Only finitely many genus $g$ smooth projective curves over a finite field

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In this tag Exercise 101.56.7 says:

Let $k$ be a finite field. Let $g > 1$. Sketch a proof of the following: there are only a finite number of isomorphism classes of smooth projective curves over $k$ of genus $g$.

I wonder whether there are some approaches to this problem other than theory of moduli space.

In this answer, the approach is to use some version of the canonical embedding to show that the 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 $k$.

However, I feel confused about some points for such approach. First of all, is the embedding of curve to the projective space related to the divisor in the finite field case in the same way as the algebraically closed field case? To find a divisor of degree approximately $2g+1$, I think there should be some assumption of $k$-rational point of $X$. Secondly, is the number of polynomials needed to define $X$ in the projective space is bounded only by $g$? If X is not completely intersection, there might be large degree hypersurface needed in the definition of $X$. Again, I am a little confused about degree in the non-algebraically closed field.

Thanks for any detailed explaination for the proof of the result of above exercise.

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This started out as a comment and became too long...

Here's a sketch of how I would do it (there are several details that always pop-up when dealing with non-algebraically closed fields and are not special to this problem and so I have left them out of this answer).

Let's split into cases:


Non-Hyperelliptic case:

Here the idea is to use that the canonical divisor $K_C$ is very ample (which can be shown by Riemann-Roch type arguments similar to the complex case) and embeds $C$ as a degree $2g-2$-curve in $\mathbb{P}^{g-1}$.

This embedding depends only on the isomorphism class of the curve and so defines a map from the set of isomorphism classes of curves to the set of $k$-rational points of the hilbert space of genus $g$ degree $2g-2$ curves in $\mathbb{P}^{g-1}$ (modulo reparametrization):

$$\{ \text{genus $g$ non-hyperelliptic curves over $k$} \} \to Hilb_{g,2g-2}(\mathbb{P}^{g-1}_k)/PGL_{g} (k)$$

This map is clearly injective and the RHS is a finite set since it is the set of $k$-rational points of a finite type scheme (with $k$-finite).


Hyperelliptic case

In this case we know the canonical embedding of $C$ factors through a $2$-fold cover of some smooth conic curve in $\mathbb{P}^2_k$. The set of all smooth conic curves are a subset of rational points of $\mathbb{P}^5$ and so is finite. It suffices to show now that given a fixed smooth conic $X$ there are only finitely many isomorphism classes of hyperelliptic curves over it.

But the category of all such curves is equivalent to the category of all quadratic extensions of $K(X)$ (suppose now that the characteristic is not $2$ - otherwise i'm not sure what to do). In this case due to Kummer theory all quadratic extensions are given by adjoining $\sqrt f$ for some $f \in K(X)^{\times}/K(X)^{\times 2}$. The degree of $f$ constrained by the genus, This set should be finite from simple algebraic considerations which elude me up to now (sorry about that...).


I hope this was at least a little bit helpful.

For a careful discussion of algebraic curves over general fields I highly recommend Qing Liu's "Algebraic Geometry and Arithmetic Curves".