next up previous
Next: Holomorphic Functions Up: Lecture 32: Fermat's Last Previous: Lecture 32: Fermat's Last

Fermat's Last Theorem

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

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

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

$\displaystyle 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

$\displaystyle 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''

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

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

$\displaystyle 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

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

of ``level $ 2$'' such that

$\displaystyle a_p \equiv b_p \pmod{\ell}$   for all $\displaystyle 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.


next up previous
Next: Holomorphic Functions Up: Lecture 32: Fermat's Last Previous: Lecture 32: Fermat's Last
William A Stein 2001-11-30