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

\centerline{Fractional ideals}
\bigskip

In the following, let~$K$ be a number field with ring of integers~$\mathfrak{O}_{K}$.
\medskip

We have seen that addition and multiplication of ideals have nice properties:
commutativity, distributivity, there is an additive identity,~$(0)$, and a
multiplicative identity,~$(1)$.  However, nonzero ideals do not necessarily have
multiplicative inverses.  To fix that in the case of number rings, we introduce
fractional ideals.

\begin{defn}
  An $\mathfrak{O}_{K}$-submodule~$I\subseteq K$ is a \emph{fractional ideal}
  of~$\mathfrak{O}_{K}$ if there exists~$\alpha\in
  \mathfrak{O}_{K}\setminus\{0\}$ such that~$\alpha I\subseteq\mathfrak{O}_{K}$

  The product of two fractional ideals~$I,J$ in~$\mathfrak{O}_{K}$ is the
  $\mathfrak{O}_{K}$-submodule of~$K$ 
  \[ 
    IJ =\Span_{\mathfrak{O}_{K}}\{ij:i\in I, j\in J\}. 
  \]
\end{defn}

\begin{remark}
  \begin{enumerate}
    \item Let~$I$ be a fractional ideal of~$\mathfrak{O}_{K}$, and say~$\alpha\in
      \mathfrak{O}_{K}\setminus\{0\}$ is such that~$\mathfrak{a}:=\alpha I\subseteq \mathfrak{O}_{K}$.  
      Then $\mathfrak{a}$ is an ideal of~$\mathfrak{O}_{K}$ (reason: $I$
      an~$\mathfrak{O}_{K}$-submodule implies $\alpha I$ 
      an~$\mathfrak{O}_{K}$-submodule of~$\mathfrak{O}_{K}$, i.e., an ideal).  
    \item The fractional ideals are exactly the~$\mathfrak{O}_{K}$-submodules
      of~$K$ of the form~$\alpha^{-1}\mathfrak{a}$ for some ideal~$\mathfrak{a}$ of
      $\mathfrak{O}_{K}$ and nonzero~$\alpha\in\mathfrak{O}_{K}$.
    \item If $I$ is an $\mathfrak{O}_{K}$-submodule, then $I$ is a fractional
      ideal if and only if there exists some~$c\in K\setminus\{0\}$ such
      that~$cI\subseteq\mathfrak{O}_{K}$.  (In the definition, $c$ is required
      to be in $\mathfrak{O}_{K}$.)  Suppose such a $c$ exists.  Then since~$K$
      is the quotient field of~$\mathfrak{O}_{K}$, there
      exists~$\alpha,\beta\in\mathfrak{O}_{K}$ such that~$c=\alpha/\beta$.  Then
      \[
        \alpha I = (c\beta) I \subseteq cI\subseteq\mathfrak{O}_{K}.
      \]
  \end{enumerate}
\end{remark}

\begin{eg}
  In the rational integers~$\Z$, the fractional ideals have the form
  \[
    r\Z=\{ra:a\in \Z\}.
  \]
  For instance, the set of all integer multiples of~$2/3$ is a fractional ideal
  of~$\Z$.  In general, if~$\mathfrak{O}_{K}$ is a PID, then every fractional
  ideal has the form $c\mathfrak{O}_{K}$ for some $c\in K$.  
\end{eg}

\begin{prop}
  Fractional ideals of~$\mathfrak{O}_{K}$ are exactly
  finitely generated $\mathfrak{O}_{K}$-submodules of~$K$.
\end{prop}
\begin{proof}
   First, suppose that~$I$ is a fractional ideal of~$\mathfrak{O}_{K}$, and take
   $\alpha\in\mathfrak{O}_{K}\setminus\{0\}$ such that~$\alpha
   I\subseteq\mathfrak{O}_{K}$.  Then $\alpha I$ is an ideal of the Noetherian
   ring~$\mathfrak{O}_{K}$.  Hence, $\alpha I$ is finitely generated as
   an~$\mathfrak{O}_{K}$-module.  We have an isomorphism of
   $\mathfrak{O}_{K}$-modules:
   \begin{align*}
     I&\to \alpha I\\
     x&\mapsto \alpha x.
   \end{align*}
   Hence, $I$ is a finitely generated as an~$\mathfrak{O}_{K}$-module (just
   multiply the generators of~$\alpha I$ by $\alpha^{-1}$ to get generators
   for~$I$).

