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?


1 Answer 1

  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).

  • 2
    $\begingroup$ 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 $p = 2$. $\endgroup$
    – KConrad
    Jul 20, 2010 at 22:16
  • 1
    $\begingroup$ 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? $\endgroup$ Jul 20, 2010 at 22:17
  • $\begingroup$ 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 $C\to X$. $\endgroup$
    – Qing Liu
    Jul 20, 2010 at 22:33
  • $\begingroup$ Qing Liu's answer above is very helpful. The fact that normal implies regular in dimension 1 is independent of the ground field, right? So when the ground field is not perfect, the issue is really smoothness as opposed to regularity. Maybe the universal property mentioned above is sufficient for working with the curve. It would still be helpful to describe the functor of points, though. $\endgroup$
    – A. Pascal
    Jul 21, 2010 at 6:24
  • $\begingroup$ 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 $K$. $\endgroup$
    – Qing Liu
    Jul 21, 2010 at 9:31

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