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

\centerline{The class group}
\bigskip

Let~$K$ be a number field, and let~$I\subseteq K$ be an
$\mathfrak{O}_{K}$-module. Recall that $I$ is a \emph{fractional ideal}
of~$\mathfrak{O}_{K}$ if it satisfies any of the following equivalent
conditions:

\begin{enumerate}
  \item There exists~$\alpha\in K\setminus\{0\}$ such
    that~$\alpha I\subseteq\mathfrak{O}_{K}$.
  \item There exists~$\alpha\in \mathfrak{O}_{K}\setminus\{0\}$ such
    that~$\alpha I\subseteq\mathfrak{O}_{K}$.
  \item There exists an ordinary ideal~$\mathfrak{a}\subseteq\mathfrak{O}_{K}$
    and $\alpha\in K\setminus\{0\}$ such
    that~$I=\alpha\mathfrak{a}$.
  \item There exists an ordinary ideal~$\mathfrak{a}\subseteq\mathfrak{O}_{K}$
    and $\alpha\in\mathfrak{O}_{K}\setminus\{0\}$ such
    that~$I=\frac{1}{\alpha}\mathfrak{a}$.
  \item $I$ is finitely generated as an $\mathfrak{O}_{K}$-module.
\end{enumerate}

A \emph{principal fractional ideal} is defined to be a fractional ideal
generated as an~$\mathfrak{O}_{K}$-module by a single element.  Thus, a
nonzero principal fractional ideal has the form~$\alpha\mathfrak{O}_{K}$ for some
$\alpha\in K\setminus\{0\}$.

Let~$\mathcal{F}$ denote the multiplicative group of nonzero fractional ideals
of~$\mathfrak{O}_{K}$, and let~$\mathcal{P}$ denote the subgroup of nonzero
principal factional ideals.
\begin{defn}
  The \emph{class group} of~$\mathfrak{O}_{K}$ is the quotient group
  \[
    \mathcal{H} = \mathcal{F}/\mathcal{P}.
  \]
  The \emph{class number} of $\mathfrak{O}_{K}$ is the size of this group:
  \[
    h_{K} =|\mathcal{H}|.
  \]
\end{defn}

\begin{prop}
  Every element of~$\mathcal{H}$ is represented by an ordinary ideal
  of~$\mathfrak{O}_{K}$.
\end{prop}
\begin{proof}
 Let~$I=\alpha\mathfrak{a}$ where $\alpha\in K\setminus\{0\}$ and~$\mathfrak{a}$
 is an ordinary ideal.  Then $\alpha\mathfrak{O}_{K}$ is a principal fractional
 ideal.  Therefore,
 \[
   I = \alpha\mathfrak{a}=(\alpha\mathfrak{O}_{K})\mathfrak{a} =\mathfrak{a}
   \bmod \mathcal{P}.
 \]
\end{proof}

\begin{prop}
  Two ordinary ideals $\mathfrak{a}$ and $\mathfrak{b}$ represent the same
  element in~$\mathcal{H}$ if and only if there
  exist~$\alpha,\beta\in\mathfrak{O}_{K}\setminus\{0\}$ such
  $\alpha\mathfrak{a}=\beta\mathfrak{b}$. 
\end{prop}
\begin{proof}
 We have~$\mathfrak{a}=\mathfrak{b}\bmod \mathcal{P}$ if and only if there exists
 $\gamma\in K\setminus\{0\}$ such that
 $(\gamma\mathfrak{O}_{K})\mathfrak{a}=\mathfrak{b}$.
 Write~$\gamma=\alpha/\beta$ to get the result.
\end{proof}

The above proposition gives another way to define~$\mathcal{H}$.  Say two
nonzero ideals $\mathfrak{a}$ and $\mathfrak{b}$ are equivalent if there exist
nonzer $\alpha,\beta\in\mathfrak{O}_{K}$ such that
$\alpha\mathfrak{a}=\beta\mathfrak{b}$.  Let~$[\mathfrak{}]$ denote the
equivalence class of a nonzero ideal~$\mathfrak{a}$.  Then~$\mathcal{H}$ can be
defined to be the set of equivalence class of nonzero ideals
of~$\mathfrak{O}_{K}$ with multiplication defined be
$[\mathfrak{a}][\mathfrak{b}]:=[\mathfrak{a}\mathfrak{b}]$. 

