Could you please check whether the solution below is ok?
There is an exercise from Shafarevich's Basic Algebraic Geometry, vol. 1, ex. 7.7.21.
Let $o$ be a point of an smooth algebraic curve $X$ of genus $g$. Using the Riemann–Roch theorem, prove that any divisor $D$ with $\deg D=0$ is equivalent to a divisor of the form $D_0-go$, where $D_0 > 0, \deg D_0=g$.
It's equivalent to show that the space $\mathcal{L}(go+D)$ has dimension $\geqslant 1$. Suppose it has dimension $0$. By Riemann-Roch theorem, we have
$\ell(go+D)-\ell(K-go-D) = 1$. Therefore, $\ell(K-go-D)=-1$. Since $K-go+D$ and $K-go$ are of the same degree, we have
$\ell(K-go) \leqslant \ell(K-go+D)$. Now apply Riemann-Roch to $go$:
$$\ell(go)-\ell(K-go) = 1$$
But the space $\mathcal{L}(go)$ is the space of all rational functions having pole of order $\leqslant g$ at $o$, so of dimension $g+1$,
and $\ell(K-go)$ is of dimension $g$ – contradiction.
I don't like the last argument and that we didn't use the structure of a canonical divisor.
Thanks in advance.
Best Answer
I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at $$\ell(go+D)-\ell(K-go-D)=1.$$ Since $\ell(E)\ge 0$ for any divisor $E$, it follows that $\ell(go+D)\ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)