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.