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

Let $k$ be a finite field. We define $N(d,g)$ to be the number of plane curves $f(x,y)$ defined over $k$ of degree $d$ with (geometric) genus $g$. If $D(d) := (d-1)(d-2)/2$ (the maximum possible genus), I would expect that as $d$ goes to infinity that the proportion of curves of degree $d$ with genus $D(d)$ would go to 1. If, on the other hand, we're interested in curves of a fixed small genus (say $g=0$ or 1), I would expect that $N(d,g)$ would still approach infinity, albeit at a much slower rate. The question that I have is how do $N(d,0)$ or $N(d,1)$ approach infinity? Is it a polynomial in $\log d$, faster, slower?

I realize that there has been a lot done with enumerative geometry (Gromon-Witten, Caporaso-Harris), but that seems a bit different, since it always works over an algebraically closed field, and classifies curves by having them pass through some set of generic points, and possibly prescribing, tangency, etc.

share|cite|improve this question
up vote 6 down vote accepted

Fix $g$, the genus, and $q$, the order of $k$. $N(d,g)$ should be $\approx C q^{3d}$, where $C$ is some constant dependent on $q$ and $g$. (Note that my $C$ has absorbed the $q^{-4}$ in Felipe's answer.) There are some nonrigorous details here.

There are finitely many isomorphism classes of pair $(X, L)$ where $X$ is a genus $g$ curve over $k$ and $L$ is a degree $d$ line bundle. I claim that, for each such pair, there are roughly $C(X,L,q) q^{3d}$ degree $d$ curves in $\mathbb{P}^2$ such that the curve is isomorphic to $X$ and the pullback of $\mathcal{O}(1)$ is $L$, and $C(X,L,q)$ is some constant dependent only on $q$, $X$ and $L$.

Recall that such curves, more or less, come from linear maps $H^0(X, L) \to k^3$, modulo rescaling on the image. The number of such maps is $q^{3 \dim H^0(X, L)}$. By Riemman-Roch, for $d$ larger than $2g$, we have $\dim H^0(X, L) = d-g+1$. So this is where I get $q^{3d}$ from, absorbing everything else into the constant.

To be more precise, we need to (1) discard the maps that have base points (2) discard the maps that correspond to branched covers rather than immersions (3) if $(X, L)$ has nontrivial automorphism group, count orbits under that group.

I claim that (1) reduces our count by a factor of about $Z_X(3)$, where $Z_X$ is the zeta function of $X$. (2) should be negligible for large $d$ -- I get that there are about $q^{2d}$ maps coming from $d$-fold covers of a line, and fewer for every other case. And (3) I would guess just divides by $|\mathrm{Aut}(X,L)|$ -- there probably aren't a lot of cases with nontrivial stabilizers. I haven't checked the claims in this paragraph carefully, but that is what I'd expect the counts to be.

share|cite|improve this answer
@David, and @Felipe, Thanks. There's one question I have: you are implicitly assuming that the number of isomorphism classes of $(X,L)$ where $L$ has degree $d$ is bounded independently of $d$. Taking a cue from Felipe's description of $g=0$ is it proved something like the following? The semigroup of very ample divisors for $X$ is finitely generated. Then every degree $d$ $(X,L)$ is parametrized by (at least) one of the $(X,L')$ where $L'$ is one of the generators of the semigroup. – Victor Miller May 20 '11 at 15:08
$X$ must have a point defined over some finite extension of $\mathbb{F}_q$; say $x \in X(\mathbb{F}_{q^e})$. Then twisting by $\mathcal{O}(-x)$ is a bijection between line bundles of degree $e$ and line bundles of degree $d-e$. So we just need to count the number of line bundles of degrees $0$, $1$, ..., $e-1$. (One could certainly give more precise bounds than this, but this is the quickest argument I can think of.) – David Speyer May 20 '11 at 15:31
@David, Thanks. – Victor Miller May 20 '11 at 15:42
Over finite fields, every curve has a rational divisor of degree one, so the number of divisors classes of degree $d$ is independent of $d$. There is an elementary way to see this, but a quick way is to deduce from the Weil bound that there is a divisor of degree $n$ (orbit of a point of degree $n$) and $n+1$ and take the difference. – Felipe Voloch May 20 '11 at 22:40
Thanks Felipe! I had just started worrying whether or not that was true. – David Speyer May 21 '11 at 12:43

The set of singular curves of degree $d$ contains the curves with $f(0,0)=f_x(0,0)=f_y(0,0)=0$. The chance that each of these values ($f(0,0),\ldots$) is zero is $1/q$ over the field of $q$ elements, so the proportion of such curves among all curves of degree $d$ is $1/q^3$, hence the proportion of smooth curves is at most $1-1/q^3$ regardless of $d$ and doesn't go to $1$ as $d$ goes to infinity. (I believe this observation is due to Poonen).

The set of curves of degree $d$ and genus zero is the set of parametrized curves $(P_0(t):P_1(t):P_2(t))$, $P_i$ polynomials of degree $d$, modulo the action of $PGL_2(\mathbb{F}_q)$ on the variable $t$. So there are about $q^{3d-4}$ such curves. That's your $N(d,0)$. You should be able to do $N(d,1)$ using similar ideas.

share|cite|improve this answer

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.