We shall now take the first steps in class field theory. Specifically, since we are interested in groups of the form {J_k/k^* N J_L}, 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 {L/K} be a cyclic extension of local fields of degree {n} and ramification {e}. Then {Q(U_L)=1}.

 

Let {G} be the Galois group. We will start by showing that {Q(U_L)=1}.

Indeed, first of all let us choose a normal basis of {L/K}, i.e. a basis {(x_\sigma)_{\sigma \in G}} such that {\tau x_{\sigma} = x_{\sigma \tau}} for all {\tau, \sigma \in G}. 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 {G}-submodule {V_a} of the additive group {\mathcal{O}_L} isomorphic to {\mathcal{O}_K[G]}, i.e. is induced. We see that {V_a} has trivial Tate cohomology and Herbrand quotient 1 by Shapiro’s lemma: any {G}-module induced from the subgroup 1 satisfies this, because {H_T^i(1, A)= 0} for any {A}.

But if {V_a} is taken sufficiently close to zero, then there is a {G}-equivariant map {\exp: V_a \rightarrow U_L}, defined via

\displaystyle \exp(x) = \sum_k \frac{x^k}{k!}

which converges appropriately at sufficiently small {x}. (Proof omitted, but standard. Note that {k! \rightarrow 0} in the nonarchimedean case though!) In other words, the additive and multiplicative groups are locally isomorphic. This map (for {V_a} sufficiently small) is an injection, the inverse being given by the logarithm power series. Its image is an open subgroup {V} of the units, and since the units are compact, of finit index. So we have

\displaystyle 1 = Q(V_a) = Q(V) = Q(U_L).

This proves the theorem.

2. The cohomology of {L^*}

Theorem 2

Hypotheses as above (cyclic extension of local fields), we have\displaystyle Q(L^*) = (K^*:NL^*)=n .
 

 

 

The equality {Q(L^*)=n} follows because as {G}-modules, we have {L^* = \mathbb{Z} \times U_L}, where {\mathbb{Z}} has the trivial action. Since {Q(\mathbb{Z}) = n}, we find that

\displaystyle Q(L^*) = n Q(U_L).

We then use the previous computation of {Q(U_L)}. Then, recall that (by HT 90) {Q(L^*) = (K^*:NL^*)}. 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 {(U_K: NU_L) = e}, for {e} the ramification index. In particular, every unit is a norm in the unramified case (since unramifiedness implies cyclicity).

 

Indeed, we have to show that {|H_T^{-1}(G,U_L)|=e} since the Herbrand quotient is 1. But, this is equal to

\displaystyle \left( \text{units of norm 1}: \text{quotients } \sigma b/b \text{ for } b \text{ a unit in } L \right).

Fix a generator {\sigma} for the Galois group. By Hilbert’s Theorem 90, the units of norm 1 are precisely of the form {\sigma c/c}, where {c \in L^*} (not necessarily of norm 1). Let us use the notation {A^f} to denote the image of {A} under a map {f}. Then we have

\displaystyle |H_T^{-1}(G,U_L)| = ( (L^*)^{1-\sigma}: U_L^{1-\sigma} )= ( (L^*)^{1-\sigma}: (K^*U_L)^{1-\sigma} )

Let {A_f}, similarly, denote the kernel of {f}. It is easy to show that {(A^f:B^f)(A_f: B_f) = (A:B)} for {B \subset A} a subgroup. It follows that

\displaystyle |H_T^{-1}(G,U_L)| = \frac{ (L^*: K^* U_L) }{ K^*: K^*)} = e.

This completes the computation of {H_T^{-1}} and thus the proof.

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