Fermat's last theorem....

[Ted Shortliffe <ehs@camis.stanford.edu>]

Folks,
        I'm sure you'll all be as excited as I was to hear about the following.
Seriously, I wonder if technical discussions in medical informatics sound
this unintelligible to others......?
Ted

-----------

At 10:30 am today in Cambridge, England, Andrew Wiles claimed to
have proved Fermat's Last Theorem, and the Shimura-Taniyama-Weil
conjecture for semistable elliptic curves over Q. (see below)

> From K.C.Rubin@newton.cam.ac.uk Wed Jun 23 02:53:28 1993
> Date: Wed, 23 Jun 93 10:50 BST
> From: K.C.Rubin@newton.cam.ac.uk
> Subject: big news
>
> Andrew Wiles just announced, at the end of his 3rd lecture here,
> that he has proved Fermat's Last Theorem.  He did this by proving
> that every semistable elliptic curve over Q (i.e. square-free
> conductor) is modular.  The curves that Frey writes down, arising
> from counterexamples to Fermat, are semistable and by work of
> Ribet they cannot be modular, so this does it.
>
> It's an amazing piece of work.
>
> Karl
>
> From K.A.Ribet@newton.cam.ac.uk Wed Jun 23 05:40:01 1993
> Date: Wed, 23 Jun 93 13:36 BST
> From: K.A.Ribet@newton.cam.ac.uk
> To: nts_local@math.berkeley.edu
> Subject: announcement of Taniyama conjecture
>
> I imagine that many of you have heard rumours about Wiles's
> announcement a few hours ago that he can prove Taniyama's conjecture
> for semistable elliptic curves over Q.  This case of the Taniyama
> conjecture implies Fermat's Last Theorem, in view of the result
> that I proved a few years ago.  (I proved that the "Frey elliptic
> curve" constructed from a possible solution to Fermat's equation
> cannot be modular, i.e., satisfy Taniyama's Conjecture.  On the
> other hand, it is easy to see that it is semistable.)
>
> 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; you also have to use Euler systems a la
> Kolyvagin, except in some new geometric guise.
>
> 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 parametrizes 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.
>
> -ken ribet