Let be a finite set of places of a number field , containing the archimedean ones. Suppose is a cyclic extension with Galois group . Then, if keeps invariant, keeps the group of -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, is isomorphic to a lattice in , though—this is the unit theorem. This isomorphism is by the log map, and it is even a -isomorphism if is given an action on coming from the permutation of (i.e. the permutation representation). This lattice is of maximal rank in the -invariant hyperplane .

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

**1. The cohomology of a lattice **

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

Proposition 1Let be two lattices in of maximal rank. Then .

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 -modules , and -modules defined similarly. Moreover, we have

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

Now are isomorphic to each other (and to !). So fix bases of . We can construct a polynomial function on , namely the determinant, which does not always vanish. In particular, it cannot always vanish on the dense subspace . So, there is a -isomorphism of that sends to , where both are regarded as subspaces of . By multiplying by a highly divisible integer, we may assume that the image of is contained in . Then because is of finite index in .

**2. The main formula **

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

Now recall that we have the log map , where the last space has the permutation representation of on it. is the -invariant space where the sum of the coordinates is zero.

Let the image be the lattice ; the map has finite kernel, so by the properties of the Herbrand quotient we need only compute . Then the lattice , for the vector of coordinate 1 everywhere, is of full rank in , in view of the unit theorem. Also, fixes , so

But is of full rank, so we can replace the computation of with of another lattice of full rank. For this, we choose the standard basis vectors for , i.e. the basis for the permutation representation. The lattice has full rank, so . However, we can compute the Herbrand quotient directly.

Let be the set of -valuations that prolongs. As a -module, is isomorphic to . The Herbrand quotient is multiplicative, so we are reduced to computing for each . But we can use the semilocal theory: is isomorphic to for prolonging fixed and the inertia group. In particular, by Shapiro’s lemma,

If we put all this together, we find:

Theorem 2For a finite cyclic extension and a set of places as above,

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