   Conversely, suppose that~$I=\Span_{\mathfrak{O}_{K}}\{x_{1},\ldots,x_{m}\}$ is a
   finitely-generated~$\mathfrak{O}_{K}$-submodule of~$K$. Since~$K$ is the
   quotient field of $\mathfrak{O}_{K}$, we can
   write~$x_{i}=\alpha_{i}/\beta_{i}$ with $\beta_{i}\neq0$ for all~$i$.  Define~$\alpha=\prod_{i=1}^{m} \beta_{i}$. 
   Then~$\alpha I\subseteq\mathfrak{O}_{K}$.  So $I^{-1}$ is finitely generated.
\end{proof}

\begin{prop}
  The set of nonzero fractional ideals in a number field~$K$ forms an abelian group
  under multiplication.  If $I$ is a nonzero fractional ideal of~$\mathfrak{O}_{K}$,
  then its inverse is
  \[
    I^{-1}=\{x\in K: xI\subseteq\mathfrak{O}_{K}\}.
  \]
\end{prop}
\begin{proof}
  Let~$I$ and~$J$ be fractional ideals.  Say~$I=c\mathfrak{a}$
  and~$J=d\mathfrak{b}$ for some ideals $\mathfrak{a},\mathfrak{b}$
  of~$\mathfrak{O}_{K}$ and some nonzero elements~$c,d\in\mathfrak{O}_{K}$.  Then 
  \[
    IJ = (c\mathfrak{a})(d\mathfrak{b}) = (cd){\mathfrak{a}\mathfrak{b}}
  \]
  is fractional ideal since $cd\in \mathfrak{O}_{K}\setminus\{0\}$
  and~$\mathfrak{a}\mathfrak{b}$ is an ideal of~$\mathfrak{O}_{K}$.  Hence,
  the set of fractional ideals is closed under multiplication.  Multiplication
  is clearly associative, and there is an identity element, the principal
  ideal~$(1)$.  We prove that nonzero fractional ideals have inverses as stated
  as part of the next theorem.
\end{proof}

\begin{defn}
  If $I,J$ are ideals in a ring $R$, then $I$ \emph{divides} $J$, denoted~$I|J$
  if there exists an ideal $H$ such that~$J=IH$.
\end{defn}

\begin{prop}
  (\emph{To contain is to divide.})
  Let~$\mathfrak{a}$ and $\mathfrak{b}$ be ideals in $\mathfrak{O}_{K}$.  Then
  $\mathfrak{a}|\mathfrak{b}$ if and only if
  $\mathfrak{b}\subseteq\mathfrak{a}$.
\end{prop}
\begin{proof} ($\Rightarrow$) Suppose that~$\mathfrak{a}|\mathfrak{b}$, and
  take~$\mathfrak{c}$ such that~$\mathfrak{b}=\mathfrak{a}\mathfrak{c}$.  The
  result follows since~$\mathfrak{a}\mathfrak{c}\subseteq\mathfrak{a}$.
  
  ($\Leftarrow$) Now suppose that~$\mathfrak{b}\subseteq\mathfrak{a}$.
  If~$\mathfrak{a}=0$, the result is trivial.  So suppose~$\mathfrak{a}\neq0$.
  We then have
  \[
    \mathfrak{b}\subseteq\mathfrak{a}\quad\Rightarrow\quad
    \mathfrak{a}^{-1}\mathfrak{b}\subseteq
    \mathfrak{a}^{-1}\mathfrak{a}=\mathfrak{O}_{K}.
  \]
  Define $\mathfrak{c}=\mathfrak{a}^{-1}\mathfrak{b}$.
  Then $\mathfrak{c}$ is a fractional ideal contained in~$\mathfrak{O}_{K}$, so
  $\mathfrak{c}$ is an ideal.  Further, $\mathfrak{a}\mathfrak{c}=\mathfrak{b}$,
  as required.
\end{proof}

\begin{thm}
  Let $K$ be a number field.
  Every nonzero ideal of~$\mathfrak{O}_{K}$ can be factored into a product of
  prime ideals, uniquely up to the order of factors.
\end{thm}
\begin{proof}
  We follow our text, breaking down the proof to several steps.
\medskip

