We shall now consider a number field ${k}$ and an abelian extension ${L}$. Let ${S}$ be a finite set of primes (nonarchimedean valuations) of ${k}$ containing the ramified primes, and consider the group ${I(S)}$ of fractional ideals prime to the elements of ${S}$. This is a free abelian group on the primes not in ${S}$. We shall define a map, called the Artin map from ${I(S) \rightarrow G(L/k)}$.

1. How does this work?

Specifically, let ${\mathfrak{p} \notin S}$ be a prime in ${k}$. There is a prime ${\mathfrak{P}}$ of ${L}$ lying above it. If ${A,B}$ are the rings of integers in ${k,L}$, respectively, then we have a field extension ${A/\mathfrak{p} \rightarrow B/\mathfrak{P}}$. As is well-known, there is a surjective homomoprhism of the decomposition group ${G_{\mathfrak{P}}}$ of ${\mathfrak{P}}$ onto ${G(B/\mathfrak{P} / A/\mathfrak{p})}$ whose kernel, called the inertia group, is of degree ${e(\mathfrak{P}|\mathfrak{p})}$.

But, we know that the extension ${B/\mathfrak{P} / A/\mathfrak{p}}$ is cyclic, because these are finite fields. The Galois group is generated by a canonically determined Frobenius element which sends ${a \rightarrow a^{|A/\mathfrak{p}|}}$. We can lift this to an element ${\sigma_{\mathfrak{p}}}$ of ${G_{\mathfrak{P}}}$, still called the Frobenius element.

First of all, ${\sigma_{\mathfrak{p}}}$ does not depend on the choice of lifting to ${G_{\mathfrak{P}}}$—indeed, the ramification index is one, so ${G_{\mathfrak{P}} \rightarrow G(B/\mathfrak{P} / A/\mathfrak{p})}$ is even an isomorphism. Moreover, ${G_{\mathfrak{P}}}$ is independent of ${\mathfrak{P}}$, because any two decomposition groups are conjugate (since any two primes of ${L}$ lying over the same prime of ${k}$ are conjugate), and we are working with an abelian extension. It follows similarly that ${\sigma_{\mathfrak{p}}}$ is independent of the extension ${\mathfrak{P}}$.

By multiplicativity, we get a homomorphism ${I(S) \rightarrow G(L/k)}$. This is called the Artin map. We write ${(\mathfrak{a}, L/k)}$ to denote the image of the fractional ideal ${\mathfrak{a}}$ (prime to ${S}$). Eventually, we will define this as a map on the ideles, and this is how we will get the isomorphism of class field theory.

2. Basic properties

There are a few easy properties of it that we may note.

First of all, suppose ${k \subset L \subset M}$ is a tower with ${M/k}$ abelian. If ${\mathfrak{a}}$ is an ideal prime to the primes ramified in ${M/k}$, then we have ${(\mathfrak{a}, M/k)_L = (\mathfrak{a}, L/k)}$. To see this, we may assume ${\mathfrak{a} = \mathfrak{p}}$, a prime ideal. Then this is because both induce the Frobenius automorphism on the extension of residue fields and lie in appropriate decomposition groups (formally, take ${\mathfrak{P}}$ of ${L}$ extending ${\mathfrak{p}}$ and ${\mathfrak{Q}}$ extending ${\mathfrak{P}}$, and look at the actions of these on the residue class fields of the rings of integers in ${L,M}$ quotiented by ${\mathfrak{P}, \mathfrak{Q}}$—it is the same).

Next, suppose we have an abelian extension ${L/k}$ and a finite extension ${E/k}$. Then ${LE/E}$ is an abelian extension too and the Galois group is a subgroup of ${G(L/k)}$. Moreover, I claim that we have ${(\mathfrak{b}, LE/E)_L = (N^E_k \mathfrak{b}, L/k)}$. So the Artin map behaves specially with respect to the norm. To see this, we need only check on prime ideals ${\mathfrak{P}}$ of ${E}$ (whose restriction ${\mathfrak{p}}$ to ${k}$ is unramified in ${L/k}$). Then ${(\mathfrak{P}, LE/E)_L}$ induces the map ${a \rightarrow a^{\mathbf{N}\mathfrak{P}}}$ on the residue fields. By contrast, ${N^E_k \mathfrak{P} = \mathfrak{p}^f}$ for ${f}$ the residue class field degree and ${(\mathfrak{P}, L/k)}$ induces the map ${a \rightarrow a^{N \mathfrak{p}}}$ on the residue class fields. But ${\mathbf{N}\mathfrak{p}^f = \mathbf{N}\mathfrak{P}}$, so we are done.

In particular, the following important fact follows: ${(N^L_k \mathfrak{b}, L/k)= 1}$ when ${\mathfrak{b}}$ is an ideal of ${L}$ (not divisible by the ramified primes).

3. The Artin map is surjective

This is the primary thing we shall prove today, and it is far from trivial. In fact, it will use the first inequality.

So, let ${S}$ be a finite set of primes containing the ones ramified in a finite abelian extension ${L/k}$. Consider the subgroup ${H}$ of ${G(L/k)}$ generated by the Frobenius elements ${\sigma_{\mathfrak{p}}, \mathfrak{p} \notin S}$. Then the fixed field ${Z}$ of this satisfies the following: ${Z/k}$ is Galois, with group ${G(L/k)/H}$.

I claim now that the primes ${\mathfrak{p}, \mathfrak{p} \notin S}$ all split completely in ${Z}$. Indeed, they are unramified by assumption. In addition, the residue class field extension must be trivial. When ${\mathfrak{p} \notin S}$, we have by consistency ${(\mathfrak{p}, Z/k) = 1}$, because ${\sigma_{\mathfrak{p}}}$ fixes ${Z}$. But ${(\mathfrak{p}, Z/k)}$ induces on the residue class field extension the Frobenius, which must be the identity; thus ${Z/k}$ satisfies ${f_{\mathfrak{p}}=2}$ for ${\mathfrak{p} \notin S}$. Thus all these primes split completely.

So, we now prove:

Theorem 1 Let ${M/k}$ be an abelian extension. Then if all but finitely many primes of ${k}$ split completely in ${M}$, we have ${M=k}$.

First, since splitting completely is defined by ${e=f=1}$, it is preserved by subextensions. So it is enough to prove the result when ${M/k}$ is cyclic, in view of Galois theory.

I claim that if all but finitely many primes split completely, then we have ${J_k = k^* NJ_M}$. To see this, first note that complete splitting implies that ${M_w = k_v}$ for ${w \mid v, v}$ outside a finite set ${S}$ of bad ${v}$‘s.

Now pick some idele ${x \in J_k}$. We can multiply it by ${c \in k^*}$ so that ${xc}$ has component near 1 at all ${v \in S}$, because of the approximation theorem. Locally, any element of ${k_v}$ close enough to 1 is a norm, because the norms form an open subgroup of finite index (which we have computed!). So if ${x}$ is chosen appropriately, ${xc}$ will be a local norm at all ${ v\in S}$, and it is so at all ${v \notin S}$ by complete splitting. So ${xc \in NJ_M}$.

We find in particular that ${(J_k: k^* NJ_M) = 1}$, which means that ${M=k}$ by the first inequality.

Corollary 2 The Artin map ${I(S) \rightarrow G(L/k)}$ is surjective for ${L/k}$ an arbitrary abelian extension.