I found this printout in a computer lab on the campus of Oklahoma University a year and a half ago. It's a bit old, and I couldn't make out all of it. Plus, my knowledge of mathematical notation is way the hell off, and I have no chance in hell of understanding this, so the transcription of this will be more than a bit off. So, in mangled form, here's the outline of the proof of Fermat's Last Theorem.

### Groundwork : Frey Curves

Suppose there were a nontrivial solution of the Fermat equation for some number N, i. e. nonzero integers A, B, C, N such that A^N + B^N = C^N.

Then we recall that around 1982 Frey called attention to the elliptic curve Y^2 = X(X - A^N)(X + B^N).

Call this curve E. Frey noted it had some very unusual properties, and guessed it might be so unusual it could not actually exist.

To begin with, various routine calculations enable us to make some useful simplifying assumptions, without loss of generality. For instance, N may be supposed to be prime and is greater than or equal to 5. B can be assumed to be even, A is equal to 3 (mod 4), and C is equal to 1 mod 4. A, B, and C can be assumed relatively prime.

The "minimal discriminant" of E, can be computed to be ((ABC)^(2N)) = (2^8) - a power of 2 times a perfect prime power. One unusual thing about E is how large the discriminant is.

The conductor is a product of primes at which E has bad reduction, which is the same as the set of primes that divide the minimal discriminant. However, the exact power of each prime occurring in the conductor depends on what type of singularity the curve possesses modulo the primes of bad reduction. The definition of the conductor provides that P divides the conductor only to the first power if X(X-A)(X+B) has only a double root rather than a triple root mod P. Now, any prime can divide only A or B but not both, since otherwise it would also divide C, and we have assumed A, B, and C are relatively prime. Hence the the polynomial will have the form X(X+D) mod P, where (P,D) = 1. Hence there is only at most a double root modulo any prime, and therefore the conductor is square-free. In other words, E is semistable.

There are other odd things about E, which have to do with specific properties of its Galois representations. Because of these, Ribet's results allow us to conclude that E cannot be modular.

### Proof of Fermat's Last Theorem from the Taniyama-Shimura conjecture

After Frey drew attention to the unusual elliptic curve which would result if there were actually a nontrivial solution to the Fermat equation, Jean-Pierre Serre (who has made many contributions to modern number theory and algebraic geometry) formulated various conjectures which, sometimes alone and sometimes together with the Taniyama-Shimura conjecture, could be used to prove Fermat's Last Theorem.

Kenneth Ribet quickly found a way to prove one of these conjectures. The conjecture itself doesn't really talk about either Frey curves or FLT. Instead, it simply states that if the Galois representation associated with an elliptic curve E has certain properties, then E cannot be modular. Specifically, it cannot be modular in the sense that there exists a modular form which gives rise to the same Galois representation.

We need to introduce a little additional notation and terminology to explain this more precisely. Let S(N) be the (vector) space of cusp forms for Theta(N) of weight 2. "Classical" theory of modular forms shows that S(N) can be identified with the space of "holomorphic differentials" on the Riemann surface X(N). Furthermore, the dimension of S(N) is finite and equal to the "genus" of X(N). "Genus" is a standard topological property of surfaces, which is intuitively the number of holes in the surface. (E. g. a torus, such as an elliptic curve, has genus 1.)

But there are relatively simple explicit formulas for the genus of X(N). These formulas, developed long ago by Hurwitz in the theory of Riemann surfaces, involve the index of Theta(N) in G. A fact of crucial importance is that for N <. 11, the genus of X(N), and hence the dimension of S(N), is zero. In other words, S(N) contains only the constant form 0 in that case. We shall use this fact about S(2) very soon.

There are certain operators called Hecke operators, after Erich Hecke, on spaces of modular forms, and for the subspace S(N) in particular, since they preserve the weight of a form. Hecke operators can be defined concretely in various ways. There is a Hecke operator T(N) for all N grater than or equal to 1. There are formulas that relate T(N) for composite N to T(P) where is a prime dividing N, so T(P) for prime P determine all T(N).

All T(N) are linear operators on S(N). If there is an F in S(N) that is a simultaneous eigenvector of all T(N), i. e. T(N)(F) = Rho(N)F, where Rho(N) is the set of (C, F) is called an eigenfor. (Nontrivial eigenforms need not exist, e. g. if S(N) has dimension 0.) F is said to be normalized if its leading Fourier series coefficient is 1. In that case, the eigenvalues (n) turn out to be the Fourier series coefficients in the expansion F(N) = The sum of (from n = 0 to infinity) ANe^(2 * Pi * I * N * (undecipherable)).

It can be shown that if F(Z) is a cusp form which is a normalized eigenfunction for all T(P), then there is an Euler product decomposition for the L-function L(F,S). This is obviously of great technical usefulness in relating L-functions of forms and those of elliptic curves (which are Euler products by definition).

If F is in the set of all S(N) is a normalized eigenform of all Hecke operators, it can in fact be shown that the coefficients in the Fourier expansion are all algebraic numbers and that they generate a finite extension K of Q.

