Number Fields

Next we attempt to show $x^n+y^n=z^n$ has no solutions over the integers for $n>2$. There is a truly remarkable proof of this, but unfortunately this site is too small to contain it. We can make a start though.

Using the result on Pythagorean triples, it can be shown that the case $n=4$ has no solutions, so we need only consider when $n$ is some odd prime $p$.

There are two cases. The first, case 1, is that $p$ divides none of $x,y,z$. Case 2 is that $p$ divides exactly one of $x,y,z$. We only consider the first case. The second is messier, but the results are similar.

For $p=3$, considering the equation modulo 9 immediately shows there are no case 1 solutions.

For $p>3$, we may attempt to mimic the process that used to find Pythagorean triples, as Kummer did. First we rewrite the equations as

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

where $\omega =e^{2\pi i/p}$.

Then if $\mathbb{Z}[\omega ]$ is a UFD, one can show that $x+y\omega =u{a}^p$ for some unit $u$ and $a\in \mathbb{Z}[\omega ]$. This in turn implies $x=y(\mathrm{mod}p)$. We also have $x^p+(-z{)}^p=(-y{)}^p$, which by the same reasoning implies $x=-z(\mathrm{mod}p)$. Thus

$$2x^p=x^p+y^p=z^p=-x^p(\mathrm{mod}p)$$

so $p$ divides $3x^p$, a contradiction since $p>3$ and $p$ does not divide $x$ because we are in case 1.

Unfortunately $\mathbb{Z}[\omega ]$ is not always a UFD. However, it turns out it is always a Dedekind domain, which means its ideals factor uniquely into prime ideals. Using this fact, one can show that the ideal $⟨x+y\omega ⟩=I^p$ for some ideal $I$.

If $p$ is regular, defined below, this implies that $I$ is also a principal ideal thus $⟨x+y\omega ⟩=I^p=⟨a{⟩}^p=⟨a^p⟩$ for some element $a$. In other words, $x+y\omega =u{a}^p$ for some unit $u$, and we can continue as before to prove case 1 of Fermat's Last Theorem for such primes.

Define a relation $~$ on the set of ideals of $\mathbb{Z}[\omega ]$ as follows. For ideals $A,B$, say $A~B$ if $aA=bB$ for some $a,b\in \mathbb{Z}[\omega ]$. It turns out this is an equivalence relation, and furthermore, there are finitely many equivalence classes. The number of classes is the class number, and denoted by $h$.

We call a prime $p$ regular if $p$ does not divide $h$.

The ideal classes form a group under multiplication. The identity element is the class of all principal ideals. This explains why regularity is important. If $p$ is regular, then its ideal class group cannot contain an element of order $p$, and hence if $I^p$ is principal, then so is $I$.