[Andrew Granville's message.]
Dear All,
There is great excitement here at the Newton Institute in Cambridge. Andrew
Wiles, from Princeton, has just announced his proof of Fermat's Last Theorem.
The work is extremely deep involving the latest ideas from a score of
different fields, including the theories of group schemes, crystalline
cohomology, Galois representations, deformation theory, Gorenstein rings,
(geometric) Euler systems and many others.
What he actually proves is a significant part of the Shimura-Taniyama-Weil
conjecture, namely that every semistable elliptic curve over Q is modular
(both reducible and irreducible). The Frey-Serre-Ribet theorem tells us that
this implies Fermat's Last Theorem.
The experts (Birch, Blasius, etc.) seem confident that it should not be
difficult to obtain all of the Shimura-Taniyama-Weil conjecture.
Because Wiles's proof rests on so much theory, one cannot be a hundred percent
confident that this is correct. There is a lot of skepticism over Faltings'
p-adic Hodge-Tate stuff (whatever that means), but the experts feel that it
can all be patched up.
So it does seem very likely that FLT is proved (and I can rip up ten of my
papers that are now entirely redundant).
All the best,
Andrew Granville.
------------------------------------------------
[Un messaggio di Ken Ribet, diffuso da Silverman.]
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 lecutures.
The method of Wiles borrows results and techniques from lots and lots of
people. To mention a few: Mazur, Hida, Flach, Kolyvagin, yours truly
[Silverman?], 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/3).
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_O(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 rationalpoint of X over which the covering remains ireducible.
You take E' to be the curve corresponding to this chosen rationalpoint of X.
Ken Ribet