $K_i$ such that $x\in X_i$. Furthermore, we know ([1], p. 4-06, th. 2) that the dimension of a local integral $k$-algebra of finite type is equal to the transcendence degree over $k$ of its field of fractions. Finally, the dimension of ${\mathcal{O}}_x$ is the upper bound of the dimensions of the ${\mathcal{O}}_x/{\mathfrak{p}}_{\alpha }$; now, condition (a) implies that these dimensions are equal to 1, so (a) implies (b); conversely, if ${\mathcal{O}}_x$ is of dimension 1, none of the ${\mathfrak{p}}_{\alpha }$ can be equal to the maximal ideal of ${\mathcal{O}}_x$, otherwise ${\mathcal{O}}_x$ would be of dimension 0; therefore each of the ${\mathcal{O}}_x/{\mathfrak{p}}_{\alpha }$ is of dimension 1, which shows that (b) entails (a).

We note that under the conditions of (7.4.1) the set $X$ is empty or infinite, as follows immediately from (I, 6.4.4).

Definition.

We call an algebraic curve over $k$ a nonempty algebraic scheme over $k$ satisfying the conditions of (7.4.1).

In the language of dimension, which will be introduced in chapter IV, this is expressed by saying that an algebraic curve over $k$ is a nonempty algebraic $k$-scheme all of whose irreducible components are of dimension 1.

We note that if $X$ is an algebraic curve over $k$, the closed reduced subpreschemes $X_i$ ($1\le i\le n$) of $X$ having for underlying spaces the irreducible components of $X$ are algebraic curves over $k$.

Corollary.

Let $X$ be an irreducible algebraic curve. The only non-closed point of $X$ is its generic point. The closed subsets of $X$ distinct from $X$ are the finite sets of closed points; these are also the only subsets of $X$ that are not everywhere dense.

Proof. If a point $x\in X$ is not closed, its closure in $X$ is an irreducible closed subset of $X$, hence necessarily the whole of $X$ by virtue of (7.4.1), and consequently

$x$ is the generic point of $X$. A closed subset $F$ of $X$ distinct from $X$ cannot contain the generic point of $X$, so all its points are closed (in $X$, and a fortiori in $F$); by considering the closed reduced subprescheme of $X$ having $F$ as underlying space (I, 5.2.1), it follows therefore from (I, 6.2.2) that $F$ is finite and discrete. The closure in $X$ of an infinite subset of $X$ is therefore necessarily equal to $X$.

When $X$ is an arbitrary algebraic curve, applying (7.4.3) to the irreducible components of $X$ shows that the only non-closed points of $X$ are the generic points of those components.

Corollary.

Let $X$ and $Y$ be two irreducible algebraic curves over $k$, and $f:X\to Y$ a $k$-morphism. For $f$ to be dominant, it is necessary and sufficient that $f^{-1}(y)$ be finite for every $y\in Y$.

Proof. Indeed, if $f$ is not dominant, $f(X)$ is necessarily a finite subset of $Y$ by virtue of (7.4.3), so it is not possible that $f^{-1}(y)$ be finite for every point of $Y$, since otherwise $X$ would be finite, which is absurd (7.4.1). Conversely, if $f$ is dominant, for every $y\in Y$ distinct from the generic point $\eta$ of $Y$, $f^{-1}(y)$ is closed in $X$ since $y$ is closed in $Y$ (7.4.3); on the other hand, by hypothesis, $f^{-1}(y)$ does not contain the generic point $\xi$ of $X$, hence is finite by virtue of (7.4.3). Finally, to see that when $f$ is dominant, $f^{-1}(\eta )$ is finite, one notes that the fibre $f^{-1}(\eta )$ is an irreducible scheme of finite type over $\kappa (\eta )$, with generic point $\xi$ (I, 6.3.9 and 6.4.11). Since $\kappa (\xi )$ and $\kappa (\eta )$ are extensions of finite type of $k$, of the same transcendence degree 1, $\kappa (\xi )$ is necessarily an extension of finite degree of $\kappa (\eta )$, hence $\xi$ is closed in $f^{-1}(\eta )$ (I, 6.4.2), and $f^{-1}(\eta )$ is consequently reduced to the point $\xi$.

