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\begin{document}
\centerline{Math 361 lecture for Wednesday, Week 7}
\bigskip

\centerline{Dedekind domains}
\bigskip

\textbf{Field of fractions of a domain.}  The rational numbers~$\Q$ form a field
that is just big enough to contain the inverses of all nonzero integers.  Here
we review and generalize the construction of the rationals from the integers.
Let~$R$ be a domain.  Define an equivalence relation on $R\times
R\setminus\{0\}$ by~$(a,b)\sim (c,d)$ if $ad=bc$, and then let fraction~$a/b$ denote the
equivalence class of~$(a,b)$.  Thus,
\[
  \frac{a}{b}=\frac{c}{d}\ \Longleftrightarrow\ ad=bc.
\]
Define addition and multiplication as usual:
\[
  \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\quad\text{and}\quad
  \frac{a}{b}\cdot\frac{c}{d}= \frac{ac}{bd}.
\]
The collection of these fractions along with these two operations give the \emph{field of fractions} of~$R$, denoted~$Q(R)$.
The field of fractions is also called the \emph{quotient field} of~$R$.
(Exercise: check that~$Q(R)$ is actually a field).
We identify~$r\in R$ with $r/1\in Q(R)$.

\begin{prop}
  Let~$K$ be a number field, and let $\mathfrak{O}_{K}$ be its ring of integers.
  Then $K$ is the field of fractions of~$\mathfrak{O}_{K}$.
\end{prop}
\begin{proof}
  We have seen (in homework) that if $\alpha\in K$ then there exists a nonzero integer~$c\in\Z$ such
   that~$c\alpha=\beta\in\mathfrak{O}_{K}$.  Thus, $\alpha=\beta/c$ with
   $\beta,c\in\mathfrak{O}_{K}$.  Thus, every element of~$K$ is in the field of
   fractions of~$\mathfrak{O}_{K}$.  Conversely, since~$K$ is a field and
   contains~$\mathfrak{O}_{K}$, it contains the field of fractions
   of~$\mathfrak{O}_{K}$.
\end{proof}


\textbf{Dedekind domains.}
We define a Dedekind domain below.  The definition captures some of the most
consequential properties of a number ring.  There are several equivalent
formulations of the definition, some of which we will get to later.  (For
instance: our definition is equivalent to the property that every
nonzero prime ideal in a domain factors into prime ideals!)

Recall that an element $\alpha\in Q(R)$ is \emph{integral over $R$} is there
exists a monic polynomial $f\in R[x]$ such that~$f(\alpha)=0$.

\begin{defn}
 A domain $R$ is \emph{integrally closed} if the only elements of its field of
 fractions $Q(R)$ that are integral over~$R$ are the elements of~$R$, itself.
\end{defn}

Recall that we denote the ring of all complex numbers that are integral over~$\Z$
by~$\mathfrak{O}$, and the ring of integers in a number field~$K$
by~$\mathfrak{O}_{K}:=K\cap\mathfrak{O}$.  One of our goals today is to show
that~$\mathfrak{O}_{K}$ is integrally closed.

\begin{eg}
  The quotient field of~$\Z$ is~$\Q$, and we has seen that~$\Z$ is integrally
  closed: the elements of~$\Q$ integral over~$\Z$ are exactly the elements
  of~$\Z$.
\end{eg}

\begin{defn}
 A \emph{Dedekind domain} is an integrally closed Noetherian domain in which
 every nonzero prime ideal is maximal.
\end{defn}

\begin{thm}\label{thm:07wedthm}
  Let~$K$ be a number field, and let $\mathfrak{O}_{K}$ be its ring of integers.
  Then $\mathfrak{O}_{K}$ is a Dedekind domain.
\end{thm}

We give some preliminary results that will be used in proving the theorem.
\medskip

\textbf{From now on, we take $K$ to be a number field with ring of
	integers~$\mathfrak{O}_{K}$.}
\medskip

\begin{thm} (Structure theorem for finitely-generated $\Z$-modules.)
  Let~$M$ be a finitely generated $\Z$-module.  Then there exists a nonnegative
  integer~$r$ and a list (possibly empty) of integers~$n_{1},\ldots,n_{k}$ with
  $n_{i}>1$ for all~$i$ such that~$M$ is isomorphic as a $\Z$-module to
  \[
    \Z^{r}\times \Z/{n_1}\Z\times\dots\times\Z/{n_k}\Z.
  \]
  It is possible to take the~$n_{i}$ so that~$n_{i}|n_{i+1}$ for all~$i$, in
  which case, the above representation of~$M$ as a product of cyclic groups is
  unique.
\end{thm}
\begin{proof}
   We will give a constructive proof later in the course. (Probably.)
\end{proof}

\begin{prop}
  Let $R$ be a finite domain.  Then~$R$ is a field.
\end{prop}
\begin{proof}
  Homework.
\end{proof}

