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\begin{document}
\centerline{Math 361 lecture for Monday, Week 12}
\bigskip

\centerline{Dirichlet's unit theorem}
\bigskip

Let~$K$ be a number field, and let~$\mathfrak{O}_{K}^{*}$ denote the units
in~$\mathfrak{O}_{K}$.

\begin{eg}\ 
  \begin{itemize}
    \item We have $\pm1\in \mathfrak{O}_{K}^{*}$ for all~$K$.
    \item If $K=\Q(i)$, then $\mathfrak{O}_{K}^{*}=\{\pm1,\pm i\}$.
    \item If $K=\Q(\sqrt{2})$, then $1+\sqrt{2}\in\mathfrak{O}_{K}^{*}$,
      and~$|1+\sqrt{2}|>1$.  So $\{(1+\sqrt{2})^{k}:k\in\Z_{>0}\}$ is an
      infinite collection of units in $\mathfrak{O}_{K}^{*}$.
  \end{itemize}
\end{eg}

\begin{remarks}\
  \begin{enumerate}[(a)]
    \item $\mathfrak{O}_{K}^{*}$ is a multiplicative group.
    \item An element~$u\in \mathfrak{O}_{K}$ is a unit if and only if
      $N(u)=\pm1$. (Reminder: If $N(u)=\prod_{i=1}^{n} \sigma_{i}(u)=\pm1$ where
      $\sigma_{1}=\id$, then let~$v:=\prod_{i=2}^{n} \sigma^{i}(u)$.  We
      have~$v\in\mathfrak{O}_{K}$ (let $\sigma_{i}$ act on the minimal
      polynomial for $u$), and $\pm v$ is the multiplicative inverse of~$u$.)
    \item Elements of finite order in~$\mathfrak{O}_{K}^{*}$ are roots of unity,
      and every root of unity in~$K$ is in $\mathfrak{O}_{K}^{*}$. (If~$\zeta\in K$
      and~$\zeta^{m}=1$, them~$\zeta$ satisfies $x^{m}-1$, and hence is an
      algebraic integer in~$K$.)
    \item The elements of~$\mathfrak{O}_{K}$ with finite order form a finite
      cyclic subgroup of~$\mathfrak{O}_{K}$ of even order.  
      \begin{proof}
         These elements clearly form a subgroup.  For finiteness, note that the
	 mapping $\sigma: K\to\mathbb{L}^{s,t}\simeq\R^{n}$ maps
	 $\mathfrak{O}_{K}$ to a lattice in~$\R^{n}$, and the image of~$\{\zeta\in
	 K: |\zeta|=1\}$ maps to a compact set.  Finally,
	 since~$-1\in\mathfrak{O}_{K}$ and has order~$2$, it follows that $2$
	 divides the order of the subgroup. The proof that any finite subgroup
	 of~$K^{*}$ must be cyclic is similar to that given in Step~1 of the two
	 square theorem the in the notes for Friday, Week 10.
      \end{proof}
  \end{enumerate}
\end{remarks}

\begin{eg}
  Let~$K$ be a cubic number field, i.e., $[K\mathbin{:}\Q]=3$.  Let~$\zeta$ be a
  root of unity in~$K$.  We claim that~$\zeta=\pm1$. Since~$\Q\subseteq \Q(\zeta)\subseteq K$, we have
  \[
    3=[K\mathbin{:}\Q]=[\Q\mathbin{:}\Q(\zeta)][\Q(\zeta)\mathbin{:}K].
  \]
  Therefore, either $\Q(\zeta)=\Q$, in which case~$\zeta=\pm1$,
  or~$K=\Q(\zeta)$. We now show that the case $K=\Q(\zeta)$ is not possible.
  This is due to that fact (not proven here) that if~$\zeta$ is a $k$-th root of
  unity, then~$[\Q(\zeta)\mathbin{:}\Q]=\phi(k)$, where $\phi$ is the Euler
  phi-function. However, $\phi$ never takes the value~$3$.  For~$k\geq 3$, we
  have that~$\phi(k)$ is even, and $\phi(1)=\phi(2)=1$.   
  So the only units in a cubic number are $\pm1$.
\end{eg}