We shall see in chapter III that a proper morphism $f:X\to Y$ of Noetherian preschemes such that $f^{-1}(y)$ is finite for every $y\in Y$ is necessarily finite; it will then follow from (7.4.4) that a proper dominant morphism from an irreducible algebraic curve to an algebraic curve is finite.

Corollary.

Let $X$ be an algebraic curve over $k$. For $X$ to be regular, it is necessary and sufficient that $X$ be normal, or again that the local rings of its closed points be discrete valuation rings.

Proof. This follows immediately from condition (b) of (7.4.1) and from (7.1.6).

Corollary.

Let $X$ be a reduced algebraic curve, and $\mathcal{A}$ a reduced coherent $\mathcal{O}(X)$-algebra; then the integral closure $X^{\prime }$ of $X$ relative to $\mathcal{A}$ (6.3.4) is a normal algebraic curve, and the canonical morphism $X^{\prime }\to X$ is finite.

Proof. The fact that $X^{\prime }\to X$ is finite follows from (6.3.10); $X^{\prime }$ is therefore an algebraic $k$-scheme; moreover, if $x_i$ ($1\le i\le n$) are the generic points of the irreducible components of $X$, $x_j^{\prime }$ ($1\le j\le m$) those of the irreducible components of $X^{\prime }$, each of the $\kappa (x_j^{\prime })$ is a finite algebraic extension of one of the $\kappa (x_i)$ (6.3.6), hence has transcendence degree 1 over $k$. $X^{\prime }$ is thus indeed an algebraic curve over $k$, and moreover one knows that $X^{\prime }$ is a sum of a finite number of integral and normal schemes (6.3.6 and 6.3.7).

(7.4.7)

We say that an algebraic curve $X$ over $k$ is complete if it is proper over $k$.

Corollary.

For a reduced algebraic curve $X$ over $k$ to be complete, it is necessary and sufficient that its normalisation $X^{\prime }$ be so.

Proof. Indeed, the canonical morphism $f:X^{\prime }\to X$ is then finite (7.4.6), hence proper (6.1.11) and surjective (6.3.8); if $g:X\to \mathrm{Spec}(k)$ is the structure morphism, $g$ and $g\circ f$ are therefore proper simultaneously, as follows from (5.4.2, (ii)) and (5.4.3, (ii)), $g$ being separated by hypothesis.

Proposition.

Let $X$ be a normal algebraic curve over $k$, and $Y$ a proper algebraic $k$-scheme over $k$. Then every $k$-rational map from $X$ to $Y$ is everywhere defined, in other words is a morphism.

Proof. Indeed, it follows from (7.3.7) that at the points $x\in X$ where such a map is not defined, the dimension of ${\mathcal{O}}_x$ would have to be $\ge 2$, so the set of such points is empty; the last assertion comes from (I, 7.2.3).

Corollary.

A normal algebraic curve over $k$ is quasi-projective over $k$.

Proof. Since $X$ is a sum of a finite number of integral and normal algebraic curves (6.3.8), one may restrict to the case where $X$ is integral (5.3.6). Since $X$ is quasi-compact, it is covered by a finite number of affine opens $U_i$ ($1\le i\le n$), and since each of these is of finite type over $k$, there exist an integer $n_i$ and a $k$-immersion $f_i:U_i\to {\mathbb{P}}_k^{n_i}$ (5.3.3 and 5.3.4, (i)). Since $U_i$ is dense in $X$, it follows from (7.4.9) that $f_i$ extends to a $k$-morphism $g_i:X\to {\mathbb{P}}_k^{n_i}$, whence a $k$-morphism $g=(g_1,\cdots ,g_n{)}_k$ of $X$ into the product $P$ of the ${\mathbb{P}}_k^{n_i}$ over $k$. Moreover, for each index $i$, since the restriction of $g_i$ to $U_i$ is an immersion, so too is the restriction of $g$ to $U_i$ (I, 5.3.14). Since the $U_i$ cover $X$ and $g$ is separated (I, 5.5.1, (v)), $g$ is an immersion of $X$ into $P$ (I, 8.2.8). Since the Segre morphism (4.3.3) gives an immersion of $P$ into a ${\mathbb{P}}_k^N$, this completes the proof that $X$ is quasi-projective.

Corollary.

A normal algebraic curve $X$ is isomorphic to the scheme induced on an everywhere dense open subset of a normal complete algebraic curve $\hat{X}$, determined up to a unique isomorphism.

