\chapter{Dedekind domains}

The notion of a Dedekind domain allows one to generalize the usual unique
factorization in principal ideal domains as in $\mathbb{Z}$ to settings such
as the ring of integers in an algebraic number field. In general, a Dedekind
domain does not have unique factorization, but the \emph{ideals} in a Dedekind
domain do factor uniquely into a product of prime ideals. 
We shall see that Dedekind domains have a short characterization in terms of
the characteristics we have developed.

After this, we shall study the case of an \emph{extension} of Dedekind domains $A \subset B$. It will be of
interest to determine how a prime ideal of $A$ factors in $B$. This should
provide background for the study of basic algebraic number theory, e.g. a
rough equivalent of the first chapter of 
\cite{La94} or \cite{Se79}.




\section{Discrete valuation rings}


\subsection{Definition}

We start with the simplest case of a \emph{discrete valuation ring,} which is
the local version of a Dedekind domain.
Among the one-dimensional local noetherian rings, these will be the nicest.

\begin{theorem} \label{DVRthm}
Let $R$ be a noetherian local domain whose only prime ideals are $(0)$ and the maximal
ideal 
$\mathfrak{m} \neq 0$. 
Then, the following are equivalent:
\begin{enumerate}
\item $R$ is factorial.
\item $\mathfrak{m}$ is principal.
\item  $R$ is integrally closed.
\item $R$ is a valuation ring with value group $\mathbb{Z}$. 
\end{enumerate}
\end{theorem} 
\begin{definition} 
A ring satisfying these conditions is called a \textbf{discrete valuation
ring} (\textbf{DVR}).
A discrete valuation ring necessarily has only two prime ideals, namely
$\mathfrak{m}$ and $(0)$. 

Alternatively, we can say that a noetherian local domain is a DVR if and only
if it is of dimension one and integrally closed.
\end{definition} 



\begin{proof}
Assume 1: that is, suppose $R$ is factorial. Then every prime ideal of height one is principal
by \cref{heightonefactoriality}.
But $\mathfrak{m}$ is the only prime that can be height one: it is minimal over
any nonzero nonunit of $R$, so $\mathfrak{m}$ is principal. Thus 1 implies 2, and similarly 2 implies 1 by
\cref{heightonefactoriality}.

1 implies 3 is true for any $R$: a factorial ring is always integrally
closed, by \cref{factorialimpliesnormal}.

4 implies 2 is easy as well. Indeed, suppose $R$ is a valuation ring with
value group $\mathbb{Z}$. Then, one chooses an element $x \in R$ such that the valuation of
$x$ is one. It is easy to see that  $x$ generates $\mathfrak{m}$: if $y
\in \mathfrak{m}$ is a non-unit, then the valuation of $y$ is at least one,
so $y/x \in R$ and $y \in (x)$. 

The proof that 2 implies 4 is also straightforward. Suppose
$\mathfrak{m}$ is principal, generated by $t$. 
In this case, we claim that any $x \in R$ is associate (i.e. differs by a
unit from) a power of $t$.
Indeed, since $\bigcap \mathfrak{m}^n = 0$ by the Krull intersection theorem
(\cref{krullintersection}), it follows that there exists $n$ such that $x$ is
divisible by $t^n$ but not by $t^{n+1}$. 
In particular, if we write $x = u t^n$, then $u \notin (t)$ is a unit. This
proves the claim.

With this in mind, we need to show that $R$ is a valuation ring with value
group $\mathbb{Z}$.
If $x
\in R$, we define the valuation of $x$ to be the nonnegative integer $n$ such
that  $(x) = (t^n)$. One can
easily check that this is a valuation on $R$, which extends to the quotient
field by additivity.

The interesting part of the argument is the claim that 3
implies 2. Suppose $R$ is integrally closed, noetherian, and of dimension one; we claim that $\mathfrak{m}$ is
principal. Choose $x \in \mathfrak{m} - \left\{0\right\}$. If $(x) =
\mathfrak{m}$, we are done. 

Otherwise, we can look at $\mathfrak{m}/(x) \neq
0$.  The module $\mathfrak{m}/(x)$ is finitely generated module a noetherian ring which is
nonzero, so it has an associated prime. That associated prime is either zero or
$\mathfrak{m}$ because $R$ has dimension one. But $0$ is not an associated prime because every element in the
module is killed by $x$. So $\mathfrak{m}$ is an associated prime of
$\mathfrak{m}/(x)$.

There is $\overline{y} \in \mathfrak{m}/(x)$ whose annihilator is
$\mathfrak{m}$.
Thus, there is $y \in \mathfrak{m}$ such that $y \notin (x)$ and $\mathfrak{m}y
\subset (x)$. In particular, $y/x \in K(R) - R$, but
\[ (y/x) \mathfrak{m} \subset R.  \]
There are two cases:
\begin{enumerate}
\item Suppose $(y/x) \mathfrak{m}  = R$. Then we can write $\mathfrak{m} =
R(x/y)$. So $\mathfrak{m}$ is principal. (This argument shows that $x/y \in R$.)	
\item The other possibility is that $y/x \mathfrak{m} \subsetneq R$. In this
case, $(y/x)\mathfrak{m}$ is an ideal, so 
\[ (y/x) \mathfrak{m} \subset \mathfrak{m}.  \]
In particular, multiplication by $y/x$ carries $\mathfrak{m}$ to itself, and
stabilizes the finitely generated \emph{faithful} module
$\mathfrak{m}$. By \cref{thirdintegralitycriterion}, we see that
$y/x$ is integral over $R$. In particular, we find that $y/x \in R$, as $R$
was integrally closed, a contradiction as $y \notin (x)$. 
\end{enumerate}
\end{proof} 

Let us give several examples of DVRs.
\begin{example} 
The localization $\mathbb{Z}_{(p)}$ at any prime ideal $(p) \neq 0$ is a DVR.
The associated valuation is the $p$-adic valuation.
\end{example} 

\begin{example} 
Although we shall not prove (or define) this, the local ring of an
algebraic curve at a smooth point is a DVR. The associated valuation  measures the
extent to which a function (or germ thereof) has a zero (or pole) at  that
point.
\end{example} 

\begin{example} 
The formal power series ring $\mathbb{C}[[T]]$ is a discrete valuation ring,
with maximal ideal $(T)$. 
\end{example} 

\subsection{Another approach}

In the proof of \cref{DVRthm}, we freely used the notion of associated primes,
and thus some of the results of \cref{noetherian}.
However, we can avoid all that and give a more ``elementary approach,'' as in
\cite{CaFr67}.


