In this post we shall sketch the reduction of Fermat’s Last Theorem to Wiles’ Theorem that every semistable elliptic curve over {{\mathbb Q}} is modular. I believe the ideas are due mainly to Frey, with a helpful big black box courtesy of Serre and Ribet. My main reference is the article of Stephens from the 1995 conference.

Suppose, for a fixed prime {p \geq 5}, that there are integers {a,b,c} with {a^p+b^p+c^p = 0}, {abc \not=0} and, assuming wlog that {a \equiv 3 \mod 4} and {b} is even, form the elliptic curve

\displaystyle E_0: y^2 = x(x-a^p)(x+b^p).

Then, by a direct computation, {E_0} is semistable with minimal discriminant {\Delta_{E_0} = 2^{-8} (abc)^{2p}} and conductor {N_{E_0}=\prod_{l|abc} l}.

Our strategy is to study the representation {\rho_E: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)} coming from the action of Galois on the {p}-torsion points of {E} over {\bar{{\mathbb Q}}} (note that this is the same {p} as in the previous paragraph).

In general, such Galois representations can be restricted to {G_{{\mathbb Q}_l}}, considered as the decomposition group of the prime {l} in {G_{\mathbb Q}} (that this is only defined up to conjugacy, determined by a choice of embedding {\bar{{\mathbb Q}} \rightarrow \bar{{\mathbb Q}}_l} is not important for us – we’ve already picked a basis for the torsion points anyway). The typical `good’ local behaviour at these primes are, for {l \not= p}, that {\rho_E} be unramified at {l} (in the sense that the absolute inertia group acts trivially), and for {l=p} a more general rather more technical condition which we describe by saying that {\rho_E} is flat at {p}.

Let us (nonstandardly) call a prime {l} bad (for a representation {\rho_E}) if {l \not= p} and {\rho_E} is ramified at {l}, or {l=p} and {\rho_E} fails to be flat at {p}. The important fact we need is that bad primes for such representations coming from elliptic curves are all flagged up by the minimal discriminant.

Proposition 1 For {E} a semistable elliptic curve, {l} is a bad prime for {\rho_E} iff {p} does not divide {ord_l(\Delta_E)}.

In particular, this implies for our curve {E_0} that the only bad prime for {\rho_{E_0}} is {2}. People suspected that this possibility of such an `arithmetically simple’ representation coming from an elliptic curve should be pretty unlikely.

Now, as well as these representations on {\mathbb{F}_p} coming from torsion points on an elliptic curve, it is possible to construct similar representations coming from certain modular forms. Indeed, from a newform {f} of weight 2, whose Fourier coefficients and character are rational, it is possible to form a representation {\rho_f: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)}. Serre made a series of conjectures about which representations could be obtained from newforms, one of which was the following result, a hard theorem proved by Ribet.

Theorem 2 (Serre’s Epsilon Conjecture) If {f} is a weight two newform of conductor {N} and {\rho_f} is absolutely irreducible, then letting {N'} be the factor of {N} obtained by throwing away all the non-bad primes which divide {N} at most once, it is possible to find a different weight two newform {g} of conductor {N'} such that {\rho_f \cong \rho_g}.

Finally, we need some results about modular curves. In fact, we just need the following basic fact. If {E} is a modular curve, then the representation {\rho_E} is equal to {\rho_f} for some newform {f} of conductor {N_E}.

Proposition 3 The curve {E_0} above (constructed from a contradiction of Fermat’s Last Theorem) is not modular.

First, one shows that {\rho_{E_0}} defined above is absolutely irreducible. Now, if {E_0} it were modular, {\rho_{E_0}} would be isomorphic to a representation {\rho_f} coming from a newform of conductor {N_{E_0}}. But looking at {N_{E_0}} and the fact that {2} is the only bad prime of {\rho_f}, Ribet’s theorem implies that there is a newform {g} of weight {2} and conductor {2} such that {\rho_f \cong \rho_g}. But the space {S_2(\Gamma_0(2))} has dimension {0}, by the basic theory of modular forms, so no such newform exists.

Having proved proposition 3, recalling that {E_0} was a semistable elliptic curve over {{\mathbb Q}}, Wiles’ theorem that every semistable elliptic curve over {{\mathbb Q}} is modular gives a contradiction and thus proves that there are no integers {a,b,c} with {a^p+b^p+c^p=0}, {abc \not= 0}.

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