# From function field to curve: non-algebraically closed ground field and functor of points

This question concerns the (re)construction of a smooth projective curve $C$ over a field $k$, using the function field $K=K(C)$ of $C$. When $k$ is algebraically closed, this is described for instance in Hartshorne, I.6.

My questions are the following:

1. At a few points in the construction, Hartshorne uses that $k$ is algebraically closed, at other points at least that $k$ is infinite. Can one get around this, and use non-algebraically closed or even finite ground fields for reconstructing a curve from its function field?

2. In constructing a smooth projective curve from a finitely generated field $K$ of transcendence degree $1$ over $k$, one takes as the underlying points of the curve the discrete valuations on $K$, defines a topology on this set by declaring closed sets to be fintie or the whole set, and then building an appropriate sheaf of rings. Then one checks that the result is a smooth projective curve. Is there a slick way to describe the functor of points for this curve, instead of the associated locally ringed space?

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1. The construction holds for any base field $k$. But if $k$ is not perfect, you get a projective curve which is normal but not necessarily smooth. For example, if $k$ has characteristic $p>2$ and $t\in k$ is a not a $p$-th power in $k$, consider the function field $K=k(x,y)$ defined by the relation $y^2=x^P-t$. The curve you get is not smooth, and there is no projective smooth curve over $k$ whose function field is $K$. Note that smooth curves are always normal, and the converse is true if $k$ is perfect.
2. Pick any transcendental element $x\in K$. Then $K$ is finite over $k(x)$. Let $A$ be the integral closure of $k[x]$ in $K$ and let $B$ be the integral closure of $k[1/x]$ in $K$. Then the localizations $A_x$ and $B_{1/x}$ are both equal to the integral closure of $k[x, 1/x]$ in $K$. Therefore we can glue the affine curves ${\rm Spec} A$ and ${\rm Spec} B$ along ${\rm Spec} A_x$ and get a curve $C$. By constuction $C$ is normal and integral, with field of functions $K$, and there is finite morphism $C\to \mathbb P^1={\rm Spec} k[x] \cup {\rm Spec} k[1/x]$ (obtained by glueing ${\rm Spec} A\to {\rm Spec} k[x]$ and ${\rm Spec} B\to {\rm Spec} (k[1/x])$). Hence $C$ is its projective, and it is the projective normal curve associated to $K$.
3. As a bonus, one also has a nice correspondance betweeen finite morphisms of curves and extensions of function fields of one variable. If $f : C\to D$ is a finite morphism of projective normal integral curves over $k$, then it induces a finite extension $k(D)\to k(C)$. One can show that this establises a anti-equivalence from the category of integral normal projective curves over $k$ (where morphisms are finite morphisms of $k$-curves) to the category of function fields of one variable over $k$ (where morphisms are morphisms of $k$-extensions).
In (1), let $l$ be any prime different from $p$ (so $l = 2$ or $l = 3$ is sufficient) and use $y^l = x^p - t$. Then (1) goes through for $p = 2$. –  KConrad Jul 20 '10 at 22:16
Is it also true that $C$ can be characterized by a universal property? Something like the universal example of a scheme $X$ with a proper morphism to $Spec k$ and a morphism from $Spec K$ such that the composed morphism $Spec K\to Spec k$ is what it should be? –  Tom Goodwillie Jul 20 '10 at 22:17
I agree. Any morphism $f$ of $k$-schemes from ${\rm Spec } K$ to a proper $k$-scheme $X$ factorizes uniquely as the canonical morphism ${\rm Spec} K\to C$ followed by a $k$-morphism $C\to X$. –  Qing Liu Jul 20 '10 at 22:33
A regular scheme or variety is always normal. A locally noetherian normal scheme of dimension 1 (e.g. normal curve over any field) is always regular. To describe the points of $C$, (2) gives you a concrete method. Of course, the valuation theory as in Hartshorne also decribes the points of $C$. But I don't know how to decribe the functor of points of $C$ directly from the field $K$. If $X$ and $Y$ are birational integral varieties over $k$, you can not distinguish the dominant morphisms for $X$ and $Y$ to $C$ in terms of $K$. –  Qing Liu Jul 21 '10 at 9:31