Interesting People mailing list archives
Fermat's last theorem....
From: Ted Shortliffe <ehs () camis stanford edu>
Date: Thu, 24 Jun 93 17:34:09 -0800
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
Current thread:
- Fermat's last theorem.... Ted Shortliffe (Jun 24)
- <Possible follow-ups>
- Re: Fermat's last theorem.... Tom Lincoln (Jun 24)