Re: x^5+y^5=31

From: JoeS (jhs_at_math.brown.edu)
Date: 03/22/05 

Date: 21 Mar 2005 18:01:31 -0800

Dave Rusin wrote:
> In article <1111110944.724288.27010@f14g2000cwb.googlegroups.com>,
> JoeS <jhs@math.brown.edu> wrote:
> >Sorry, David, that's not quite right about integral points on
elliptic
> >curves.
>
> Thanks, Joe, for the corrections. I don't know why my memory stopped
at
> the point of Siegel's proofs.
>
> >Finally, Elkies pointed out that if one had an effective version of
the
> >ABC conjecture, then using Belyi's theorem, one could turn it into
an
> >effective version of Faltings' theorem.
>
> You lost me there. What is Belyi's theorem?
>
> dave

It's a very beautiful, but very strange, theorem. It says that a
(smooth projective) algebraic curve X defined over the complex numbers
has a model defined over the algebraic numbers (i.e., over an algebraic
closure of the rationals) if and only if there is a rational map from X
onto the projective line P^1(C) (i.e, the Riemann sphere) that is
ramified over exactly three points.

The theorem is strange (at least, to me), because, in other language,
it says that a Riemann surface X has a model in projective space given
by equations with algebraic coefficients if and only if X admits a
holomorphic covering to the Riemann sphere ramified at exactly three
points.

The connection with ABC is roughly as follows. Let X be a curve defined
over a number field. Then by Belyi's theorem, it admits a map f:X -->
P^1 ramified over exactly three points, which we may take to be the
points 0, 1, and infinity. Then, very roughly speaking, if P is a
rational point on X, then one applies the ABC conjecture (really in the
form A/C + B/C = 1) to the equation
     f(P) + (1-f(P)) = 1.

Joe