Curve of genus $g$ with a point removed Let $C$ be a smooth projective curve of genus $g$. If we pick two distinct points $p,q\in C$, when are $C\setminus\{p\}$ and $C\setminus\{q\}$ isomorphic or not isomorphic? When $g = 0$, they are always isomorphic. But for $g\ge 1$, I wasn't sure how to approach this problem. 
Since curves of genus $g\ge 2$ have finite automorphism groups, I would expect $C\setminus\{p\}$ and $C\setminus\{q\}$ to be not isomorphic to each other. But I wasn't sure if this is a correct approach to the problem.
 A: (I assume we work over an algebraically closed field, hence an infinite field.) 
An isomorphism $C \backslash \{p\} \to C \backslash \{q\}$ induces a map $C \backslash \{p\} \to C$, and since $C$ is smooth, this extends canonically to a map $C \to C$ (this is proven in Hartshorne, Chapter 1). This map $C \to C$ is a regular map between projective varieties, hence is proper ; since it has closed and dense image, $p$ must be sent to $q$ (reason the other way around to notice that this extension to $C$ has to be an automorphism, i.e. biregular). We just established a bijection between the set of isomorphisms $C \backslash \{p\} \to C \backslash \{q\}$ and the set of isomorphisms $C \to C$ which maps $p$ to $q$. Therefore such an automorphism exists for each $p$ and $q$ if and only if the automorphism group is transitive. 
For $g = 0$ this is elementary, and for $g \ge 2$ this follows from Hurwitz, as you stated (the curve has infinitely many closed points). Since elliptic curves admit a group structure, their automorphism group is transitive (the automorphism sending $p$ to $q$ is just translation by $q-p$). 
Hope that helps,
