FOM: faltings
Lou van den Dries
vddries at math.uiuc.edu
Thu Nov 6 20:05:15 EST 1997
As promised earlier, and for what it's worth, I will state below
Faltings' theorem in elementary terms, but just for the field of rational
numbers, and just the part that solves the Mordell conjecture.
(Elementary: no talk of complex numbers, projective space, or genus).
Disclaimer: taking f.o.m. in Simpson's sense,
no f.o.m significance is claimed for this, nothing is new here, but
there is a certain interest in having results stated at this level.
Also, contrary to remarks by Harvey (Oct. 28) and Steve (Oct. 29),
I never said I would explain in "general intellectual terms" the
fundamental ideas behind Faltings' proof or explicate the interest of
Faltings' theorem. What I do claim is that Faltings'
theorem can be *stated* in a reasonably natural and striking way
assuming nothing more than the elementary algebra of polynomials in
two variables with rational coefficients, including the notions of
divisibility and irreducibility for such polynomials.
I don't think one can do much better with MRDP in about the same space.
In fact, the full statement of MRDP (not some watered down version)
can only be appreciated by those who are familiar with the somewhat technical
notion of recursive set, and its motivation by the Turing-Church analysis.
Roughly speaking, Faltings' theorem says:
A polynomial equation f(x,y) = 0 with rational coefficients can only have
finitely many rational solutions, except possibly in the very special
case where by a change of variables one of the irreducible factors of
f(x,y) becomes *linear* or *cubic*.
That's about the same level as the popular statement of a corollary of MRDP:
No computer program, even with unlimited time and space available,
can tell if an arbitrary given polynomial equation in several
variables with integer coefficients has a solution in integers.
When going into a little more detail, it makes sense to
first discuss older theorems of Mordell and Mazur on rational
solutions to cubic equations, which lead up to the Mordell conjecture
that Faltings' theorem solved. These results are remarkable in
themselves, so nothing is lost in spending first some time on them. In
discussing this I will use the language of cartesian
analytic geometry. (Objections?)
Now, Mordell's theorem. Consider a cubic equation
(*) y^2 = x^3 + ax + b
with real coefficients a and b such that 4a^3 + 27b^2 is not 0.
(This last condition means that the curve defined by (*) is smooth,
but never mind if that sounds too arcane.)
Let C be the curve in the real xy-plane defined by (*).
Take any two points P=(p_1,p_2) and Q = (q_1,q_2) on C and draw the
straight line L through them. Then L meets the curve in one more point
R = (r_1,r_2), and r_1 and r_2 are given by rational expressions in
a, b , p_1, p_2, q_1 and q_2. Explanation: intersecting L with C amounts
to solving a cubic equation in one unknown of which already two
real roots are known (corresponding to the points P and Q), so there
is exactly one more root which can be expressed as indicated. This is
just "Descartes". (If P=Q the line L is to be taken as the tangent
line to C at P, and when in addition P=Q lies on the x-axis, then this
tangent line is vertical and the point R is to be viewed as the "
point at infinity on C". This gives a first idea why it is helpful to
"complete" the curve by one extra "ideal" point "at infinity", to be
thought of as the point where all vertical lines meet.
Suppose now in addition that the coefficients a and b are rational,
and that the points P and Q are also rational. Then the point R is
rational as well, since its coordinates r_1 and r_2 are given by
rational expressions in a, b , p_1, p_2, q_1 and q_2. (By the way,
Harvey might like the following general fact: definable functions in
algebraically closed fields of characteristic 0 are necessarily
"piecewise" rational. The above rationality phenomenon can be viewed
as a special case.) We say that the point R is obtained from P and Q by the
"chord and tangent process". Below we always assume that a,b are rational.
MORDELL'S THEOREM. There are finitely many rational points on C
such that all other rational points can be obtained from those by
iteratively applying the chord and tangent process.
(The point at infinity is to be counted as a rational point, and the
lines connecting it to the other points should be taken as vertical
lines.)
By the way, this result from 1922 was suggested by Poincare in 1901.
A theorem of Mazur implies that if you have more than 16 rational
points to start with, then you can get infinitely many rational points by
iteration of this "chord and tangent process".
Translation into the language of "college algebra" :
What's really going on here is that there is a hidden abelian group
structure on "C union the point at infinity", where we have P + Q + R = O
in the notation above, where O, the zero of the group, is the
point at infinity. The rational points form a subgroup, and Mordell's
theorem says that this subgroup of rational points is finitely
generated. The largest finite subgroup of the group of rational
points can be effectively determined from a and b. Mazur's theorem
says that this finite subgroup can have at most 16 elements, and it describes
all possible 15 isomorphism types of these finite groups that can
occur as the coefficients a and b vary. For certain values of a,b the
group of rational points is finite, for other values it is infinite,
and no decision procedure is presently known to distinguish between
these two possibilities (though the Birch & Swinnerton-Dyer conjecture
would provide a decision procedure if I am not mistaken).
