\documentclass[12pt]{book}
\usepackage{amsthm}
\theoremstyle{definition}
\newtheorem{theorem}{Theorem}
\newtheorem{defin}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{sublemma}[theorem]{Sublemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{algorithm}[theorem]{Algorithm}
\newcommand{\Ex}{\begin{example}}
\newcommand{\Eex}{\end{example}}
\newcommand{\Z}{{\bf Z}}
\newcommand{\Q}{{\bf Q}}
\newcommand{\C}{{\bf C}}
\newcommand{\R}{{\bf R}}
\newcommand{\F}{{\bf F}}
\newcommand{\Fp}{{\bf F}_p}
\newcommand{\chart}{{\rm char}}
\newcommand{\degg}{{\rm deg}}
\newcommand{\disc}{{\rm disc}}
\newcommand{\Hom}{{\rm Hom}}
\newcommand{\m}{{\rm m}}
\newcommand{\spn}{{\rm span}}
\newcommand{\trr}{Tr}
\newcommand{\ZF}{\Z_F}
\newcommand{\defn}[1]{{\bf #1}}
\begin{document}

\chapter*{October 12:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Ideal lemmas
\item 
Invertible ideals
\item
Main Theorems
\end{itemize}

\section{Ideal lemmas}

The main goal today is to prove that every ideal in a Dedekind
domain is invertible, that ideals factor uniquely into prime
ideals, and that the ideal ``classes'' form a group.
The main lines of the proof here are due to van der Waerden.

We start with two lemmas.  Throughout let $R$ be a Dedekind
domain and $F$ its fraction field.  (There are some minor
ways in which the proof could be simplified if $R = \ZF$ for
some number field~$F$.)  Also, in one-dimensional rings we
will follow convention and reserve the word ``ideal'' for
nonzero ideals.

\begin{lemma}
Any ideal contains a product of prime ideals.
\end{lemma}

\begin{proof}
The set of ideals not containing such a product has a maximal
element under inclusion, since~$R$ is Noetherian.  Let $I$ 
be such an ideal, i.e., an ideal such that any ideal that properly
contains~$I$ also contains a producdt of prime ideals.

Then $I$ isn't a prime ideal itself, so there are elements $x$ and~$y$
of~$R$ such that $xy \in i$ but $x$ and~$y$ are not in~$I$.
Then $I+(x)$ is an ideal that is strictly larger than~$I$, so it
contains a product $\prod P_i$ of prime ideals, and similarly
$I+(y)$ contains a product $\prod Q_j$ of prime ideals.  But then
$$I \supset (I+(x))(I + (y)) \supset \prod_i P_i \prod_j Q_j$$
so that $I$ in fact contains a product of prime ideals.
\end{proof}

If $I$ is an ideal, then an element $x$ of the fraction field~$F$ is
said to be a \defn{multiplier} for~$I$ if $xI \subset R$, i.e.,
$x$ multiplies anything in~$I$ into~$R$.  Very roughly, this means that
the denominator of $x$ is an element of~$I$.

\begin{lemma}
Any proper ideal (i.e., not equal too the entire ring) has a
multiplier that is not in the ring.
\end{lemma}

\begin{proof}
Let $I$ be a proper ideal, and choose
a nonzero element $y$ in~$I$.
The ideal $(y)$ contains a product $\prod_{i=1}^m P_i$ of prime ideals
by the previous lemma; we choose such a product in which the number
$m$ of ideals is as small as possible.
Also, choose a prime ideal~$P$ that contains~$I$, so that we have
$$ P \subset I \subset (y) \subset \prod_{i=1}^m P_i.$$
Then $P = P_i$ for some $i$ (if not, then for each~$i$ choose $x_i$
in $P_i$ by not in~$P$; one finds that the product of the $x_i$, and
hence one of the $x_i$, is in~$P$ since $P$ is a prime ideal).
Let $P = P_1$, without loss of generality.

Now choose $z$ that is in the product $P_2 \cdots P_m$ that is not
in~$(y)$ ($z$ exists because of the minimality of~$m$).

Then the claim is that $x = z/y$ is a multiplier for~$I$; note that
$x$ is clearly not in~$R$ since $z$ is not in the principal ideal~$(y)$.

On the other hand
$$xI = (z/y)I \subset (z/y)P \subset (1/y)P_1 P_2 \cdots P_m 
\subset R$$
by our various choices.
\end{proof}

\section{Invertible ideals}

An ideal $I$ in~$R$ is said to be \defn{invertible} if there is
an ideal~$J$ such that $IJ$ is principal.  Any ideal in a
Dedekind domain is invertible and, though we won't prove it,
any one-dimensional integral domain with this property is in fact
a Dedekind domain.

\begin{theorem}
If $I$ is an ideal in~$R$ and $y$ is a nonzero element of~$I$, then
the set 
$$ J := \{ x \in R : xI \subset (y) \}$$
is an ideal, and $IJ = (y)$.
\end{theorem}

\begin{proof}
It is easy to check that $J$ is an ideal, and that $IJ$ is contained
in~$(y)$.  So it suffices to check that $IJ = (y)$.  In order to do
this, let
$$K = (1/y)IJ.$$
The set $K$ is contained in~$R$ by our definitions, and it is then
easy to check that it is an ideal.
We are trying to prove that $K = R$.

