# 4 Fermat’s Last Theorem for regular primes

Let \(p \geq 5\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(x, y \in \mathbb {Z}\) coprime.

For \(i \neq j\) with \(i,j \in {0,\dots ,p-1}\) we can write

with \(u\) a unit in \(\mathbb {Z}[\zeta _p]\). From this it follows that the ideals

are pairwise coprime.

Lemma 3.8 gives that \(u\) is a unit. So all that needs to be proved is that the ideals are coprime. Assume not, then for some \(i \neq j\) we have some prime ideal \(\mathfrak {p}\) dividing by \((x+y\zeta _p^i)\) and \((x+y\zeta _p^j)\). It must then also divide their sum and their difference, so we must have \(\mathfrak {p}| (1-\zeta _p)\) or \(\mathfrak {p}| y\). Similarly, \(\mathfrak {p}\) divides \(\zeta _p^j(x+y\zeta _p^i)-\zeta _p^i(x+y\zeta _p^j)\) so \(\mathfrak {p}\) divides \(x\) or \((1-\zeta _p)\). We can’t have \(\mathfrak {p}\) dividing \(x,y\) since they are coprime, therefore \(\mathfrak {p}|(1-\zeta _p)\). We know that since \((1-\zeta _p)\) has norm \(p\) it must be a prime ideal, so \(\mathfrak {p}=(1-\zeta _p)\). Now, note that \(x+y \equiv x+y\zeta _p^i \equiv 0 \mod \mathfrak {p}\). But since \(x,y \in \mathbb {Z}\) this means we would have \(x+y \equiv 0 \pmod p\), which implies \(z^p \equiv 0 \pmod p\) which contradicts our assumptions.

Let \(p\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(\alpha \in \mathbb {Z}[\zeta _p]\). Then \(\alpha ^p\) is congruent to an integer modulo \(p\).

Just use \((x+y)^p \equiv x^p + y^p \pmod p\) and that \(\zeta _p\) is a \(p\)-th root of unity.

Let \(p\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\) with \(\alpha =\sum _i a_i \zeta _p^i\). Let us suppose that there is \(i\) such that \(a_i = 0\). If \(n\) is an integer that divides \(\alpha \) in \(\mathbb {Z}[\zeta _p]\), then \(n\) divides each \(a_i\).

Looking at \(\alpha =a_0+a_1\zeta _p+\cdots +a_{p-1}\zeta _p^{p-1}\), if one of the \(a_i\)’s is zero and \(\alpha /n \in \mathbb {Z}[\zeta _p]\), then \(\alpha /n=\sum _i a_i/n \zeta _p^i\). Now, as \(\alpha /n \in \mathbb {Z}[\zeta _p]\), pick the basis of \(\mathbb {Z}[\zeta _p]\) which does not contain \(\zeta _p\) (which is possible as any subset of \(\{ 1,\zeta _p,\dots ,\zeta _p^{p-1}\} \) with \(p-1\) elements forms a basis of \(\mathbb {Z}[\zeta _p]\).). Then \(\alpha =\sum _i b_i \zeta _p^i\) where \(b_i \in \mathbb {Z}\). Therefore comparing coefficients, we get the result.

Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\). Let \(x\) and \(y\) be integers such that \(x+y\zeta _p^i=u \alpha ^p\) with \(u \in \mathbb {Z}[\zeta _p]^\times \) and \(\alpha \in \mathbb {Z}[\zeta _p]\). Then there is an integer \(k\) such that

Using lemma 3.7 we have \((x+y\zeta _p^i) =\beta \zeta _p^k \alpha ^p\) which is congruent modulo \(p\) to \(\beta \zeta _p^k a \pmod p\) for some integer \(a\) by 4.2. Now, if we consider the complex conjugate we have \(\overline{(x+y\zeta _p^i) }\equiv \beta \zeta _p^{-k}a \pmod p\). Looking at \((x+y\zeta _p^i)-\zeta _p^{2k}\overline{(x+y\zeta _p^i) }\) then gives the result.

Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(K=\mathbb {Q}(\zeta _p)\). Assume that we have \(x,y,z \in \mathbb {Z}\) with \(\gcd (xyz,p)=1\) and such that

This is the so called "case I". To prove Fermat’s last theorem, we may assume that:

\(p \geq 5\);

\(x,y,z\) are pairwise coprime;

\(x \not\equiv y \mod p\).

The first part is easy.

Reducing modulo \(p\), using Fermat’s little theorem, you get that if \(x \equiv y \equiv -z \pmod p\) then \(3z \equiv 0 \pmod p\). But since \(p {\gt}3\) this means \(p |z\) but this contradicts \(\gcd (xyz,p)=1\). Now, if \(x \equiv y \pmod p\) then \(x \not\equiv -z \pmod p\) we can relabel \(y,z\) so that wlog \(x \not\equiv y\) (this uses that \(p\) is odd).

