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?)

  • $\begingroup$ I don't know enough about this to say something more concrete, but these people that study moduli of stable vector bundles have ways of telling when the moduli space is non-empty. Maybe they resort to one of the above in this type of situation though? $\endgroup$ – Matt Sep 22 '12 at 2:24
  • $\begingroup$ @Matt: ok, but I know nothing about moduli of stable vector bundles! Are there any key results on non-emptiness that you think might be relevant? $\endgroup$ – Akhil Mathew Sep 22 '12 at 11:27
  • $\begingroup$ Dear Akhil, My answer below was incomplete, and I added a brief sketch of how to complete it. Cheers, $\endgroup$ – Matt E May 9 '13 at 5:06

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.

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    $\begingroup$ If $K$ is a field with non-trivial Brauer group, then there are conics without rational points. For conics, having a $0$-cycle of degree $1$ is equvialent to having a rational point by Riemann-Roch. So the required property can't hold over such fields. This include global fields and local fields. Of course finite fields have trivial Brauer groups, so there is no contradiction. $\endgroup$ – user18119 Nov 1 '12 at 22:25
  • $\begingroup$ I'll have a look at this. Thanks! $\endgroup$ – Akhil Mathew Nov 2 '12 at 23:28
  • $\begingroup$ Incidentally, (since you also explained this argument to me), you might be interested in Matt E's edit below. There's an extra step to get from "points of $\mathrm{Pic}^0$" to "line bundles on $C$" (which works in the case of a finite field). $\endgroup$ – Akhil Mathew May 10 '13 at 13:52

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.

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    $\begingroup$ Thank you! This is a very nice argument. $\endgroup$ – Akhil Mathew Mar 3 '13 at 13:44
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    $\begingroup$ @AkhilMathew: Dear Akhil, You're welcome. I think this method of Lang for killing $H^1$ over a finite field goes back to a paper he wrote maybe in the 60's, and has lots of other applications, e.g. in the theory of algebraic groups over finite fields. Best wishes, $\endgroup$ – Matt E Mar 3 '13 at 23:10
  • $\begingroup$ Doesn't the fact that the line bundle is defined over $\mathbb{F}_q$ mean that the (or rather, a) divisor one could associate to it has to be defined over $\mathbb{F}_q$? Perhaps I'm missing something... $\endgroup$ – Akhil Mathew May 9 '13 at 22:16
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    $\begingroup$ @AkhilMathew: Dear Akhil, Unfortunately not. There is an obstruction to lifting an element of $Pic^0$ that is defined over the ground field $k$ to an actual divisor defined over $k$, which lies in the Galois $H^1$ of the group of principal divisors. By HT90, this $H^1$ ebmeds into the Galois $H^2$ of $\overline{k}^{\times}$, and so one can reinterpret this obstruction as a certain element of the Brauer group of $k$, which need not be trivial. I guess the point is that there is a difference between line bundles defined over $k$, and line bundles over $\overline{k}$ whose isomorphism class $\endgroup$ – Matt E May 10 '13 at 1:26
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    $\begingroup$ Thanks for clarifying! I had forgotten that in defining the Picard scheme, one needs to use either a rational point or a sheafification process. So that's apparently where the $H^2$ of $\mathbb{G}_m$ comes from. $\endgroup$ – Akhil Mathew May 10 '13 at 13:51

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