Prime ideals of the ring of integers of K are the analogues of prime numbers of Q. In the case that f is a normalized eigenform it is possible to carry out the construction of a Galois representation Phi(F,Gamma) of Gal(Lambda/Q) for any prime ideal Gamma of the ring of integers of K.

At last we can describe what Ribet proved. Suppose E is a semistable elliptic curve with conductor N and that its associated Galois representation Phi(E,P) for some prime P has certain properties. Suppose 2 divides N (which is true for Frey curves). If E is modular, then there is a normalized eigenform f and a prime ideal Gamma lying over P (i. e. one of the prime factors of P in the extension field generated by the Fourier coefficients of F) such that the Galois representation Phi(F,Gamma) is Phi(E,P). Ribet showed that it is possible to find an odd prime q is not equal to p which divides N such that there is another F' is in the set of all S(N/Q), and a corresponding prime ideal Gamma' of the ring of integers in the field generated by the coefficients of F' such that Phi(F',Gamma') gives essentially the same Galois representation. This is known as the "level lowering" conjecture since it asserts that under the right conditions there is an eigenform of a lower level that gives essentially the same representation.

But this process can be repeated as long as N has any odd prime factors. It is important that the curve E is semistable so that N is square-free. This means all that all odd prime factors of N can been eliminated, so there must be a nontrivial eigenform of level 2, i. e. in S(2), that gives essentially the same Galois representation. And that is a contradiction, since S(2) has dimension 0, hence contains no non-trivial forms. The contradiction means that E can't be modular.

Now we invoke the "unusual" properties of the Frey curve resulting from a solution of FLT. These properties allow it to be shown that the associated Galois representation has the properties required to apply Ribet's result. Hence the Frey curve can't be modular.

But the Frey curve is semistable, so the semistable case of the Taniyama-Shimura conjecture, which Wiles proved, implies the curve is modular. This contradiction means that the assumption of the existence of a nontrivial solution of the Fermat equation must be wrong, and so Fermat's Last Theorem is proved.

### Proof of the semistable case of the Taniyama-Shimura conjecture

Not very surprisingly (since it was such hard work), the proof is quite technical. However, the outline of it is relatively simple. In the following, we assume that E is a semistable elliptic curve with conductor N. We have to prove E is modular.

We know we can construct a Galois representation Phi(E,P^Infinity) : G->GL(Z) (I think... undecipherable) for any prime P. To show that E is modular, we have to show this representation is modular in a suitable sense. The wonderful thing is, this needs to be done for only one prime P, and we can "shop around" for whatever prime is easiest to work with.

To show Phi(E,P^Infinity) is modular involves finding a normalized eigenform F in S(N) with appropiate properties. The properties required are that the eigenvalues of F, which are its Fourier series coefficients, should be congruent mod Q to trace(Phi(E,P^Infinity) (something, undecipherable)) for all but a finite number of prime Q. ((something, undecipherable) G is the "Frobenius element".) We know that the trace is, for Q prime to PN, the coefficient A = Q + 1 - #(E(F)) of the Dirichlet series of L(E,S).

The longest and hardest part of Wiles' work was to prove a general result which is roughly that if Phi(E,P) is modular then so is Phi(E,P^Infinity). In other words, to show that E is modular, it is actually sufficient just to show that Phi(E,P) : G->GL(Z/pZ) is modular. This is called the "modular lifting problem".

The problem boils down to assuming that Phi(E,P) is modular and attempting to "lift" the representation to Phi(E,P). This is done mainly by working with the theory of representations as much as possible, without specific reference to the curve E. The proof uses a concept called "deformation", which suggests intuitively what goes on in the process of lifting.

The outcome of this part of Wiles work is:

Theorem: Suppose that E is a semistable elliptic curve over Q. Let P be an odd prime. Assume that the representation Phi(E,P) is both irreducible and modular. Then E is a modular elliptic curve.

At this point, all we have to do is find a single prime P such that Phi(E,P) is irreducible and modular. But Langlands and Tunnell had already proven in 1980-81 that Phi(E,3) is modular.

Unfortunately, this isn't quite enough. If Phi(E,3) is irreducible, we are done. But otherwise, one more step is required. So suppose Phi(E,3) is reducible. Wiles then considered Phi(E,5). That may be either reducible or irreducible as well. If it is reducible, Wiles proved directly that E is modular.

So the last case is if Phi(E,5) is irreducible. Wiles showed that there is another semistable curve E' such that Phi(E',3) is irreducible, and hence E' is modular by the above theorem. But Wiles could also arrange that the representations Phi(E',5) and Phi(E,5) are isomorphic. Hence Phi(E,5) is irreducible and modular, so E is modular by the theorem.

#### Copyright © 1996 by Charles Daney, All Rights Reserved (Reprinted without Permission)

I've e-mailed the last known address of the guy asking for permission. No reply. I'm leaving this in hopes that he or someone close to him reads this someday - just tell someone to delete this, give proof of who you are, and it will be done, no messy legal action needed.