\textbf{Step 1.} Claim: Let $\mathfrak{a}\neq0$ be an ideal of~$\mathfrak{O}_{K}$.
Then there there exists nonzero prime
ideals~$\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}$ such
\[
  \mathfrak{p}_{1}\cdots\mathfrak{p}_{r}\subseteq\mathfrak{a}.
\]
\emph{Proof of claim.} Let $\mathcal{A}$ be the set of all nonzero ideals that
do not have the desired property.  We would like to show that~$\mathcal{A}$ is
empty.  For the sake of contradiction, suppose it is not. Then since~$\mathfrak{O}_{K}$ is Noetherian, $\mathcal{A}$ has a
maximal element~$\mathfrak{a}$.  Since~$\mathfrak{a}$ does not have the desired
property, it cannot be  prime.  So there exist $\beta,\gamma\in\mathfrak{O}_{K}$ such
that~$\beta \gamma\in\mathfrak{a}$, yet~$\beta\not\in\mathfrak{a}$ and $\gamma\not\in\mathfrak{a}$.
We have
\[
  \mathfrak{a}\subsetneq \mathfrak{a}+(\beta)\quad\text{and}\quad
  \mathfrak{a}\subsetneq \mathfrak{a}+(\gamma).
\]
By maximality of~$\mathfrak{a}$, the ideals $\mathfrak{a}+(\beta)$ and
$\mathfrak{a}+(\gamma)$ are not in~$\mathcal{A}$.  Hence, there exist prime ideals
$\mathfrak{p}_{i}$ and $\mathfrak{q}_{j}$ such that
\[
 \prod_{i=1}^{k} \mathfrak{p}_{i}\subseteq\mathfrak{a}+(\beta)
 \quad\text{and}\quad
  \prod_{i=j}^{\ell} \mathfrak{q}_{i}\subseteq\mathfrak{a}+(\gamma).
\]
It follows that
\[
  \left(\prod_{i=1}^{k} \mathfrak{p}_{i}\right)\left(\prod_{i=j}^{\ell}
  \mathfrak{q}_{i}\right)
  \subseteq \left(\mathfrak{a}+(\beta)\right)\left(\mathfrak{a}+(\gamma)\right)
  \subseteq \mathfrak{a}+(\beta \gamma)
  \subseteq \mathfrak{a},
\]
which yields that contradiction that~$\mathfrak{a}\not\in\mathcal{A}$.
\bigskip

\textbf{Step 2.} Given a nonzero fractional ideal~$I$, define
\[
  I^{-1} := \{x\in K: xI\subseteq\mathfrak{O}_{K}\}.
\]
Since~$I$ is a fractional ideal, there exists~$\alpha\in
\mathfrak{O}_{K}\setminus\{0\}$ such
that~$\alpha I\subseteq\mathfrak{O}_{K}$.  Hence, $\alpha\in I^{-1}$.  So
$I^{-1}\neq\emptyset$ (and $I^{-1}\neq0$). It is straightforward to check that~$I^{-1}$ is
an~$\mathfrak{O}_{K}$-submodule. Letting~$y\in I\neq0$, we have~$y\in
K\setminus\{0\}$ and~$yI^{-1}\subseteq\mathfrak{O}_{K}$ (recall that it suffices
to find such an element in~$K$, not necessarily in~$\mathfrak{O}_{K}$).
Hence,~$I^{-1}$ is a fractional ideal.
\medskip

Claim: $I^{-1}$ is the multiplicative inverse of~$I$, i.e.,
$II^{-1}=(1)=\mathfrak{O}_{K}$.  We now prove this in several steps.
\medskip

\textbf{Step 2.1.} Let $\mathfrak{a}\subseteq\mathfrak{O}_{K}$ be a proper
nonzero ideal. (By ``proper'' we mean $\mathfrak{a}\subsetneq\mathfrak{O}_{K}$.)
Claim: $\mathfrak{O}_{K}\subsetneq\mathfrak{a^{-1}}$.
\medskip

