Why are Riemann surfaces algebraic curves?

Solution 1:

As Akhil writes in his comment, to see that some given Riemann surface is an algebraic curve, you usually need Riemann-Roch. A prototypical example of its use is the usual proof of existence of a Weierstrass equation for an elliptic curves (see e.g. Silverman).

But for the modular curves, things are actually simpler. It is easy to see that a modular curve is a covering of $\Gamma(1)\backslash \mathbb{H}^*$. Now, the latter is the simplest Riemann surface there is, the Riemann sphere, and that is clearly an algebraic curve. There are lots of ways of seeing that $\Gamma(1)\backslash \mathbb{H}^*$ is the Riemann sphere, some of which are sketched in Milne's notes, Proposition 2.21. An explicit isomorphism of $\Gamma(1)\backslash \mathbb{H}$ with $\mathbb{C}$ is provided by the $j$-function.

You can now use the covering to obtain an equation for a modular curve. For the minute details of this calculation for the case $\Gamma_0(N)$, see Milne's notes, Theorem 6.1. This does not need Riemann-Roch, and only relies on explicit computations with the $j$-function.

A more powerful approach to modular curves, which will also give you more information about fields of definition, is by interpreting the modular curves are moduli varieties. Milne also sketches some of this in section 8, and much more can be found in his notes on Shimura varieties.