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.