\emph{Proof of claim.}  Since~$\mathfrak{a}$ is an ideal, it is clear from the definition of~$\mathfrak{a}^{-1}$
that~$\mathfrak{O}_{K}\subseteq \mathfrak{a}^{-1}$. Let~$\mathcal{A}$ now be the
set of proper ideals of~$\mathfrak{O}_{K}$.  Since $\mathfrak{O}_{K}$ is a
Noetherian ring and~$\mathcal{A}\neq\emptyset$, it follows that~$\mathcal{A}$
has a maximal element~$\mathfrak{p}$.  Since~$\mathfrak{p}$ is maximal, it is
prime.

Since~$\mathfrak{a}\subseteq\mathfrak{p}$, it follows
that~$\mathfrak{p}^{-1}\subseteq\mathfrak{a}^{-1}$.  So it suffices to show that
$\mathfrak{p}^{-1}\neq\mathfrak{O}_{K}$.  In other words, we must show that
$\mathfrak{p}^{-1}$ contains an element that is not integral over~$\Z$.
Pick~$0\neq \alpha\in\mathfrak{p}$.  Using Step~$1$, we may pick a minimal~$r$ such that
\[
  \mathfrak{p}_{1}\cdots\mathfrak{p}_{r}\subseteq (\alpha)\subseteq\mathfrak{p}
\]
for some nonzero prime ideals~$\mathfrak{p}_{i}$.  Since~$\mathfrak{p}$ is
prime~$\mathfrak{p}_{i}\subseteq \mathfrak{p}$ for some $i$ (this follows from a
homework problem).  Without loss of
generality, say~$\mathfrak{p}_{1}\subseteq\mathfrak{p}$.
Since~$\mathfrak{O}_{K}$ is Dedekind, nonzero primes are maximal.  Hence,
$\mathfrak{p}_{1}=\mathfrak{p}$.  By minimality of~$r$,
$\mathfrak{p}_{2}\cdots\mathfrak{p}_{r}$ is not contained in $(\alpha)$.  
Take $\beta\in\mathfrak{p}_{2}\cdots\mathfrak{p}_{r}\setminus\{(\alpha)\}$.   Our
goal is to show that~$\alpha^{-1}\beta\in\mathfrak{p}^{-1}$ but $\alpha^{-1}\beta
\not\in\mathfrak{O}_{K}$.  
We have
\[
  \beta\mathfrak{p}
  =
  \beta\mathfrak{p}_{1}
  \subseteq
  \mathfrak{p}_{1}\mathfrak{p}_{2}\cdots\mathfrak{p}_{r}
  \subseteq(\alpha).
\]
So $\alpha^{-1}\beta\mathfrak{p}\subseteq \alpha^{-1}(\alpha)=(1)=\mathfrak{O}_{K}$, and
thus~$\alpha^{-1}\beta\in \mathfrak{p}^{-1}$.  However, $\beta\not\in
(\alpha)=\alpha\mathfrak{O}_{K}$.
So $\alpha^{-1}\beta\not\in\mathfrak{O}_{K}$.
\medskip

\textbf{Step 2.2.} Claim: if $\mathfrak{a}$ is a nonzero ideal and
$\mathfrak{a}S\subseteq\mathfrak{a}$ for any subset~$S\subseteq K$,
then~$S\subseteq\mathfrak{O}_{K}$.
\medskip

\emph{Proof of claim.} Let~$\theta\in S$.  To show~$\theta\in\mathfrak{O}_{K}$,
we must show that~$\theta$ is integral over~$\Z$.  For that, it suffices to find
a finitely generated $\Z$-module $M\subset K$ such that~$\theta M\subseteq M$
(recall the determinant trick).  We know that~$\mathfrak{O}_{K}$ is a Noetherian
$\Z$-module (since is finitely generated as a module over the Noetherian
ring~$\Z$).  The ideal~$\mathfrak{a}\subseteq \mathfrak{O}_{K}$ is thus not only
finitely generated as an ideal (i.e., as an $\mathfrak{O}_{K}$-submodule
of~$\mathfrak{O}_{K}$), it is finitely generated as a $\Z$-module.  So we can
let~$M=\mathfrak{a}$.
\medskip