On last preliminary result:

\begin{prop}
  A prime ideal in a number ring contains the norm of each of its elements:
  if~$\mathfrak{p}$ is a prime ideal in~$\mathfrak{O}_{K}$, and~$\alpha\in
  \mathfrak{p}$, then~$\Z\ni N(\alpha)\in\mathfrak{p}$. 
\end{prop}
\begin{proof}
  We have already seen that~$N(\alpha)\in\Z$ since the norm is a coefficient of
  the field polynomial for~$\alpha$, which in turn is a power of the minimal
  polynomial for~$\alpha$.

  If $\alpha=0$, the result is obvious.  So assume that~$\alpha\neq0$.
  Let~$\sigma_{1},\ldots,\sigma_{n}$ be the embeddings of~$K$ with
  $\sigma_{1}=\mathrm{id}$, as usual.  Then, define~$\beta$ by
  \[
    N(\alpha)=\prod_{i=1}^{n} \sigma_{i}(n)=\alpha\cdot
    \underbrace{\sigma_{2}(\alpha)\cdots\sigma_{n}(\alpha)}_{\beta}.
  \]
  Since~$N(\alpha)\in\Z\subset K$ and~$0\neq\alpha\in K$, it follows
  that~$\beta=N(\alpha)/\alpha\in K$.  Let $\mathfrak{O}$ denote the set of all
  algebraic integers (in~$\C$).  We have seen that
  each~$\sigma_{i}(\alpha)$ is an algebraic integer (apply~$\sigma_{i}$ to the
  minimal polynomial for~$\alpha$). Since~$\mathfrak{O}$ is a ring, and
  each~$\sigma_{i}\in\mathfrak{O}$, it follows that
  $\beta\in\mathfrak{O}$.  So we have~$\beta\in K\cap
  \mathfrak{O}=:\mathfrak{O}_{K}$.  Therefore, $N(\alpha)=\alpha \beta$ with
  $\alpha\in \mathfrak{p}$ and~$\beta\in \mathfrak{O}_{K}$.  It follows
  that~$N(\alpha)\in\mathfrak{p}$.
\end{proof}

We now prove our main theorem:

\begin{proof}[Proof of Theorem~\ref{thm:07wedthm}]
We have seen that~$\mathfrak{O}_{K}$ is a Noetherian $\Z$-module since it is a
finitely generated~$\Z$-module and~$\Z$ is Noetherian.
If~$\mathfrak{a}\subseteq\mathfrak{O}_{K}$ is an ideal, it follows that~$\mathfrak{a}$ is a
finitely generated~$\Z$-module.  Any set of generators over~$\Z$ will
generate~$\mathfrak{a}$ as an ideal.  Thus, $\mathfrak{O}_{K}$ is a Noetherian
ring (i.e., $\mathfrak{O}_{K}$ is Noetherian as an~$\mathfrak{O}_{K}$-module.) Alternatively, we could use the Hilbert
basis theorem: since~$\mathfrak{O}_{K}$ is finitely generated as a~$\Z$-module,
it is certainly finitely generated as a ring over~$\Z$.  Then, since~$\Z$ is a
Noetherian domain, so is $\mathfrak{O}_{K}$.

