MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Castelnuovo bound says that if we have a function field(algebraic curve) $F$ and a divisor on it $D$ then: $g\leq c\frac{\deg(D)^2}{\ell(D)}$(where $c$ is some global constant say 2 and $g$ is a genus of the curve). I would like to ask if the converse is true? My question is if the converse is true for every $\ell(D)$? Formally the question is the following:

Does there exists a constant $c$ such that for every function field $F$ and for every integer $2\leq l \leq g$ there exists a divisor $D$ with $\ell(D)= l$ and $g\geq c\frac{\deg(D)^2}{\ell(D)}$?

share|cite|improve this question
I removed the "l-functions" tag, since $\ell(D)$ has nothing to do with L-functions. The implication $\deg(D) \le \sqrt{g} \Longrightarrow \ell(D) \le 1$ is wrong in general: for a hyperelliptic curve, with $D$ twice a Weierstrass point, we have $\ell(D) = 2$ and $\deg(D) = 2$, which is $\le \sqrt{g}$ for $g$ sufficiently large. – Michael Stoll Apr 29 '12 at 16:43

Over an algebraically closed field, the general curve of genus $g$ has a divisor of degree $d$ and (projective) dimension $r = l(D) - 1$ if and only if $r(d − r + 1) − (r − 1)g \ge 0$. This is the main result of Brill-Noether theory.

Over non-algebraically closed fields the answer to your question is probably no, in general.

share|cite|improve this answer
Thanks for the answer. I agree that if $r(d−r+1)−(r−1)g \geq 0$ then there exists a divisor but I think that it the converse is not true. It may happen that $r(d−r+1)−(r−1)g <0$ and still you will have divisor of degree $d$ and dimenion $r+1$. Over general curve we usually have divisor of degree $2\sqrt{g}$ with r=2. Taking $g$ large we will get that your equation is neggative – Klim Efremenko Apr 29 '12 at 15:34
If $r(d−r+1)−(r−1)g < 0$ then there exists a curve of that genus that has no divisor with those parameters. In fact this will be true for "most" curves. – Felipe Voloch Apr 29 '12 at 17:12
What you said about $r=2$ is false. You may be thinking of smooth plane curves, but most curves are not like that. – Felipe Voloch Apr 29 '12 at 17:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.