\begin{thm}[Dirichlet's unit theorem]
  Let~$K$ have~$s$ real embeddings and~$t$ complex embeddings. Then, we have a
  group isomorphism
  \[
    \mathfrak{O}_{K}^{*}\simeq W\times \Z^{s+t-1}
  \]
  where~$W$ is the finite cyclic group of roots of unity in~$K$ (the subgroup
  of~$\mathfrak{O}_{K}$ of elements of finite order).
\end{thm}
\begin{proof}
  See our text, Appendix~B for a full proof.  For the idea of the proof,
  consider the mapping~$\ell\colon \mathfrak{O}_{K}^{*}\to \R^{s+t}$ which is our usual mapping
  $\sigma\colon K\to \mathbb{L}^{s,t}=\R^{s}\times \C^{t}$, restricted
  to~$\mathfrak{O}_{K}^{*}$, followed by the
  mapping
  \begin{align*}
    \R^{s}\times \C^{t} &\to \R^{s+t} \\
  (x_{1},\ldots,x_{s},z_{1},\ldots,z_{t})  &\mapsto 
  (\ln|x_{1}|,\ldots,\ln|x_{s}|,\ln|z_{1}|^{2},\ldots,\ln|z_{t}|^{2}).
  \end{align*}
  Note that due to the logs, the mapping~$\ell$ takes a multiplicative group
  into and additive subgroup of~$\R^{n}$.  One may show that the image of~$\ell$
  is a lattice of dimension~$s+t-1$, and the kernel of~$\ell$ is the set of elements
  of~$\mathfrak{O}_{K}^{*}$ of finite order. Thus,~$\ell$ induces
  an isomorphism of~$\mathfrak{O}_{K}^{*}/W$ with a lattice of rank~$s+t-1$.  It
  follows from the structure theorem for finitely generated abelian groups that
  \[
    \mathfrak{O}_{K}^{*}\simeq W\times \Z^{s+t-1}.
  \]
  
  To see that the $W\subseteq\ker(\ell)$, first take~$\zeta\in W$.  Say~$\zeta^{m}=1$
  with $m\geq1$.  Then
  \[
    0=\ell(1)=\ell(\zeta^{m})=m\ell(\zeta).
  \]
  Hence,~$\zeta\in\ker(\ell)$.  The opposite inclusion requires a bit more work
  (see our text).

  It is also straightforward to see that the image of $\ell$ lies in a lattice
  of rank at most~$s+t-1$.  Let $\alpha\in\mathfrak{O}_{K}^{*}$.
  Then
  \[
    \ell(\alpha)=(\ell_{1}(\alpha),\ldots,\ell_{s+t}(\alpha))
  \]
  where 
  \[
    \ell_{j}=
    \begin{cases}
      \ln(|\sigma_{j}(\alpha)|) &\text{for $1\leq j\leq s$}\\
      \ln(|\sigma_{j}(\alpha)|^{2}) &\text{for $s+1\leq j\leq s+t$}.
    \end{cases}
  \]
  Therefore, 
  \begin{align*}
    0&= \sum_{j=1}^{s+t} \ell_{j}(\alpha)\\
    &=\sum_{j=1}^{s} \ln|\sigma_{j}(\alpha)| + \sum_{j=s+1}^{s+t}
    \ln|\sigma_{j}(\alpha)|^{2}\\
    &=\ln|N(\alpha)|\\
    &=0,
  \end{align*}	
  since the norm of a unit is~$\pm 1$.  We see that the image of~$\ell$ sits in
  the set~$\{x\in\R^{s+t}: x_{1}+\cdots+x_{s+t}=0\}$.  
\end{proof}

\begin{remark} The finite group~$W$ consists of the roots of unity of~$K$.  The
  only real roots of unity are~$\pm 1$.  Since field embeddings preserve roots
  of $1$, including their orders, it follows that if~$K$ has any real
  embeddings, then~$W=\{-1,1\}$.
\end{remark}
\end{document}