\begin{prop}
  $\mathfrak{O}_{K}$ is a UFD if and only if $h_{K}=1$, i.e., if and only if the
  class group is trivial.
\end{prop}
\begin{proof}
   ($\Rightarrow$) Suppose that~$\mathfrak{O}_{K}$ is a UFD, and let~$I$ be a
   fractional ideal.  We can write $I=\frac{1}{\alpha}\mathfrak{a}$ for
   some nonzero~$\alpha\in\mathfrak{O}_{K}$ and some nonzero ordinary ideal~$\mathfrak{a}$.
   We saw earlier that~$\mathfrak{O}_{K}$ is a UFD if and only if it is a PID.
   Hence,~$\mathfrak{a}=(\beta)$ for some nonzero~$\beta\in\mathfrak{O}_{K}$.
   Therefore~$I$ is generated as an $\mathfrak{O}_{K}$-module by the single
   element~$\beta/\alpha$.  So $I$ is principal.  It follows that~$\mathcal{H}$
   is trivial.

   ($\Leftarrow$)  Suppose that $\mathcal{H}$ is trivial, and let~$\mathfrak{a}$
   be a nonzero ideal of~$\mathfrak{O}_{K}$.  We may regard~$\mathfrak{a}$ as a
   fractional ideal, and since~$\mathcal{H}$ is trivial, it follows
   that~$\mathfrak{a}$ is a principal fractional ideal.  Thus,
   $\mathfrak{a}=\alpha\mathfrak{O}_{K}$ for some nonzero element~$\alpha\in K$.
   Since~$\mathfrak{a}\subseteq\mathfrak{O}_{K}$, it follows that~$\alpha\in
   \mathfrak{O}_{K}$, and thus, $\mathfrak{a}$ is the principal
   ideal~$(\alpha)\subseteq\mathfrak{O}_{K}$.  We have shown
   that~$\mathfrak{O}_{K}$ is a PID, from which it follows
   that~$\mathfrak{O}_{K}$ is a UFD.
\end{proof}

\begin{thm}\label{thm:11wed}
  Every element of $\mathcal{H}$ is represented by an ideal with norm at most 
  \[
    \left(\frac{2}{\pi}\right)^{t}\sqrt{|\Delta|}
  \]
  where~$2t$ is the number of complex embeddings of~$K$ and~$\Delta$ is the
  discriminant of~$K$.
\end{thm}
\begin{proof}
 We will prove this in an upcoming lecture.
\end{proof}


\begin{cor}
  The class group~$\mathcal{H}$ is finite.
\end{cor}
\begin{proof}
  Recall that there are finitely many ideals with a given norm.  (To recall the
  proof: Let $a\in\N$ and suppose~$\mathfrak{a}$ is an ideal with
  $N(\mathfrak{a})=a$.  In $\mathfrak{O}_{K}$ we factor the ideal~$(a)$:
  \[
    (a)=\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_{i}}.
  \]
  We know that and ideal contains the norm of each of its element.  In
  particular, $a\in\mathfrak{a}$.  This implies $(a)\subseteq\mathfrak{a}$, and
  hence, $\mathfrak{a}|(a)$.  It follows from unique factorization of ideals
  that
  \[
    \mathfrak{a} = \prod_{i=1}^{k} \mathfrak{p}_{i}^{\ell_{i}}
  \]
  where $0\leq\ell_{i}\leq e_{i}$ for $i=1,\ldots,k$.  There are only finitely
  many possibilities for the~$\ell_{i}$, and hence, only finitely many
  possibilities for~$\mathfrak{a}$.)