A prime number \(p\) is called regular if it does not divide the class number of \(\mathbb {Q}(\zeta _p)\).

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(\gcd (xyz,p)=1\).

For \(p=3\) use the standard elementary arguments, so assume \(p \geq 5\).

First thing is to note that if \(x^p+y^p=z^p\) then

as ideals. Then since by 4.1 we know the ideals are coprime, then by lemma 3.9 we have that each \((x+y\zeta _p^i)=\mathfrak {a}^p\), for \(\mathfrak {a}\) some ideal. Note that, \([\mathfrak {a}^p]=1\) in the class group. Now, since \(p\) does not divide the size of the class group we have that \([\mathfrak {a}]=1\) in the class group, so its principal. So we have \(x+y\zeta _p^i=u_i\alpha _i^p\) with \(u_i\) a unit. So by 4.4 we have some \(k\) such that \(x+y\zeta _p-\zeta _p^{2k}x-\zeta _p^{2k-1}y \equiv 0 \pmod p\). If \(1,\zeta _p,\zeta _p^{2k},\zeta _p^{2k-1}\) are distinct, then 4.3 says that (since \(p \geq 5\)) \(p\) divides \(x,y\), contrary to our assumption. So they cannot be distinct, but checking each case leads to a contradiction, therefore there cannot be any such solutions.

Let \(p\) be a regular prime and let \(u \in \mathbb {Z}[\zeta _p]^\times \). If \(u \equiv a \mod p\) for some \(a \in \mathbb {Z}\), then there exists \(v \in \mathbb {Z}[\zeta _p]^\times \) such that \(u=v^p\).

See the Lean proof.

In these next few lemmas we are following [ BS66 , SD01 , HLA\(^{+}\)98 ] .

Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\) and that \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\). Then each of \((x+\zeta _p^i y)\) is divisible by \((1-\zeta _p)\) and there is a unique \(i_0\) such that \((x+\zeta _p^{i_0})\) is divisible by \((1-\zeta _p)^2\).

By our assumptions we have the following equality of ideals in \(\mathbb {Z}[\zeta _p]\):

where \(\mathfrak {a}=(z)\), \(\mathfrak {p}=(1-\zeta _p)\) (which is prime) and \(m=n(p-1)\). Now as \(mp \geq p\) we must have that at least one of the terms on the lhs is divisible by \(\mathfrak {p}\).

Note that since

it follows every \(x+\zeta _p^k\) is divisible by \(\mathfrak {p}\) for \(0 \le k \le p-1\). This proves the first claim.

For the second claim, we begin by observing that if \(x+\zeta _p^ky \equiv x + \zeta _p^i y \mod \mathfrak {p}^2\) (for \(0 \leq k {\lt} i \leq p-1\)) then \(\zeta _p^ky(1-\zeta _p^{i-k}) \equiv 0 \mod \mathfrak {p}^2\) which cannot happen as \(y\) is coprime to \(\mathfrak {p}\). Therefore since, for \(0 \leq k \leq p-1\), \(x+\zeta _p^ky\) are all distinct modulo \(\mathfrak {p}^2\) we must have that \(\frac{x+\zeta _p^k}{1-\zeta _p}\) are non-congruent modulo \(\mathfrak {p}\). The second claim now follows by noting that (since \(N(\mathfrak {p})=p\)), the numbers \(\frac{x+\zeta _p^k}{1-\zeta _p}\) form a complete set of residues modulo \(\mathfrak {p}\), so one must be divisible by \(\mathfrak {p}\).

Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\), \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) and \(x+y\) is divisible by \(\mathfrak {p}^2\) and \(x+ \zeta _p^ky\) is only divisible by \(\mathfrak {p}=(1-\zeta _p)\) (for \(0 {\lt} k \leq p-1\)). Let \(\mathfrak {m}= \mathrm{gcd}((x),(y))\). Then:

We can write

\[ (x+y)=\mathfrak {p}^{p(m-1)+1}\mathfrak {m}\mathfrak {c}_0 \]and

\[ (x+\zeta _p^k y)=\mathfrak {p}\mathfrak {m}\mathfrak {c}_k \](for \(0 {\lt} k \leq p-1\)) where \(m=n(p-1)\) and with \(\mathfrak {c}_i\) pairwise coprime.

Each \(\mathfrak {c}_k=\mathfrak {a}_k^p\) and \(\mathfrak {a}_k\mathfrak {a}_0^{-1}\) is principal (as a fractional ideal).

Let \(p\) be a regular odd prime, \(\epsilon \in \mathbb {Z}[\zeta _p]^\times \) and \(n \in \mathbb {Z}_{\geq 1}\). Then the equation \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) has no solutions with \(x,y,z \in \mathbb {Z}[\zeta _p]\), all non-zero and \(xyz\) coprime to \((1-\zeta _p)\).

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(p | xyz\).

See the Lean proof.

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(xyz \ne 0\).

See the Lean proof.