The genus of a curve with a group structure I'm reading Milne's Elliptic Curves and came across this statement: If a nonsingular projective curve has a group structure defined by polynomial maps, then it has genus 1.
In this question a similar question was asked, and an answer given there (the one that was not accepted) someone backs this up using machinery/notation that I do not understand. Can someone give or direct me to a proof of this statement? It doesn't have to be very formal; this isn't homework. Also, in the affine case this statement clearly breaks down - the affine line has the group structure of addition, for instance. Is there some sort of correction or does the group structure bear no relation to the genus in this case?
 A: For complex algebraic curves, this is actually very easy. Suppose we have an algebraic group law on a complex algebraic curve. Then it is necessarily continuous in a classical topology, so we get a topological group. Now, a fundamental group of a topological group has to be abelian (this is an interesting and not that hard exercise). But, the curves with genus > 1 don't have abelian fundamental groups!
A: I accepted another answer but I'm going to post a more elementary (from my perspective) way to prove this. The Riemann-Hurwitz Formula gives the nice bound $|\text{Aut}(\mathcal{C})| < \infty$ for a $g(\mathcal{C}) > 1$ (we'll assume characteristic zero so all maps are tamely ramified). But each point gives the translation automorphism, and these are all clearly distinct, whence we obtain a contradiction for $g(\mathcal{C}) > 1$. So the question becomes: does the projective line have a group law? Of course not, for then a hyperelliptic curve $E$ ($g(E) > 1$) would have one too. The gist is that if we select a degree-2 map $f \colon E \to \mathbb{P}^1$, totally ramified, then it must be injective, so we can pull back the group law.
A: Just to throw another way to formalize this fact, if $X$ is a variety that is a group, then the canonical sheaf $\omega_{X}$ must be trivial. The intuitive reason is that there is a canonical way to identify the tangent space $T_{x}$ at any point $x \in X$ with the tangent space $T_{e}$ at identity. (Namely, the map $y \leadsto y * x^{-1}$). 
If $X$ is a curve, then Riemann-Roch implies that $deg(K) = 2g-2$, where $K$ is the divisor corresponding to $\omega_{X}$. Since $\omega_{X}$ is trivial, $deg(K) = 0$. This implies that $g = 1$. 