  By the previous theorem, each element of~$\mathcal{H}$ is represented by an
  ideal of norm at most $(2/\pi)^{t}\sqrt{|\Delta|}$.  There are only finitely
  many positive integers less than this bound.  Coupled with the fact that there
  are only finitely many ideals with a given norm, the result follows.
\end{proof}

\begin{eg}
  Let $K=\Q(\sqrt{-5})$.  Then~$\mathfrak{O}_{K}=\Span_{\Z}\{1,\sqrt{-5}\}$.  The discriminant of~$K$ is
  \[
    \Delta = \det
    \left(\begin{array}{rr}
	1 & \sqrt{-5} \\
	1 & -\sqrt{-5}
    \end{array}\right)^{2}
  = (-2\sqrt{-5})^{2}=-20.
  \]
  Then~$K$ has~$2$ complex embeddings.  So according to
  Theorem~\ref{thm:11wed}, each element of~$\mathcal{H}$ is represented by an
  ideal with norm at most
  \[
    \left(\frac{2}{\pi}\right)\sqrt{|\Delta|}<2.9.
  \]
  So each element of~$\mathcal{H}$ is represented by an ideal with norm
  either~$1$ or~$2$.  What are the ideals with these norms?  The only ideal with
  norm~$1$ is $(1)=\mathfrak{O}_{K}$. Next, suppose that~$\mathfrak{a}$ is an
  ideal with norm~$2$.  We know that~$2\in \mathfrak{a}$, and hence,
  $\mathfrak{a}|(2)$.

  To see how~$(2)$ factors, we factor the minimal polynomial for~$\sqrt{-5}$
  modulo~$2$:
  \[
    x^2+5=x^2+1=(x+1)^{2}\bmod 2.
  \]
  Hence, $(2)=(2,1+\sqrt{-5})^{2}$.  Taking norms, we
  have~$N(2)=4=N(2,1+\sqrt{-5})^{2}$.  Hence, $N(2,1+\sqrt{-5})=2$, and we see
  that~$(2,1+\sqrt{-5})$ is the only ideal with norm equal to~$2$.

  So far, we have seen that every element of~$\mathcal{H}$ is represented by
  $(1)$ or~$(2,1+\sqrt{-5})$.  To show these ideals are distinct in~$H$, we must
  show that $(2,1+\sqrt{5})$ is not principal.  For sake of contradiction,
  suppose it is.  Then there exist $a,b\in\Z$ such
  that~$(a+b\sqrt{-5})=(2,1+\sqrt{-5})$.  Taking norms we have
  \[
    N(((a+b\sqrt{-5})))=|N(a+b\sqrt{-5}))|=a^2+5b^2=N((2,1+\sqrt{-5}))=2.
  \]
  However, there are no solutions in~$\Z$ to this equation.
\end{eg}

\begin{cor}
  Let~$\mathfrak{a}$ be an ideal of~$\mathfrak{O}_{K}$, and let~$h=|\mathcal{H}|$ be the class
  number of~$K$.  Then
  \begin{enumerate}
    \item $\mathfrak{a}^{h}$ is principal, and
    \item If $u\in\N$ is relatively prime to~$h$, and~$\mathfrak{a}^{u}$ is
      principal, then $\mathfrak{a}^{u}$ is
      principal.  
  \end{enumerate}
\end{cor}
\begin{proof}
  Let~$[\mathfrak{a}]$ denote the equivalence class of~$\mathfrak{a}$
  modulo~$\mathcal{P}$.  In a finite group, raising any element to the order of
  the group yields the identity of the group.  Hence,
  $[\mathfrak{a}]^{h}=[\mathfrak{a}^{h}]=(1) \bmod \mathcal{P}$.  So it follows
  that there is a principal fractional ideal~$(\alpha)\mathfrak{O}_{K}$ such
  that~$\mathfrak{a}^{h}=(1)(\alpha)=(\alpha)$.  We see that~$\alpha$ must be
  in~$\mathfrak{O}_{K}$ since $\mathfrak{a}$ is in $\mathfrak{O}_{K}$.

  If $u\in \N$ is relatively prime to~$h$, then there exist~$a,b\in\Z$ such
  that~$au+bh=1$.  Supposing~$\mathfrak{a}^{u}$ is principal, it follows that
  \[
    \mathfrak{a}=\mathfrak{a}^{1}=\mathfrak{a}^{au+bh}=\mathfrak{a}^{au}\mathfrak{a}^{h}=\mathfrak{a}^{au}=(\mathfrak{a}^{u})^{a}=(1)^{a}=(1)\bmod\mathcal{P}.
  \]
  Hence, $\mathfrak{a}$ is principal.
\end{proof}
\end{document}
