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.

Let ${S}$ be a finite set of places of a number field ${L}$, containing the archimedean ones. Suppose ${L/k}$ is a cyclic extension with Galois group ${G}$. Then, if ${G}$ keeps ${S}$ invariant, ${G}$ keeps the group ${L_S}$ of ${S}$-units invariant. We will need to compute its Herbrand quotient in order to do the same for the idele classes (next time), and that is the purpose of this post.

Up to a finite group, ${L_S}$ is isomorphic to a lattice in ${\mathbb{R}^{|S|}}$, though—this is the unit theorem. This isomorphism is by the log map, and it is even a ${G}$-isomorphism if ${G}$ is given an action on ${\mathbb{R}^{|S|}}$ coming from the permutation of ${S}$ (i.e. the permutation representation). This lattice is of maximal rank in the ${G}$-invariant hyperplane ${W \subset \mathbb{R}^{|S|}}$.

Motivated by this, we study the Herbrand quotient on lattices next.

1. The cohomology of a lattice

Fix a cyclic group ${G}$. We will suppose given a ${G}$-lattice ${L}$, that is to say a ${\mathbb{Z}}$-free module of finite rank on which ${G}$ acts. One way to get a ${G}$-lattice is to consider a representation of ${G}$ on some real vector space ${V}$, and choose a lattice in ${V}$ that is invariant under the action of ${G}$.

Proposition 1 Let ${L, L'}$ be two lattices in ${V}$ of maximal rank. Then ${Q(L)=Q(L')}$.

The way Lang (following Artin-Tate’s Class Field Theory) approaches this result seems a little unwieldy to me. I will follow Cassels-Frohlich (actually, Atiyah-Wall in their article on group cohomology in that excellent conference volume).  We have ${\mathbb{Q}[G]}$-modules ${L_{{\mathbb Q}} = L \otimes \mathbb{Q},L'_{{\mathbb Q}}= L' \otimes \mathbb{Q}}$, and ${\mathbb{R}}$-modules ${L_{{\mathbb R}}, L'_{{\mathbb R}}}$ defined similarly. Moreover, we have

$\displaystyle \mathbb{R} \otimes_{{\mathbb Q}} \hom_{\mathbb{Q}[G]}( L_{{\mathbb Q}}, L_{{\mathbb Q}}) \simeq \hom_{{\mathbb R}[G]}(L_{{\mathbb R}}, L'_{{\mathbb R}}).$ (more…)

We continue our quest to climb Mount Takagi-Artin.

In class field theory, it will be important to compute and keep track of the orders of groups such as ${(K^*:NL^*)}$, where ${L/K}$ is a Galois extension of local fields. A convenient piece of machinery for doing this is the Herbrand quotient, which we discuss today. I only sketch the proofs though, and a little familiarity with the Tate cohomology groups will be useful (but is not strictly necessary if one accepts the essentially combinatorial results without proof or proves them directly).

1. Definition

Let ${G}$ be a cyclic group generated by ${\sigma}$ and ${A}$ a ${G}$-module. It is well-known that the Tate cohomology groups ${H^i_T(G, A)}$ are periodic with period two and thus determined by ${H^0}$ and ${H^{-1}}$. By definition,

$\displaystyle H^0(G,A) = A^G/ NA ,$

where ${A^G}$ consists of the elements of ${A}$ fixed by ${G}$, and ${N: A \rightarrow A}$ is the norm map, ${a \rightarrow \sum_g ga}$. Moreover,

$\displaystyle H^{-1}(G, A) = \mathrm{ker} N/ (\sigma -1) A.$

(Normally, for ${G}$ only assumed finite, we would quotient by the sum of ${(\sigma - 1)A}$ for ${\sigma \in G}$ arbitrary, but here it is enough to do it for a generator—easy exercise.)

If both cohomology groups are finite, define the Herbrand quotient ${Q(A)}$ as

$\displaystyle Q(A) = \frac{ |H_T^0(G,A)|}{|H_T^{-1}(G,A)|}.$ (more…)