Lemma 50.2.1. Let $k$ be a field. Let $X$ be a curve and $Y$ a proper variety. Let $U \subset X$ be a nonempty open and let $f : U \to Y$ be a morphism. If $x \in X$ is a closed point such that $\mathcal{O}_{X, x}$ is a discrete valuation ring, then there exists an open $U \subset U' \subset X$ containing $x$ and a morphism of varieties $f' : U' \to Y$ extending $f$.

## 50.2 Curves and function fields

In this section we elaborate on the results of Varieties, Section 32.4 in the case of curves.

**Proof.**
This is a special case of Morphisms, Lemma 28.40.5.
$\square$

Lemma 50.2.2. Let $k$ be a field. Let $X$ be a normal curve and $Y$ a proper variety. The set of rational maps from $X$ to $Y$ is the same as the set of morphisms $X \to Y$.

**Proof.**
A rational map from $X$ to $Y$ can be extended to a morphism $X \to Y$ by Lemma 50.2.1 as every local ring is a discrete valuation ring (for example by Varieties, Lemma 32.43.16). Conversely, if two morphisms $f,g: X \to Y$ are equivalent as rational maps, then $f = g$ by Morphisms, Lemma 28.7.10.
$\square$

Lemma 50.2.3. Let $k$ be a field. Let $f : X \to Y$ be a nonconstant morphism of curves over $k$. If $Y$ is normal, then $f$ is flat.

**Proof.**
Pick $x \in X$ mapping to $y \in Y$. Then $\mathcal{O}_{Y, y}$ is either a field or a discrete valuation ring (Varieties, Lemma 32.43.16). Since $f$ is nonconstant it is dominant (as it must map the generic point of $X$ to the generic point of $Y$). This implies that $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective (Morphisms, Lemma 28.8.6). Hence $\mathcal{O}_{X, x}$ is torsion free as a $\mathcal{O}_{Y, y}$-module and therefore $\mathcal{O}_{X, x}$ is flat as a $\mathcal{O}_{Y, y}$-module by More on Algebra, Lemma 15.22.10.
$\square$

Lemma 50.2.4. Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes over $k$. Assume

$Y$ is separated over $k$,

$X$ is proper of dimension $\leq 1$ over $k$,

$f(Z)$ has at least two points for every irreducible component $Z \subset X$ of dimension $1$.

Then $f$ is finite.

**Proof.**
The morphism $f$ is proper by Morphisms, Lemma 28.39.7. Thus $f(X)$ is closed and images of closed points are closed. Let $y \in Y$ be the image of a closed point in $X$. Then $f^{-1}(\{ y\} )$ is a closed subset of $X$ not containing any of the generic points of irreducible components of dimension $1$ by condition (3). It follows that $f^{-1}(\{ y\} )$ is finite. Hence $f$ is finite over an open neighbourhood of $y$ by More on Morphisms, Lemma 36.39.2 (if $Y$ is Noetherian, then you can use the easier Cohomology of Schemes, Lemma 29.21.2). Since we've seen above that there are enough of these points $y$, the proof is complete.
$\square$

Lemma 50.2.5. Let $k$ be a field. Let $X \to Y$ be a morphism of varieties with $Y$ proper and $X$ a curve. There exists a factorization $X \to \overline{X} \to Y$ where $X \to \overline{X}$ is an open immersion and $\overline{X}$ is a projective curve.

**Proof.**
This is clear from Lemma 50.2.1 and Varieties, Lemma 32.42.6.
$\square$

Here is the main theorem of this section. We will say a morphism $f : X \to Y$ of varieties is *constant* if the image $f(X)$ consists of a single point $y$ of $Y$. If this happens then $y$ is a closed point of $Y$ (since the image of a closed point of $X$ will be a closed point of $Y$).

Theorem 50.2.6. Let $k$ be a field. The following categories are canonically equivalent

The category of finitely generated field extensions $K/k$ of transcendence degree $1$.

The category of curves and dominant rational maps.

The category of normal projective curves and nonconstant morphisms.

The category of nonsingular projective curves and nonconstant morphisms.

The category of regular projective curves and nonconstant morphisms.

The category of normal proper curves and nonconstant morphisms.

**Proof.**
The equivalence between categories (1) and (2) is the restriction of the equivalence of Varieties, Theorem 32.4.1. Namely, a variety is a curve if and only if its function field has transcendence degree $1$, see for example Varieties, Lemma 32.20.3.

The categories in (3), (4), (5), and (6) are the same. First of all, the terms “regular” and “nonsingular” are synonyms, see Properties, Definition 27.9.1. Being normal and regular are the same thing for Noetherian $1$-dimensional schemes (Properties, Lemmas 27.9.4 and 27.12.6). See Varieties, Lemma 32.43.16 for the case of curves. Thus (3) is the same as (5). Finally, (6) is the same as (3) by Varieties, Lemma 32.42.4.

