I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.


I think it should be true, but I want to prove the following concretely: 


"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."


(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)


Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $\deg D = g-1$.


Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.


More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?



Proving existence of non-special divisors of a given degree d on compact Riemann surfaces
Take $g+1$ general points $p\_1, \dots, p\_{g+1}$ on your curve. The divisor $p\_1+ \cdots +p\_g - p\_{g+1}$ is non-special. The proof is easy from the following lemma: if $D$ is a divisor such that $\mathrm{h}^0(D) > 0$, then $\mathrm{h}^0(D - p) = \mathrm{h}^0(D) - 1$ for all but finitely many points $p$. First you use Riemann-Roch to deduce that $\mathrm{h}^0(p\_1+ \cdots +p\_g) = 1$, then you apply the lemma once again to $p\_1+ \cdots +p\_g$. This also works for $g = 0$ and $g = 1$.


