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From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
Subject: sci.math FAQ: Wiles attack
Summary: Part 6 of many, New version,
Originator: alopez-o@neumann.uwaterloo.ca
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Date: Fri, 17 Nov 1995 17:14:06 GMT
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Archive-Name: sci-math-faq/FLT/Wiles
Last-modified: December 8, 1994
Version: 6.2
Wiles' line of attack
Here is an outline of the first, incorrect proposed proof. The bits
about Euler system are
From Ken Ribet:
Here is a brief summary of what Wiles said in his three
lectures.
The method of Wiles borrows results and techniques from lots
and lots of people. To mention a few: Mazur, Hida, Flach,
Kolyvagin, yours truly, Wiles himself (older papers by Wiles),
Rubin... The way he does it is roughly as follows. Start with a
mod p representation of the Galois group of Q which is known to
be modular. You want to prove that all its lifts with a certain
property are modular. This means that the canonical map from
Mazur's universal deformation ring to its maximal Hecke algebra
quotient is an isomorphism. To prove a map like this is an
isomorphism, you can give some sufficient conditions based on
commutative algebra. Most notably, you have to bound the order
of a cohomology group which looks like a Selmer group for Sym^2
of the representation attached to a modular form. The
techniques for doing this come from Flach; and then the proof
went on to use Euler systems a la Kolyvagin, except in some new
geometric guise. [This part turned out to be wrong and
unnecessary].
If you take an elliptic curve over Q , you can look at the
representation of Gal on the 3-division points of the curve. If
you're lucky, this will be known to be modular, because of
results of Jerry Tunnell (on base change). Thus, if you're
lucky, the problem I described above can be solved (there are
most definitely some hypotheses to check), and then the curve
is modular. Basically, being lucky means that the image of the
representation of Galois on 3-division points is GL(2,Z/3Z) .
Suppose that you are unlucky, i.e., that your curve E has a
rational subgroup of order 3. Basically by inspection, you can
prove that if it has a rational subgroup of order 5 as well,
then it can't be semistable. (You look at the four non-cuspidal
rational points of X_0(15) .) So you can assume that E[5] is
``nice''. Then the idea is to find an E' with the same
5-division structure, for which E'[3] is modular. (Then E' is
modular, so E'[5] = E[5] is modular.) You consider the modular
curve X which parameterizes elliptic curves whose 5-division
points look like E[5] . This is a twist of X(5) . It's
therefore of genus 0, and it has a rational point (namely, E ),
so it's a projective line. Over that you look at the
irreducible covering which corresponds to some desired
3-division structure. You use Hilbert irreducibility and the
Cebotarev density theorem (in some way that hasn't yet sunk in)
to produce a non-cuspidal rational point of X over which the
covering remains irreducible. You take E' to be the curve
corresponding to this chosen rational point of X .
From the previous version of the FAQ:
(b) conjectures arising from the study of elliptic curves and
modular forms. - The Taniyama-Weil-Shmimura conjecture.
There is a very important and well known conjecture known as
the Taniyama-Weil-Shimura conjecture that concerns elliptic
curves. This conjecture has been shown by the work of Frey,
Serre, Ribet, et. al. to imply FLT uniformly, not just
asymptotically as with the ABC conjecture.
The conjecture basically states that all elliptic curves can be
parameterized in terms of modular forms.
There is new work on the arithmetic of elliptic curves. Sha,
the Tate-Shafarevich group on elliptic curves of rank 0 or 1.
By the way an interesting aspect of this work is that there is
a close connection between Sha, and some of the classical work
on FLT. For example, there is a classical proof that uses
infinite descent to prove FLT for n = 4 . It can be shown that
there is an elliptic curve associated with FLT and that for n =
4 , Sha is trivial. It can also be shown that in the cases
where Sha is non-trivial, that infinite-descent arguments do
not work; that in some sense ``Sha blocks the descent''.
Somewhat more technically, Sha is an obstruction to the
local-global principle [e.g. the Hasse-Minkowski theorem].
From Karl Rubin:
It has been known for some time, by work of Frey and Ribet,
that Fermat follows from this. If u^q + v^q + w^q = 0 , then
Frey had the idea of looking at the (semistable) elliptic curve
y^2 = x(x - a^q)(x + b^q) . If this elliptic curve comes from a
modular form, then the work of Ribet on Serre's conjecture
shows that there would have to exist a modular form of weight 2
on Gamma_0(2) . But there are no such forms.
