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\begin{document}
\centerline{Math 361 lecture for Friday, Week 11}
\bigskip

\centerline{The class group II}
\bigskip

Let $K$ be a number field of degree~$n$ with real embeddings
$\sigma_{1},\ldots,\sigma_{s}$, and complex embeddings
$\sigma_{s+1},\overline{\sigma}_{s+1},\ldots,\sigma_{s+t},\overline{\sigma}_{s+t}$.
Recall our $\Q$-algebra embedding:
\begin{align*}
  \sigma_{K}=\sigma\colon K&\to \mathbb{L}^{s,t}:=\R^{s}\times \C^{t}\\
  \alpha&\mapsto
  (\sigma_{1}(\alpha),\ldots,\sigma_{s}(\alpha),\sigma_{s+1}(\alpha),\ldots,\sigma_{s+t}(\alpha)),
\end{align*}
and our identification
\begin{align*}
  \R^{s}\times \C^{t}&\simeq \R^{n}\\
 (x_{1},\ldots,x_{s},z_{1},\ldots,z_{t}) &\mapsto
 (x_{1},\ldots,x_{s},u_{1},v_{1},\ldots,u_{t},v_{t}) 
\end{align*}
Define the \emph{norm} of $q=(x_{1},\ldots,x_{s},z_{1},\ldots,z_{t})\in
\R^{s}\times \C^{t} $ to be 
\[
  N(q) = x_{1}\cdots x_{s}z_{1}\overline{z_{1}}\cdots z_{t}\overline{z_{t}}
  =x_{1}\cdots x_{s}|z_{1}|^{2}\cdots|z_{t}|^{2}
\]
and note that it is consistent with our earlier definition: 
for $\alpha\in K$,
\[
  N(\alpha)=
\sigma_{1}(\alpha)\cdots\sigma_{s}(\alpha)\sigma_{s+1}(\alpha)\overline{\sigma}_{s+1}(\alpha)\cdots\sigma_{s+1}(\alpha)\overline{\sigma}_{s+t}(\alpha)
=N(\sigma(\alpha)).
\]
We proved a theorem giving the volume of a fundamental domain of the image of a
lattice under~$\sigma$ and derived the following:

\textbf{Corollary.}
Let~$\mathfrak{a}$ be a nonzero ideal in~$\mathfrak{O}_{K}$.
Identifying~$\mathbb{L}^{s,t}$ with~$\R^{n}$,
regard~$\sigma(\mathfrak{a})\subset\R^{n}$.  Then~$\sigma(\mathfrak{a})$ is a
lattice with fundamental domain of volume
\[
  2^{-t}N(\mathfrak{a})\sqrt{|\Delta|}
\]
where~$\Delta$ is the discriminant of~$K$.
\bigskip

Our goal now it two prove two theorems, one of which was used in the last
lecture to show that the class group is finite.

\begin{thm}
  If $\mathfrak{a}$ is a nonzero ideal of~$\mathfrak{O}_{K}$, then there
  exists~$0\neq\alpha\in\mathfrak{O}_{K}$ such that
  \[
    |N(\alpha)|\leq
    \left(\frac{2}{\pi}\right)^{t}N(\mathfrak{a})\sqrt{|\Delta|}.
  \]
\end{thm}
\begin{proof}
 Fix a real number $\varepsilon>0$, and select positive real numbers
 $c_{1},\ldots,c_{s+t}$ such that
 \[
   c_{1}\cdots c_{n} =
   \left(\frac{2}{\pi}\right)^{t}N(\mathfrak{a})\sqrt{|\Delta|}.
 \]
 Define $X_{\varepsilon}\in\R^{n}$ to be
 those~$x=(x_{1},\ldots,x_{s},x_{s+1},y_{s+1},\ldots,x_{s+t},y_{s+t})\in\R^{n}$ such that
 \begin{itemize}
   \item $|x_{1}|<c_{1}+\varepsilon$,
   \item $|x_{2}|<c_2,\ldots, |x_{s}|<c_s$,
   \item
     $|x_{s+1}^{2}+y_{s+1}^{2}|<c_{s+1},\ldots,|x_{s+t}^{2}+y_{s+t}^{2}|<c_{s+t}$.
 \end{itemize}
 Then $X_{\varepsilon}$ is centrally symmetric about the origin and convex (exercise).  We have
 \begin{align*}
   \vol(X_{\varepsilon})&>[(2c_{1})\cdots {2c_{s}}][(\pi c_{s+1})\cdots(\pi c_{s+t})]\\
   &= 2^{s}\pi^{t}(c_{1}\cdots c_{s+t})\\
   &=2^{s}\pi^{t} \left(\frac{2}{\pi}\right)^{t}N(\mathfrak{a})\sqrt{|\Delta|}\\
   &=2^{s+t}N(\mathfrak{a})\sqrt{|\Delta|}\\
   &=2^{s+2t}\cdot 2^{-t}N(\mathfrak{a})\sqrt{|\Delta|}\\
   &=2^{n}\vol(F)
 \end{align*}
 where~$F$ is a fundamental domain for~$\sigma(\mathfrak{a})$.

