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 be a finite cyclic extension of global fields of degree . In the following, the Herbrand quotient will always be respect to the Galois group .

Theorem 1We have . In particular,

**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 acts on the ideles , and also on (clearly). As a result, we get an action on the **idele classes** . There is a map ; I claim that it is an injection, and the fixed points of in are precisely the points of . This can be proved using group cohomology. We have an exact sequence , and consequently one has a long exact sequence

by Hilbert’s Theorem 90, and where is the ordinary (non-Tate) cohomology groups, that is to say just the -stable points. Since we know that and , we find that , q.e.d.

So, anyhow, this Big Theorem today computes the Herbrand quotient . It in particular implies that , and since this group is none other than

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 is really, really big, and we want to cut it down. Hence, we need:

Lemma 2Let be a number field. There is a finite set of places of such that , where denotes the -ideles (units outside of ).

Indeed, first consider the map , where is the ideal class group, and the map proceeds by taking an idele , 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 , so is finite by the finiteness of the ideal class group. Since , the lemma follows.

**3. Reduction **

Now choose so large such that , as in the previous section, and enlarge it if necessary to contain all ramified primes and so that is stable under (so that is a -module). We use this to simply the computation of . We have

We have already computed . We need to compute . 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 . Now we first compute . Let be the set of places of that extends. Then is

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

Next we compute . But for each place of whose prolongations lie outside , we have for because the ramified primes are contained in . It follows that since commutes with arbitrary direct products for : indeed, for .

So .

**5. The final computation **

We have , so if we put everything together, we find

Bingo.

June 3, 2010 at 1:13 pm

[…] field theory using only ideal theory, but now the language of the ideles is convenient too (and as we saw, the ideles lend themselves very nicely to computing Herbrand quotients). But they are not as […]

June 5, 2010 at 7:18 am

[…] any -th power is a norm. It is also finite; this is true because the Herbrand quotient is defined (and has been computed!). So the order is a power of […]