[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