We continue our quest to climb Mount Takagi-Artin.

In class field theory, it will be important to compute and keep track of the orders of groups such as ${(K^*:NL^*)}$, where ${L/K}$ is a Galois extension of local fields. A convenient piece of machinery for doing this is the Herbrand quotient, which we discuss today. I only sketch the proofs though, and a little familiarity with the Tate cohomology groups will be useful (but is not strictly necessary if one accepts the essentially combinatorial results without proof or proves them directly).

1. Definition

Let ${G}$ be a cyclic group generated by ${\sigma}$ and ${A}$ a ${G}$-module. It is well-known that the Tate cohomology groups ${H^i_T(G, A)}$ are periodic with period two and thus determined by ${H^0}$ and ${H^{-1}}$. By definition, $\displaystyle H^0(G,A) = A^G/ NA ,$

where ${A^G}$ consists of the elements of ${A}$ fixed by ${G}$, and ${N: A \rightarrow A}$ is the norm map, ${a \rightarrow \sum_g ga}$. Moreover, $\displaystyle H^{-1}(G, A) = \mathrm{ker} N/ (\sigma -1) A.$

(Normally, for ${G}$ only assumed finite, we would quotient by the sum of ${(\sigma - 1)A}$ for ${\sigma \in G}$ arbitrary, but here it is enough to do it for a generator—easy exercise.)

If both cohomology groups are finite, define the Herbrand quotient ${Q(A)}$ as $\displaystyle Q(A) = \frac{ |H_T^0(G,A)|}{|H_T^{-1}(G,A)|}.$

Why is this so useful? Well, local class field theory gives an isomorphism ${K^*/NL^* \rightarrow G(L/K)}$ for ${L/K}$ a finite abelian extension of local fields. To prove this, one needs first to compute the order of ${K^*/NL^*}$. I claim that when ${G=G(L/K)}$ is cyclic, this is in fact ${Q(L^*)}$ (where ${L^*}$ is regarded as ${G}$-module). Indeed, $\displaystyle Q(L^*) = \frac{ (K^*: NL^*)}{ |H^{-1}_T(G, L^*)|} = (K^*:NL^* )$

since ${H^{-1}_T(G, L^*) =0}$ by Hilbert’s Theorem 90! (I never proved HT 90 or really covered any of this cohomology business in much detail, but it basically states that if you have a cyclic extension of fields ${L/K}$ and ${a \in L}$ has norm 1 to ${K}$, then it can be written as ${\sigma b/b}$ for ${\sigma}$ a generator of the Galois group and ${b \in L}$. This precisely means that the cohomology group at -1 vanishes.)

So instead of computing actual orders of cohomology groups, one just has to compute the Herbrand quotient. And the Herbrand quotient has many nice properties that make its computation a lot easier, as we will see below.

2. Proporties of the Herbrand quotient

1. ${Q}$ is an Euler-Poincaré function. In other words, if ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$ is an exact sequence of ${G}$-modules, and if ${Q}$ is defined on two of ${A,B,C}$, then it is defined on all three and $\displaystyle Q(B) = Q(A) Q(C).$

This follows because there is always a long exact sequence of Tate cohomology groups associated to the short exact sequence, which becomes an exact hexagon since ${G}$ is cyclic and we have periodicity. When one has an exact hexagon of abelian groups, the orders satisfy a well-known multiplicative relation (essentially an extension of ${|A| = (A:B)|B|}$ for ${B \subset A}$ a subgroup) This is how one proves multiplicativity.

2. ${Q(A)= 1}$ if ${A}$ is finite. This is somewhat counterintuitive, but is is incredibly useful.

To prove it, we shall write some exact sequences. Define ${A_G = A/(\sigma - 1)A}$, ${A^G}$ the fixed points of ${\sigma}$ (i.e. those of ${G}$). Then there is an exact sequence of abelian groups: $\displaystyle 0 \rightarrow A^G \rightarrow A \rightarrow^{1 - \sigma} A \rightarrow A_G \rightarrow 0,$

which proves that ${|A^G| = |A_G|}$, if ${A}$ is finite. Then $\displaystyle 0 \rightarrow \mathrm{ker}(N)/(\sigma - 1)A \rightarrow A_G \rightarrow A^G \rightarrow A^G/NA \rightarrow 0$

is exact, for ${A_G \rightarrow A^G}$ the norm map ${N}$. It now follows that ${|H_T^{-1}(G,A)| = |H_T^0(G,A)|}$, since these are precisely the extreme terms of the last exact sequence, and the two middle terms have the same order.

3. If ${A =\mathbb{Z}}$ with trivial action, then ${Q(A) = |G|}$. Indeed, in this case the norm is just multiplication by ${G}$, so that ${H_T^{-1}}$ is trivial. ${A^G = \mathbb{Z}}$ and ${NA = |G| \mathbb{Z}}$, so ${H_T^0}$ has order ${|G|}$.

Next time, we’re going to use just these three properties to compute the local norm index of a cyclic extension.