\textbf{Step 2.3} Let~$\mathfrak{p}$ be a maximal ideal of~$\mathfrak{O}_{K}$.
Claim:~$\mathfrak{p}^{-1}\mathfrak{p}=(1)=\mathfrak{O}_{K}$.
So~$\mathfrak{p}^{-1}$ is the multiplicative inverse of~$\mathfrak{p}$.
\medskip

\emph{Proof of claim.} From the definition
of~$\mathfrak{p}^{-1}$, it immediately follows that~$\mathfrak{p}\subseteq
\mathfrak{p}\mathfrak{p}^{-1}\subseteq\mathfrak{O}_{K}$.
Since~$\mathfrak{p}\mathfrak{p}^{-1}$ is a product of fractional ideals, it is a
fractional ideal.  Hence~$\mathfrak{p}\mathfrak{p}^{-1}$ is
an~$\mathfrak{O}_{K}$-submodule of~$\mathfrak{O}_{K}$, i.e., an ideal.
Since~$\mathfrak{p}$ is maximal, 
$\mathfrak{p}\mathfrak{p}^{-1}$ is either~$\mathfrak{p}$ or $\mathfrak{O}_{K}$.
If~$\mathfrak{p}\mathfrak{p}^{-1}=\mathfrak{p}$, then Step~2.2 implies
that~$\mathfrak{p}^{-1}\subseteq\mathfrak{O}_{K}$, in contradiction to Step~2.1.
Hence, $\mathfrak{p}\mathfrak{p}^{-1}=\mathfrak{O}_{K}$, as claimed.
\medskip

\textbf{Step 2.4.} For every nonzero
ideal~$\mathfrak{a}\subseteq\mathfrak{O}_{K}$, we
have~$\mathfrak{a}\mathfrak{a}^{-1}=(1)=\mathfrak{O}_{K}$.
\medskip

\emph{Proof of claim.} If not, since~$\mathfrak{O}_{K}$ is a Noetherian ring, we
can choose an ideal~$\mathfrak{a}$ that is maximal with respect to the property
that $\mathfrak{a}\mathfrak{a}^{-1}\neq\mathfrak{O}_{K}$.  We can then choose a
maximal ideal~$\mathfrak{p}$ such
that~$\mathfrak{a}\subset\mathfrak{p}\subsetneq\mathfrak{O}_{K}$.
Hence,~$\mathfrak{O}_{K}\subseteq\mathfrak{p}^{-1}\subseteq\mathfrak{a}^{-1}$.
Multiplying this string of subset inclusions through by $\mathfrak{a}$,
\[
  \mathfrak{a}\subseteq\mathfrak{a}\mathfrak{p}^{-1}\subseteq\mathfrak{a}\mathfrak{a}^{-1}\subseteq\mathfrak{O}_{K}.
\]
Since~$\mathfrak{a}\mathfrak{p}^{-1}\subseteq\mathfrak{O}_{K}$, it is an ideal.
It cannot be that~$\mathfrak{a}=\mathfrak{a}\mathfrak{p}^{-1}$ since, otherwise,
$\mathfrak{p}^{-1}\subseteq\mathfrak{O}_{K}$ by Step~2.2, contradicting Step~2.1.
Therefore~$\mathfrak{a}\subsetneq \mathfrak{a}\mathfrak{p}^{-1}$.
By maximality of~$\mathfrak{a}$, we have
\[
  (\mathfrak{a}\mathfrak{p}^{-1})(\mathfrak{a}\mathfrak{p}^{-1})^{-1}=\mathfrak{O}_{K}.
\]
It then follows from the definition of~$\mathfrak{a}^{-1}$ that
\[
  \mathfrak{p}(\mathfrak{a}\mathfrak{p}^{-1})^{-1}\subseteq \mathfrak{a}^{-1},
\]
but then
\[
  \mathfrak{O}_{K}=\mathfrak{a}\mathfrak{p}(\mathfrak{a}\mathfrak{p}^{-1})^{-1}\subseteq
  \mathfrak{a}\mathfrak{a}^{-1}\subseteq\mathfrak{O}_{K}.
\]
This forces~$\mathfrak{a}\mathfrak{a}^{-1}=\mathfrak{O}_{K}$, contradicting our
choice of~$\mathfrak{a}$.
\medskip