Let us suppose that $R$ is an integrally closed, local noetherian domain of
dimension one. We shall prove that the maximal ideal $\mathfrak{m} \subset R$
is principal. This was the hard part of \cref{DVRthm}, and the only part
where we used associated primes earlier.
\begin{proof}
We will show that $\mathfrak{m}$ is principal, by showing it is \emph{invertible} (as will be seen below).   We divide the proof into steps:

\paragraph{Step one}
For a nonzero ideal $I \subset R$, let $I^{-1} := \{ x \in K(R): xI \subset R \}$,
where $K(R)$ is the quotient field of $R$.  Then clearly $I^{-1} \supset R$
and $I^{-1}$ is an $R$-module, but in general we cannot say that $I^{-1} \neq
R$ even if $I$ is proper.  
Nevertheless, we claim that in the present situation, we have  \[
{\mathfrak{m}^{-1} \neq R.}\] This is the conclusion of Step one.

The proof runs across a familiar line: we show that any maximal element in the
set of ideals $I \subset R$ with $I^{-1} \neq R$ is prime. 
The set of such ideals is nonempty: it contains   any $(a)$ for $a \in \mathfrak{m}$ (in which case $(a)^{-1} = Ra^{-1} \neq R$).  
There must be a maximal element in this set of ideals by noetherianness, which
as we will see is prime; thus,  that maximal element must be $\mathfrak{m}$, which proves our claim.

So to fill in the missing link, we must prove:
\begin{lemma} If $S$ is a noetherian domain, any maximal element in the set of ideals $I \subset S$ with $I^{-1} \neq S$ is prime.
\end{lemma}

\begin{proof}
Let $J$ be a maximal element, and suppose we have $ab \in J$, with  $a,b \notin J$.  I claim that if $z \in J^{-1} - S$, then $za, zb \in J^{-1} - S$.  The $J^{-1}$ part follows since $J^{-1}$ is a $S$-module.

By symmetry it is enough to prove the other half for $a$, namely that $za \notin
S$; but then if $za \in S$, we would have $z( (a) + J ) \subset S$, which implies $ ( (a) + J)^{-1} \neq S$, contradiction, for $J$ was maximal.

Then it follows that $z(ab) = (za) b \in J^{-1} - S$, by applying the claim  just made twice.  But $ab \in J$, so $z(ab) \in S$, contradiction.
\end{proof}



\paragraph{Step two} In the previous step, we have established that
$\mathfrak{m}^{-1} \neq R$.

We now claim that $\mathfrak{m}\mathfrak{m}^{-1} = R$.  First, we know of course
that $\mathfrak{m}\mathfrak{m}^{-1} \subset R$ by definition of inverses,
and equally $\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1}$ too.  So $\mathfrak{m}\mathfrak{m}^{-1}$ is an ideal sandwiched between $\mathfrak{m}$ and $R$.  
Thus we only need to prove that $\mathfrak{m} \mathfrak{m}^{-1} = \mathfrak{m}$
is impossible.  If this were the case, we could choose some $a \in
\mathfrak{m}^{-1} - R$ which must satisfy $a \mathfrak{m} \subset \mathfrak{m}$.
Then $a$ would integral over $R$. 
As $R$ is integrally closed, this is impossible.

\paragraph{Step three}

Finally, we  claim that $\mathfrak{m}$ is principal, which is the final step of
the proof.
In fact, let us prove a more general claim.

\begin{proposition} 
Let $(R, \mathfrak{m})$ be a local noetherian domain such that $\mathfrak{m}
\mathfrak{m}^{-1} = R$. Then $\mathfrak{m}$ is principal.
\end{proposition} 
\begin{proof} 
Indeed, since $\mathfrak{m} \mathfrak{m}^{-1} = R$, write
\[ 1 = \sum m_i n_i, \quad m_i \in \mathfrak{m}, \ n_i \in \mathfrak{m}^{-1}.\]
At least one $m_j n_j$ is invertible, since $R$ is local.
It follows that there are $x \in \mathfrak{m}$ and $y \in \mathfrak{m}^{-1}$
whose product $xy$ is a unit in $R$.
We may even assume $xy = 1$.

Then we claim $\mathfrak{m} = (x)$.  
Indeed, we need only prove $\mathfrak{m} \subset (x)$.  For this, if $q \in
\mathfrak{m}$, then $qy \in R$ by definition of $\mathfrak{m}^{-1}$, so  \[ q =
x(qy)  \in ( x).\]
\end{proof}

\end{proof} 

So we are done in this case too.  Taking stock, we have an effective way to say whether a ring is a DVR.  These three conditions are much easier to check in practice (noetherianness is usually easy, integral closure is usually automatic, and the last one is not too hard either for reasons that will follow) than the existence of an absolute value.


\section{Dedekind rings}

\subsection{Definition}
We now introduce a closely related notion.  
\begin{definition} 
A \textbf{Dedekind ring} is a noetherian domain $R$ such that 
\begin{enumerate}
\item $R$ is integrally closed.
\item Every nonzero prime ideal of $R$ is maximal. 
\end{enumerate}
\end{definition} 


\begin{remark} 
If $R$ is Dedekind, then any nonzero element is height one. This is evident
since every nonzero prime is maximal. 

If $R$ is Dedekind, then $R$ is locally factorial. In fact, the localization of
$R$ at a nonzero prime $\mathfrak{p}$ is a DVR.
\begin{proof} 
$R_{\mathfrak{p}}$ has precisely two prime ideals: $(0)$ and
$\mathfrak{p}R_{\mathfrak{p}}$. As a localization of an integrally closed
domain, it is integrally closed. So $R_{\mathfrak{p}}$ is a DVR by the above
result (hence
factorial).
\end{proof} 
\end{remark} 


Assume $R$ is Dedekind now.
We have an exact sequence
\[ 0 \to R^* \to K(R)^* \to \cart(R) \to \pic(R) \to 0.  \]
Here $\cart(R) \simeq \weil(R)$. But $\weil(R)$ is free on the nonzero
primes, or equivalently maximal ideals, $R$ being Dedekind. 
In fact, however, $\cart(R)$ has a simpler description.

\begin{proposition} 
Suppose $R$ is Dedekind. Then $\cart(R)$ consists of all nonzero finitely generated
submodules of $K(R)$ (i.e. \textbf{fractional ideals}). 
\end{proposition} 

This is the same thing as saying as every nonzero finitely generated submodule of $K(R)$ is
invertible.
\begin{proof} 
Suppose $M \subset K(R)$ is nonzero and finitely generated It suffices to check that $M$ is
invertible after localizing at every prime, i.e. that $M_{\mathfrak{p}}$ is
an invertible---or equivalently, trivial, $R_{\mathfrak{p}}$-module. At the
zero prime, there is nothing to check. We might as well assume that
$\mathfrak{p}$ is maximal. Then $R_{\mathfrak{p}}$ is a DVR and
$M_{\mathfrak{p}}$ is a finitely generated submodule of $K(R_{\mathfrak{p}}) = K(R)$. 

