TITLE: Existence of divisors of degree one on a curve over a finite field
QUESTION [18 upvotes]: 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?)

REPLY [8 votes]: 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.