If $f : X \to Y$ is a nonconstant morphism of nonsingular projective curves, then $f$ sends the generic point $\eta $ of $X$ to the generic point $\xi $ of $Y$. Hence we obtain a morphism $k(Y) = \mathcal{O}_{Y, \xi } \to \mathcal{O}_{X, \eta } = k(X)$ in the category (1). If two morphisms $f,g: X \to Y$ gives the same morphism $k(Y) \to k(X)$, then by the equivalence between (1) and (2), $f$ and $g$ are equivalent as rational maps, so $f=g$ by Lemma 50.2.2. Conversely, suppose that we have a map $k(Y) \to k(X)$ in the category (1). Then we obtain a morphism $U \to Y$ for some nonempty open $U \subset X$. By Lemma 50.2.1 this extends to all of $X$ and we obtain a morphism in the category (5). Thus we see that there is a fully faithful functor (5)$\to $(1).

To finish the proof we have to show that every $K/k$ in (1) is the function field of a normal projective curve. We already know that $K = k(X)$ for some curve $X$. After replacing $X$ by its normalization (which is a variety birational to $X$) we may assume $X$ is normal (Varieties, Lemma 32.27.1). Then we choose $X \to \overline{X}$ with $\overline{X} \setminus X = \{ x_1, \ldots , x_ n\} $ as in Varieties, Lemma 32.42.6. Since $X$ is normal and since each of the local rings $\mathcal{O}_{\overline{X}, x_ i}$ is normal we conclude that $\overline{X}$ is a normal projective curve as desired. (Remark: We can also first compactify using Varieties, Lemma 32.42.5 and then normalize using Varieties, Lemma 32.27.1. Doing it this way we avoid using the somewhat tricky Morphisms, Lemma 28.51.16.) $\square$

Definition 50.2.7. Let $k$ be a field. Let $X$ be a curve. A *nonsingular projective model of $X$* is a pair $(Y, \varphi )$ where $Y$ is a nonsingular projective curve and $\varphi : k(X) \to k(Y)$ is an isomorphism of function fields.

A nonsingular projective model is determined up to unique isomorphism by Theorem 50.2.6. Thus we often say “the nonsingular projective model”. We usually drop $\varphi $ from the notation. Warning: it needn't be the case that $Y$ is smooth over $k$ but Lemma 50.2.8 shows this can only happen in positive characteristic.

Lemma 50.2.8. Let $k$ be a field. Let $X$ be a curve and let $Y$ be the nonsingular projective model of $X$. If $k$ is perfect, then $Y$ is a smooth projective curve.

**Proof.**
See Varieties, Lemma 32.43.16 for example.
$\square$

Lemma 50.2.9. Let $k$ be a field. Let $X$ be a geometrically irreducible curve over $k$. For a field extension $K/k$ denote $Y_ K$ a nonsingular projective model of $(X_ K)_{red}$.

If $X$ is proper, then $Y_ K$ is the normalization of $X_ K$.

There exists $K/k$ finite purely inseparable such that $Y_ K$ is smooth.

Whenever $Y_ K$ is smooth

^{1}we have $H^0(Y_ K, \mathcal{O}_{Y_ K}) = K$.Given a commutative diagram

\[ \xymatrix{ \Omega & K' \ar[l] \\ K \ar[u] & k \ar[l] \ar[u] } \]of fields such that $Y_ K$ and $Y_{K'}$ are smooth, then $Y_\Omega = (Y_ K)_\Omega = (Y_{K'})_\Omega $.

**Proof.**
Let $X'$ be a nonsingular projective model of $X$. Then $X'$ and $X$ have isomorphic nonempty open subschemes. In particular $X'$ is geometrically irreducible as $X$ is (some details omitted). Thus we may assume that $X$ is projective.

Assume $X$ is proper. Then $X_ K$ is proper and hence the normalization $(X_ K)^\nu $ is proper as a scheme finite over a proper scheme (Varieties, Lemma 32.27.1 and Morphisms, Lemmas 28.42.11 and 28.39.4). On the other hand, $X_ K$ is irreducible as $X$ is geometrically irreducible. Hence $X_ K^\nu $ is proper, normal, irreducible, and birational to $(X_ K)_{red}$. This proves (1) because a proper curve is projective (Varieties, Lemma 32.42.4).

Proof of (2). As $X$ is proper and we have (1), we can apply Varieties, Lemma 32.27.3 to find $K/k$ finite purely inseparable such that $Y_ K$ is geometrically normal. Then $Y_ K$ is geometrically regular as normal and regular are the same for curves (Properties, Lemma 27.12.6). Then $Y$ is a smooth variety by Varieties, Lemma 32.12.6.

If $Y_ K$ is geometrically reduced, then $Y_ K$ is geometrically integral (Varieties, Lemma 32.9.2) and we see that $H^0(Y_ K, \mathcal{O}_{Y_ K}) = K$ by Varieties, Lemma 32.26.2. This proves (3) because a smooth variety is geometrically reduced (even geometrically regular, see Varieties, Lemma 32.12.6).

If $Y_ K$ is smooth, then for every extension $\Omega /K$ the base change $(Y_ K)_\Omega $ is smooth over $\Omega $ (Morphisms, Lemma 28.32.5). Hence it is clear that $Y_\Omega = (Y_ K)_\Omega $. This proves (4). $\square$

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