Existence of divisors of degree one on a curve over a finite field

Let $C$ be a smooth, geometrically irreducible projective curve defined over a finite field $\mathbb{F}_q$. Given a (scheme-theoretic) point $x \in C$, define the degree of $x$ to be the degree of the extension $[k(x): \mathbb{F}_q]$. The degree of a divisor on $C$ can thus be defined by linearity.

We proved today in my algebraic number theory class that there is always a divisor of degree one. The argument was to suppose that all the degrees $[k(x): \mathbb{F}_q]$ were divisible by some $m >1$, and then to apply (a very weak form of) the Cebotarev density theorem to the extension of function fields $k(C) \to k(C) \otimes_{\mathbb{F}_q} \mathbb{F}_{q^m}$. All places would have to split completely if $\mathbb{F}_{q^m}$ is contained in every residue field, which is a contradiction.

I also learned another proof from a comment of Felipe Voloch on MO: for large $n$, the Weil bound on the number of $\mathbb{F}_{q^n}$-rational points implies that there is a point of degree $n$ and a point of degree $n+1$. Taking the difference gives a divisor of degree one.

Is there an elementary geometric way of seeing this? (Related question: Are there other fields for which this is true?)

Solution 1:

This is not an elementary proof, but it is a little different to the others mentioned here:

The space of degree $1$ divisors on $C$ is a torsor over Jac$(C)$, and you are asking for a proof that it is the trivial torsor.

The space of such torsors is computed by $H^1(Gal(\overline{k}/k), \mathrm{Jac}(C))$. Now when $C$ is finite, the absolute Galois group is procyclic, generated by Frobenius, and there is an argument of Lang that shows for any connected smooth commutative group scheme $G$ over $k$ that $H^1(Gal(\overline{k}/k),G)$ vanishes in this case.

Namely, the endomorphism $\mathrm{Frob} - 1$ of $G$ is surjective, because it is smooth (take its deritative, and use that fact that the derivative of Frob vanishes), hence has open image (which is also closed, being a subgroup) and $G$ is connected.

This allows you to trivialize any 1-cocycle.

In your particular case, we should be able to make this concrete: choose a point of $C$ defined over some extension of $\mathbb F_q$, say $P$, then consider the degree zero divisor $\mathrm{Frob}_q(P) - P$. Applying Lang's argument to $\mathrm{Jac}(C)$, we find another degree zero divisor $D$ on $C$ (defined over some extension of $\mathbb F_q$) such that $\mathrm{Frob}_q(D) - D = \mathrm{Frob}_q (P) - P.$ Then $P - D$ is a degree one divisor that is fixed by $\mathrm{Frob}_q$, and so defined over $\mathbb F_q$.

So there is not really any cohomology involved (not that the above Galois cohomology is particular difficult), but you need to know that degree zero divisors are parameterized by a connected commutative smooth group scheme (the Jacobian).

Edit: The above argument actually shows that there is a degree one divisor whose linear equivalence class is defined over $\mathbb F_q$. An extra argument is needed to actually get a degree one divisor defined over $\mathbb F_q$. One writes downs the various obvious exacts sequences involving non-zero functions, principal divisors, all divisors, and Pic, and takes Galois cohomology. Using Hilbert Thm. 90 plus the vanishing of the Brauer group $H^2(G_{\mathbb F_q}, \overline{\mathbb F}_q^{\times})$, one finds that a Galois invariant linear equiv. class does indeed lift to an invariant divisor.

Solution 2:

There is a fairly elementary proof outlined in Exercise 6 here. The fact in question appears as a step in a proof of Weil's Riemann hypothesis for curves. I don't know if this counts as "geometric": it has an analytic flavor, but that's not too surprising since Professor Rabinoff's proof used Cebotarev density.

As for the related question, this is pretty obvious for algebraically closed fields. It fails for $\mathbb{R}$, since there are conics like $x^2+y^2+z^2=0$ with no rational point. I would be very interested in seeing an answer for global and non-Archimedean local fields.