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Note that for $I \subset \O_K$, there exists $k >
0$ such that $I^k$ is principal if and only if the
order of $I$ in $\Cl(K)$ is finite. But we now
that $\Cl(K)$ is finite, hence the order is always
finite, so there always exists some $k > 0$ such
that $I^k$ is principal.

\textbf{Units:} $\alpha \in \O_K$ is a unit if
$\alpha^{-1} \in \O_K$. Notation:
\[ \O_K^\times \defeq \{u \in \O_K \st \text{$u$
is a unit}\} .\]

\begin{flashcard}[units-lemma]
\begin{lemma*}
  The following are equivalent for $\alpha \in
  \O_K$:
  \begin{enumerate}[(1)]
    \item $\alpha \in \O_K^\times$.
    \item \cloze{$N(\alpha) = \pm 1$.}
    \item \cloze{$\langle \alpha \rangle = \O_K$.}
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1) $\Rightarrow$ (2)]
    \item[(1) $\Rightarrow$ (2)] \cloze{$N(\alpha) \in
      \ZZ$ and
      \[ N(\alpha) N(\alpha^{-1}) = N(\alpha
      \alpha^{-1}) = N(1) = 1 \]
      with both $N(\alpha), N(\alpha^{-1}) \in
      \ZZ$ since $\alpha, \alpha^{-1} \in \O_K$.
      Hence $N(\alpha) = \pm 1$.}
    \item[(2) $\Rightarrow$ (3)] \cloze{Note:
      \[ N(\langle \alpha \rangle) = |N(\alpha)| =
      1 \implies |\O_K / \langle \alpha \rangle| =
      1 \implies \langle \alpha \rangle = \O_K
      .\]}
    \item[(3) $\Rightarrow$ (1)] \cloze{If $\langle
      \alpha \rangle = \O_K$, then $1 = \alpha
      \cdot \beta$ for some $\beta \in \O_K$.
      Hence $\alpha \in \O_K^\times$.}
  \end{enumerate}
\end{proof}
\end{flashcard}

\subsubsection*{Quadratic fields}

Let $m \neq 0, 1$, $m$ square-free, $K =
\QQ(\sqrt{m})$. Recall:
\[ \O_K = \begin{cases}
  a + b\sqrt{m} : a, b \in \ZZ & \text{if $m \equiv
  2, 3 \pmod{4}$} \\
  \frac{a + b \sqrt{m}}{2} : a, b \in \ZZ, 2 \mid
  a + b & \text{if $m \equiv 1 \pmod{4}$}
\end{cases} \]
We have
\[ N(a + b\sqrt{m}) = (a + b\sqrt{m})(a -
b\sqrt{m}) = a^2 - mb^2 .\]
There are $2$ cases:
\begin{itemize}
  \item $m \equiv 2, 3 \pmod 4$: $\O_K^\times$ is
    the elements $u = a + b\sqrt{m}$ with $a, b \in \ZZ$
    such that
    \[ a^2 - mb^2 = \pm 1 \tag{$*$}
    \label{lec13_l70} \]
  \item $m \equiv 1 \pmod{4}$: $\O_K^\times$ is
    the elements $u = \frac{a + b\sqrt{m}}{2}$
    with $a, b \in \ZZ$ such that
    \[ a^2 - mb^2 = \pm 4 \tag{$**$}
    \label{lec13_l75} \]
\end{itemize}

First consider $m < 0$. If $m \le -5$, then
\[ -mb^2 = \pm 4 - a^2 \le 4 \implies |b| \le
\frac{4}{5} \implies b = 0 .\]
Then $u = \pm 1$. We can go over the cases $m =
-1, -2, -3, -4$ by hand:
\begin{itemize}
  \item $m = -1$, the units are $\pm 1$, $\pm
    \sqrt{-1}$.
  \item $m = -2, -4$ the units are $\pm 1$.
  \item $m = -3$, the units are $\pm 1, \frac{\pm
    1 \pm \sqrt{-3}}{2}$.
\end{itemize}

Now move onto $m \ge 2$.