Let $S$ be the set of integers $n$ such that there exists $ x \in
M_{\mathfrak{p}}$ with $v(x) = n$, for $v$ the valuation of $R_{\mathfrak{p}}$.
By finite generation of $M$, $S$ is bounded below. Thus $S$ has a least element
$k$. There is an element of $M_{\mathfrak{p}}$, call it $x$, with valuation $k$.

It is easy to check that $M_{\mathfrak{p}}$ is generated by $x$, and is in fact free with
generator $x$. The reason is simply that $x$ has the smallest valuation of
anything in $M_{\mathfrak{p}}$.
\end{proof} 

What's the upshot of this?

\begin{theorem}
If $R$ is a Dedekind ring, then any nonzero ideal $I \subset R$ is invertible,
and therefore uniquely described as a product of powers of (nonzero) prime ideals, $I =
\prod \mathfrak{p}_i^{n_i}$.
\end{theorem} 
\begin{proof} 
This is simply because $I$ is in $\cart(R) = \weil(R)$ by the above result. 
\end{proof} 

This is Dedekind's generalization of unique factorization.

We now give the standard examples:
\begin{example} 
\begin{enumerate}
\item Any PID (in particular, any DVR) is Dedekind.  
\item If $K$ is a finite extension of $\mathbb{Q}$, and set $R $ to be the
integral closure of $\mathbb{Z}$ in $K$, then $R$ is a Dedekind ring. The ring
of integers in any number field is a Dedekind ring. 
\item If $R$ is the coordinate ring of an algebraic variety which is smooth and
irreducible of dimension one, then $R$ is Dedekind. 
\item  Let $X$ be a compact Riemann surface, and let $S \subset X$ be a
nonempty finite subset. Then the ring of meromorphic functions on $X$ with
poles only in $S$ is
Dedekind. The maximal ideals in this ring are precisely those corresponding to
points of $X-S$. 
\end{enumerate}
\end{example} 



\subsection{A more elementary approach}

We would now like to give  a more elementary approach to the unique
factorization of ideals in Dedekind domains, one which does not use the heavy
machinery of Weil and Cartier divisors.

In particular, we can encapsulate what has already been proved as:
\begin{theorem} Let $A$ be a Dedekind domain with quotient field $K$. Then there is a bijection between the discrete valuations of $K$ that assign nonnegative orders to elements of $A$ and the nonzero prime ideals of $A$.
\end{theorem}
\begin{proof} Indeed, every valuation gives a prime ideal of elements of positive order; every prime ideal $\mathfrak{p}$ gives a discrete valuation on $A_{\mathfrak{p}}$, hence on $K$. \end{proof}


This result, however trivial to prove, is the main reason we can work essentially interchangeably with prime ideals in Dedekind domains and discrete valuations.

Now assume  $A$ is Dedekind. A finitely generated $A$-submodule of the quotient field $F$ is called a \textbf{fractional ideal}; by multiplying by some element of $A$, we can always pull a fractional ideal into $A$, when it becomes an ordinary ideal.  The sum and product of two fractional ideals are fractional ideals.

\begin{theorem}[Invertibility]  If $I$ is a nonzero fractional ideal and $ I^{-1} := \{ x \in F: xI \subset A \}$, then $I^{-1}$ is a fractional ideal and $I I^{-1} = A$.
\end{theorem}

Thus, the nonzero fractional ideals are an \emph{abelian group} under multiplication.

\begin{proof}
To see this, note that invertibility is preserved under localization: for a multiplicative set $S$, we have $S^{-1} ( I^{-1} ) = (S^{-1} I)^{-1}$, where the second ideal inverse is with respect to $S^{-1}A$; this follows from the fact that $I$ is finitely generated.  Note also that invertibility is true for discrete valuation rings: this is because the only ideals are principal, and principal ideals (in any integral domain) are obviously invertible.

So for all primes $\mathfrak{p}$, we have $(I I^{-1})_{\mathfrak{p}} = A_{\mathfrak{p}}$, which means the inclusion of $A$-modules $I I^{-1} \to A$ is an isomorphism at each localization.  Therefore it is an isomorphism, by general algebra.
\end{proof} 

The next result says we have unique factorization of \textbf{ideals}:
\begin{theorem}[Factorization] Each ideal $I \subset A$ can be written uniquely as a product of powers of prime ideals.
\end{theorem}
\begin{proof}
Let's use the pseudo-inductive argument to obtain existence of a prime factorization.  Let $I$ be the maximal ideal which can't be written in such a manner, which exists since $A$ is Noetherian.  Then $I$ isn't prime (evidently), so it's contained in some prime $\mathfrak{p}$.  But $I = (I\mathfrak{p}^{-1})\mathfrak{p}$, and $I\mathfrak{p}^{-1} \neq I$ can be written as a product of primes, by the inductive assumption. Whence so can $I$, contradiction.

Uniqueness of factorization follows by localizing at each prime.
\end{proof}

\begin{definition} Let $P$ be the subgroup of nonzero principal ideals in the group $I$ of nonzero ideals.  The quotient $I/P$ is called the \textbf{ideal class group}.
\end{definition}

The ideal class group of the integers, for instance (or any principal ideal domain) is clearly trivial.  In general, this is not the case, because Dedekind domains do not generally admit unique factorization.
\begin{proposition} Let $A$ be a Dedekind domain. Then $A$ is a UFD if and only if its ideal class group is trivial.
\end{proposition}
\begin{proof} If the ideal class group is trivial, then $A$ is a principal ideal domain, hence a UFD by elementary algebra.  Conversely, suppose $A$ admits unique factorization.  
Then, by the following lemma, every prime ideal is principal.  Hence every ideal is principal, in view of the unique factorization of ideals.
\end{proof}
\begin{lemma} Let $R$ be a UFD, and let $\mathfrak{p}$ be a prime ideal which contains no proper prime sub-ideal except for $0$.  Then $\mathfrak{p}$ is principal.
\end{lemma}
The converse holds as well; a domain is a UFD if and only if every prime ideal
of height one is principal, by \rref{heightonefactoriality}.
\begin{proof}
First, $\mathfrak{p}$ contains an element $x \neq 0$, which we factor into irreducibles $\pi_1 \dots \pi_k$.  One of these, say $\pi_j$, belongs to $\mathfrak{p}$, so $\mathfrak{p} \supset (\pi_j)$. Since $\mathfrak{p}$ is minimal among nonzero prime ideals, we have $\mathfrak{p} = (\pi_j)$.  (Note that $(\pi_j)$ is prime by unique factorization.)
\end{proof}

\begin{exercise} 
This exercise is from \cite{Li02}. If $A$ is the integral closure of
$\mathbb{Z}$ in a number field (so that $A$ is a Dedekind domain), then it is
known (cf. \cite{La94} for a proof) that the ideal class group of $A$ is
\emph{finite}. From this, show that every open subset of $\spec A$ is a
principal open set $D(f)$. Scheme-theoretically, this means that every open
subscheme of $\spec A$ is affine (which is not true for general rings).
\end{exercise} 

