# Proving existence of non-special divisors of a given degree d on compact Riemann surfaces

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?

• Divisors of degree d modulo rational equivalence form a g dimensional algebraic variety, Pic^d(C), and there is a natural morphism C^d to Pic^d(C) given by sending an n-tuple (p_1,...p_d) to the divisor $\sum_i [p_i]$. The image consists exactly of the effective divisors. The morphism C^d to Pic^0(C) is generically finite (for d $\leq$ g) so there always exist non-special divisors for d < g. (The canonical reference for special divisors is the book of Arbarello, Cornalba, Griffiths, Harris.)
– naf
Commented Apr 19, 2010 at 9:22
• @unknown: For answering the original question, it is not even important that the morphism is generically finite, since we just need an upper bound on the dimension of image. Commented Apr 19, 2010 at 12:57

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 $g = 1$.