Let $C$ be a smooth projective curve of genus $g$ over complex numbers.
Is it true that any degree one divisor is linearly equivalent to $1.p$ for some point $p$?
I know it's true for $\mathbb P^1$. I am wondering if something like $p+q-r$ will be linearly equivalent to $p'$ on higher genus curves.
Any help is appreciated.
Here is a proof that $C$ is of genus $\le1$. Suppose for a contradiction that $g_C\ge2$. For any $x\ne y\in C$ we have $2x-y$ is a degree one divisor, so there is some point $p\in C$ such that $2x-y\sim p$. Thus there is a degree two rational function $f\colon C\to\mathbb P^1$ with zeroes $p$ and $y$ and a pole $x$ of multiplicity two. Thus, $C$ is a hyper-elliptic curve. Moreover, there are infinitely many rational functions $f\colon C\to\mathbb P^1$ giving a double cover.
But by Proposition 19.5.7 of Vakil, this is a contradiction.