Let ${S}$ be a finite set of places of a number field ${L}$, containing the archimedean ones. Suppose ${L/k}$ is a cyclic extension with Galois group ${G}$. Then, if ${G}$ keeps ${S}$ invariant, ${G}$ keeps the group ${L_S}$ of ${S}$-units invariant. We will need to compute its Herbrand quotient in order to do the same for the idele classes (next time), and that is the purpose of this post.

Up to a finite group, ${L_S}$ is isomorphic to a lattice in ${\mathbb{R}^{|S|}}$, though—this is the unit theorem. This isomorphism is by the log map, and it is even a ${G}$-isomorphism if ${G}$ is given an action on ${\mathbb{R}^{|S|}}$ coming from the permutation of ${S}$ (i.e. the permutation representation). This lattice is of maximal rank in the ${G}$-invariant hyperplane ${W \subset \mathbb{R}^{|S|}}$.

Motivated by this, we study the Herbrand quotient on lattices next.

1. The cohomology of a lattice

Fix a cyclic group ${G}$. We will suppose given a ${G}$-lattice ${L}$, that is to say a ${\mathbb{Z}}$-free module of finite rank on which ${G}$ acts. One way to get a ${G}$-lattice is to consider a representation of ${G}$ on some real vector space ${V}$, and choose a lattice in ${V}$ that is invariant under the action of ${G}$.

Proposition 1 Let ${L, L'}$ be two lattices in ${V}$ of maximal rank. Then ${Q(L)=Q(L')}$.

The way Lang (following Artin-Tate’s Class Field Theory) approaches this result seems a little unwieldy to me. I will follow Cassels-Frohlich (actually, Atiyah-Wall in their article on group cohomology in that excellent conference volume).  We have ${\mathbb{Q}[G]}$-modules ${L_{{\mathbb Q}} = L \otimes \mathbb{Q},L'_{{\mathbb Q}}= L' \otimes \mathbb{Q}}$, and ${\mathbb{R}}$-modules ${L_{{\mathbb R}}, L'_{{\mathbb R}}}$ defined similarly. Moreover, we have $\displaystyle \mathbb{R} \otimes_{{\mathbb Q}} \hom_{\mathbb{Q}[G]}( L_{{\mathbb Q}}, L_{{\mathbb Q}}) \simeq \hom_{{\mathbb R}[G]}(L_{{\mathbb R}}, L'_{{\mathbb R}}).$

This follows by basic properties of flat base extension. Cf. Chapter 1 of Bourbaki’s Commutative Algebra, for instance.

Now ${L_{{\mathbb R}}, L'_{{\mathbb R}}}$ are isomorphic to each other (and to ${V}$!). So fix bases of ${L, L'}$. We can construct a polynomial function on ${\hom_{{\mathbb R}[G]}(L_{{\mathbb R}}, L'_{{\mathbb R}})}$, namely the determinant, which does not always vanish. In particular, it cannot always vanish on the dense subspace ${\hom_{\mathbb{Q}[G]}( L_{{\mathbb Q}}, L_{{\mathbb Q}})}$. So, there is a ${G}$-isomorphism ${f}$ of ${V}$ that sends ${L_{\mathbb Q}}$ to ${L'_{\mathbb Q}}$, where both are regarded as subspaces of ${V}$. By multiplying by a highly divisible integer, we may assume that the image of ${L}$ is contained in ${L'}$. Then ${Q(L) = Q(f(L)) = Q(L')}$ because ${f(L)}$ is of finite index in ${L'}$.

2. The main formula

Now there is a bit of confusion since ${L}$ is a field, yet I have used ${L}$ in the last section for a lattice. I will no longer do that. Now, ${L}$ denotes just the field, and a lattice will be denoted by ${M}$. We are now ready to resume our problem as in the beginning of this post. Let ${n=[L:k]}$.

Now recall that we have the log map ${h: L_S \rightarrow W \subset \mathbb{R}^S}$, where the last space has the permutation representation of ${G}$ on it. ${W}$ is the ${G}$-invariant space where the sum of the coordinates is zero.

Let the image ${h(L_S)}$ be the lattice ${M}$; the map ${h}$ has finite kernel, so by the properties of the Herbrand quotient we need only compute ${Q(h(L_S))}$. Then the lattice ${M' = M \oplus \mathbb{Z}v}$, for ${v}$ the vector of coordinate 1 everywhere, is of full rank in ${\mathbb{R}^S}$, in view of the unit theorem. Also, ${G}$ fixes ${v}$, so ${Q(M) = \frac{1}{n} Q(M')}$

But ${M'}$ is of full rank, so we can replace the computation of ${Q(M')}$ with ${Q}$ of another lattice of full rank. For this, we choose the standard basis vectors ${(e_w)_{w \in S}}$ for ${\mathbb{R}^S}$, i.e. the basis for the permutation representation. The lattice ${M'' = \bigoplus \mathbb{Z} e_w}$ has full rank, so ${Q(M')=Q(M'')}$. However, we can compute the Herbrand quotient directly.

Let ${T}$ be the set of ${k}$-valuations that ${S}$ prolongs. As a ${G}$-module, ${M''}$ is isomorphic to ${\bigoplus_{v \in T} \bigoplus_{w | v} \mathbb{Z} e_w}$. The Herbrand quotient is multiplicative, so we are reduced to computing ${Q\left(\bigoplus_{w | v} \mathbb{Z} e_w\right)}$ for each ${v \in T}$. But we can use the semilocal theory: ${ \left(\bigoplus_{w | v} \mathbb{Z} e_w\right)}$ is isomorphic to ${\mathrm{Ind}_{G_{w_0}}^{G} (\mathbb{Z})}$ for ${w_0 \in S}$ prolonging ${v}$ fixed and ${G_{w_0}}$ the inertia group. In particular, by Shapiro’s lemma, $\displaystyle Q\left(\bigoplus_{w | v} \mathbb{Z} e_w\right) = n_v = [L_{w_0}: k_v].$

If we put all this together, we find:

Theorem 2 For a finite cyclic extension ${L/k}$ and a set ${S}$ of places as above, $\displaystyle Q(L_S) = \frac{1}{n} \prod_{v \in T} n_v.$