Number Fields

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 $⟨x+y\omega ⟩=I^p$ 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 $yp$. This is a contradiction because $z,yp$ 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

$$⟨x+y⟩⟨x+y\omega ⟩...⟨x+y{\omega }^{p-1}⟩=⟨z{⟩}^p$$

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

Lemma: Suppose $x^p+y^p=z^p$ for integers $x,y,z$ for some prime $p\ge 5$. Then $x=y(\mathrm{mod}p)$.

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

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