Threefold category equivalence: algebraic curves, Riemann surfaces and fields of transcendence degree 1

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.

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).
• 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, – Matt E Jan 12 '14 at 2:05
• 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 $\mathbb{P}^3$. – rfauffar Jan 12 '14 at 12:10