\subsection{Modules over Dedekind domains}

Let us now consider some properties of Dedekind domains.

\begin{proposition} Let $A$ be a Dedekind domain, and let $M$ be a finitely generated $A$ module. Then
$M$ is projective (or equivalently flat, or locally free) if and only if it is torsion-free.
\label{Dedekind means projective=tors free}
\end{proposition}
\begin{proof}
If $M$ is projective, then it is a direct summand of a free module, so it is torsion-free. So we need to show that if $M$ is torsion-free, then it is projective. Recall that to show $M$ is projective, it suffices to show that $M_\mathfrak{p}$ is projective for any prime $\mathfrak{p} \subset M$. But note that $A_\mathfrak{p}$ is a PID so a module over it is torsion free if and only if it is flat, by Lemma \ref{PID means flat=tors free}. However, it is also a local Noetherian ring, so a module is flat if and only if it is projective. So $M_\mathfrak{p}$ is projective if and only if it is torsion-free, so it now suffices to show that it is torsion-free.

However for any multiplicative set $S \subset A$, if $M$ is torsion-free then $M_S$ is also torsion-free. This is because if
\[\frac{a}{s'} \cdot \frac{m}{s}=0\]
then there is $t$ such that $tam=0$,
as desired.
\end{proof}

\begin{proposition}
Let $A$ be a Dedekind domain. Then any finitely generated module $M$ over it has (not canonically) a decomposition $M=M^{tors} \oplus M^{tors-free}$.
\end{proposition}
\begin{proof}
Note that by Lemma \ref{tors tors-free ses}, we have a short exact sequence
\[ 0 \to M^{tors} \to M \to M^{tors-free} \to 0\]
but by proposition \ref{Dedekind means projective=tors free} the torsion free part is projective, so $M$ can be split, not necessarily canonically as  $M^{tors} \oplus M^{tors-free}$, as desired.
\end{proof}

Note that we may give further information about the torsion free part of the module:
\[M^{tors}=\bigoplus_{\mathfrak{p}} M_{\mathfrak{p}}^{tors}\]
First note that there is a map
\[M^{tors} \to \bigoplus_{\mathfrak{p}} M_{\mathfrak{p}}^{tors}\]
because $M$ is torsion, every element is supported at finitely many points, so the image of $f$ in $M^{tors}_\mathfrak{p}$ is only nonzero for finitely many $\mathfrak{p}$.
It is an isomorphism, because it is an isomorphism after every localization.

So we have pretty much specified what the torsion part is. We can in fact also classify the torsion free part;  in particular, we have
\[M^{tors-free} \simeq \oplus \mathcal{L}\]
where $\mathcal{L}$ are locally free modules of rank 1.
This is because we know from above that the torsion free module is projective, we may apply Problem Set 10, Problem 12, and then since $L$ is a line bundle, and $I_{-D}$ is also, $L \otimes I_{-D}$ is a line bundle, and then $M/L \otimes I_{-D}$ is flat, so it is projective, so we may split it off.



\begin{lemma} For $A$ a Dedekind Domain, and $I \subset A$ an ideal, then $I$ is a locally free module of rank 1.
\end{lemma}
\begin{proof}
First note that $I$ is torsion-free and therefore projective by \ref{Dedekind means projective=tors free}, and it is also finitely generated, because $A$ is Noetherian. But for a finitely generated module over a Noetherian ring, we know that it is projective if and only if it is locally free, so we have shown that it is locally free.

Also recall that for a module which is locally free, the rank is well defined,
i.e, any localization which makes it free makes it free of the same rank. So
to test the rank, it suffices to show that if we tensor with the field of
fractions $K$, it is free of rank 1. But note that since $K$, being a localization of $A$ is flat over $A$ so we have short exact sequence
\[0 \to I \otimes_A K \to A \otimes_A K \to (A/I) \otimes_A K \to 0\]

However, note that $\supp(A/I)=V(\Ann(A/I))=V(I)$, and the prime $(0)$ is not
in $V(I)$, so $A/I \otimes_{A} K$, which is the localization of $A/I$ at $(0)$ vanishes, so we have
\[I\otimes_A K \simeq A\otimes_A K\]
but this is one-dimensional as a free $K$ module, so the rank is 1, as desired.
\end{proof}

We close by listing a collection of useful facts about Dedekind domains.
 \underline{A dozen things every Good Algebraist should know about Dedekind domains}. $R$
 is a Dedekind domain.
 \begin{enumerate}
   \item $R$ is local $\Longleftrightarrow$ $R$ is a field or a DVR.
   \item $R$ semi-local $\Longrightarrow$ it is a PID.
   \item $R$ is a PID $\Longleftrightarrow$ it is a UFD $\Longleftrightarrow$ $C(R)=\{1\}$
   \item $R$ is the full ring of integers of a number field $K$ $\Longrightarrow$ $|C(R)|<
   \infty$, and this number is the \emph{class number} of $K$.
   \item $C(R)$ can be any abelian group. This is Clayborn's Theorem.
   \item For any non-zero prime $\mathfrak{p}\in \spec R$, $\mathfrak{p}^n/\mathfrak{p}^{n+1}\cong R/\mathfrak{p}$ as an
   $R$-module.
   \item ``To contain is to divide'', i.e.~if $A,B\subset R$, then $A\subset B$
   $\Longleftrightarrow$ $A=BC$ for some $C\subset R$.
   \item (Generation of ideals) Every non-zero ideal $B\subset R$ is generated by two elements.
   Moreover, one of the generators can be taken to be any non-zero element of $B$.
   \item (Factor rings) If $A\subset R$ is non-zero, then $R/A$ is a PIR
	(\textbf{principal ideal ring}).
   \item (Steinitz Isomorphism Theorem) If $A,B\subset R$ are non-zero ideals, then $A\oplus
   B\cong {}_RR\oplus AB$ as $R$-modules.
   \item If $M$ is a finitely generated torsion-free $R$-module of rank $n$,\footnote{The
   rank is defined as $rk(M)=\dim_{K(R)} M\otimes_R K(R)$ where $K(R)$ is the
	quotient field.} then it is of the
   form $M\cong R^{n-1}\oplus A$, where $A$ is a non-zero ideal, determined up to
   isomorphism.
   \item If $M$ is a finitely generated torsion $R$-module, then $M$ is uniquely of the
   form $M\cong R/A_1\oplus
   \cdots \oplus R/A_n$ with $A_1\subsetneq A_2\subsetneq \cdots \subsetneq
   A_n\subsetneq R$.
    \end{enumerate}

