ArchiveName: scimathfaq/FLT/Wiles
Lastmodified: December 8, 1994 Version: 6.2 See reader questions & answers on this topic!  Help others by sharing your knowledge 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 3division 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 3division 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 noncuspidal 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 5division 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 5division 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 3division structure. You use Hilbert irreducibility and the Cebotarev density theorem (in some way that hasn't yet sunk in) to produce a noncuspidal 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 TaniyamaWeilShmimura conjecture. There is a very important and well known conjecture known as the TaniyamaWeilShimura 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 TateShafarevich 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 nontrivial, that infinitedescent arguments do not work; that in some sense ``Sha blocks the descent''. Somewhat more technically, Sha is an obstruction to the localglobal principle [e.g. the HasseMinkowski 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 BlochKato 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 Kolyvagintype 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 nonsemistable 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 mod5 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 5adically close to E . Then it is semistable at 5, and at other primes because rho_(E',5) = rho_(E,5) .) _________________________________________________________________ alopezo@barrow.uwaterloo.ca Tue Apr 04 17:26:57 EDT 1995 User Contributions:Comment about this article, ask questions, or add new information about this topic:
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