We shall now take the first steps in class field theory. Specifically, since we are interested in groups of the form , we will need their orders. And the first place to begin is with a local analog.
1. The cohomology of the units
Theorem 1 Let be a cyclic extension of local fields of degree and ramification . Then .
Let be the Galois group. We will start by showing that .
Indeed, first of all let us choose a normal basis of , i.e. a basis such that for all . It is known (the normal basis theorem) that this is possible. By multiplying by a high power of a uniformizer, we find that there is a -submodule of the additive group isomorphic to , i.e. is induced. We see that has trivial Tate cohomology and Herbrand quotient 1 by Shapiro’s lemma: any -module induced from the subgroup 1 satisfies this, because for any .
But if is taken sufficiently close to zero, then there is a -equivariant map , defined via
which converges appropriately at sufficiently small . (Proof omitted, but standard. Note that in the nonarchimedean case though!) In other words, the additive and multiplicative groups are locally isomorphic. This map (for sufficiently small) is an injection, the inverse being given by the logarithm power series. Its image is an open subgroup of the units, and since the units are compact, of finit index. So we have
This proves the theorem.
2. The cohomology of
Theorem 2Hypotheses as above (cyclic extension of local fields), we have
The equality follows because as -modules, we have , where has the trivial action. Since , we find that
We then use the previous computation of . Then, recall that (by HT 90) . This proves the theorem.
3. The norm index of the units
The following will also be a useful bit of the story:
Theorem 3 Hypotheses as above, we have , for the ramification index. In particular, every unit is a norm in the unramified case (since unramifiedness implies cyclicity).
Indeed, we have to show that since the Herbrand quotient is 1. But, this is equal to
Fix a generator for the Galois group. By Hilbert’s Theorem 90, the units of norm 1 are precisely of the form , where (not necessarily of norm 1). Let us use the notation to denote the image of under a map . Then we have
Let , similarly, denote the kernel of . It is easy to show that for a subgroup. It follows that
This completes the computation of and thus the proof.