Proof. If X_1, X_2 are two normal complete curves, it follows immediately from (7.4.9) that every isomorphism of an open U_1 dense in X_1 onto an open U_2 dense in X_2 extends in a unique way to an isomorphism of X_1 onto X_2; whence the uniqueness assertion. To prove the existence of $\hat{X}$, it suffices to remark that one may regard $X$ as a subscheme of a projective bundle ${\mathbb{P}}_k^n$ (7.4.10). Let $\bar{X}$ be the closure of $X$ in ${\mathbb{P}}_k^n$ (I, 9.5.11); since $X$ is induced by $\bar{X}$ on an open dense in $\bar{X}$ (I, 9.5.10), the generic points $x_i$ of the irreducible components of $\bar{X}$ are also those of the irreducible components of $X$, and the $\kappa (x_i)$ are the same for these two schemes, so (7.4.1) $\bar{X}$ is an algebraic curve over $k$, which is reduced (I, 9.5.9) and projective over $k$ (5.5.1), hence complete (5.5.3). Let us then take for $\hat{X}$ the normalisation of $\bar{X}$, which is again complete (7.4.8); moreover, if $h:\hat{X}\to \bar{X}$ is the canonical morphism, the restriction of $h$ to $h^{-1}(X)$ is an isomorphism onto $X$ since $X$ is normal (6.3.4), and since $h^{-1}(X)$ contains the generic points of the irreducible components of $\hat{X}$ (6.3.8), it is dense in $\hat{X}$, which concludes the proof.

Remark.

We shall show in chapter V that the conclusion of (7.4.10) still holds without supposing the curve normal (nor even reduced); we shall also show that for an algebraic curve (reduced or not) to be affine, it is necessary and sufficient that its (reduced) irreducible components be not complete.

Corollary.

Let $X$ be a normal irreducible curve with field $R(X)=K$, $Y$ an integral complete curve with field $L=R(Y)$. There is a canonical bijective correspondence between dominant $k$-morphisms $X\to Y$ and $k$-monomorphisms $L\to K$.

Proof. By virtue of (7.4.9), the $k$-rational maps from $X$ to $Y$ are identified with the $k$-morphisms $u:X\to Y$. The dominant morphisms $u$ being characterised by the fact that $u(x)=y$ (denoting by $x$ and $y$ the respective generic points of $X$ and $Y$), the corollary follows from these remarks and from (I, 7.1.13).

(7.4.14)

One may sharpen the result of (7.4.13) when one takes for $Y$ the projective line ${\mathbb{P}}_k^1=\mathrm{Proj}(k[T_0,T])$, $T_0$ and $T$ being two indeterminates. This is indeed an integral scheme (2.4.4), and the scheme induced on the open $D_{+}(T_0)$ of $Y$ is isomorphic to $\mathrm{Spec}(k[T])$ (2.3.6), so the generic point of $Y$ is the ideal (0) of k[T] and its field of rational functions $k(T)$, which shows that $Y$ is a complete algebraic curve over $k$. Moreover, the only graded prime ideal of $S=k[T_0,T]$ containing $T_0$ and distinct from $S_{+}$ is the principal ideal $(T_0)$, so the complement of $D_{+}(T_0)$ in $Y={\mathbb{P}}_k^1$ reduces to a single closed point, called the "point at infinity", which we denote $\infty$ (for a general study of the relations between vector bundles and projective bundles, see 8.4). With these notations:

Corollary.

Let $X$ be a normal irreducible curve with field $R(X)=K$. There exists a canonical bijective map of $K$ onto the set of morphisms $u$ of $X$ into ${\mathbb{P}}_k^1$ distinct from the constant morphism of value $\infty$. For $u$ to be dominant, it is necessary and sufficient that the corresponding element of $K$ be transcendental over $k$.

Proof. This statement follows immediately from (7.4.9) and from the

Lemma.

Let $X$ be an integral prescheme over $k$, and let $K=R(X)$ be its field of rational functions. There exists a canonical bijective map of the set $K$ onto the set of rational maps $u$ of $X$ into ${\mathbb{P}}_k^1$ distinct from the constant morphism of value $\infty$. For such a rational map to be dominant, it is necessary and sufficient that the corresponding element of $K$ be transcendental over $k$.

