## Fermat’s Last Theorem: Regular Primes

Here we fill in details for proving Fermat’s Last Theorem for regular primes (for case 1 solutions).

**Lemma:** Let \(p\) be an odd prime, and let \(\omega = e^{2 \pi i / p}\).
Suppose \(x^p + y^p = z^p\) for integers \(x, y, z\).

Then \(\langle x + y \omega \rangle = I^p\) for some ideal \(I\) of \(\mathbb{Z}[\omega]\).

**Proof:** The \(p\)th roots of unity give rise to the identity

which yields

and

For now assume \(\mathbb{Z}[\omega]\) is a UFD.

Suppose a prime \(\pi\) divides \((x+y\omega)\). Then \(\pi\) must also divide \(z\). Now if \(\pi\) divides \((x + y\omega^k)\) for some \(k \ne 1\), then \(\pi\) must also divide the difference \(y(\omega - \omega^k)\). From one of the identities above, this implies \(\pi\) also divides \(y p\). This is a contradiction because \(z, y p\) are coprime (recall we are only considering case 1 solutions).

Thus we deduce \(x + y\omega = u a^p\) for some unit \(u\) and \(a \in \mathbb{Z}[\omega]\).

Now we relax the assumption that \(\mathbb{Z}[\omega]\) is a UFD. It turns out that ideals of \(\mathbb{Z}[\omega]\) factor uniquely into prime ideals, and we proceed with a similar argument. Consider the ideal equation

Suppose a prime ideal \(I\) divides \(\langle x+y\omega \rangle\), that is, \(\langle x+y\omega\rangle \subset I\). If for some \(k \ne 1\) we also have \(\langle x+y\omega^k\rangle \subset I\), this would imply \(y(\omega - \omega^k) \in I\), thus \(y p \in I\). But we also have \(z \in I\), and since \(z, y p\) are coprime, we have that \(1 \in I\) which is a contradiction. Hence \(\langle x + y\omega\rangle = I^p\) for some ideal \(I\).∎

**Lemma:** Suppose \(x^p + y^p = z^p\) for integers \(x, y, z\) for some prime
\(p \ge 5\). Then \(x = y \pmod p\).

**Proof:** The polynomial \(f(t) = t^{p-1} + ...+t + 1\) is irreducible:
use Eisenstein’s criterion on \(f(t+1) = ((t+1)^p - 1)/t\).
Thus every element of \(\mathbb{Z}[\omega]\) may be written in the form
\(a_0 + a_1 \omega +... + a_{p-2}\omega^{p-2}\) where \(a_i \in \mathbb{Z}\)
(thus \(\mathbb{Q}[\omega]\) is a degree \(p-1\) extension of \(\mathbb{Q}\)).

Then \(p\) divides \(\alpha = a_0 + a_1 \omega +... + a_{p-2}\omega^{p-2}\) if and only if \(p\) divides all the \(a_i\), and we may easily do computations modulo \(p\mathbb{Z}[\omega]\), which we denote by \(mod p\). We have \(\beta = \gamma \pmod p\) implies \(\bar{\beta} = \bar{\gamma} \pmod p\) where the overbar denotes conjugation. \((\beta + \gamma)^p = \beta^p + \gamma^p \pmod p\) and \(\alpha^p =a \pmod p\) for some \(a \in \mathbb{Z}\).

Now suppose \(x + y\omega = u \alpha ^p \pmod p\). Note \(\alpha^p \in \mathbb{Z} \pmod p\) and \(\bar{\omega} = \omega^{-1}\). By a lemma due to Kummer, dividing both sides by their conjugates gives \(x + y\omega = (x + y \omega^{-1})\omega^k \pmod p\) for some integer \(k\). Since the minimal polynomial of \(\omega\) has degree \(p-1 \gt 3\), we must have \(k = 1 \pmod p\) (recall we are considering only case 1 solutions so \(p\) does not divide \(x\) or \(y\)). We may as well pick \(k = 1\) since \(\omega^p = 1\). Thus we see \((x -y) + (y -x)\omega = 0 \pmod p\), which implies \(x = y \pmod p\).∎

*blynn@cs.stanford.edu*💡