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.

2. A lemma

First of all, the whole idele group {J_L} is really, really big, and we want to cut it down. Hence, we need:

Lemma 2 Let {L} be a number field. There is a finite set {S} of places of {L} such that {J_L = L^* J_S}, where {J_S} denotes the {S}-ideles (units outside of {S}).

Indeed, first consider the map {J_L \rightarrow  \mathbf{Cl}_L}, where {\mathbf{Cl}} is the ideal class group, and the map proceeds by taking an idele {(x_v)_v}, finding the orders at all the nonarchimedean places, and taking the product of the associated prime ideals of the ring of integers raised to those orders. The kernel is clearly {L^*  J_{S_\infty}}, so {J_L/L^* J_{S_\infty}} is finite by the finiteness of the ideal class group. Since {J_L = \bigcup_S  J_S}, the lemma follows.

3. Reduction

Now choose {S} so large such that {J_L = L^*  J_S}, as in the previous section, and enlarge it if necessary to contain all ramified primes and so that {S} is stable under {G} (so that {J_S} is a {G}-module). We use this to simply the computation of {Q(C_L)}. We have

\displaystyle Q(C_L) = Q(L^* J_S/L^*) = Q(J_S/(L \cap  J_S)) = Q(J_S/L_S) = Q(J_S)/Q(L_S) .

We have already computed {Q(L_S)}. We need to compute {Q(J_S)}. Once we do so (and show that it is defined!), we will have completed our goal of computing the cohomology of the idele classes and thereby ensured that the fate of humanity is secure.

4. The cohomology of the ideles

We have {J_S = \prod_{w \in S} L_w^* \times \prod_{v \notin S} U_w = A  \times B}. Now we first compute {Q(A)}. Let {T} be the set of places of {k} that {S} extends. Then {Q(A)} is

\displaystyle Q( \prod_{v \in S_k} \prod_{w \mid v}  L_w^*) = \prod_{v \in S_k} n_v

in view of Shapiro’s lemma and also the computation of the local Herbrand quotient.

Next we compute {Q(B)}. But for each place {v} of {k} whose prolongations lie outside {S}, we have {H_T^i(G, \prod_{w \mid v} U_w) =  H_T^i(G_{w_0}, U_{w_0}) = 0} for {i = 0,  -1} because the ramified primes are contained in {S}. It follows that {Q(B) = 1} since {H_T^i} commutes with arbitrary direct products for {i=0,-1}: indeed, {H_T^i(G, B) = 0} for {i=0,-1}.

So {Q(J_S) = \prod_{v \in S_k} n_v}.

5. The final computation

We have {Q(L_S) = \frac{1}{n} \prod_{v \in S_k} n_v}, so if we put everything together, we find

\displaystyle Q(C_L) = n.

Bingo.

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