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

Then let be the stabilizer of for some fixed , i.e. Then, we have . This is what I meant about being induced.

I claim that

In particular, we get an equality of the Herbrand quotients .

(This is actually true for all , 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. **

We compute ; I claim it is isomorphic to .

Indeed, we shall define maps , which are inverses when restricted to the fixed points.

Suppose . Then for , i.e. . So, set .

Conversely, given , we can define an extension by selecting coset representatives (note that is abelian!) for , and noting that these coset representatives map bijectively onto . In detail, we choose such that for each . Then we set , and in this way get a map . This map sends into . Indeed, because is fixed by in this case, this is -invariant.

So, the map induces an isomorphism .

Now the norm map can be computed as follows: it is

In particular, the norm maps are compatible with the maps . As a result, we get the claimed isomorphism .

**2. **

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

where are the *augmentation ideals* (generated by terms of the form ). This isomorphism is defined by sending to . 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.

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