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. (more…)

We are now (finally) ready to start handling the cohomology of the idele classes.  The previous few posts in this series contain important background computations of the Herbrand quotients of local fields and units, which should be read before this.  Let ${L/k}$ be a finite cyclic extension of global fields of degree ${n}$. In the following, the Herbrand quotient will always be respect to the Galois group ${G=G(L/k)}$.

Theorem 1 We have ${Q(J_L/L^*) = n}$. In particular, $\displaystyle (J_k: k^* NJ_L) \geq n.$

1. Some remarks

The point of class field theory, of course, is that there is exactly an equality in the above statement, which is induced by an isomorphism between the two groups, and which holds for an arbitrary abelian extension of number fields.

Before we prove this theorem, let’s review a little. We know that ${G}$ acts on the ideles ${J_L}$, and also on ${L^*}$ (clearly). As a result, we get an action on the idele classes ${C_L=J_L/L^*}$. There is a map ${C_k = J_k/k^* \rightarrow C_L}$; I claim that it is an injection, and the fixed points of ${G}$ in ${C_L}$ are precisely the points of ${C_k}$. This can be proved using group cohomology. We have an exact sequence ${0 \rightarrow L^* \rightarrow J_L \rightarrow C_L \rightarrow 0}$, and consequently one has a long exact sequence $\displaystyle 0 \rightarrow H^0(G, L^*) \rightarrow H^0(G, J_L) \rightarrow H^0(G, C_L) \rightarrow H^1(G, L^8) = 0$

by Hilbert’s Theorem 90, and where ${H^0}$ is the ordinary (non-Tate) cohomology groups, that is to say just the ${G}$-stable points. Since we know that ${H^0(G, L^*)=k^*}$ and ${H^0(G, J_L)=J_k}$, we find that ${(C_L)^G = C_k}$, q.e.d.

So, anyhow, this Big Theorem today computes the Herbrand quotient ${Q(C_L)}$. It in particular implies that ${H_T^0(G, C_L) \geq n}$, and since this group is none other than $\displaystyle C_k/NC_L = J_k/k^* NC_L$

we get the other claim of the theorem. We are reduced to computing this messy Herbrand quotient, and it will use all the tools that we have developed up to now.