## 53.2 Curves and function fields

In this section we elaborate on the results of Varieties, Section 33.4 in the case of curves.

Lemma 53.2.1. 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 exist an open $U \subset U' \subset X$ containing $x$ and a morphism of varieties $f' : U' \to Y$ extending $f$.

Proof. This is a special case of Morphisms, Lemma 29.42.5. $\square$

Lemma 53.2.2. 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 53.2.1 as every local ring is a discrete valuation ring (for example by Varieties, Lemma 33.43.8). Conversely, if two morphisms $f,g: X \to Y$ are equivalent as rational maps, then $f = g$ by Morphisms, Lemma 29.7.10. $\square$

Lemma 53.2.3. 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 33.43.8). 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 29.8.7). 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 53.2.4. Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes over $k$. Assume

1. $Y$ is separated over $k$,

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

3. $f(Z)$ 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 29.41.7. 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 37.44.2 (if $Y$ is Noetherian, then you can use the easier Cohomology of Schemes, Lemma 30.21.2). Since we've seen above that there are enough of these points $y$, the proof is complete. $\square$

Lemma 53.2.5. 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 53.2.1 and Varieties, Lemma 33.43.6. $\square$

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

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

2. The category of curves and dominant rational maps.

3. The category of normal projective curves and nonconstant morphisms.

4. The category of nonsingular projective curves and nonconstant morphisms.

5. The category of regular projective curves and nonconstant morphisms.

6. 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 33.4.1. Namely, a variety is a curve if and only if its function field has transcendence degree $1$, see for example Varieties, Lemma 33.20.3.

The categories in (3), (4), (5), and (6) are the same. First of all, the terms “regular” and “nonsingular” are synonyms, see Properties, Definition 28.9.1. Being normal and regular are the same thing for Noetherian $1$-dimensional schemes (Properties, Lemmas 28.9.4 and 28.12.6). See Varieties, Lemma 33.43.8 for the case of curves. Thus (3) is the same as (5). Finally, (6) is the same as (3) by Varieties, Lemma 33.43.4.

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 53.2.2. 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 53.2.1 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 33.27.1). Then we choose $X \to \overline{X}$ with $\overline{X} \setminus X = \{ x_1, \ldots , x_ n\}$ as in Varieties, Lemma 33.43.6. 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 33.43.5 and then normalize using Varieties, Lemma 33.27.1. Doing it this way we avoid using the somewhat tricky Morphisms, Lemma 29.53.16.) $\square$

Definition 53.2.7. 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 53.2.6. 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 53.2.8 shows this can only happen in positive characteristic.

Lemma 53.2.8. 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 33.43.8 for example. $\square$

Lemma 53.2.9. 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 $(X_ K)_{red}$.

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

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

3. Whenever $Y_ K$ is smooth1 we have $H^0(Y_ K, \mathcal{O}_{Y_ K}) = K$.

4. 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 33.27.1 and Morphisms, Lemmas 29.44.11 and 29.41.4). 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 33.43.4).

Proof of (2). As $X$ is proper and we have (1), we can apply Varieties, Lemma 33.27.4 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 28.12.6). Then $Y$ is a smooth variety by Varieties, Lemma 33.12.6.

If $Y_ K$ is geometrically reduced, then $Y_ K$ is geometrically integral (Varieties, Lemma 33.9.2) and we see that $H^0(Y_ K, \mathcal{O}_{Y_ K}) = K$ by Varieties, Lemma 33.26.2. This proves (3) because a smooth variety is geometrically reduced (even geometrically regular, see Varieties, Lemma 33.12.6).

If $Y_ K$ is smooth, then for every extension $\Omega /K$ the base change $(Y_ K)_\Omega$ is smooth over $\Omega$ (Morphisms, Lemma 29.34.5). Hence it is clear that $Y_\Omega = (Y_ K)_\Omega$. This proves (4). $\square$

[1] Or even geometrically reduced.

Comment #2057 by David Hansen on

In Lemma 46.2.5, you should probably say that X is a curve!

Comment #4169 by Che Shen on

In the third paragraph of theorem 0BY1, we conclude that "there is a fully faithful functor (5)→(1)". We only proved it's full, but did not prove it's faithful.

To prove faithfulness, we can add the following before "Conversely, suppose ...":

Suppose $f,g: X \to Y$ gives the same field homomorphism $K(X)\to K(Y)$. By 01RK, $f,g$ are dominant morphisms. So by 0BXN, $f$ and $g$ are equal as ratinal maps. So they agree on an open subset $U \subset X$. Since $Y$ is separated, $f = g$ as morphisms.

(For the last sentence above, I think there should be a lemma in stacks project saying "If $f,g: X \to Y$ agree on an open subset $U \subset X$ and $Y$ is separated, then $f=g$", but I did not find it.)

Comment #5048 by Raphael Rouquier on

Maybe an irreducibility assumption is needed for the curves in Theorem 0BY1?

Comment #5049 by Raphael Rouquier on

Apologies, by definition a curve has the required irreducibility property.

Comment #8811 by Thomas Manopulo on

I think in Lemma 0BXY it should be mentioned that the point x lies in U, no?

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