Perhaps this is so obvious a question that no one has asked it before, but can someone provide a simple example of a canonical divisor which is not a principal divisor or conversely on a Riemann Surface?
I know that if $g(X)=1$ then $\operatorname{KDiv}(X)=\operatorname{PDiv}(X)$, but I imagine that this does not always hold. Here is my presumably false proof that $"\operatorname{KDiv}(X)=\operatorname{PDiv}(X)"$ in general:
If $\omega$ is a meromorphic $1-$form on $X$, then $\operatorname{div}(\omega)=\sum_{p\in X} \operatorname{ord}_p(\omega)\cdot p$, and for $f\in \mathcal{M}(X)$, we have $\operatorname{div}(f)=\sum_{p\in X} \operatorname{ord}_p(f)\cdot p$. However, $\operatorname{ord}_p(\omega)=\operatorname{ord}_0(g_p(z) dz)$ where $z$ is a local parameter near $p$, centered at $p$. I believe one should be able to glue together the $g_p$ to form a global meromorphic function $g\in \mathcal{M}(X)$ with $\operatorname{div}(g)=\operatorname{div}(\omega)$. In this case, we would get that $\operatorname{KDiv}(X)=\operatorname{PDiv}(X)$.
I think my mistake is that the gluing procedure produces something undesirable because of the method by which differential forms transform under a change of variables.
A very simple example is to take $X$ to be the Riemann sphere $\Bbb C_\infty$. A differential is $\omega=dz$ which has a double pole at $\infty$. So its divisor is $-2(\infty)$ which is not principal, since its degree is nonzero.
In general, a canonical divisor has degree $2g-2$ where $g$ is the genus, so it can only be principal if $g=1$, that is on an elliptic curve.