basic application of riemann roch applied to genus 1 curve On a genus $1$ smooth projective curve $C$ (over $\mathbb{C}$), if $p, q$ are two points and there is a rational function $f$ on $C$ such that $(f)=p-q$, then must $p=q$? 
I'm sure the answer is well known but I don't have any books with me at the moment. I can't seem to extract anything from Riemann Roch.
 A: This doesn't need Riemann-Roch (and I don't think Riemann-Roch says anything useful here). 
The theorem that you need is that for any (let's say smooth, but I don't think this is necessary) curve $C$, if $p - q = 0$ in the class group of $C$, i.e. if there is a meromorphic function $f$ such that $p - q = (f)$, then $C \simeq \mathbb{P}^1$. (equivalently, when $C$ is smooth, $g(C) = 0$). 
This is because any global meromorphic function $f$ defined on $C$ defines a rational map from $C$ to $\mathbb{P}^1$ such that the poles of $f$ map to $\infty$ (this is familiar from complex analysis: an entire meromorphic function defines an entire regular map to the Riemann sphere). This map is given by $[f : 1]$ on the locus where $f$ is regular and $[1 : \frac{1}{f}]$ on the locus where $\frac{1}{f}$ is regular. Now, if $(f) = p - q$, it follows that $f$ has a unique simple pole at $q$ and a unique zero at $p$. It is injective, since the function $f - a$ also has a unique simple pole at $q$ for any $a \in k$. Since the degree of a meromorphic function must be zero, this means it has a unique simple zero. In particular, it is non-constant, so finite, and the degree of the map is $1$. This shows that it is birational, and a birational map to a smooth curve is an isomorphism. 
