## 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 $x = y (mod p)$.

**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 $\beta = \gamma (mod p)$ implies $\bar{\beta} = \bar{\gamma} (mod p)$ where the overbar denotes conjugation. $(\beta + \gamma)^p = \beta^p + \gamma^p (mod p)$ and $\alpha^p =a (mod p)$ for some $a \in \mathbb{Z}$.

Now suppose $x + y\omega = u \alpha ^p (mod p)$. Note $\alpha^p \in \mathbb{Z} (mod p)$ and $\bar{\omega} = \omega^{-1}$. By a lemma due to Kummer, dividing both sides by their conjugates gives $x + y\omega = (x + y \omega^{-1})\omega^k (mod p)$ for some integer $k$. Since the minimal polynomial of $\omega$ has degree $p-1 \gt 3$, we must have $k = 1 (mod p)$ (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 $(x -y) + (y -x)\omega = 0 (mod p)$, which implies $x = y (mod p)$.∎