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The logarithmic embedding is
\[ \log : K \to \RR^{r + s}; \alpha \mapsto
(\log|\sigma_1(\alpha)|, \ldots,
\log|\sigma_r(\alpha), 2\log|\tau_1(\alpha)|,
\ldots, 2\log|\tau_s(\alpha)|)^\top ,\]
which is a homomorphism from $(K, \cdot)$ to
$(\RR^{r + s}, +)$. Observe that
\[ \log|N(\alpha)| = \sum_{i = 1}^{r + s}
(\log(\alpha))_j .\]
We write $V \subset \RR^{r + s}$ for $\{x : x + 1
\cdots + x_{r + s} = 0\}$. If $\alpha \in
\O_K^\times$, then $N(\alpha) = \pm 1$, and hence
$\log \alpha \in V$.

\begin{proposition}
  $\ker(\log) = W$ and $|W| < \infty$.
\end{proposition}

\begin{proposition}
  $\log(\O_K^\times)$ is a \gls{latt} in $V$.
\end{proposition}

\begin{proof}[Proof of
  \nameref{dirichlet_unit_thm} (non-examinable)]
  Let $x_1, \ldots, x_{r + s - 1}$ be a basis for
  $\log(\O_K^\times)$. We can choose $u_j$ such
  that $\log(u_j) = x_j$. Easy to check that the
  theorem holds with this choice.
\end{proof}

\begin{proof}[Proof of Proposition 1]
  If $\log \alpha = 0$, then $|\sigma_j(\alpha)| =
  1$, $|\tau_j(\alpha)| = 1$ for all $j$. This
  means that
  \[ \| \Sigma(\alpha) \| \le \sqrt{d} ,\]
  and $\Sigma(\O_K)$ is a \gls{latt}, so it has a
  finite intersection with $B(0, \sqrt{d}) = \{v
  \in \RR^d \st \|v\| < \sqrt{d}\}$. Then
  $|\ker(\log)| < \infty$. $\ker(\log)$ is a group
  under $\cdot$. So $\alpha \in \ker \log$ has
  finite, i.e. $\alpha^m = 1$ for some $n \in
  \ZZ_{> 0}$. Thus $\alpha \in W$.
\end{proof}

\begin{lemma*}
  Let $\Lambda \subset V$ be an additive subgroup.
  Then $\Lambda$ is a \gls{latt} if and only if there
  is $R \in \RR_{> 0}$ such that $\Lambda \cap
  B(x, R)$ is finite and non-empty for all $x \in
  V$.
\end{lemma*}

\begin{proof}
  Omitted.
\end{proof}

\begin{proof}[Proof of Proposition 2]
  To prove Proposition 2, we need the following:
  Given $x \in \RR^{r + s}$ with $\sum_j x_j = 0$,
  we need to show that the set of units $u \in
  \O_K^\times$ that satisfy
  \[ \|\log(u) - x\| < R \]
  is finite and non-empty. For simplicity assume $s
  = 0$. The above inequality is equivalent to
  \[ e^{x_j} e^{-\tilde{R}} \le |\sigma_j(u)| \le
  e^{x_j} \cdot e^{\tilde{R}} \]
  for all $i$. Finiteness follows from
  $\Sigma(\O_K)$ being a lattice.

  Non-empty is more difficult. Observe: enough to
  show $\exists u \in \O_K^\times$ with
  \[ |\sigma_j(u)| \le C_0 e^{x_j} . \tag{$*$}
  \label{lec14_l77} \]
  This is
  because: $|N(u)| = 1$, so
  \[ \prod |\sigma_j(u)| = 1 \implies
  |\sigma_j(u)| \ge \left( \prod_{k \neq j}
  |\sigma_k(u)| \right)^{-1} \ge C_0^{d - 1}
  e^{\sum_{k \neq j} x_k}  = C_0^{d - 1} e^{-x_j} \]
  By \nameref{mink_thm} applied to the \gls{latt}
  $\Sigma(\O_K)$ and the convex set
  \[ \{v : |v_j| < C_0 e^{x_j}\} \]
  gives $\alpha \in \O_K$ that satisfies
  \eqref{lec14_l17} provided $C_0$ is large
  enough. Now the problem is that $\alpha$ may not
  be a unit. However:
  \[ |N(\alpha)| \le C_0^d \prod_i e^{x_i} = C_0^d \]
  where $C_0^d$ is some constant which depends
  only on $K$. There are only finitely many
  principal ideals in $\O_K$ with norm $\le
  C_0^d$. Fix a generator in each of them, say
  $\alpha_I$ for the generator of $I$. Let $\alpha
  \in \O_K$ that the argument gives, so it
  satifies \eqref{lec14_l77} and $|N(\alpha)| <
  C_0^d$. Then $\langle \alpha \rangle = \langle
  \alpha_{\langle \alpha \rangle} \rangle$.
  Therefore $\alpha \cdot \alpha_{\langle \alpha
  \rangle}^{-1} \in \O_K^\times$.
\end{proof}

\subsubsection*{Cyclotomic Fields}

\begin{notation*}
  $k \in \ZZ_{> 0}$, then $\theta_k = 2^{2\pi i /
  k}$. This is a primitive $k$-th root of unity.
\end{notation*}

\begin{flashcard}[cyclotomic-field-roots-of-unity-lemma]
\begin{lemma*}
  Fix $p \in \ZZ$ a prime. Let $K = \QQ(\theta_p)$.
  Let $W$ be the roots of unity in $\O_K$. Then
  \[ W = \{\pm \theta_p^k : k = 0, \ldots, p - 1\}
  = \{\theta_{2p}^k : k = 0, \ldots, 2p - 1\} .\]
\end{lemma*}

\begin{proof}
  \cloze{Let $t \in \RR_{> 0}$ minimal with the property
  that $e^{2\pi i t} \in W$. Recall that $W$ is
  finite. Recall that $W$ is finite, so this
  minimum exists. Claim: if $e^{2\pi i s} \in W$,
  then $s / t \in \ZZ$. If not then $e^{2\pi i (s
  - (s / t) t} \in W$. This contradicts
  minimality. I know $e^{2\pi i / 2p} \in W$. So
  $t = \frac{1}{k2p}$ for some $k \in \ZZ_{> 0}$.}
\end{proof}
\end{flashcard}