 By Minkowski's theorem, $X_{\varepsilon}$ contains a nonzero point in the
 lattice $\sigma(\mathfrak{a})$, i.e., there exists $0\neq\beta\in\mathfrak{a}$ such
 that $\sigma(\beta)\in X_{\varepsilon}$.  For each $\varepsilon>0$, define
 \[
   A_{\varepsilon}=\{\beta\in\mathfrak{a}:\beta\neq0, \sigma(\beta)\in
   X_{\varepsilon}\}.
 \]
 For each~$\beta\in A_{\varepsilon}$, we have
 \[
   |N(\beta)|<(c_{1}+\varepsilon)c_{2}\cdots c_{s+t}.
 \]
 We have seen that each~$A_{\varepsilon}$ is nonempty.  Further,
 since~$\sigma(\mathfrak{a})$ is a lattice, each $A_{\varepsilon}$ is finite.
 We have
 \[
   A_{1}\supseteq A_{1/2}\supseteq A_{1/3}\supseteq \cdots \supseteq
   A_{1/k}\supseteq \cdots.
 \]
 Hence, $\bigcap_{k\geq 1}A_{1/k}\neq\emptyset$.  Let~$\alpha\in A$, then
 since~$\alpha\in A_{1/k}$ for all~$k\geq1$, we have
 \[
   |N(\alpha)|<(c_{1}+1/k)c_{2}\cdots c_{s+t}.
 \]
 It follows that
 \[
   |N(\alpha)|\leq c_{1}\cdots c_{s+t} = 
   \left(\frac{2}{\pi}\right)^{t}N(\mathfrak{a})\sqrt{|\Delta|}.
 \]
\end{proof}

\begin{thm}
  Every element of the class group~$\mathcal{H}$ is represented by an ideal with
  norm at most 
  \[
    \left(\frac{2}{\pi}\right)^{t}\sqrt{|\Delta|}.
  \]
\end{thm}
\begin{proof}
  Consider an arbitrary equivalence class $[\mathfrak{c}]\in \mathcal{H}$
  where~$\mathfrak{c}$ is an ordinary ideal.  (We know that every element
  of~$\mathcal{H}$ has this form.) Then~$[\mathfrak{c}^{-1}]\in \mathcal{H}$, so
  there exists an ordinary ideal~$\mathfrak{b}$ representing~$[\mathfrak{c}^{-1}]$. 
  
  Take $0\neq \beta\in\mathfrak{b}$ with
  \[
    |N(\beta)| \leq \left(\frac{2}{\pi}\right)^{t}N(\mathfrak{b})\sqrt{|\Delta|}.
  \]
  Since~$N(\beta)\in \mathfrak{b}$, it follows
  that~$(\beta)\subseteq\mathfrak{b}$.  Multiply this inclusion through by
  $\mathfrak{b}^{-1}$ to define
  \[
    \mathfrak{a}:= (\beta)\mathfrak{b}^{-1}\subseteq
    \mathfrak{b}\mathfrak{b}^{-1}=\mathfrak{O}_{K}.
  \]
  Hence, $\mathfrak{a}$ is an ideal.  Further,
  \[
    [\mathfrak{a}]=[\mathfrak{b^{-1}}]=[\mathfrak{c}],
  \]
  since~$\mathfrak{a}$ differs from~$\mathfrak{b}^{-1}$ by a factor of a
  principal ideal.
   Finally,
  \[
    N(\mathfrak{a})=N((\beta))N(\mathfrak{b}^{-1})=\frac{|N(\beta)|}{N(\mathfrak{b})}
    \leq\left(\frac{2}{\pi}\right)^{t}\sqrt{|\Delta|}.
  \]
\end{proof}

\begin{eg}
  Let $K=\Q(\sqrt{-13})$.  Every ideal class in $\mathcal{H}$ is represented by
  an ideal with norm at most
  \[
    \left(\frac{2}{\pi}\right)\sqrt{4\cdot 13}<4.6.
  \]
  So we must consider ideals with norms $1,2,3,4$.  If an ideal~$\mathfrak{a}$
  contains a rational integer $a$, then $(a)\subseteq\mathfrak{a}$ implies
  that~$\mathfrak{a}$ divides $(a)$.   So to find the ideals with
  norms~$a$, we look for divisors of the ideal~$(a)$.  In our case, where
  $a\in\{1,2,3,4\}$, we factor the minimal polynomial $x^{2}+13$ modulo $p$ for
  $p=2,3$ to find 
  \[
  (2)=(2,\sqrt{-13})^{2},\quad (3) = (3).
  \]
  An ideal with norm~$4$ divides
  \[
    (4)=(2)^{2}=(2,\sqrt{-13})^{4}.
  \]
  Therefore, $h=|\mathcal{H}|=1$ or~$2$, depending on whether~$(2,\sqrt{-13})$
  is principal.  

  If $(2,\sqrt{-13})=(a+b\sqrt{-13})$ for some $a,b\in\Z$, taking norms, we find
  \[
    2 = a^{2}+13b^2,
  \]
  for which there are no solutions.  Therefore,~$\mathcal{H}$ is a group with
  two elements: $[(1)]$ and~$[(2,\sqrt{-13})]$.
\end{eg}
\end{document}