\add{eventually, proofs of these should be added}

\section{Extensions}


In this section, we will essentially consider the following question: if $A$
is a Dedekind domain, and $L$ a finite extension of the quotient field of $A$,
is the integral closure of $A$ in $L$ a Dedekind domain? The general answer
turns out to be yes, but the result is somewhat simpler for the case of a
separable extension, with which we begin.

\subsection{Integral closure in a finite separable extension}

One of the reasons Dedekind domains are so important is
\begin{theorem} \label{intclosdedekind} Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable extension of $K$, and $B$ the integral closure of $A$ in $L$.  Then $B$ is Dedekind.
\end{theorem}

This can be generalized to the Krull-Akizuki theorem below (\cref{}).

First let us give an alternate definition of ``separable''.
For a finite field extension $k'$ of $k$, we may consider the bilinear pairing
$k' \otimes_k k' \to k$ given by $x,y \mapsto \Tr_{k'/k}(xy)$. Which is to say $xy \in k'$ can be seen as a $k$-linear map of finite dimensional vector spaces $k' \to k'$, and we are considering the trace of this map. Then we claim that $k'$ is separable if and only if the bilinear pairing $k' \times k' \to k$ is non-degenerate.

To show the above claim, first note that the pairing is non-degenerate if and
only if it is non-degenerate after tensoring with the algebraic closure. This
is because if $\Tr(xy)=0$ for all $y \in k'$, then $\Tr((x\otimes
1_{\overline{k}})y)=0$ for all $y \in k' \otimes_k \overline{k}$, which we may see to be true by decomposing into pure tensors. The other direction is obtained by selecting a basis of $\overline{k}$ over $k$, and then noting that for $y_i$ basis elements, if $\Tr(\sum x y_i)=0$ then $\Tr(xy_i)=0$ for each $i$. 

So now we just need to show that $X=k'  \otimes_k \overline{k}$ is reduced if
and only if the map $X \otimes_{\overline{k}} X \to \overline{k}$ given by $a \otimes b \mapsto \Tr(ab)$ is non-degenerate. To do this, we show that elements of the kernel of the bilinear map are exactly the nilpotents. But note that $X$ is a finite dimensional algebra over $\overline{k}$, and we may elements as matrices. Then if $\Tr(AB)=0$ for all $B$ if and only if $\Tr(PAP^{-1}PBP^{-1})=0$ for all $B$, so we may assume $A$ is in Upper Triangular Form. From this, the claim becomes clear.


\begin{proof}  We need to check that $B$ is Noetherian, integrally, closed, and of dimension 1.

\begin{itemize}
\item Noetherian.  Indeed, $B$ is a finitely generated $A$-module, which
obviously implies Noetherianness. To see this, note that the $K$-linear map
$(.,.): L \times L \to K$, $a,b \to \mathrm{Tr}(ab)$ is nondegenerate since
$L$ is separable over $K$ (\rref{}).  Let $F \subset B$ be a free module spanned by a $K$-basis for $L$.  Then since traces preserve integrality and $A$ is integrally closed, we have $B \subset F^*$, where $F^* := \{ x \in K: (x,F) \subset A \}$.  Now $F^*$ is $A$-free on the dual basis of $F$ though, so $B$ is a submodule of a finitely generated $A$ module, hence a finitely generated $A$-module.
\item Integrally closed.  $B$ is the integral closure of $A$ in $L$, so it
is integrally closed (integrality being transitive).
\item Dimension 1.  Indeed, if $A \subset B$ is an integral extension of
domains, then $\dim A = \dim B$. This follows essentially from the theorems of
``lying over'' and ``going up.'' Cf. \cite{Ei95}.
\end{itemize}

So, consequently the ring of algebraic integers (integral over $\mathbb{Z}$) in a number field (finite extension of $\mathbb{Q}$) is Dedekind.
\end{proof}



Note that the above proof actually implied (by the argument about traces) the following useful fact:
\begin{proposition}\label{intclosurefgen} Let $A$ be a noetherian integrally closed domain with quotient field $K$.  Let $L$ be a finite separable extension and $B$ the ring of integers. Then $B$ is a finitely generated $A$-module.
\end{proposition}

We shall give another, more explicit proof of \rref{intclosurefgen} whose technique will be useful in the sequel.
Let $\alpha  \in B$ be a generator of $L/K$.  Let $n=[L:K]$ and $\sigma_1, \dots, \sigma_n$ the distinct embeddings of $L$ into the algebraic closure of $K$.
Define the \textbf{discriminant} of $\alpha$ to be
\[ D(\alpha) = \left(\det \begin{bmatrix} 
1 & \sigma_1\alpha & (\sigma_1 \alpha)^2 & \dots \\
1 & \sigma_2\alpha & (\sigma_2 \alpha)^2 & \dots \\
\vdots & \vdots & \vdots & \ddots \end{bmatrix}\right)^2 .\]
This maps to the same element under each $\sigma_i$, so is in $K^*$ (and even
$A^*$ by integrality); it is nonzero by basic facts about vanderMonde
determinants since each $\sigma_i$ maps $\alpha$ to a different element.  The  next lemma clearly implies that $B$ is contained in a finitely generated $A$-module, hence is finitely generated (since $A$ is noetherian).
\begin{lemma} We have $B \subset D(\alpha)^{-1} A[\alpha]$. 
\end{lemma}
\begin{proof}
Indeed, suppose $x \in B$.  We can write $x = c_0 (1) + c_1 (\alpha) + \dots c_{n-1}(\alpha^{n-1})$ where each $c_i \in K$.  We will show that in fact, each $c_i \in D(\alpha)^{-1}A$, which will prove the lemma.  Applying each $\sigma_i$, we have for each $i$, $\sigma_i x = c_0 (1) + c_1 (\sigma_i \alpha) + \dots + c_{n-1} ( \sigma_i \alpha^{n-1})$.
Now by Cramer's lemma, each $c_i$ can be written as a quotient of determinants of matrices involving $\sigma_jx $ and the $\alpha^{j}$.  The denominator determinant is in fact $D(\alpha)$.  The numerator is in $K$ and must be integral, hence is in $A$.  This proves the claim and the lemma.
\end{proof}