Suppose that $K$ is a proper ideal.  Let $x$ be a multiplier for~$K$
that does not lie in~$R$.  

First note that $J \subset K$.  Indeed, $y \in I$ so $yJ \subset IJ$
and therefore $J \subset (1/y)IJ$.

Next, note that since $xK \subset R$ we have
$$xIJ \subset (y).$$
This means that anything in $xJ$ (which, by the first remark is a subset
of~$R$) multiplies $I$ into $(y)$, i.e.,, that $xJ$ is contained in~$J$.

This is actually a contradiction since it implies that $x$ is the root
of a monic polynomial with coefficients in~$R$, i.e., that (by the
definition of a Dedekind domain) $x$ is in~$R$.  Indeed, if 
$\alpha_1, \cdots , \alpha_k$ is a generating set for~$J$, then
each $x\alpha_i$ can be written as a linear combination of the $\alpha_i$,
with coefficients in~$R$.  In matrix form this gives
$$xA = MA, \qquad (xI-M)A = 0$$
where $A$ the column vector of $\alpha_i$'s.  A singular matrix has zero
determinant, so $\det(xI-M) = 0$ and $x$ is the root of a monic polynomial
in $R[x]$, as claimed.

\end{proof}

\section{The Main Theorems}

Finally, we are in a position to smoothly prove the major results about
ideals in Dedekind domains.

\begin{theorem} If $I$ and $J$ are ideals in~$R$ then $I \subset J$ if
and only if $I$ is divisible by~$J$.
\end{theorem}

\begin{remark}
A smaller ideal will be divisible by more primes, so the statement is
in fact not counterintuitive.
\end{remark}

\begin{proof} If $I = JK$ then anything in~$J$ is certainly in~$I$
(this implication is of course true in any ring).

On the other hand, if $I \subset J$ then find an ideal $K$ such that
$JK = (y)$ is principal.  Then $I' := (1/y)IK$ is easily checked to be
an ideal and
$$I'J = (1/y)IKJ = I.$$
So $I$ is indeed the product of $J$ and another ideal, i.e., $I$ is
divisible by~$J$ as claimed.
\end{proof}

\begin{theorem} Cancellation holds for multiplication of ideals in a
Dedekind domain, i.e., $IJ = IK$ implies that $J = K$.
\end{theorem}

\begin{proof} Find an ideal $I'$ such that $II' = (y)$.  Multiplying
the given equation by $I'$ gives $(y)J = (y)K$ from which $J = K$
follows. 
\end{proof}

\begin{theorem}
Every ideal in $R$ can be uniquely represented as a product of prime
ideals.
\end{theorem}

\begin{proof}
To show existence, we consider an ideal $I$ that is as large as possible
that is not a product of prime ideals.  Then $I \ne R$ (since we
regard $R$ as an empty product of prime ideals).  Therefore $I$ is
contained in a maximal (prime) ideal~$P$, $I \subset P$.
Clearly $I$ is not equal to~$P$ (since we regard a prime ideal as
a product of a single prime ideal) so, by the preceding result, there
is a proper ideal~$J$ such that $PJ = I$.  Then $I$ is a proper subideal
of~$J$, so $J$ is a product of prime ideals, and so is~$I$.

To prove uniqueness we observe that if 
$$\prod P_i = \prod Q_j$$
then $P_1 \supset \prod Q_j$ so that some $Q_j$, say $Q_1$ is equal
to~$P_1$.  Applying cancellation enables us to prove uniqueness
by induction.
\end{proof}

Say that two ideals $I$ and $J$ in~$R$ are \defn{equivalent} if there
are nonzero elements $x$ and $y$ of the ring such that
$$xI = yJ.$$
It is easy to check that this is an equivalence relation.  Multiplication
on ideals induces a multiplication on equivalence classes, and it is
easily checked that this operation has an identity (the entire ring $(1)$),
is commutative, and is associative.

\begin{theorem}
Equivalence classes form a group under the above operation.
\end{theorem}

\begin{proof}
If $IJ = (y)$ then
$$[I][J] = [R].$$
\end{proof}

The ideal class group of a Dedekind domain $R$ will be denoted
$Cl(R)$.  Later we will see that this group is finite if $R$ is the
ring of integers in a number field.


\end{document}


