The following situation—namely, the cohomology of induced objects—occurs very frequently, and we will devote a post to its analysis. Let {G} be a cyclic group acting on an abelian group {A}. Suppose we have a decomposition {A = \bigoplus_{i \in I} A_i} such that any two {A_i} are isomorphic and {G} permutes the {A_i} with each other. It turns out that the computation of the cohomology of {A} can often be simplified.

Then let {G_0} be the stabilizer of {A_{i_0}} for some fixed {i_0 \in I}, i.e. {G_0 = \{g: gA_{i_0} = A_{i_0} \}.} Then, we have {A = \mathrm{Ind}_{G_0}^G A_{i_0}}. This is what I meant about {A} being induced.

I claim that

\displaystyle \boxed{ H_T^i(G, A) \simeq H_T^i(G_0, A_{i_0}) , \quad i = -1, 0. }

In particular, we get an equality of the Herbrand quotients {Q(A), Q(A_{i_0})}.

(This is actually true for all {i}, and is a general fact about the Tate cohomology groups. It can be proved using abstract nonsense.)

We shall give a direct elementary (partial) proof.

1. {i=0}

We compute {A^G}; I claim it is isomorphic to {A_{i_0}^{G_0}}.

Indeed, we shall define maps {A_0 \rightarrow A}, {A \rightarrow A_0} which are inverses when restricted to the fixed points.

Suppose {(a_i)_{i \in I} \in A^G}. Then {ga_{i_0} = a_{i_0}} for {g \in G_{0}}, i.e. {a_{i_0} \in A_{i_0}^{G_0}}. So, set {f((a_i)_{i \in I}) = a_{i_0}}.

Conversely, given {a_{i_0} \in A_{i_0}^{G_0}}, we can define an extension {(a_i)_{i \in I} \in A^G} by selecting coset representatives (note that {G} is abelian!) for {G_0}, and noting that these coset representatives map bijectively onto {I}. In detail, we choose {g_i \in G} such that {g_i A_{i_0} = A_i} for each {i \in I}. Then we set {a_i = g_i a_{i_0}}, and in this way get a map {f': A_{i_0} \rightarrow A}. This map sends {A_{i_0}^{G_0}} into {A^G}. Indeed, because {a_{i_0}} is fixed by {G_0} in this case, this is {G}-invariant.

So, the map {(a_i) \rightarrow a_{i_0}} induces an isomorphism {A^G \rightarrow A_{i_0}^{G_0}}.

Now the norm map {N_G: A \rightarrow A} can be computed as follows: it is

\displaystyle \sum_I g_i N_{G_0}.

In particular, the norm maps {N_G, N_{G_0}} are compatible with the maps {f, f'}. As a result, we get the claimed isomorphism {A_{i_0}^{G_0}/NA_{i_0} \simeq A^G/NA}.

2. {i=-1}

To be honest, I’m not terribly inclined to go through the details here, especially as I’m short on time; if anyone has a burning urge to see the proof, I refer you to Lang’s Algebraic Number Theory. The idea is to prove an isomorphism between

\displaystyle \mathrm{ker} N_G/ I_G A \simeq \mathrm{ker} N_{G_0}/ I_{G_0} A_{i_0},

where {I_G, I_{G_0}} are the augmentation ideals (generated by terms of the form {g-1}). This isomorphism is defined by sending {(a_i)_{i \in I}} to {\sum a_i}. It is not too bad to check that everything works out…

Class field theory is more fun, and I want to get to it ASAP.