From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Density of rational points
Date: 3 Nov 1998 17:15:16 GMT

George Baloglou <baloglou@oswego.edu> wrote:

>My question: what is known on the density (as opposed to "mere" infinitude)
>of rational points in planar curves, and how often are we lucky enough to
>have proofs as elementary and/or easy as the one I provided for the circle? 

Well, you have to decide what a "curve" is. The existence of rational points
on curves described by various functions (e.g. exp and log) is quite a 
difficult part of transcendental number theory. But if you intend the
curve to be the zero locus of a rational polynomial in two variables, you can 
use techniques from algebraic geometry.

First off, observe that the curve  x(x^4+y^4-1)=0 consists of a line and a
closed curve; the former has rational points dense in the line, which the
latter has no nontrivial rational points (Fermat's Theorem). So it's possible
to have infinitely many points but not have them dense in the real locus.

But this example is "cheating": the curve is the union of two curves, each
itself an algebraic curve. Really you should ask your question about
irreducible curves (those which cannot be so decomposed).

By a theorem of Faltings, an irreducible curve with infinitely many rational 
points must be of genus 0 or 1, that is, it is possible to make substitutions 
with rational functions  (x,y) = (f(X,Y), g(X,Y))  for which the resulting
constraint is either
	Y = 0  (the "genus zero" case)   or
	Y^2 = X^3 + A X + B  for some  A  and  B  (the "genus one" case)
By continuity arguments, the density of rational points in the original curve
is equivalent to density of rational points in these transformed curves.

Clearly in the genus-0 case, the rational points are dense. This includes the
unit circle and indeed any conic section with a rational point.

The genus 1 case is harder. It is possible for such a curve to have only a 
finite number of rational points, but such cases can be easily checked for; 
in any other case, having _a_ rational point implies having infinitely many.
Moreover, if there are infinitely many rational points, they are dense in the
curve. Now, unlike the genus zero case (more or less by definition) the 
rational points are _not_ given as the image of any rational function, so 
"elementary proofs"  you might envision are not possible here; on the other
hand, it is possible to describe a process for generating infinitely many 
points from one (excluding the checkable cases with finitely many points.)
Unfortunately, for genus-1 curves, 
(a) it's not always clear whether or not there are any rational points
on the curve (we don't have an infallible algorithm for this)
(b) even if we can show there is a point on the curve we can't necessarily
find it easily (there's no useful upper bound on the heights of the points).
So "mere infinitude" is actually quite difficult to ascertain in practice.

dave