Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed? - MathOverflow

Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed?

2010-06-18T03:38:50Z

My question is a doubt I had in the last point to the first answer to this MO question - http://mathoverflow.net/questions/14314/algebraic-topologies-like-the-zariski-topology

Can one associate a Riemann surface to any arbitrary field extension? The statement there says its true, it is the (equivalence class of) valuations which fix the base field. The result seems too fascinating to be true. Are there any extra hypotheses?

I am aware of the result that the valuations of C(x)/C form the Riemann sphere (counting the exponent of any (x-a) in each rational function gives a correspondence between points on the complex plane (minus infinity) and valuations.)

Any references?

Answer by Charles Siegel for Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed?

2010-06-18T04:08:56Z

You need the field extension to have transcendence degree 1 over $\mathbb{C}$ to get a Riemann surface. More generally, you can get an algebraic curve at least for every transcendence degree 1 extension over an algebraically closed field. Chapter I of Hartshorne does this in detail, I don't know an analytic approach.

Answer by Tom Goodwillie for Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed?

2010-06-18T05:13:54Z

Zariski introduced an abstract notion of Riemann surface associated to, for example, a finitely generated field extension $K/k$. It's a topological space whose points are equivalence classes of valuations of $K$ that are trivial on $k$, or equivalently valuation rings satisfying $k\subset R_v\subset K$. If $A$ is a finitely generated $k$-algebra inside $K$ then those $R_v$ which contain $A$ form an open set.

In the case of a (finitely generated and) transcendence degree 1 extension all of these valuation rings are the familiar DVRs -- local Dedekind domains -- and they serve to identify the points in the unique complete nonsingular curve with this function field. (There is also the trivial valuation with $R_v=K$, which corresponds to the generic point of that curve.)

In higher dimensions there are lots of complete varieties to contend with -- you can keep blowing up. Also there are more possibilities for valuations. Most of the valuation rings are not Noetherian. A curve in a surface gives you a discrete valuation ring, consisting of those rational functions which can meaningfully be restricted to rational functions on the curve: those which do not have a pole there. A point on a curve on a surface gives you a valuation whose ring consists of those functions which do not have a pole all along the curve, and which when restricted to the curve do not have a pole at the given point. The value group is $\mathbb Z\times \mathbb Z$ lexicographically ordered. A point on a transcendental curve in a complex surface, or more generally a formal (power series) curve in a surface gives you a valuation by looking at the order of vanishing; the value group is a subgroup of $\mathbb R$.

This space of valuations has something of the flavor of Zariski's space of prime ideals in a ring: it is compact but not Hausdorff, for example. It can be thought of as the inverse limit, over all complete surfaces $S$ with this function field, of the space (Zariski topology) of points $S$.

Answer by stjc for Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed?

2012-02-24T13:19:45Z

For compact Riemann surface, the algebraic approach and the analytic approach is the same(Chow's lemma), so the algebraic answer is sufficient for you. i.e. For every extension finitely generated over \mathbb{C} which has transcendental degree 1(up to isomorphism), there is a unique Riemann surface you want.