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According to this article,

The category of curves over the complex numbers is equivalent to two other categories: Riemann surfaces and fields of transcendence degree 1 over $\mathbb{C}$.

But I can not find details of this equivalence:

  1. Is it correct only for projective algebraic curves and compact Riemann surfaces?
  2. Is the category of curves considered up to birational equivalence?
  3. Is this field equal both to field of meromorpic functions of surface and coordinate field of curve?

And the most general question, is there an online material on this topic which intended for beginners? It was mentioned as something widely-known in all articles I found.

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  • $\begingroup$ A small but important note: on either the curve or Riemann surface side you need to restrict to the dominant, or equivalently nonconstant, maps. The constant maps don't induce maps on function fields (which go the other way, so a second small but important note is that the equivalence with function fields is contravariant). $\endgroup$
    – Qiaochu Yuan
    Commented Jul 17 at 19:30

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  1. Yes, this is for smooth projective algebraic curves and compact Riemann surfaces. Smoothness is required so that the curve is a complex manifold, and since it is projective then it is a compact complex manifold of dimension 1. Vice versa, a Riemann surface can be embedded into projective space with a very ample divisor. A priori, this embedding is as a complex submanifold. However, a theorem of Chow assures us that a closed submanifold of projective space is actually algebraic.

  2. You may forget smooth projective curves and just consider curves, since in every birational class of curves there is only one non-singular curve. This is because if $f:C\to C'$ is a rational morphism between non-singular curves, then it is a morphism. This comes from a general result saying that if $f:X\to\mathbb{P}^n$ is rational and $X$ is a smooth projective variety, then the place where $f$ is not defined is of codimension at least $2$ (see Shafarevich for example).

  3. Yes, the field of meromorphic functions is equal to the coordinate field. Miranda proves in his book, for example, that if there is a subfield of meromorphic functions that separates points and tangents, then it is actually equal to the whole meromorphic function field.

Miranda is a good book to start with that talks about this kind of equivalence. Also look for a book that proves Chow's Theorem on submanifolds of projective space.

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  • $\begingroup$ Dear Robert, As you likely know, there is another way to prove projectivity (which I think was Riemann's original method): you use the Dirichilet principle to construct an appropriate harmonic function on $C$, promote it a mermorphic function, and so realize your Riemann surface as a branched cover of $\mathbb P^1$. You then show that such a branched cover is necessarily an algebraic curve (Riemann's original form of the Riemann existence theorem). Regards, $\endgroup$
    – Matt E
    Commented Jan 12, 2014 at 2:05
  • $\begingroup$ Dear Matt, I am aware of Riemann's construction, and I thank you for pointing out the weakness in what I said above: The hard part of all this is showing that Riemann surfaces have enough meromorphic functions! $\endgroup$
    – rfauffar
    Commented Jan 12, 2014 at 12:07
  • $\begingroup$ Assuming that R. surfaces do have enough meromorphic functions, one can prove by hand that they are projective, without resorting to Chow's theorem. This can be done by embedding the surface in projective space, and projecting to $\mathbb{P}^2$ (we can allow nodes, etc.). Then it is not hard to show by hand that the projection to $\mathbb{P}^2$ of the surface must be algebraic, and we can then lift this equation to $\mathbb{P}^3$, where we know the R. surface can live. Playing around with these equations it is not hard to show that the surface is algebraic in $\mathbb{P}^3$. $\endgroup$
    – rfauffar
    Commented Jan 12, 2014 at 12:10

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