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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 $g\geq c\frac{\deg(D)^2}{\ell(D)}$?

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    $\begingroup$ 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. $\endgroup$
    – Michael Stoll
    Apr 29, 2012 at 16:43

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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.

http://en.wikipedia.org/wiki/Brill%E2%80%93Noether_theory

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

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  • $\begingroup$ 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 $\endgroup$
    – Klim Efremenko
    Apr 29, 2012 at 15:34
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    $\begingroup$ 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. $\endgroup$
    – Felipe Voloch
    Apr 29, 2012 at 17:12
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    $\begingroup$ What you said about $r=2$ is false. You may be thinking of smooth plane curves, but most curves are not like that. $\endgroup$
    – Felipe Voloch
    Apr 29, 2012 at 17:13

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