# Zombies Ate My Neighbors

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1Q: What is the current status of Fermat's last theorem?

and

Did Fermat prove this theorem? Fermat's Last Theorem:

There are no positive integers x,y,z, and n > 2 such that x^n + y^n = z^n.

I heard that <insert name here> claimed to have proved it but later on the proof was found to be wrong. ...

A: The status of FLT has remained remarkably constant. Every few

years, someone claims to have a proof ... but oh, wait, not quite.

UPDATE... UPDATE... UPDATE

Andrew Wiles, a researcher at Princeton, Cambridge claims to have found a proof.

SECOND UPDATE...

A mistake has been found. Wiles is working on it. People remain mildly optimistic about his chances of fixing the error. The proposed proof goes like this:

The proof was presented in Cambridge, UK during a three day seminar to an audience including some of the leading experts in the field.

The manuscript has been submitted to INVENTIONES MATHEMATICAE, and is currently under review. Preprints are not available until the proof checks out. Wiles is giving a full seminar on the proof this spring.

The proof is long and cumbersome, but here are some of the first few details:

*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; you also have to use Euler systems a la Kolyvagin, except in some new geometric guise.

CLARIFICATION: This step in Wiles' manuscript, the Selmer group bound, is currently considered to be incomplete by the reviewers. Yet the reviewers (or at least those who have gone public) have confidence that Wiles will fill it in. (Note that such gaps are quite common in long proofs. In this particular case, just such a bound was expected to be provable using Kolyvagin's techniques, independently of anyone thinking of modularity. In the worst of cases, and the gap is for real, what remains has to be recast, but it is still extremely important number theory breakthrough work.)

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 parameterizes 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. *From the previous version of the FAQ: (b) conjectures arising from the study of elliptic curves and modular forms. -- The Taniyama-Weil-Shmimura conjecture. There is a very important and well known conjecture known as the Taniyama-Weil-Shimura 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 conj. 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 Tate-Shafarevich 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 non-trivial, that infinite-descent arguments do not work; that in some sense 'Sha blocks the descent'. Somewhat more technically, Sha is an obstruction to the local-global principle [e.g. the Hasse-Minkowski theorem].

*From Karl Rubin:

Theorem. If E is a semistable elliptic curve defined over Q, then E is modular.

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(Q^bar/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 Bloch-Kato conjecture, is what Wiles needs to verify. He does this using a Kolyvagin-type 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. 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 non-semistable 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 mod-5 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) not GL_2(Z/3)). In other words, an E' of the type we need exists. (To make sure E' is semistable, choose it 5-adically close to E. Then it is semistable at 5, and at other primes because rho_{E',5} = rho_{E,5}.)

Referencesm:

American Mathematical Monthly January 1994.

Notices of the AMS, Februrary 1994.