\textbf{Step 2.5.} If $I$ is a nonzero fractional ideal,
then~$II^{-1}=\mathfrak{O}_{K}$.
\medskip

\emph{Proof of claim.} Suppose~$I$ is a nonzero fractional ideal.
Pick~$\alpha\in\mathfrak{O}_{K}\setminus\{0\}$ such that~$\alpha
I\subseteq\mathfrak{O}_{K}$.  By Step~2.4, we have $(\alpha I)(\alpha
I)^{-1}=\mathfrak{O}_{K}$.  We have
\[
  (\alpha I)^{-1} = \{ x\in K: x(\alpha I)\subseteq\mathfrak{O}_{K}\}.
\]
So~$x\in (\alpha I)^{-1}$ if and only if~$\alpha x\in I^{-1}$.  Therefore,
$(\alpha I)^{-1}=(1/\alpha)I^{-1}$.  We have
\[
  \mathfrak{O}_{K}=(\alpha I)(\alpha I)^{-1} = (\alpha
  I)\left(\frac{1}{\alpha}I^{-1}\right) = I I^{-1}.
\]
\medskip

\textbf{Step 3.} Claim: Every nonzero ideal $\mathfrak{a}\subseteq\mathfrak{O}_{K}$ is
a product of prime ideals.
\medskip

\emph{Proof of claim.}  If not, since~$\mathfrak{O}_{K}$ is Noetherian, we can
take an ideal~$\mathfrak{a}$ maximal with respect to the property of not having
a prime factorization.  In particular, $\mathfrak{a}$ is not prime.  It is also
the case that~$\mathfrak{a}\neq\mathfrak{O}_{K}=(1)$.  That's
because~$\mathfrak{O}_{K}$ does have a prime factorization---the empty
factorization. (This is just like the case of ordinary prime factorization
in~$\Z$: every nonzero integer has a prime factorization, including $\pm1$.)  Pick
a maximal ideal~$\mathfrak{p}$ containing~$\mathfrak{a}$.  In Step~2.4, we
showed that
\[
  \mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{p}^{-1}.
\]
By maximality of~$\mathfrak{a}$, the ideal~$\mathfrak{a}\mathfrak{p}^{-1}$ has a
prime factorization.  So 
\[
  \mathfrak{a}\mathfrak{p}^{-1} = \mathfrak{p}_{2}\cdots\mathfrak{p}_{r}
\]
for some primes~$\mathfrak{p}_{i}$.  Multiplying through by $\mathfrak{p}$, we
get
\[
  \mathfrak{a} = \mathfrak{p}_{1}\mathfrak{p}_{2}\cdots \mathfrak{p}_{r}.
\]
That contradicts the fact that~$\mathfrak{a}$ does not factor into primes.  The
result follows.
\medskip

\textbf{Step 4.} Claim: prime factorization of ideals in~$\mathfrak{O}_{K}$ is
unique.
\medskip

\emph{Proof of claim.} Suppose that there are nonzero prime ideals
$\mathfrak{p}_{i}$ and $\mathfrak{q}_{j}$ such that
\[
  \mathfrak{p}_{1}\cdots\mathfrak{p}_{r} =
  \mathfrak{q}_{1}\cdots\mathfrak{q}_{s}.
\]
Since~$\mathfrak{p}_{1}$ divides~$\mathfrak{q}_{1}\cdots\mathfrak{q}_{s}$
and~$\mathfrak{p}_{1}$ is prime, it follows that divides
some~$\mathfrak{q}_{i}$.  Without loss of generality, say
$\mathfrak{p}_{1}|\mathfrak{q}_{1}$.  Therefore,
$\mathfrak{q}_{1}\subseteq\mathfrak{p}_{1}$.  Since~$\mathfrak{q}_{1}$ is a
nonzero prime ideal in~$\mathfrak{O}_{K}$, it is maximal.
Therefore~$\mathfrak{p}_{1}=\mathfrak{q}_{1}$.  By induction,~$r=s$ and the set
of $\mathfrak{p}_{i}$s equals the set of~$\mathfrak{q}_{j}$s.
\end{proof}
\end{document}