First of all, the rational maps of $X$ into ${\mathbb{P}}_k^1$ correspond bijectively to the points of ${\mathbb{P}}_k^1$ with values in the extension $K$ of $k$ (I, 7.1.12). If such a point is localised (I, 3.4.5) at the generic point of ${\mathbb{P}}_k^1$, the corresponding rational map is evidently dominant. In the contrary case, since every point of ${\mathbb{P}}_k^1$ distinct from the generic point is closed (7.4.3), the image of the domain of definition $U$ of $u$ by the unique morphism $U\to {\mathbb{P}}_k^1$ of the class $u$ (I, 7.2.2) reduces to a closed point $y$ of ${\mathbb{P}}_k^1$, and this morphism (which is not necessarily everywhere defined in $X$) is therefore not dominant; by abuse of language, one then says that the rational map $u$ is "constant, of value $y$". It remains to put in bijective correspondence the points of ${\mathbb{P}}_k^1$ with values in $K$ of locality (I, 3.4.5) distinct from $\infty$ and the elements of $K$, and to verify that the locality of such a point is the generic point of ${\mathbb{P}}_k^1$ if and only if it corresponds to an

element transcendental over $k$. Now, this verification is immediate from (4.2.6, example 1°).

Corollary.

Let $X$, $Y$ be two algebraic curves over $k$, normal, complete and irreducible; let $K=R(X)$, $L=R(Y)$ be their fields. There is a canonical bijective correspondence between the set of $k$-isomorphisms $X\stackrel{\sim }{\to }Y$ and the set of $k$-isomorphisms $L\stackrel{\sim }{\to }K$.

Proof. This is an evident consequence of (7.4.13).

(7.4.17)

The corollary (7.4.16) shows that an algebraic curve over $k$, normal, complete and irreducible, is determined by its field of rational functions $K$ up to a unique isomorphism; by definition, $K$ is an extension of finite type of $k$, of transcendence degree 1 (what is classically called a field of algebraic functions of one variable). Moreover:

Proposition.

For every extension $K$ of $k$, of finite type and of transcendence degree 1, there exists a normal, complete and irreducible algebraic curve $X$ such that $R(X)=K$ (determined up to unique isomorphism). The set of local rings of $X$ is identified (I, 8.2.1) with the set consisting of $K$ and the valuation rings containing $k$ and having $K$ as field of fractions.

Proof. Indeed, $K$ is an extension of finite degree of a pure transcendental extension $k(T)$ of $k$, which is identified, as we have seen, with the field of rational functions of the projective line $Y={\mathbb{P}}_k^1$. Let $X$ be the integral closure of $Y$ relative to $K$ (6.3.4); $X$ is a normal algebraic curve with field $K$ (6.3.7), and it is complete since the morphism $X\to Y$ is finite (7.4.6). The local rings ${\mathcal{O}}_x$ of $X$ are: the field $K$ when $x$ is the generic point; if $x$ is distinct from the generic point, ${\mathcal{O}}_x$ is a discrete valuation ring containing $k$ and having $K$ as field of fractions (7.4.5). Conversely, let $A$ be such a ring; since the morphism $X\to \mathrm{Spec}(k)$ is proper and $A$ dominates $k$, $A$ dominates a local ring ${\mathcal{O}}_x$ of $X$ (7.3.10); the latter being a valuation ring having $K$ as field of fractions, is therefore necessarily equal to $A$.

Remarks.

It follows from (7.4.16) and (7.4.18) that the data of an algebraic curve over $k$, normal, complete and irreducible, is essentially equivalent to the data of an extension $K$ of $k$, of finite type and of transcendence degree 1. We note that if $k^{\prime }$ is an extension of the base field $k$, $X{\otimes }_k{k}^{\prime }$ will again be a complete algebraic curve over $k^{\prime }$ (5.4.2, (iii)), but in general it will be neither reduced nor irreducible. It will be so, however, if $K$ is a separable extension of $k$ and $k$ is algebraically closed in $K$ (which is expressed, in a classical terminology that we shall not follow, by saying that $K$ is a "regular extension" of $k$). But even in this case, it can happen that $X{\otimes }_k{k}^{\prime }$ is not normal. The reader will find details on these questions in chapter IV.