The above technique may be illustrated with an example.
\begin{example} Let $p^i$ be a power of a prime $p$ and consider the extension
$\mathbb{Q}(\zeta_{p^i})/\mathbb{Q}$ for $\zeta_{p^i}$ a primitive $p^i$-th
root of unity.  This is a special case of a cyclotomic extension, an important
example in the subject.  We claim that the ring of integers (that is, the
integral closure of $\mathbb{Z}$)
in $\mathbb{Q}(\zeta_{p^i})$ is precisely $\mathbb{Z}[\zeta_{p^i}]$.  This is true in fact for all cyclotomic extensions, but we will not be able to prove it here.

First of all, $\zeta_{p^i}$ satisfies the equation $X^{p^{i-1}(p-1)} +
X^{p^{i-1}(p-2)} + \dots + 1 = 0$.  This is because if $\zeta_p$ is a $p$-th
root of unity, $(\zeta_p-1)(1+\zeta_p + \dots + \zeta_p^{p-1}) = \zeta_p^p - 1 =
0$.  In particular, $X - \zeta_{p^i} \mid X^{p^{i-1}(p-1)} +
X^{p^{i-1}(p-2)} + \dots + 1 $, and consequently (taking $X=1$), we find that
$1 - \zeta_{p^i}$ divides $p$ in the ring of integers in
$\mathbb{Q}(\zeta_{p^i})/\mathbb{Q}$.  This is true for \emph{any} primitive
$p^i$-th root of unity for \emph{any} $p^i$.  Thus the norm to $\mathbb{Q}$ of $1 - \zeta_{p^i}^j$ for any $j$ is a power of $p$.

I claim that this implies that the discriminant $D(\zeta_{p^i})$ is a power of
$p$, up to sign. But by the vanderMonde formula, this discriminant is a
product of terms of the form $\prod (1 - \zeta_{p^i}^{j})$ up to roots of
unity.  The norm to $\mathbb{Q}$ of each factor is thus a power of $p$, and the discriminant itself plus or minus a power of $p$.

By the lemma, it follows that the ring of integers is contained in
$\mathbb{Z}[p^{-1}, \zeta_{p^i}]$. To get down further to
$\mathbb{Z}[\zeta_{p^i}]$ requires a bit more work. \add{this proof}
\end{example}


\subsection{The Krull-Akizuki theorem}

We are now going to prove a general theorem that will allow us to remove the
separability hypothesis in \cref{}. Let us say that a noetherian domain has
\textbf{dimension at most one} if every nonzero prime ideal is maximal; we shall later
generalize this notion of ``dimension.''

\begin{theorem}[Krull-Akizuki] Suppose $A$ is a noetherian domain of dimension
at most one. Let $L$ be a finite extension of the quotient field $K(A)$, and suppose
$B \subset L$ is a domain containing $A$. Then $B$ is noetherian of  dimension
at most one.
\end{theorem} 


From this, it is clear:

\begin{theorem} 
The integral closure of a Dedekind domain in any finite extension of the
quotient field is a Dedekind domain.
\end{theorem} 
\begin{proof} 
Indeed, by Krull-Akizuki, this integral closure is noetherian and of dimension
$\leq 1$; it is obviously integrally closed as well, hence a Dedekind domain.
\end{proof} 

Now let us prove Krull-Akizuki.
\add{we need to introduce material about length}
\begin{proof} 
We are going to show that for any $a \in A - \left\{0\right\}$, the $A$-module
$B/a B$ has finite length. (This is quite nontrivial, since $B$ need not even
be finitely \emph{generated} as an $A$-module.) 
From this it will be relatively easy to deduce the result.

Indeed, if $I \subset B$ is any nonzero ideal, then $I$ contains a nonzero  element of $A$; to
see this, we need only choose an element $b \in I$ and consider an irreducible polynomial
\[ a_0 X^n + \dots + a_n  \in K[X]  \]
that it satisfies. We can assume that all the $a_i \in A$ by clearing
denominators. It then follows that $a_n \in A \cap I$. 
So choose some $a \in (A \cap I) - \left\{0\right\}$. 
We then know by the previous paragraph (though we have not proved it yet) that
$B/aB$ has finite length as an $A$-module (and a fortiori as a $B$-module); in
particular, the submodule $I/aB$ is finitely generated as a
$B$-module. The
exact sequence
\[ 0 \to a B \to I \to I/aB \to 0  \]
shows that $I$ must be finitely generated as a $B$-module, since the two outer
terms are.
Thus any ideal of $B$ is finitely generated, so $B$ is noetherian.

\add{$B$ has dimension at most one}

To prove the Krull-Akizuki theorem, we are going to prove:
\begin{lemma}[Finite length lemma]
If $A$ is a noetherian domain of dimension at most one, then for any
torsion-free
$A$-module $M$ such that $K(A) \otimes_A M$ is finite-dimensional
(alternatively: $M$ has finite rank) and $a \neq 0$, $M/aM$ has finite length.
\end{lemma} 
\begin{proof} 
We are going to prove something stronger. If $M$ has rank $n$ and is
torsion-free, then will show
\begin{equation} \label{boundka} \ell(M/aM) \leq n \ell(A/aA).   \end{equation}
Note that $A/aA$ has finite length. This follows because there is a filtration of
$A/aA$ whose quotients are of the form $A/\mathfrak{p}$ for $\mathfrak{p}$
prime; but these $\mathfrak{p}$ cannot be zero as $A/aA$ is torsion. So these
primes are maximal, and $A/aA$ has a filtration whose quotients are
\emph{simple}. Thus $\ell(A/aA) < \infty$. In fact, we see thus that \emph{any
torsion, finitely-generated module has finite length}; this will be used in
the sequel.


There are two cases:
\begin{enumerate}
\item $M$ is finitely generated.
We can choose generators $m_1, \dots, m_n$ in $M$ of $K(A) \otimes_A M$; we
then from these generators get a map
\[ A^n \to M  \]
which becomes an isomorphism after localizing at $A - \left\{0\right\} $. In
particular, the kernel and cokernel are torsion modules. The kernel must be trivial
($A$ being a domain), and $A^n \to M$ is thus injective. 
Thus we have found a finite free submodule $F \subset M$ such that $M/F$ is a
torsion module $T$, which is also finitely generated.



