From function field to curve: non-algebraically closed ground field and functor of points - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:05:11Z http://mathoverflow.net/feeds/question/32692 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32692/from-function-field-to-curve-non-algebraically-closed-ground-field-and-functor-o From function field to curve: non-algebraically closed ground field and functor of points A. Pascal 2010-07-20T20:35:56Z 2010-07-20T21:54:35Z <p>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.</p> <p>My questions are the following:</p> <ol> <li><p>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?</p></li> <li><p>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?</p></li> </ol> http://mathoverflow.net/questions/32692/from-function-field-to-curve-non-algebraically-closed-ground-field-and-functor-o/32702#32702 Answer by Qing Liu for From function field to curve: non-algebraically closed ground field and functor of points Qing Liu 2010-07-20T21:54:35Z 2010-07-20T21:54:35Z <ol> <li><p>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. </p></li> <li><p>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 $K$. </p></li> <li><p>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). </p></li> </ol>