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.