Next, we would like to show that every nonzero prime ideal of~$\mathfrak{O}_{K}$
is maximal.  Let~$0\neq\mathfrak{p}$ ve a prime ideal. Take any nonzero
$\alpha\in\mathfrak{p}$.  Let~$N$ denote the integer $N(\alpha)\in\Z$.  We have seen that~$N\in\mathfrak{p}$.
Consider the (surjective) quotient mapping
\begin{align*}
   \mathfrak{O}_{K}&\to \mathfrak{O}_{K}/\mathfrak{p} \\
  \beta&\mapsto \overline{\beta}. 
\end{align*}
Since~$N\in\mathfrak{p}$, it is in the kernel of the quotient mapping.
Therefore, we get a well-defined surjective mapping
\begin{align}\label{surj:07wed}
   \mathfrak{O}_{K}/(N)&\to \mathfrak{O}_{K}/\mathfrak{p} \\
  \beta&\mapsto \overline{\beta}. \nonumber
\end{align}
where~$(N)$ is the principal ideal generated by $N$ in $\mathfrak{O}_{K}$.
Now, $\mathfrak{O}_{K}/(N)$ is a finitely generated $\Z$-module (indeed,
even~$\mathfrak{O}_{K}$, itself, is a finitely generated $\Z$-module).  Thus, we
have an isomorphism of~$\Z$-modules:
\[
 \mathfrak{O}_{K}/(N) \simeq \Z^{r}\times \Z/{n_1}\Z\times\dots\times\Z/{n_k}.
\]
for some $r,n_{1},\ldots,n_{k}$.  If~$r\neq0$, then~$\mathfrak{O}_{K}/(N)$ would
have elements of infinite order.  However, for each $\gamma\in
\mathfrak{O}_{K}$, we have
\[
  \underbrace{\gamma+\cdots+\gamma}_{\text{$N$ times}}=N\gamma =0\in
  \mathfrak{O}_{K}/(N).
\]
So
\[
 \mathfrak{O}_{K}/(N) \simeq \Z/{n_1}\Z\times\dots\times\Z/{n_k}.
\]
for some list of integers~$n_{i}$ with $n_{i}>1$.  This means
that~$\mathfrak{O}_{K}/(N)$ is finite.  The surjection~(\ref{surj:07wed}), then
implies that~$\mathfrak{O}_{K}/\mathfrak{p}$ is finite.

Since~$\mathfrak{p}$ is prime, $\mathfrak{O}_{K}/\mathfrak{p}$ is a domain.
Since a finite domain is a field (from a proposition, above), it follows
that~$\mathfrak{O}_{K}/\mathfrak{p}$ is a field.  Hence,~$\mathfrak{p}$ is
maximal, as we wanted to show.

It remains to show that~$\mathfrak{O}_{K}$ is integrally closed.  Take
$\alpha\in K$ and suppose that~$\alpha$ is integral over~$\mathfrak{O}_{K}$.  We
must show that~$\alpha$ is integral over~$\Z$, i.e., that~$\alpha\in
\mathfrak{O}$.  We then have that~$\alpha\in
K\cap\mathfrak{O}=\mathfrak{O}_{K}$.  We do so by showing there exists a
finitely generated $\Z$-module~$M\subset K$ such that~$\alpha M\subseteq M$.
(This was one of the equivalent conditions established for integrality.  You may
remember that it involved multiplying each element of a generating set by
$\alpha$ and then evaluating a certain determinant.)

Since~$\alpha$ is integral over~$\mathfrak{O}_{K}$, there exists a monic
polynomial~$f\in \mathfrak{O}_{K}[x]$ such that~$f(\alpha)=0$.  Say
\[
  f(x)= x^{k}+b_{k-1}x^{k-1}+\cdots+b_{1}x+b_{0}
\]
with~$b_{i}\in\mathfrak{O}_{K}$ for all~$i$.  Let~$B$ be the subring of~$\mathfrak{O}_{K}$
generated~$b_{i}$ over~$\Z$:
\[
  B = \Z[b_{0},\ldots,b_{k}]:=\{g(b_{0},\ldots,b_{k}): g\in
  \Z[x_{0},\ldots,x_{k}]\}.
\]
The ring~$B$ is a $\Z$-submodule of~$\mathfrak{O}_{K}$.
Since~$\mathfrak{O}_{K}$ is a Noetherian~$\Z$-module, it follows that $B$ is
finitely generated as a~$\Z$-module.  Since~$\Z$ is Noetherian, it follows
that~$B$ is a Noetherian~$\Z$-module.  Next consider the ring
\[
  B[\alpha] :=\left\{g(\alpha): g\in B[x]\right\}
\]
Since
\[
  0=f(\alpha)=\alpha^{k}+b_{k-1}\alpha^{k-2}+\cdots+b_{1}\alpha+b_{0},
\]
it follows that~$B[\alpha]$ is finitely generated as a $B$-module by
$\{1,\alpha,\ldots,\alpha^{k}\}$.  

We have~$\Z\subseteq B\subseteq B[\alpha]$ with~$B$ finitely generated as a
$\Z$-module and~$B[\alpha]$ finitely generated as a~$B$-module.  We have seen
that this implies~$B[\alpha]$ is finitely generated as a $\Z$-module (by the set
of products of generators of~$B$ over~$\Z$ with generators of~$B[\alpha]$
over~$B$).

So $B[\alpha]$ is a finitely generated $\Z$-module, and it is clear
that~$\alpha B[\alpha]\subseteq B[\alpha]$.  Therefore, $\alpha$ is integral
over~$\Z$, completing the proof.
\end{proof}

\end{document}
