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\begin{document}
\centerline{Math 361 lecture for Monday, Week 9}
\bigskip

\centerline{The norm of an ideal}
\bigskip

Let~$K$ be a number field, and let~$\mathfrak{O}_{K}$ be its ring of integers.
Recall the proof that if~$\mathfrak{a}$ is a nonzero ideal of~$\mathfrak{O}_{K}$, then
$\mathfrak{O}_{K}/\mathfrak{a}$ is finite.  Pick~$0\neq \alpha\in \mathfrak{a}$.
Then we saw that~$N(\alpha)\in\mathfrak{a}$, and hence the usual quotient map
$\mathfrak{O}_{K}\to\mathfrak{O}_{K}/\mathfrak{a}$ induces a surjection
\[
  \mathfrak{O}_{K}/(N(\alpha)) \to \mathfrak{O}_{K}/\mathfrak{a}.
\]
We then appeal to the structure theorem of finite abelian groups and use the
fact that no element of~$\mathfrak{O}_{K}/(N(\alpha))$ has infinite
order to conclude that~$\mathfrak{O}_{K}/(N(\alpha))$ is a finite
product of finite cyclic groups, hence, finite.  The surjection above then
implies that~$\mathfrak{O}_{K}/\mathfrak{\alpha}$ is also finite.

\begin{defn}
  Let~$\mathfrak{a}$ be a nonzero ideal of~$\mathfrak{O}_{K}$.  Then the \emph{norm}
  of $\mathfrak{a}$ is
  \[
    N(\mathfrak{a})=\left|\mathfrak{O}_{K}/\mathfrak{a}\right|.
  \]
\end{defn}

\begin{eg} Let~$K=\Q(\sqrt{-14})$, and consider the
  ideal~$\mathfrak{a}=(6,1+\sqrt{-14})\subset \mathfrak{O}_{K}$.  Since~$-14\neq
  1\bmod 4$, each element of $\mathfrak{O}_{K}$ has the form~$a+b\sqrt{-14}$ for
  some $a,b\in\Z$.  Working modulo~$\mathfrak{a}$, we have
  \begin{equation}\label{eqn:09moneq1}
    a+b\sqrt{-14}=(a+b\sqrt{-14})-b(1+\sqrt{-14l})=(a-b) \bmod \mathfrak{a}.
  \end{equation}
  Since~$6\in\mathfrak{a}$, one might be tempted to jump to the conclusion
  that~$\mathfrak{O}_{K}/\mathfrak{a}$ is isomorphic to~$\Z/6\Z$.  However, that
  reasoning assumes that $0,1,2,3,4,5$ are distinct modulo~$\mathfrak{a}$.
  The above reasoning actually says that we get a well-defined surjection
  \begin{align*}
   \Z/6\Z&\to \mathfrak{O}_{K}/\mathfrak{a}\\
    x&\mapsto x.
  \end{align*}
  It is well-defined since~$6\in\mathfrak{a}$, and it is surjective by
  equation~(\ref{eqn:09moneq1}).  The kernel of this mapping is one of the
  following: $(0)$, $(1)$, $(2)$, or~$(3)$, since these are the only ideals
  of~$\Z/6\Z$. If the kernel is $(0)$, the mapping is an isomorphism.
  Otherwise, we will have~$1,2$ or~$3$ in~$\mathfrak{a}$.  So our problem is
  solved by finding the smallest positive integer in~$\mathfrak{a}$.  

  An arbitrary element of~$\mathfrak{a}$ has the form
  \[
    \alpha=(a+b\sqrt{-14})6+(c+d\sqrt{-14})(1+\sqrt{-14})
    =(6a+c-14d)+(6b+c+d)\sqrt{-14}.
  \]
  Then~$\alpha$ is a rational integer if and only if $d=-6b-c$.  In that case,
  \[
    \alpha = 6a+c-14d = 6a+c-14(-6b-c)=6a+84b+15c=3(2a+28b+5c).
  \]
  The possible values for~$2a+28b+5c$ are the elements in the
  ideal~$(2,28,5)=(1)=\Z$.  So the smallest positive integer in~$\mathfrak{a}$
  is $3$.  We can get this by letting~$a=-2,b=0, c=1$, and $d=-6b-c=-1$:
  \[
    (-2)6+(1-\sqrt{-14})(1+\sqrt{-14}) = -12+1+14=3.
  \]
  Therefore,
  \[
    \Z/3\Z\simeq\mathfrak{O}_{K}/\mathfrak{a},
  \]
  and
  \[
    N(\mathfrak{a})=\left|\mathfrak{O}_{K}/\mathfrak{a}\right|=3.
  \]
\end{eg}