We have an exact sequence
\[ 0 \to F/(aM \cap F) \to M/aM \to T/aT \to 0.  \]
Here the former has length at most $\ell(F/aF) = n \ell(A/aA)$, and we get
the bound $\ell(M/aM) \leq n \ell(A/aA) + \ell(T/aT)$. However, we
have the annoying last term to contend with, which makes things somewhat
inconvenient. Thus, we use a trick: for each $t > 0$, we consider the exact
sequence
\[ 0 \to F/(a^tM \cap F) \to M/a^tM \to T/a^tT \to 0.  \]
This gives 
\[ \ell(M/a^t M) \leq t n \ell(A/aA) + \ell(T/a^t T) \leq  t n \ell(A/aA) +
\ell(T). \]
However, $\ell(T) < \infty$ as $T$ is torsion (cf. the first paragraph).
If we divide by $t$, we get the inequality
\begin{equation} \label{almostboundka} \frac{1}{t} \ell(M/a^t M) \leq n
\ell(A/aA) + \frac{\ell(T)}{t}.  \end{equation}
However, the filtration $a^t M \subset a^{t-1}M \subset \dots \subset a M
\subset M$ whose quotients are all isomorphic to $M/aM$ ($M$ being
torsion-free) shows that $\ell(M/a^t M ) = t \ell(M/aM)$
In particular, letting $t \to \infty$ in \eqref{almostboundka} gives
\eqref{boundka} in the case where $M$ is finitely generated.
\item  $M$ is not finitely generated. Now we can use a rather cheeky argument. 
$M$ is the inductive limit of its finitely generated submodules $M_F \subset
M$, each of which
is itself torsion free and of rank at most $n$. Thus $M/aM $ is the inductive limit
of its submodules $M_F/(aM \cap M_F)$ as $M_F$ ranges over 
We know that $\ell(M_F/(aM \cap M_F)) \leq n \ell(A/aA)$ for each finitely
generated $M_F \subset M$ by the first case above (and the fact that
$M_F/(aM \cap M_F)$ is a quotient of $M_F/aM_F$).

But if $M/aM$ is the inductive limit of \emph{submodules} of length at most $n
\ell(A/aA)$, then it itself can have length at most $n \ell(A/aA)$. For $M/aM$
must be in fact equal to the submodule $M_F/(aM \cap M_F)$ that has the
largest length (no other submodule $M_{F'}/(aM \cap M_{F'})$ can properly
contain this).
\end{enumerate}


\end{proof} 

With this lemma proved, it is now clear that Krull-Akizuki is proved as well.

\end{proof} 

\subsection{Extensions of discrete valuations}

As a result, we find: 
\begin{theorem} 
Let $K$ be a field, $L$ a finite separable extension.  Then a discrete valuation on $K$ can be extended to one on $L$.
\end{theorem}
\add{This should be clarified --- what is a discrete valuation?}
\begin{proof}
Indeed, let $R \subset K$ be the ring of integers of the valuation, that is
the subset of elements of nonnegative valuation.  Then $R$ is a DVR, hence Dedekind, so the integral closure $S \subset L$ is Dedekind too (though in general it is not   a DVR---it may have several non-zero prime ideals) by \rref{intclosdedekind}.  Now as above, $S$ is a finitely generated $R$-module, so if $\mathfrak{m} \subset R$ is the maximal ideal, then 
\[ \mathfrak{m} S \neq S \]
by Nakayama's lemma (cf. for instance \cite{Ei95}).  So $\mathfrak{m} S$ is contained in a maximal ideal $\mathfrak{M}$ of $S$ with, therefore, $\mathfrak{M} \cap R = \mathfrak{m}$.  (This is indeed the basic argument behind lying over, which I could have just invoked.) Now $S_{\mathfrak{M}} \supset R_{\mathfrak{m}}$ is a DVR as it is the localization of a Dedekind domain at a prime ideal, and one can appeal to \rref{niupmeansdvr}.  So there is a discrete valuation on $S_{\mathfrak{M}}$.  Restricted to $R$, it will be a power of the given $R$-valuation, because its value on a uniformizer $\pi$ is $<1$.  However, a power of a discrete valuation is a discrete valuation too.  So we can adjust the discrete valuation on $S_{\mathfrak{M}}$ if necessary to make it an extension.

This completes the proof.  
\end{proof}

Note that there is a one-to-one correspondence between extensions of the valuation on $K$ and primes of $S$ lying above $\mathfrak{m}$.   Indeed, the above proof indicated a way of getting valuations on $L$ from primes of $S$.  For an extension of the valuation on $K$ to $L$, let $\mathfrak{M} := \{ x \in S: \left| x \right| < 1\}$.
\section{Action of the Galois group}

 Suppose we have an integral domain (we don't even have to assume it Dedekind) $A$ with quotient field $K$, a finite Galois extension $L/K$, with $B$ the integral closure in $L$.  Then the Galois group $G = G(L/K)$ acts on $B$; it preserves $B$ because it preserves equations in $A[X]$.
In particular, if $\mathfrak{P} \subset B$ is a prime ideal, so is $\sigma \mathfrak{P}$, and the set $\spec B$ of prime ideals in $B$ becomes a $G$-set.

\subsection{The orbits of the Galois group} It is of interest to determine the orbits; this question has  a very clean answer.

\begin{proposition} The orbits of $G$ on the prime ideals of $B$ are in bijection with the primes of $A$, where a prime ideal $\mathfrak{p} \subset A$ corresponds to the set of primes of $B$ lying over $A$.\footnote{It is useful to note here that the lying over theorem works for arbitrary integral extensions.}  Alternatively, any two primes $\mathfrak{P}, \mathfrak{Q} \subset B$ lying over $A$ are conjugate by some element of $G$.
\end{proposition}

