Fermat's Last Theorem

Theorem 1.1   Let $n>2$ be an integer. If $a, b, c\in\mathbb{Z}$ and

$$a^n + b^n = c^n,$$

then $abc=0$.

Proof. [Proof (sketch)] First reduce to the case when $n=\ell$ is a prime greater than $3$ (see Exercise 1.2). Suppose that

$$a^\ell + b^\ell = c^\ell$$

with $a, b, c\in\mathbb{Z}$ and $abc\neq 0$. Permuting $(a,b,c)$, we may suppose that $b$ is even and that we have $a\equiv 3\pmod{4}$. Following Gerhard Frey, consider the elliptic curve $E$ defined by

$$y^2 = x(x-a^{\ell})(x+b^{\ell}).$$

The discriminant of $E$ is  $2^4(abc)^{2\ell}$ (see Exercise 1.3 below).

Andrew Wiles and Richard Taylor [Annals of Math., May 1995] proved that $E$ must be ``modular''. This means that there is a ``modular form''

$$f(z) = \sum_{n=1}^{\infty} a_n e^{2\pi i n z}$$

of ``level $N=abc$'' such that for all primes $p\nmid abc$,

$$a_p = p+1-\char93 E(\mathbb{Z}/p\mathbb{Z}).$$

Ken Ribet [Inventiones Math., 1991] used that the discriminant of $E$ is a perfect $\ell$th power (away from $2$) to prove that there is a cuspidal modular form

$$g(z) = \sum_{n=1}^{\infty} b_n e^{2\pi i n z}$$

of ``level $2$'' such that

$$a_p \equiv b_p \pmod{\ell}$$   for all $$p\nmid abc.$$

This is a contradiction because the space of ``cuspidal modular forms'' of level $2$ has dimension 0 (see Section 3.1). $\qedsymbol$

Exercise 1.2   Reduce to the prime case. That is, show that if Fermat's last theorem is true for prime exponents, then it is true.

Exercise 1.3   Prove that $y^2 = x(x-a^{\ell})(x+b^{\ell})$ has discriminant $2^4(abc)^{2\ell}$.

The rest of this lecture is about the words in the proof that are in quotes.