\begin{prop}\label{prop:09mon1}
  Let~$\mathfrak{a}$ be a nonzero ideal of~$\mathfrak{O}_{K}$ and pick
  a~$\Z$-module basis $\{\alpha_{1},\ldots,\alpha_{n}\}$
  for~$\mathfrak{a}$.\footnote{We have seen that~$\mathfrak{a}$ if a free
  $\Z$-module of rank~$n=[K\mathbin{:}\Q]$.}  Then
  \[
    N(\mathfrak{a})=\left|\frac{\Delta[\alpha_{1},\ldots,\alpha_{n}]}{\Delta}\right|^{1/2}
  \]
  where $\Delta$ is the discriminant of~$K$ (i.e., the discriminant of any
  $\Z$-basis for~$\mathfrak{O}_{K}$).
\end{prop}
\begin{proof}
 Let~$\{\omega_{1},\ldots,\omega_{n}\}$ be a $\Z$-basis for $\mathfrak{O}_{K}$.
 Each $\alpha_{i}$ is a $\Z$-linear combination of the $\omega_{i}$s.  Hence,
 there is an integer matrix~$C$ such that 
 \[
   \left(\alpha_{1},\ldots,\alpha_{n}\right)^{t} = C
   \left(\omega_{1},\ldots,\omega_{n}\right)^{t}.
 \]
 By the change of basis formula for the discriminant, we have
 \[
   \Delta[\alpha_{1},\ldots,\alpha_{n}]
   =\det(C)^{2}\Delta[\omega_{1},\ldots,\omega_{n}] 
   =\det(C)^{2}\Delta.
 \]
 On the other hand, we have the commutative diagram:
 \begin{center}
   \begin{tikzcd}
     0\ar[r] &   \mathfrak{a} \ar[r] \ar[d,"{\rotatebox{270}{$\approx$}}"]
     &\mathfrak{O}_{K} \ar[r] \ar[d,"{\rotatebox{90}{$\approx$}}"]
     &\mathfrak{O}_{K}/\mathfrak{a} \ar[r] \ar[d,"{\rotatebox{90}{$\approx$}}"]&0\\
   0\ar[r] &  \Z^n \ar[r,"C"] &\Z^n \ar[r] &\cok{C} \ar[r] &0,
   \end{tikzcd}
 \end{center}
and we have seen that~$|\cok(C)|=|\det(C)|$.  (Recall that by an integer change
of coordinates, i.e., by applying integer row and column operations to~$C$, we
can replace~$C$ be a diagonal matrix~$D$.  It is then easy to see that
$|\det(C)|=|\det(D)|=|\cok(D)|=|\cok(C)|$.) It follows that
$|\det(C)|=\left|\mathfrak{O}_{K}/\mathfrak{a}\right|=N(\mathfrak{a})$.

We then have
\[
   \Delta[\alpha_{1},\ldots,\alpha_{n}]=\det(C)^{2}\Delta=N(\mathfrak{a})^{2}\Delta,
\]
and the result follows by taking square roots.
\end{proof}

\begin{cor}
  Let~$0\neq \alpha\in\mathfrak{O}_{K}$, and consider the principal
  ideal~$(\alpha)$.  Then
  \[
    N((\alpha))=|N(\alpha)|
  \]
  where~$N(\alpha)$ is the norm we defined previously for elements of~$K$.
\end{cor}
\begin{proof}
 Let~$\{\omega_{1},\ldots,\omega_{n}\}$ be a $\Z$-basis for~$\mathfrak{O}_{K}$.
 Then $\{\alpha \omega_{1},\ldots,\alpha \omega_{n}\}$ is a $\Z$-basis for the
 principal ideal~$(\alpha)$.  Say~$\sigma_{1},\ldots,\sigma_{n}$ are the
 embeddings of~$K$ in~$\C$.  By definition of the discriminant,
 \[
   \Delta[\alpha \omega_{1},\ldots,\alpha \omega_{n}]
   = \prod_{i=1}^{n} \sigma_{i}(\alpha \omega_{j})^{2}
   = \left(\prod_{i=1}^{n}\sigma_{i}(\alpha)\right)^{2} \left(\prod_{i=1}^{n} \sigma_{i}(\omega_{j})^{2}\right)
   = N(\alpha)^{2}\Delta.
 \]
 The result now follows from Proposition~\ref{prop:09mon1}.
\end{proof}

\begin{eg}
  Let~$d$ be a square-free integer not equal to $0$ or~$1$.
  Let~$a,b\in\Z$ and consider the principal ideal~$\mathfrak{a}=(a+b\sqrt{d})$ in
  $\mathfrak{O}_{\Q(\sqrt{d})}$.  Then
  \[
   \left|\mathfrak{O}_{\Q(\sqrt{d})}/\mathfrak{a}\right| 
   = N(\mathfrak{a}) = |N(a+b\sqrt{d})| = |(a+b\sqrt{d})(a-b\sqrt{d})|=|a^2-db^{2}|.
  \]
\end{eg}

