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Riemann showed (not proved rigorously) that there is a correspondence between compact riemann surfaces and algebraic function fields in one variable (does anyone know the year?).

To construct the algebraic function field of the compact riemann surface take the field of meromorphic functions M(M). Main question: Seen as an antiequivalence between categories, what is the inverse functor? Or explicitly, how to construct a riemann surface X from its function field M(X), preferably as a polynomial in one variable with coefficients depending on another variable. If there are singularities on the recovered surface, is there a systematic way to remove them? A simpler but useful question is: how many generators do M(X) have, and how does the genus g(X) depend on M(X)? (notice that i am not restricting X to sit in any particular space).

Here is the part i am not so good at but it shows a possible solution: I think the function field K(X) can always be written as K[x](y)/< P(x,y) > where P is an irreducible polynomial. This would be fine if any function field of X has uniquely this form. Then the equation i am looking for should be the extracted P(x,y)=0. Did this make the problem any simpler? And still, what is the functor inverse to M()?

If this is really impossible maby a parametric representation of X from M(X) is possible.

Reading another post on mathoverflow i think i am asking for higher genus Weierstrass Pe-functions related to the generators, that should be constructed in terms of Riemann theta-functions. As i understand it, for X in the g(X)=1 case K=C(Pe,Pe') (for a given lattice in C), and X is parametrized by Weierstrass Pe and Pe'-functions. But that didn't help me since there were no formulas or references for higher genera. Here is one reference for something that looks like higher genus Pe-functions.

Google books: Symmetries and integrability of difference equations p68-70

Aside: since Riemann showed there is this correspondence, birational equivalence should be the same as biholomorphic equivalence for compact riemann surfaces. I have never even seen a theorem about it. How is the situation for general riemann surfaces?

Note: i will from now on write trdeg(F) for the transcenden degree of a field extension F over K.

A generalization to higher dimensions (background: infered from wikipedia). Is this true (even for singular algebraic varieties or only smooth algebraic varieties)?

Every algebraic variety X over a field K has a function field K(X) that is the field of rational functions on X, is a field extension of K that is finitely generated, and has trdeg(K(X))=dim(X) (both over K). And, all such field extensions of K with finite trdeg(K(X)) are the function field of some algebraic variety X over K.

And, question: is there a categorical antiequivalence between

the the category of algebraic function fields of trdeg(K(X)) variables that are extensions of K, and ring homomorphism as morphisms, and the category of algebraic varieties over K of dimension trdeg(K(X)) and rational functions between them as morphisms?

Finally, does this correspondence hold locally for schemes over a field?

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There is no unique equation P(x,y) = 0. Simple reason: you can always replace x with x + 1. It's like in algebra, where a field extension is an intrinsic thing but it can be generated by many particular elements. –  KConrad May 18 '10 at 3:33
As far as the genus, see mathoverflow.net/questions/152/… . –  Qiaochu Yuan May 18 '10 at 3:37
Corona, this is specific to dim 1 and trdeg 1. Learn alg. curves over an alg. closed field (with singularities, and without, via normalization). Then passage in reverse direction is easier to grock. Yes, Riemann et al. worked in days before algebraic geometry. But honestly, it is easier to pass from algebraic data of function field to algebro-geometric data of plane curve with singularities on to smooth possibly non-planar proj. alg. curve, which can be "analytified" than to follow the historical route and rigorously "desingularize" an "analytic singularity" by elementary analytic tools. –  BCnrd May 18 '10 at 4:29
If it is correct to attribute this correspondence to Riemann (which I'm not completely sure about) then the year would be 1857, in his paper Theorie der Abel'schen Functionen, Journal für die reine und angewandte Mathematik, vol. 54 (1857), pp. 101-155. –  John Stillwell May 18 '10 at 8:22

3 Answers 3

In his paper cited above on Abelian functions, and appealing to his earlier thesis results, Riemann sketches a functor from the category of irreducible plane algebraic curves with rational maps and rational functions, to compact connected complex one manifolds with holomorphic maps and meromorphic functions. [One compactifies the curve as a projective curve, and then desingularizes it as a manifold. Rational maps become holomorphic on the desingularization because of the Riemann extension theorem. Much of this occurs in the book of Miranda.] Riemann then proves that the field of rational functions on the plane curve equals the field of meromorphic functions on the manifold.

Indeed, he shows that if there is a single non constant meromorphic function on the manifold say of degree n, such as one of the plane variables gives, then the entire field of meromorphic functions on the manifold is algebraic of degree at most n over the field obtained by adjoining this one function to the constants. It follows that all holomorphic maps of the manifolds arise from rational maps of the curves and that in particular holomorphic equivalence of the manifolds is the same as birational equivalence of the curves.

This implies that your construction of a (non unique) plane curve from a function field, always yields birationally equivalent curves, hence isomorphic Riemann surfaces. Riemann himself considers the problem of birational equivalence in his paper and determines the lowest degree of a plane polynomial representation for a given Riemann surface in terms of the lowest degree of a map from that surface to the Riemann sphere.

All this is actually proved essentially rigorously in his paper, appealing only to his extension theorem for holomorphic functions. What has been criticized as to rigor is the inverse correspondence that all compact connected complex one manifolds arise from plane curves. The method was to produce harmonic functions in plane regions by the Dirichlet principle, which method was justified by Hilbert and others later, as recorded in the books of Weyl and Siegel and Springer. More modern approaches occur in Gunning, and the article by Cornalba in his Trieste lectures.