Now consider more generally an equation
(**) f(x,y) = 0
where f(x,y) is a nonconstant polynomial with rational coefficients,
To avoid repetition of the phrase "with rational coefficients",
all polynomials and rational functions mentioned below are assumed
with rational coefficients.
We want to know how it can happen that (**) has infinitely many
rational solutions. First of all, we can factor f(x,y) effectively
into irreducible factors, so for this question we may as well assume
that f is irreducible. If we now make an "invertible change of
variables", say we substitute x= uv, y= uv^2 in (**) (so that
u=x^2/y and v= y/x) we get an equation
(***) g(u,v) = 0
which in some sense is equivalent to the original equation, in particular
the (rational, real, complex) solutions of (**) are in
1-1 correspondence with the (rational, real, complex)
solutions of (***), with at most finitely many exceptions on both
sides, which have to do with certain denominators vanishing. In any case,
(**) will have infinitely many rational solutions if and only if (***)
has infinitely many rational solutions. The above is not 100% correct
but gives a preliminary idea. The only tedious part in stating
Faltings' theorem correctly in an elementary way is to give an
airtight definition of when two irreducible equations (**) and (***)
are "equivalent via an invertible change of variables", or,
more briefly, "equivalent". In elementary terms this definition looks
convoluted, but its motivation as explained above should be crystal clear.
Definition. The irreducible polynomial equations (**) and (***) are
equivalent if
(1) there are rational functions x(u,v) = x_1(u,v)/x_2(u,v), and
y(u,v) = y_1(u,v)/y_2(u,v), where the x_i(u,v) and y_i(u,v) are
polynomials and the denominators x_2(u,v) and y_2(u,v) are
relatively prime to their respective numerators, and not
divisible by g(u,v), such that
f(x(u,v), y(u,v)) = g(u,v).a(u,v)/b(u,v),
where a(u,v) and b(u,v) are polynomials with b(u,v) not divisible
by g(u,v).
(2) there are rational functions u(x,y) and v(x,y) such that (1)
holds with u(x,y) and v(x,y) instead of x(u,v) and y(u,v), and
with the roles of f(x,y) and g(u,v) interchanged.
(3) x(u(x,y),v(x,y)) is well defined (i.e. x_2(u(x,y),v(x,y)) is
not 0) and x(u(x,y),v(x,y))= x + p(x,y).f(x,y)/q(x,y) where
p and q are polynomials, and q is not divisible by f.
Also y(u(x,y),v(x,y)) is well defined (i.e. y_2(u(x,y),v(x,y)) is
not 0) and y(u(x,y),v(x,y))= x + r(x,y).f(x,y)/s(x,y) where
r and s are polynomials, and s is not divisible by f.
(4) same as (3) with the roles of x(u,v), y(u,v), f(x,y) interchanged with
u(x,y), v(x,y), g(u,v).
Comments: (1) implies that each (rational, real, complex) solution
(u',v') of (***) (with finitely many exceptions) gives a solution
(x(u',v'),y(u',v')) of (***), (2) implies that each solution (x',y') of (**)
(with finitely many exceptions) gives a solution (u(x',y'), v(x',y')) of
(***), and (3) and (4) say that these two maps (from solutions of
(**) to (***) and back) are each others inverse (modulo a finite
set of exceptions).
FALTINGS' THEOREM. Given an irreducible polynomial g(u,v) the equation
g(u,v) = 0 has only finitely many rational solutions, except possibly
in the case that the equation is equivalent to the equation y=0, or to
an equation of the form y^2 - (x^3 + ax + b) = 0 as in (*) above.
If g has degree > 3 then these two cases are indeed exceptional, as
has been known for a long time. The first exceptional case
(equivalence to y=0) was studied first by Hilbert & Hurwitz, the
second exceptional case by Poincare. These people seem largely
responsible for initiating the modern view of looking at diophantine
equations from an algebraic-geometric point of view. By the way,
equivalence to y=0 can be effectively decided, and equivalence to an
equation of the form (*) can be effectively decided if at least one
rational solution is known.
The above formulation of Faltings' theorem is a bit unnatural from
the geometric viewpoint: I took unfair advantage of the fact that
having infinitely many rational solutions and being irreducible
implies absolute irreducibility. Nevertheless, this formulation is
clearly equivalent from the loftier algebraic-geometric point of view
to the usual formulation in terms of the genus of the compact Riemann surface
associated to the equation.
Theorems like Faltings' are in the following tradition: a polynomial
equation can only have "many" solutions of a certain kind if there is a
good reason for it. Roughly speaking, the "good reason" in the case of
Faltings' is that there is a group operation on the set of (complex) solutions
which is given by rational functions. (Thus, given "enough" rational
solutions to start with and applying the group operation iteratively, one
obtains "many" rational solutions.) I believe the Lang conjectures are
suggestions to make this vague idea ("good reasons for having many
solutions") more concrete in higher dimensions.
Question to Harvey and Steve: is the fact that x^2=2
has no rational solution of foundational significance? (Just curious.)
Lou van den Dries
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