Just like the norm we defined for algebraic numbers, the norm for ideals is
multiplicative:

\begin{prop}
 Let~$\mathfrak{a}$ and~$\mathfrak{b}$ be nonzero ideals of~$\mathfrak{O}_{K}$.
 Then
 \[
   N(\mathfrak{a}\mathfrak{b})=N(\mathfrak{a})N(\mathfrak{b}).
 \]
\end{prop}
\begin{proof}
 See Theorem~5.12 in our text.
\end{proof}

\begin{prop} 
  Let~$\mathfrak{a}$ be a nonzero ideal of~$\mathfrak{O}_{K}$.  Then
  \begin{enumerate}
    \item If~$\alpha\in \mathfrak{a}$, then $N(\mathfrak{a})|N(\alpha)$.
    \item $N(\mathfrak{a})=1$ if and only if
      $\mathfrak{a}=(1)=\mathfrak{O}_{K}$.
    \item If $N(\mathfrak{a})$ is prime, $\mathfrak{a}$ is prime.
    \item $N(\mathfrak{a})\in \mathfrak{a}$.
    \item If $\mathfrak{a}$ is prime, then~$\mathfrak{a}$ contains a unique
      rational prime~$p$ and $N(\mathfrak{a})=p^{m}$ for some $1\leq
      m\leq n:=[K\mathbin{:}\Q]$.
  \end{enumerate}
\end{prop}
\begin{proof}\
 \begin{enumerate}
   \item If~$\alpha\in\mathfrak{a}$, then the principal ideal $(\alpha)$ is
     contained in~$\mathfrak{a}$.  Therefore~$\mathfrak{a}|(\alpha)$, i.e.,
     there exists an ideal~$\mathfrak{b}$ such
     that~$(\alpha)=\mathfrak{a}\mathfrak{b}$.  Taking norms yiels
     \[
       N((\alpha))=|N(\alpha)|=N(\mathfrak{a})N(\mathfrak{b}).
     \]
     The result follows.
   \item This part is immediate from the definition of the norm.
   \item Factor~$\mathfrak{a}$ into primes:
     \[
      \mathfrak{a}=\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_{i}}.
     \]
     Taking norms:
     \begin{equation}\label{eqn:09monnorm}
       N(\mathfrak{a})=\prod_{i=1}^{k} N(\mathfrak{p}_{i})^{e_{i}}.
     \end{equation}
     If~$\mathfrak{p}$ is prime, then $\mathfrak{p}\neq \mathfrak{O}_{K}$, and
     hence~$N(\mathfrak{p})>1$.  Therefore, $N(\mathfrak{a})$ is prime if and
     only if~$\mathfrak{a}$ is prime.
   \item Since $N(\mathfrak{a}) = |\mathfrak{O}_{K}/\mathfrak{a}|$, it follows
     that for any~$\alpha\in\mathfrak{O}_{K}$, we have
     $N(\mathfrak{a})\alpha=0\in\mathfrak{O}_{K}/\mathfrak{a}$, i.e.,
     $N(\mathfrak{a})\alpha\in\mathfrak{a}$. Letting $\alpha=1$ gives the
     result.
   \item Suppose that~$\mathfrak{p}$ is prime.
     Let~$N(\mathfrak{a})=\prod_{i=1}^{k} p_{i}^{e_{i}}$ be the prime
     factorization of~$N(\mathfrak{a})$. Since
     $N(\mathfrak{a})\in\mathfrak{a}$, on the level of ideals, we have
     \[
       \prod_{i=1}^{k} (p_{i})^{e_{i}}\subseteq \mathfrak{a},
     \]
     and, hence,
     \[
       \mathfrak{a}|\prod_{i=1}^{k} (p_{i})^{e_{i}}.
     \]
     Since~$\mathfrak{a}$ is prime, there exists~$i$ such
     that~$\mathfrak{a}|(p_{i})$, which means $(p_{i})\subseteq \mathfrak{a}$
     or, equivalently,~$p_{i}\in\mathfrak{a}$.
     If there exists an rational prime~$q\neq p_{i}$ in~$\mathfrak{a}$, we
     would have 
     \[
       1\in(p_{i},q)=(p_{i})+(q)\subseteq\mathfrak{a}
     \]
     However, since~$\mathfrak{a}$ is prime, it does not contain~$1$.  So there
     exists a unique prime~$p=p_{i}\in\mathfrak{a}$.  From the first part of
     this problem, we have~$N(\mathfrak{a})|N(p)$.  Since~$N(p)=p^{n}$, the
     result follows.
 \end{enumerate}
\end{proof}

