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\section{Introduction}

If $L \supset K$ are fields, then $L$ is an
extension of $K$. Notation $L / K$. We can think
of $L$ as a vector space over $K$. The dimension
of $L / K$ is called the degree of the field
extension, and is written as $[L : K]$.

\begin{flashcard}[number-field-defn]
\begin{definition*}[Number field]
  \glsnoundefn{nf}{number field}{number fields}
  \cloze{A \emph{number field} is a subfield $K$ of $\CC$ with
  $[K : \QQ] < \infty$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\QQ$.
    \item Let $\alpha \in \CC$ be algebraic, i.e.
      a root of a polynomial with integer
      coefficients. Then $\QQ(\alpha)$ (this
      notation means the smallest subfield of
      $\CC$ containing $\alpha$) is a
      \gls{nf}. $[\QQ(\alpha) : \QQ] = \deg
      f_\alpha$, where $f_\alpha$ is the unique
      monic minimal polynomial of $\alpha$ over
      $\QQ$. By the Primitive Element Theorem (see
      \courseref[Galois Theory]{Galois}), all
      number fields are of this form.
    \item Quadratic fields: $K$ with $[K : \QQ] =
      2$. $K = \QQ(\sqrt{m})$ where $m \in \ZZ$,
      $m \neq 0, \pm 1$ and square-free.
    \item Cyclotomic fields. Let $n \in \ZZ_{\ge
      3}$. Let $\theta_n = e^{2\pi i / n}$. This
      is an $n$-th root of unity, i.e. $\theta_n^n
      = 1$. Then $K = \QQ(\theta_n)$ is a
      \gls{nf}, with $[\QQ(\theta_n) : \QQ] =
      \varphi(n)$, where $\varphi(n)$ is the
      number of residue classes modulo $n$ that
      are coprime to $n$.
  \end{enumerate}
\end{example*}

\vspace{-1em}
Why study Number Fields?

Consider Fermat equation:
\[ x^n + y^n = z^n, \qquad x, y, z \in \ZZ .\]
Consider the $n = 2$ case. We are interseted in primitive
solutions (solutions with $\gcd(x, y, z) =
1$). Furthermore we assume $x, y, z \ge 0$.

Assume $2 \nmid y$. Note that $(z - x)(z + x) =
z^2 - x^2 = y^2$.

Claim: $\gcd(z - x, z + x) = 1$.
Indeed let $p \mid z - x, z + x$. Then $p \mid 2z,
2x, y^2$. But $\gcd(2x, 2z, y^2) = 1$ (since we
assumed $2 \nmid y$ and $\gcd(x, y, z) = 1$), so
no such $p$ exists.

$y^2$ has all prime factors
with even multiplicities, and these factors must
go to either $(z - x)$ or $(z + x)$ with the
multiplicity they occur in $y^2$. Conclusion: $z -
x = n^2$, $z + x = m^2$ for some $0 \le n \le m \in
\ZZ$ and coprime and odd. We now have:
\[ x = \frac{m^2 - n^2}{2}, \qquad z = \frac{m^2 +
n^2}{2}, \qquad y = mn \]
All solutions must be of this form. Easy to check
that these are all solutions. More customary to
write
\[ x = 2mn, \qquad y = m^2 - n^2, \qquad z = m^2 +
n^2 ,\]
$m > n$, $\gcd(m, n) = 1$, and exactly one of them
is even.

Fermat claimed: No solutions for $n \ge 3$ and $x,
y, z \in \ZZ_{> 0}$. First step is to factorize
the equation. For $n = 2$, we used $X^2 - 1 = (X -
1)(X + 1)$. For general $n$, we have $X^n - 1 =
\prod_{j = 0}^{n - 1} (X - \theta_n^j)$. Assume
$n$ is odd, then consider $X \to -X$: $X^n + 1 =
\prod_{j = 0}^{n - 1} (X + \theta_n^j)$. Now
substitute $X \leftarrow \frac{x}{y}$ to get
\[ z^n = x^n + y^n = \prod_{j = 0}^{n - 1} (x +
y\theta_n^j). \]
Next step: show that $(x + y\theta_n^j)$ is an
$n$-th power.

Issues:
\begin{itemize}
  \item Unique factorisation may fail. In fact,
    $\ZZ[\theta_n]$ is not a UFD for any prime $n
    \ge 23$.
  \item Even if it is a UFD, if $\alpha$ has all
    prime factors with multiplicity divisible by
    $n$, we can conclude only that $\alpha = u
    \beta^n$ for some $\beta \in \ZZ[\theta_n]$
    and some unit $u \in \ZZ[\theta_n]^\times$
    (reminder: $u \in R$ is a unit if there exists
    $u^{-1} \in R$ such that $uu^{-1} = 1$, and
    $R^\times$ denotes the set of units in $R$).
\end{itemize}

\begin{theorem*}[Kummer 1850]
  If $p$ is a regular prime (not defined here),
  then
  \[ x^p + y^p = z^p \]
  has no solutions with $x, y, z \in \ZZ_{\ge 1}$.
\end{theorem*}

\vspace{-1em}
Aims of the course:
\begin{itemize}
  \item Ring of integers in \glspl{nf}
  \item Unique factorisation of ideals
  \item Units
  \item Fermat equation: prove Kummer's Theorem in
    the case $p \nmid xyz$
\end{itemize}
