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$p \in \ZZ_{\ge 3}$ a prime, $\theta_p = e^{2\pi i
/ p}$, $K = \QQ(\theta_p)$. $\forall i, j \in \ZZ$
with $I \not\equiv j \pmod{p}$, there exists
$u_{i, j} \in \ZZ[\theta_p]^\times$ such that $p =
u_{i, j} (1 - \theta_p)^{p - 1}$.

Proof of $\O_K = \ZZ[\theta_p]$. We made an
indirect assumption, and we want to get a
contradiction. We found $\beta \in \O_K \setminus
z\ZZ[\theta_p]$ and $\gamma \in \ZZ[\theta_p]$ and
$\alpha \in \ZZ$ such that
\[ (1 - \theta_p) \beta = a + (1 - \theta_p)\gamma
.\]
We have $p \nmid a$, for otherwise
\[ \beta = \frac{a}{1 - \theta_p} + \gamma ,\]
and if $a = pa'$, then
\[ \frac{a}{1 - \theta_p} = \frac{a'u (1 -
\theta_p)^{p - 1}}{1 - \theta_p} \in \ZZ[\theta_p]
.\]
So $\beta \in \ZZ[\theta_p]$, which is not the
case. This proves $p \nmid a$. On the other hand,
\[ \frac{a}{1 - \theta_p} = \beta - \gamma \in
\O_K .\]
Then
\[ \ub{\frac{1}{a} \left( \frac{a}{1 - \theta_p}
\right)^{p - 1}}_{\in \O_K} = \ub{\frac{a^{p -
1}}{p}}_{\in \QQ} \]
hence
\[ \frac{a^{p - 1}}{p} \in \ZZ ,\]
a contradiction to $p \nmid a$.

Proof of the claim that: $\langle p \rangle = P^{p
- 1}$ for a prime $P \subset \O_K$, and
\[ P = \langle \theta_p^i - \theta_p^j \rangle \]
for any $i, j \in \ZZ$ such that $i \not\equiv j
\pmod{p}$.

Let $P_{ij} = \langle \theta_p^i -
\theta_p^j \rangle$, then $\langle p \rangle =
P_{ij}^{p - 1}$. $N(\langle p \rangle) = p^{p -
1}$, hence $N(P_{ij}) = p$. So $P_{ij}$ must be a
prime ideal. By uniqueness of factorisation,
$P_{ij}$ does not depend on $i$ and $j$.

\begin{flashcard}[regular-prime-defn]
\begin{definition*}[Regular prime]
  \glsadjdefn{regp}{regular}{prime}
  \cloze{A prime $p \in \ZZ$ is regular if $p \nmid
  \cln(\QQ(\theta_p))$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[regular-flt-thm]
\begin{theorem*}[Regular Fermat's Last Theorem]
  \cloze{Let $p \ge 5$ ba a \emph{\gls{regp} prime}. Then
  there are no solutions of
  \[ x^p + y^p = z^p \]
  with $x, y, z \in \ZZ$. such that $p \nmid xyz$
  (the case $p \nmid xyz$ is known as ``Case
  I'').}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[flt-subprop]
\begin{proposition*}
  Assume that $x, y, z$ is a solution of $x^p +
  y^p = z^p$ and assume $\gcd(x, y, z) = 1$ and $p
  \nmid xyz$. Then
  \cloze{
  \[ x + \theta_p y = u \alpha^p \]
  where $u \in \O_K^\times$, and $\alpha \in
  \O_K$.}
\end{proposition*}

\begin{proof}
  \cloze{Recall:
  \[ (x + y)(x + \theta_p y) \cdots (x +
  \theta_p^{p - 1} y) = z^p .\]
  Claim: there is no prime $Q \subset \O_K$ such
  that $Q \mid \langle x + \theta_p^i y \rangle,
  \langle x + \theta_p^j y \rangle$ for $i \not
  \equiv j \pmod{p}$.

  Suppose the contrary. Then
  \[ Q \mid \ub{\langle \theta_p^i y - \theta_p^j y
  \rangle}_{P\langle y \rangle}, \ub{\langle \theta_p^{-i} x
  - \theta_p^{-j} x \rangle}_{P\langle x \rangle}
  .\]
  If $Q = P$, then $P \mid \langle z \rangle^p$,
  so $P \mid \langle z \rangle$, so $z \in P \cap
  \ZZ = p\ZZ$, hence $p \mid z$, cotnradicting our
  assumption of being in Case I. So $Q \neq P$.
  Then $Q \mid \langle x \rangle, \langle y
  \rangle$, so $x, y \in Q$. We must have $\gcd(x,
  y) = 1$, for any common prime factor would also
  divide $z$ by $z^p = x^p + y^p$, and we assume
  $\gcd(x, y, z) = 1$. So we can find $a, b \in
  \ZZ$ such that $1 = ax + by$. Then $1 \in Q$,
  which is not possible. So we have proved the
  claim (that there is no prime $Q$ dividing
  more than one of the ideals $\langle x + \theta_p^i y
  \rangle$).

  Then $\langle x + \theta_p y \rangle = I^p$ for
  some ideal $I \subset \O_K$ (not necessarily
  prime). We assumed that $p \nmid \cln(K)$. Hence
  the only class in the class group whose $p$-th
  power is the unit element, that is the class of
  principal ideals, is the unit element itself (the class
  of principal ideals). We know that $I^p$ is
  principal because $I^p = \langle x + \theta_p y
  \rangle$, so $I$ must be principal too, and the
  proposition follows.}
\end{proof}
\end{flashcard}

\begin{flashcard}[regular-flt-prop2]
\begin{proposition*}
  Assume that $x, y, z$ is a solution of $x^p +
  y^p = z^p$ and assume $\gcd(x, y, z)$. Then we
  must have \cloze{$x \equiv y \pmod{p}$.}
\end{proposition*}

\end{flashcard}

\begin{flashcard}[proof-of-regular-flt]
\prompt{Proof of regular FLT? \\}

\begin{proof}
  \cloze{Suppose that there is a solution $x, y, z$. We may
  assume $\gcd(x, y, z) = 1$ (by dividing by any
  common factor). By a previous proposition, we
  get $x \equiv y \pmod{p}$. Applying it to $x^p -
  z^p = y^p$, we get $x \equiv -z \pmod{p}$. Then
  \[ x^p + y^p - z^p \equiv 3x^p \pmod{p} \]
  But the LHS is equal to $0$, so $p \mid 3x^p$,
  but $p \nmid 3$, because $p \ge 5$, and $p \nmid
  x$ because of Case I.}
\end{proof}
\end{flashcard}