\begin{prop}\
  \begin{enumerate}
    \item Let $\mathfrak{a}$ be an ideal of~$\mathfrak{O}_{K}$.  Then there are
      only a finite number of ideals~$\mathfrak{b}$ such that
      $\mathfrak{b}|\mathfrak{a}$,  Equivalently, there are finitely many
      ideals~$\mathfrak{b}$ such that~$\mathfrak{a}\subseteq\mathfrak{b}$.
    \item\label{prop:09monprop2} If $a\in\Z$, there are finitely many ideals~$\mathfrak{a}$ of
      $\mathfrak{O}_{K}$ containing~$a$.
    \item There are finitely many ideals with a given norm.
  \end{enumerate}
\end{prop}
\Needspace{3\baselineskip}
\begin{proof}\
  \begin{enumerate}
    \item This is an immediate consequence of prime factorization of ideals. 
    \item We have~$a\in\mathfrak{a}$ if and only if $\mathfrak{a}|(a)$.  So this
      result follows from the previous applied to the principal ideal~$(a)$.
    \item Fix~$a\in\Z_{>0}$.  If~$\mathfrak{a}$ is an ideal
      with~$N(\mathfrak{a})=a$, then from the previous proposition, we
      have~$a\in\mathfrak{a}$.  The result then follows from
      part~\ref{prop:09monprop2}.
  \end{enumerate}
\end{proof}

\begin{prop}
  The number ring~$\mathfrak{O}_{K}$ is a UFD if and only if it is a PID.
\end{prop}
\begin{proof}
We already know that a PID is a UFD and that~$\mathfrak{O}_{K}$ is a
factorization domain, i.e., every element of~$\mathfrak{O}_{K}$ has a
factorization into irreducibles.  Suppose that~$\mathfrak{O}_{K}$ is a UFD.
Since every ideal is a product of prime ideals, to show~$\mathfrak{O}_{K}$ is a
PID, it suffices to show that every prime ideal is principal.

Let~$\mathfrak{p}$ be a prime ideal of~$\mathfrak{O}_{K}$. We have
\[
  \mathfrak{p}\ni N(\mathfrak{p})=\pi_{1}\cdots\pi_{k}
\]
where the~$\pi_{i}$ are irreducibles in~$\mathfrak{O}_{K}$.
Since~$\mathfrak{p}$ is prime and divides~$\prod_{i=1}^{k} (\pi_{})$, it follows
that~$\mathfrak{p}|(\pi_{i})$ for some~$i$.  Thus,
$(\pi_{i})\subseteq\mathfrak{p}$.  In a UFD,
irreducibles are prime.  Therefore, $(\pi_{i})$ is prime.
Since~$\mathfrak{O}_{K}$ is Dedekind, nonzero primes are maximal.
Therefore~$\mathfrak{p}=(\pi_{i})$.
\end{proof}

\begin{prop}
  Suppose that~$\mathfrak{O}_{K}$ is not a UFD, and let~$\pi\in\mathfrak{O}_{K}$
  be irreducible but not prime.  Let~$(\pi)=\prod_{i=1}^{k}
  \mathfrak{p}_{i}^{e_{i}}$ be the prime factorization of~$(\pi)$.  Then no
  $\mathfrak{p}_{i}$ is principal.
\end{prop}
\begin{proof}
  For the sake of contradiction,
 suppose $\mathfrak{p}_{i}=(\alpha)$ from some~$i$ and
 some~$\alpha\in\mathfrak{O}_{K}$.  Then since~$\mathfrak{p}_{i}|(\pi)$, it
 follows that~$(\pi)\subseteq \mathfrak{p}_{i}=(\alpha)$.  Hence, $\pi=\alpha
 \beta$ from some~$\beta\in\mathfrak{O}_{K}$.  Since~$\mathfrak{p}$ is prime, so
 is~$\alpha$.  Since~$\pi$ is irreducible, $\beta$ is a unit.  Hence, $\pi$ is
 prime---a contradiction.
\end{proof}
\end{document}
