We begin with a few definitions.

Definition 1: A integral domain $R$ is called a Dedekind domain if it is noetherian, every nonzero prime ideal is maximal, and it is integrally closed in its field of fractions.

Definition 2: A ring $R$ is called a discrete valuation ring (DVR) if it is a principal ideal domain (PID) with exactly one nonzero prime ideal. (In other language, a DVR is a local PID of Krull dimension 0 or 1.)

One very important property of Dedekind domains is that ideals have unique factorizations as products of prime ideals. I used this property in the case of rings of integers in my last post to say that if $L/K$ is an extension of number fields with rings of integers $B/A$, so if $\mathfrak{p}\subset A$ is a prime ideal, then we can write $\mathfrak{p}B=\prod_{i=1}^g \mathfrak{P}_i^{e_i}$. But this result holds in more generality, for any Dedekind domain.

Also, it is very easy to check that a DVR is a Dedekind domain. But one very common occurrence of DVRs is as localizations of rings of integers. In particular, if $A$ is a Dedekind domain and $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ is a DVR.

One way to interpret a DVR is through the following filtration of ideals. Let $R$ be a DVR, and let $\mathfrak{p}$ be the unique nonzero prime ideal of $R$. Then every nonzero ideal of $R$ is of the form $\mathfrak{p}^n$ for some $n\ge 0$ (where by $\mathfrak{p}^0$ I mean $R$). Now, for any $x\in R\setminus\{0\}$, there is an integer $n$ so that $x\in\mathfrak{p}^n\setminus\mathfrak{p}^{n+1}$. We can now define a function $v:R\setminus\{0\}\to\mathbb{N}$ (where $\mathbb{N}$ includes 0 in this case) by $v(r)=n$ as above. We can extend our definition of $v$ to all of $R$ by setting $v(0)=+\infty$.

It is also possible to extend $v$ to the quotient field $K$ of $R$ by setting $v(x/y)=v(x)-v(y)$; it is easy to check that this is well-defined. Now, $v$ satisfies the following properties:

1) $v:K\setminus\{0\}\to\mathbb{Z}$ is a surjective homomorphism.

2) $v(x+y)\ge\min(v(x),v(y)$.

We call such a function $v$ a valuation of the field $K$.

Knowing $v$ is enough to reconstruct $R$, since $R=\{x\in K:v(x)\ge 0\}$. Furthermore, $\mathfrak{p}=\{x\in K:v(x)\ge 1\}$. We call $R$ the valuation ring of $K$.

Let’s look at a few places where DVRs arise naturally.

1) As we mentioned earlier, the localization of a Dedekind domain at a prime ideal is a DVR. So, for example, $\mathbb{Z}_{(p)}=\{x/y\in\mathbb{Q}:p\nmid y\}$ is a DVR if $p$ is a prime. The unique prime ideal is $p\mathbb{Z}_{(p)}$.

2) The ring $\mathbb{Z}_p$ of $p$-adic integers is a DVR with unique prime ideal $p\mathbb{Z}_p$. Also, finite extensions of the field $\mathbb{Q}_p$ of $p$-adic numbers inherit valuations from $\mathbb{Q}_p$, and so they contain DVRs as described above. In particular, if $K/\mathbb{Q}_p$ is a finite field extension, then the integral closure of $\mathbb{Z}_p$ in $K$ is a DVR.

Now, if $R$ is a DVR and $\mathfrak{p}$ is its prime ideal, then $R/\mathfrak{p}$ is a field. In the cases described above, this will always be a finite field; in what follows, we always assume that this field is finite. We call $R/\mathfrak{p}$ the residue field.

We can also put a topology on a valued field $K$ by letting the following sets be a basis for the topology: if $x\in K$ and $n\ge 0$ is an integer, then $\{y\in K:v(x-y)\ge n\}$ is an open set. These sets generate the topology. In what follows, we will assume that $K$ is complete as a topological space with this topology. Finite extensions of $\mathbb{Q}_p$ are complete with respect to this topology, so this will be our motivating example. The residue fields will also be finite.

Last post, I pointed out that if $L/K$ is a Galois extension of number fields, then $efg=[L:K]$. This holds more generally, however. If $L/K$ is a finite Galois field extension, and $A\subset K$ is a Dedekind domain so that $K$ is the quotient field of $A$, and $B$ is the integral closure of $A$ in $L$, then we still have $efg=[L:K]$.

We now interpret this in the case of $K$ a field complete with respect to a discrete valuation $v$, and $A$ the valuation ring of $K$. Let $L/K$ be a finite Galois extension, and let $B$ be the integral closure of $A$ in $L$, or, equivalently, the valuation ring of $L$. Then $L$ is also complete with respect to a discrete valuation $w$ that is very closely related to $v$, as we will see soon.

Let $\mathfrak{p}$ be the prime of $A$, and let $\mathfrak{P}$ be the prime of $B$. Since there is only one prime, $g=1$. Hence $\mathfrak{p}B=\mathfrak{P}^e$. Now, if $x\in K$, then $w(x)=ev(x)$, and if $x\in L$, then $fw(x)=v(N_{L/K}x)$. (But we won’t need these results in what follows, at least today.) The implication is the decomposition group of the extension $L/K$ is the entire Galois group.

We can put a filtration on the Galois group as follows: For $i\ge -1$, let $G_i=\{\sigma\in Gal(L/K):w(\sigma(x)-x)\ge i+1 \hbox{ for all } x\in L\}$. We call $G_i$ the $i^{\hbox{th}}$ ramification group of $L/K$. $G_{-1}=G$ is the entire Galois group (or the decomposition group; $G_0$ is the inertia group. Also, each $G_i$ is normal in $G$.

Now, I won’t prove it here, but it can be shown that if the residue field is finite of characteristic $p>0$ and $K$ is complete, then for each $i\ge 1$, $G_i/G_{i+1}$ is a direct product of copies of $\mathbb{Z}/p\mathbb{Z}$, and $G_0/G_1$ is a subgroup of the roots of unity of $B/\mathfrak{P}$ (and hence finite and cyclic of order prime to $p$). Hence, by basic group theory or otherwise, $G_0$ is a semidirect product of a normal Sylow $p$-subgroup and a cyclic group of order prime to $p$. In particular, $G_0$ is solvable. However, as shown in the last post, $G/G_0\cong Gal((B/\mathfrak{P})/(A/\mathfrak{p}))$ is cyclic since it is the Galois group of an extension of finite fields. Hence:

Theorem: $G$ is solvable.