In other words, under the natural map $\spec B \to \spec A =\spec B^G$, the latter space is the quotient under the action of $G$, while $A=B^G$ is the ring of invariants in $B$.\footnote{The reader who does not know about the $\spec$ of a ring can disregard these remarks.}

\begin{proof} 
We need only prove the second statement. 
Let $S$ be the multiplicative set $A - \mathfrak{p}$.  Then $S^{-1}B $ is the integral closure of $S^{-1}A$, and in $S^{-1}A = A_{\mathfrak{p}}$, the ideal $\mathfrak{p}$ is maximal. 
Let $\mathfrak{Q}, \mathfrak{P}$ lie over $\mathfrak{p}$; then $S^{-1}\mathfrak{Q},S^{-1} \mathfrak{P}$  lie over $S^{-1}\mathfrak{p}$ and are maximal (to be added).  If we prove that $S^{-1} \mathfrak{Q}, S^{-1} \mathfrak{P}$ are conjugate under the Galois group, then $\mathfrak{Q}, \mathfrak{P}$ must also be conjugate by the properties of localization.  \emph{In particular, we can reduce to the case of $\mathfrak{p}, \mathfrak{Q}, \mathfrak{P}$ all maximal.}

The rest of the proof is now an application of the Chinese remainder theorem.  Suppose that, for all $\sigma \in G$, we have $\sigma \mathfrak{P} \neq \mathfrak{Q}$.  Then the ideals $\sigma \mathfrak{P}, \mathfrak{Q}$ are distinct maximal ideals, so by the remainder theorem, we can find $x \equiv 1 \mod \sigma \mathfrak{P}$ for all $\sigma \in G$ and $x \equiv  0 \mod \mathfrak{Q}$.
Now, consider the norm $N^L_K(x)$; the first condition implies that it is congruent to 1 modulo $\mathfrak{p}$.  But the second implies that  the norm is in $\mathfrak{Q} \cap K = \mathfrak{p}$, contradiction.
\end{proof}

\subsection{The decomposition and inertia groups}
Now, let's zoom in on a given prime $\mathfrak{p} \subset A$.  We know that $G$ acts transitively on the set $\mathfrak{P}_1, \dots, \mathfrak{P}_g$ of primes lying above $\mathfrak{p}$; in particular, there are at most $[L:K]$ of them.  

\begin{definition} If $\mathfrak{P}$ is any one of the $\mathfrak{P}_i$, then the stabilizer in $G$ of this prime ideal is called the \textbf{decomposition group} $G_{\mathfrak{P}}$. \end{definition}
We have, clearly, $(G: G_{\mathfrak{P}}) = g$.

Now if $ \sigma \in G_{\mathfrak{P}}$, then $\sigma$ acts on the residue field $B/\mathfrak{P}$ while fixing the subfield $A/\mathfrak{p}$.  In this way, we get a homomorphism $\sigma \to \overline{\sigma}$ from $G$ into the automorphism group of $B/\mathfrak{P}$ over $A/\mathfrak{p})$ (we don't call it a Galois group because we don't yet know whether the extension is Galois).

The following result will be crucial in constructing the so-called ``Frobenius elements'' of crucial use in class field theory.

\begin{proposition} Suppose $A/\mathfrak{p}$ is perfect. Then $B/\mathfrak{P}$ is Galois over $A/\mathfrak{p}$, and the homomorphism $\sigma \to \overline{\sigma}$ is surjective from $G_{\mathfrak{P}} \to G(B/\mathfrak{P}/A/\mathfrak{p})$.
\end{proposition}
\begin{proof}
In this case, the extension $B/\mathfrak{P}/A/\mathfrak{p}$ is separable, and we can choose $\overline{x} \in B/\mathfrak{P}$ generating it by the primitive element theorem.  We will show that $\overline{x}$ satisfies a polynomial equation  $\overline{P}(X) \in A/\mathfrak{p}[X]$ all of whose roots lie in $B/\mathfrak{P}$, which will prove that the residue field extension is Galois.  Moreover, we will show that all the nonzero roots of $\overline{P}$ in $B/\mathfrak{P}$ are conjugates of $\overline{x}$ under elements of $G_{\mathfrak{P}}$.  This latter will imply surjectivity of the homomorphism $\sigma \to \overline{\sigma}$, because it shows that any conjugate of $\overline{x}$ under $G(B/\mathfrak{P}/A/\mathfrak{p})$ is a conjugate under $G_{\mathfrak{P}}$.

We now construct the aforementioned polynomial.   Let $x \in B$ lift $\overline{x}$.  Choose $y \in B$ such that $y \equiv x \mod \mathfrak{P}$ but $y \equiv 0 \mod \mathfrak{Q}$ for the other primes $\mathfrak{Q}$ lying over $\mathfrak{p}$.  We take $P(X) = \prod_{\sigma \in G} (X - \sigma(y)) \in A[X]$. Then the reduction $\overline{P}$ satisfies $\overline{P}(\overline{x})= \overline{P}(\overline{y}) = 0$, and $\overline{P}$ factors completely (via $\prod_{\sigma} (X - \overline{\sigma(t)})$) in $B/\mathfrak{P}[X]$.  This implies that the residue field extension is Galois, as already stated. 
But it is also clear that the polynomial $\overline{P}(X)$ has roots of zero and $\sigma(\overline{y}) = \sigma(\overline{x})$ for $\sigma \in G_{\mathfrak{P}}$.  This completes the proof of the other assertion, and hence the proposition.
\end{proof}

\begin{definition} 
The kernel of the map $\sigma \to \overline{\sigma}$ is called the \textbf{inertia
group} $T_{\mathfrak{P}}$.  Its fixed field is called the \textbf{inertia
field}.
\end{definition}

These groups will resurface significantly in the future.

\begin{remark} Although we shall never need this in the future, it is of
interest to see what happens when the extension $L/K$ is \emph{purely
inseparable}.\footnote{Cf. \cite{La02}, for instance.}  Suppose $A$ is integrally closed in $K$, and $B$ is the integral closure in $L$.  Let the characteristic be $p$, and the degree $[L:K] = p^i$.
In this case, $x \in B$  if and only if $x^{p^i} \in A$. Indeed, it is clear that the condition mentioned implies integrality.  Conversely, if $x$ is integral, then so is $x^{p^i}$, which belongs to $K$ (by basic facts about purely inseparable extensions).  Since $A$ is integrally closed, it follows that $x^{p^i} \in A$.

Let now $\mathfrak{p} \subset A$ be a prime ideal.  I claim that there is precisely one prime ideal $\mathfrak{P}$ of $B$ lying above $A$, and $\mathfrak{P}^{p^i} = \mathfrak{p}$.  Namely, this ideal consists of $x \in B$ with $x^{p^i} \in \mathfrak{p}$!  The proof is straightforward; if $\mathfrak{P}$ is \emph{any} prime ideal lying over $\mathfrak{p}$, then $x \in \mathfrak{P}$ iff $x^{p^i} \in L \cap \mathfrak{P} = \mathfrak{p}$.   In a terminology to be explained later, $\mathfrak{p}$ is \emph{totally ramified.}
\end{remark}