\begin{flashcard}[quadratic-units-thm]
\begin{theorem*}
  Let $K = \QQ(\sqrt{m})$, $m \ge 2$, squarefree.
  Then there is a unit $u > 1$ that is smallest,
  and all units are of the form:\cloze{
  \[ \O_K^\times = \{\pm u^n : n \in \ZZ\} .\]}
\end{theorem*}

\begin{proof}
  \cloze{We first show that all units $u > 1$ are of the
  form $u = a + b\sqrt{m}$ with $a, b > 0$. Note:
  \[ N(u) = \pm 1 = (a + b\sqrt{m})(a - b\sqrt{m}) \]
  hence
  \[ \{\pm u^{\pm 1}\} = \{\pm a \pm \sqrt{m} b\}
  .\]
  If $u > 1$, then these are distinct, and $a +
  \sqrt{m} b$ are the largest among them.
  Therefore $a, b > 0$ indeed. The fact that $u$
  with $u > 1$ exists is not examinable, but there
  are two ways to see this:
  \begin{enumerate}[(1)]
    \item The Pell equation $a^2 - mb^2 = 1$
      always has positive solutions (see
      \courseref[Part II Number Theory]{NT}).
    \item Can be proved using \nameref{mink_thm}. We will
      sketch this proof.
  \end{enumerate}
  We prove that there exists a smallest $u$ among
  those $> 1$. Suppose not. Then $\exists u_1,
  u_2, \ldots, \in \O_K^\times$ such that $u_1,
  u_2 > u_3 > \cdots > 1$. Then $\frac{u_n}{u_{n +
  1}} \to 1$, with each term lying in $\O_K^\times$ and
  greater than $1$. Then $\frac{u_n}{u_{n + 1}}
  \ge \frac{1 + \sqrt{m}}{2} > 1$, which is a
  contradiction. Let $v \in \O_K^\times$. We show
  that $v = \pm u^{\pm n}$ for some $n \in \ZZ$.
  Clearly this is true for $v$ if and only if true
  for $\pm v^{\pm 1}$. So we can assume $v \ge 1$.
  $v = 1$ is obvious, so assume $v > 1$. We cannot
  have
  \[ v \in (u^n, u^{n + 1}) \]
  for any $n \ge 0$ because then $v \cdot u^{-n}
  \in \O_K^\times$ and $1 < v \cdot u^{-n} < u$,
  contradicting the choice of $u$. So $v = u^n$
  for some $n \ge \ZZ_{\ge 1}$.}
\end{proof}
\end{flashcard}

This $u$ in the theorem is called the
\emph{fundamental unit}.

We can find the fundamental unit by searching
through the solutions of \eqref{lec13_l70} or
\eqref{lec13_l75}. For this
the following observation helps:

Let $(a_1, b_1)$ and $(a_2, b_2)$ be solutions of
\eqref{lec13_l70} with $a_1, a_2, b_1, b_2 \ge 0$.
Then $1 \le b_1 < b_2$ implies:
\[ a_1^2 = mb_1^2 \pm 1 < mb_2^2 \pm 1 = a_2^2 \]
So $a_1^2 < a_2^2$, so in fact $a_1 + b_1 \sqrt{m}
< a_2 + b_2 \sqrt{m}$. So when looking for the
fundamental solution, it suffices to find the
solution with $b$ minimal.

\begin{flashcard}[dirichlet-unit-thm]
\begin{theorem*}[Dirichlet's unit theorem]
  \label{dirichlet_unit_thm}
  \cloze{Let $K$ be a \gls{nf} with $r$ real embeddings and $s$
  pairs of complex embeddings. Let $W$ denote the
  roots of unity contained in $\O_K$, that is
  $\alpha \in \O_K$ such that $\alpha^m = 1$ for
  some $m \in \ZZ$. Then there are $r + s - 1$
  units $u_1, u_2, \ldots,u_{r + s - 1} \in
  \O_K^\times$ such that all units can be written
  uniquely as
  \[ \omega u_1^{n_1} \cdots u_{r + s - 1}^{n_{r +
  s - 1}} \]
  for some $n_1, \ldots, n_{r + s - 1} \in \ZZ$
  and $\omega \in W$. In addition, $|W| <
  \infty$.}
\end{theorem*}
\end{flashcard}