To prove the Theorem, start with an elliptic curve E , a prime
p and let rho_p : Gal(\bar(Q)/Q) --> GL_2(Z/pZ) be the
representation giving the action of Galois on the p -torsion
E[p] . We wish to show that a certain lift of this
representation to GL_2(Z_p) (namely, the p -adic representation
on the Tate module T_p(E) ) is attached to a modular form. We
will do this by using Mazur's theory of deformations, to show
that every lifting which ``looks modular'' in a certain precise
sense is attached to a modular form.
Fix certain ``lifting data'', such as the allowed ramification,
specified local behavior at p , etc. for the lift. This defines
a lifting problem, and Mazur proves that there is a universal
lift, i.e. a local ring R and a representation into GL_2(R)
such that every lift of the appropriate type factors through
this one.
Now suppose that rho_p is modular, i.e. there is some lift of
rho_p which is attached to a modular form. Then there is also a
hecke ring T , which is the maximal quotient of R with the
property that all modular lifts factor through T . It is a
conjecture of Mazur that R = T , and it would follow from this
that every lift of rho_p which ``looks modular'' (in particular
the one we are interested in) is attached to a modular form.
Thus we need to know 2 things:
(a)
rho_p is modular
(b)
R = T .
It was proved by Tunnell that rho_3 is modular for every
elliptic curve. This is because PGL_2(Z/3Z) = S_4 . So (a) will
be satisfied if we take p = 3 . This is crucial.
Wiles uses (a) to prove (b) under some restrictions on rho_p .
Using (a) and some commutative algebra (using the fact that T
is Gorenstein, basically due to Mazur) Wiles reduces the
statement T = R to checking an inequality between the sizes of
2 groups. One of these is related to the Selmer group of the
symmetric square of the given modular lifting of rho_p , and
the other is related (by work of Hida) to an L -value. The
required inequality, which everyone presumes is an instance of
the Bloch-Kato conjecture, is what Wiles needs to verify.
[This is the part that turned out to be wrong in the first
version]. He does this using a Kolyvagin-type Euler system
argument. This is the most technically difficult part of the
proof, and is responsible for most of the length of the
manuscript. He uses modular units to construct what he calls a
geometric Euler system of cohomology classes. The inspiration
for his construction comes from work of Flach, who came up with
what is essentially the bottom level of this Euler system. But
Wiles needed to go much farther than Flach did. In the end,
under certain hypotheses on rho_p he gets a workable Euler
system and proves the desired inequality. Among other things,
it is necessary that rho_p is irreducible.
[The new proof replaces the argument above with one using
commutative algebra and and some clever observations by De
Shalit to fill in the gap.]
Suppose now that E is semistable.
Case 1. rho_3 is irreducible.
Take p = 3. By Tunnell's theorem (a) above is true. Under these
hypotheses the argument above works for rho_3 , so we conclude
that E is modular.
Case 2. rho_3 is reducible. Take p = 5 . In this case rho_5
must be irreducible, or else E would correspond to a rational
point on X_0(15) . But X_0(15) has only 4 noncuspidal rational
points, and these correspond to non-semistable curves. If we
knew that rho_5 were modular, then the computation above would
apply and E would be modular.
We will find a new semistable elliptic curve E' such that
rho_(E,5) = rho_(E',5) and rho_(E',3) is irreducible. Then by
Case I, E' is modular. Therefore rho_(E,5) = rho_(E',5) does
have a modular lifting and we will be done.
We need to construct such an E' . Let X denote the modular
curve whose points correspond to pairs (A, C) where A is an
elliptic curve and C is a subgroup of A isomorphic to the group
scheme E[5] . (All such curves will have mod-5 representation
equal to rho_E .) This X is genus 0, and has one rational point
corresponding to E , so it has infinitely many. Now Wiles uses
a Hilbert Irreducibility argument to show that not all rational
points can be images of rational points on modular curves
covering X , corresponding to degenerate level 3 structure
(i.e. im(rho_3) != GL_2(Z/3) ). In other words, an E' of the
type we need exists. (To make sure E' is semistable, choose it
5-adically close to E . Then it is semistable at 5, and at
other primes because rho_(E',5) = rho_(E,5) .)
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Tue Apr 04 17:26:57 EDT 1995