## 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

$(x-1)(x-\omega)...(x-\omega^{p-1}) = x^p - 1$

which yields

$(1-\omega)...(1-\omega^{p-1}) = p$

and

$(x+y)(x+y\omega)...(x+y\omega^{p-1}) = z^p$

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

$\langle x+y \rangle\langle x+y\omega \rangle... \langle x+y\omega^{p-1}\rangle=\langle z\rangle ^p$

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 $$x = y \pmod p$$.∎

Ben Lynn blynn@cs.stanford.edu 💡