As noted above, for higher dimension there exist compact complex manifolds not arising from algebraic varieties, and there exist such examples in Shafarevich, e.g. of compact complex tori which do not have meromorphic function field of the correct transcendence degree. Manifolds which do have such meromorphic function fields, and hence could be algebraic, are called Moishezon manifolds, and he showed they can always be blown up to become algebraic, if I recall correctly.

In that famous Abelsche Functionen paper, Riemann goes on to deduce rigorously his famous inequality, by estimating the rank of a period matrix, assuming only the existence of sufficient meromorphic one forms of 1st and 2nd kinds, i.e. either holomorphic, or having zero residues at every pole. Although his proof of the existence of these forms in the manifold setting relies on his disputed use of the Dirichlet principle, he remarks in section 9 of the paper that one can simply write them down in the case of plane curves, and he actually does so for the holomorphic ones, using the "Poincare" residue principle. He says he could write down the others as well, but will not stop to do so. Such explicit expressions are given in the book on Plane Algebraic Curves of Brieskorn, by way of showing how to represent all cohomology classes on a curve by meromorphic forms of 1st and 2nd kinds. E.g. on the cubic curve y^2 = x(x-1)(x-t), the form x(x-1)dx/y^3 is an elementary form of 2nd kind with one double pole (at (t,0)) but zero residue.

If one grants that Riemann knew how to do this, as he said, then the foundation for his proof of the Riemann inequality is completely provided, and at least for plane curves, there is no need for Hilbert's analytic foundations to bolster Riemann's argument in the complex algebraic case. The 1865 paper of Roch, in which he completes Riemann's argument, rests solely on Green's theorem to compute Riemann's period matrix as a residue integral, hence is completely solid. 17 years later, Brill and Noether, using the same matrix computed by Roch, apparently showed that one can exploit the duality between divisors of form D and K-D to actually give the full proof using only the existence of the integrals of 1st kind. Since that paper was so influential, Roch's residue matrix (occurring in the middle of the second page of his paper) is now usually known as the Brill Noether matrix.

In addition to the functor from curves C to one manifolds X, Riemann also considered two more functors, the symmetric products X^(d) and the Jacobian variety J(X), as well as a natural transformation between them X^(d)--->J(X), called the Abel map. The fibers of this map are the linear series |D|, ("Abel's theorem"), and the derivative of this map is the Roch ("B-N") matrix, (by the fundamental theorem of calculus). Hence the Riemann Roch theorem becomes the assertion that the fibers of the Abel map are non singular as schemes. I.e. the fiber dimension dim |D|, equals the dimension of the kernel of the derivative, d-g+h^0(K-D). This is the formulation of Mattuck and Mayer.

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I have heard the non singularity of the fibers of the abel map attributed to Riemann as well, but have not found it. If this were true, then in conjunction with the full Abel theorem, as observed above, this would already imply the full RRT. Notice also that the full Abel theorem already implies the Riemann inequality, since the fibers of a map from X^(d) to J(X) must have dimension ≥ d-g. –  roy smith Apr 23 '11 at 21:02

Though I don't have it in front of me for the details, check out chapter I of Hartshorne...it's much less terrifying than the rest of the book is for non-algebraic geometry people. Given a function field of transcendent degree one, you can take all the DVR's with field of fractions that field, and they'll form the points of a Riemann surface (actually, smooth projective algebraic curve over an algebraically closed field).

As for polynomials, yes, any such field is the field of fractions of a field of the form $k[x,y]/(f)$ but not uniquely, as KConrad pointed out, for rather simple reasons. But also in Hartshorne chapter I, it's prove that every variety is birational to a hypersurface in affine space, and function fields are the same as birational classes of varieties. As for the singularities that occur, which will occur generically, because most curves of large genus don't embed into the plane smoothly, you can just compute the normalization of the curve.

As Qiaochu said, there's this question which talks about the function field and the genus, and as for generators, the fact that things are birational to singular plane curves will tell you that you can always choose two generators to make things work out.

Now, every variety has a function field, as for every function field having a variety, I believe it's true, but don't have a proof off the top of my head (though that might be that it's 1:30 am and it's obvious when I'm awake). As for the proposed anti-equivalence, you'll need to make it dominant rational maps, and then I believe it's true. Dominant means that the image is dense, and if it fails, then the image might be in the exceptional locus of the next morphism, and composition doesn't work out, because then you don't have a rational map, you have the empty map.

I think I covered most of your questions...but really, chapter I of Hartshorne would be a good read for you.

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@Charles: The aspect of Ch. I which is painful is exactly the step of making the smooth curve model...because there Hartshorne introduces a rather wacky (to a beginner) notion of "abstract curve", and doesn't have tools at that point to discuss normalization in a meaningful way. Getting rid of the singularities is (I think) the most serious part of the leap from the function field back to the Riemann surface. It's also subtle in a purely analytic approach if one avoids the use of algebraic methods (and doesn't have available a theory of complex-analytic spaces...). –  BCnrd May 18 '10 at 5:54

You could take a look in Lectures on Riemann surfaces by Otto Forster (Springer, Graduate Texts in Mathematics 81). The book starts at a moderate level (you just need to know basic complex analysis and the Lebesgue integral), but covers quite some topics like the correspondence between Galois groups of field extensions and covering transformations as well as how to remove the removable singularities. I don't have the book available now, but from my memory (of the German original) there should be enough in it to answer at least your first three questions.

PS: You can google for the two terms "forster" and "field of meromorphic functions" to get a link to google books for a first impression of the book.

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