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- From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
- Subject: sci.math FAQ: Wiles attack
- Summary: Part 6 of many, New version,
- Originator: alopez-o@neumann.uwaterloo.ca
- Message-ID: <DI76JI.H68@undergrad.math.uwaterloo.ca>
- Sender: news@undergrad.math.uwaterloo.ca (news spool owner)
- Approved: news-answers-request@MIT.Edu
- Date: Fri, 17 Nov 1995 17:14:06 GMT
- Expires: Fri, 8 Dec 1995 09:55:55 GMT
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- Keywords: Fermat Last Theorem
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- Xref: senator-bedfellow.mit.edu sci.math:124631 sci.answers:3459 news.answers:57910
-
- Archive-Name: sci-math-faq/FLT/Wiles
- Last-modified: December 8, 1994
- Version: 6.2
-
-
-
- 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 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 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 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:
-
-
-
-
-
-
-
- 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 Bloch-Kato 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 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.
-
- [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 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) != 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) .)
-
-
-
-
- _________________________________________________________________
-
-
-
- alopez-o@barrow.uwaterloo.ca
- Tue Apr 04 17:26:57 EDT 1995
-
-
-