# The number of curves of given genus over a field

Let $k$ be a field. Let $g\geq 0$ be an integer.

I have an elementary question.

Let $N$ be the "number" of $k$-isomorphism classes of smooth projective geometrically connected curves over $k$ of genus $g$. (Note that $N$ can also be $\infty$.)

Is $N$ finite if $k$ is finite?

When is $N$ finite in general?

I'm looking for the "most elementary" answer to this question.

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I don't know enough algebraic geometry to be sure about this, but: a suitable category of smooth projective curves over $k$ should be equivalent to the opposite of the category of function fields over $k$, and there should also be a bound on the degree of such a function field (as an extension of $k(x)$) necessary to realize any curve of genus $g$ in terms of $g$. This establishes that $N$ is finite if $k$ is finite. The converse is false, since for example if $k$ is algebraically closed and of characteristic not equal to $2$ and $g = 0$ then $N = 1$. – Qiaochu Yuan Oct 21 '12 at 23:07
@QiaochuYuan. There is a bound on the degree of the function field (as an extension of $k(t)$) depending only on the genus. This follows from Riemann-Roch. But why does $\mathbf{F}_p(t)$ only have finitely many extensions of bounded degree? The ramification isn't fixed. – Harry Oct 22 '12 at 8:39
I forgot for a second that there is only one curve of genus zero over an algebraically closed field. I changed the question accordingly. – Harry Oct 22 '12 at 8:41

Yes, this number is finite when the field $k$ is finite. Use some version of the canonical embedding to show that your curve can be realized in some projective space by equations bounded by some degree, and then observe that there are only finitely many polynomials with a given number of variables and of given degree over $k$. For an infinite field the number of isomorphism classes is infinite as soon as $g \geq 1$: one can consider a hyperelliptic curve defined by a choice of $2g+2$ points on $\mathbf P^1$ (considered up to the action of $\mathrm{Aut}(\mathbf P^1)$), and there are infinitely many such choices when the field is infinite.
A more sophisticated approach (when $g \geq 2$) uses the existence of a moduli space of curves of given genus. If $M_g$ is the coarse moduli space of curves of genus $g$ and $k$ is a finite field, then one has the formula $$\# M_g(k) = \sum_{[C]} \frac{1}{\# \mathrm{Aut}_k(C)}$$ where the sum ranges over $k$-isomorphism classes of curves $C$, and clearly $\# M_g(k)$ is finite. Unfortunately the only proof of this formula that I know uses the Grothendieck-Lefschetz trace formula on the moduli stack...
Thank you for your answer. I agree with the first part. But I have a question about the more sophisticated approach, because I don't see how this answers the question immediately. I agree that $M_g(k)$ is finite: it is the set of $k$-rational points on a $k$-variety. Now, this means that the right-hand side is a converging sum. But, this doesn't necessarily mean we are summing over finitely many elements, right? A priori, it's possible that the set of $k$-curves $C_1,C_2,\ldots$ is infinite and each curve satisfies $\#$ Aut$_k(C_n) = n^2$... – Harry Oct 22 '12 at 17:39
Now, the number of automorpshisms of a $k$-curve is bounded by the genus. I only know of an explicit bound due to Stickelberg. It's a bit bigger than the usual Hurwitz bound $84(g-1)$, but this shows the sum on the righthandside is also finite. It would be nice to have an elementary (and thus not explicit) bound on the number of elements in the automorphism group of a $k$-curve. Do you know how to prove this? – Harry Oct 22 '12 at 17:40
Oh, I hadn't thought about that, but you're absolutely right. I think the bound in positive characteristic is due to Stichtenoth, not Stickelberg, though. But you are right that there is an elementary proof for curves over a finite field: again, use the canonical embedding into some projective space. Then every automorphism of the curve extends to the ambient projective space, and the number of automorphisms of projective space is bounded by the number of $k$-points of $\mathrm{PGL}(n)$. – Dan Petersen Oct 23 '12 at 5:52