So, it turns out there’s another way to prove the second inequality, due to Chevalley in 1940. It’s purely arithmetical, where “arithmetic” is allowed to include cohomology and ideles. But the point is that no analysis is used, which was apparently seen as good for presumably the same reasons that the standard proof of the prime number theorem is occasionally shunned. I’m not going into the proof so much for the sake of number-theory triumphalism but rather because I can do it more completely, and because the ideas will resurface when we prove the existence theorem. Anyhow, the proof is somewhat involved, and I am going to split it into steps. The goal, remember, is to prove that if ${L/k}$ is a finite abelian extension of degree ${n}$, then

$\displaystyle (J_k: k^* NJ_L) \leq n.$

Here is an outline of the proof:

1. Technical abstract nonsense: Reduce to the case of ${L/k}$ cyclic of a prime degree ${p}$ and ${k}$ containing the ${p}$-th roots of unity

2. Explicitly construct a group ${E \subset J_k}$ and prove that ${NJ_L \supset E}$

3. Compute the index ${(J_k: k^* E)}$. The whole proof is too long for one blog post, so I will do step 1 (as well as some preliminary index computations—yes, these are quite fun—today). (more…)

(Argh. So, the spacing isn’t working as well as I would like on the post and it reads non-ideally (sorry). So I’ve also included a PDF of the post if it makes things better.  -AM)

So, we have defined this thing called the Artin map on the ideals prime to some set of primes. But we really care about the ideles. There has to be some way to relate ideals and ideles. In this post, we give a translation guide between the idealic and ideleic framework. In the good ol’ days, one apparently developed class field theory using only ideal theory, but now the language of the ideles is convenient too (and as we saw, the ideles lend themselves very nicely to computing Herbrand quotients).  But they are not as good for the Artin map, unless one already has local class field theory. We don’t—we could if we developed a lot of cohomological machinery and some delightful pieces of abstract nonsense—but that’s not what we’re going to do (at least not until I manage to muster some understanding of said machinery).

1. Some subgroups of the ideles

Fix a number field ${k}$. Let’s first look at the open subgroups of ${J_k}$. For this, we determine a basis of open subgroups in ${k_v}$ when ${v}$ is a place. When ${v}$ is real, ${k_v^+}$ will do. When ${v}$ is complex, ${k_v^*}$ (the full thing) is the smallest it gets. When ${v}$ is ${\mathfrak{p}}$-adic, we can use the subgroups ${U_i = 1 + \mathfrak{p}^i}$. Motivated by this, we define the notion of a cycle ${\mathfrak{c}}$: by this we mean a formal product of an ideal ${\mathfrak{a}}$ and real places ${v_1, \dots, v_l}$ induced by real embeddings ${\sigma_1, \dots, \sigma_l: k \rightarrow \mathbb{R}}$. Say that an idele ${(x_v)_v}$ is congruent to 1 modulo ${\mathfrak{c}}$ if ${x_{\mathfrak{p}} \equiv 1 \mod \mathfrak{p}^{ \mathrm{ord}_{\mathfrak{p}}(\mathfrak{a} )}}$ for all primes ${\mathfrak{p} \mid \mathfrak{a}}$ and ${x_{v_i} >0}$ for ${1 \leq i \leq l}$. We have subgroups ${J_{\mathfrak{c}}}$ consisting of ideles congruent to 1 modulo ${\mathfrak{c}}$. Note that ${k^* J_{\mathfrak{c}} = J_k}$ in view of the approximation theorem. We define the subgroup ${U(\mathfrak{c}) \subset J_{\mathfrak{c}}}$ consisting of ideles that are congruent to 1 modulo ${\mathfrak{c}}$ and units everywhere. Fix a finite Galois extension ${M/k}$. If ${\mathfrak{c}}$ is large enough (e.g. contains the ramified primes and to a high enough power), then ${U(\mathfrak{c})}$ consists of norms—this is because any unit is a local norm, and any idele in ${Y(\mathfrak{c})}$ is very close to 1 (or positive) at the ramified primes. These in fact form a basis of open subgroups of ${J_k}$. (more…)

Class field theory is about the abelian extensions of a number field ${K}$. Actually, this is strictly speaking global class field theory (there is an analog for abelian extensions of local fields), and there is a similar theory for function fields of transcendence degree 1 over finite fields, but we shall not deal with it.

Let us, however, consider the situation for local fields—which we will later investigate more—as follows. Suppose ${k}$ is a local field and ${L}$ an unramified extension. Then the Galois group ${L/k}$ is isomorphic to the Galois group of the residue field extension, i.e. is cyclic of order ${f}$ and generated by the Frobenius. But I claim that the group ${k^*/NL^*}$ is the same. Indeed, ${NU_L = U_K}$ by a basic theorem about local fields that we will prove using abstract nonsense later (but can also be easily proved using successive approximation and facts about finite fields). So ${k^*/NL^*}$ is cyclic, generated by a uniformizer of ${k}$, which has order ${f}$ in this group. Thus we get an isomorphism

$\displaystyle k^*/NL^* \simeq G(L/k)$

sending a uniformizer to the Frobenius. (more…)

As usual, let ${K}$ be a global field. Now we do the same thing that we did last time, but for the ideles.

1. Ideles

First of all, we have to define the ideles. These are only a group, and are defined as the restricted direct product

$\displaystyle J_K = \prod'_v K_v^*$

relative to the unit subgroups ${U_v}$ of ${v}$-units (which are defined to be ${K_v^*}$ if ${v}$ is archimedean). In other words, an idele ${(x_v)_v}$ is required to satisfy ${|x_v|=1}$ for almost all ${v}$.

If ${S}$ is a finite set of places containing the archimedean ones, we can define the subset ${J^S_K = \prod_{v \in S} K_v \times \prod_{v \notin S} U_v}$; this has the product topology and is an open subgroup of ${J_K}$. These are called the ${S}$-ideles. As we will see, they form an extremely useful filtration on the whole idele group.

Dangerous bend: Note incidentally that while the ideles are a subset of the adeles, the induced topology on ${J_K}$ is not the ${J_K}$-topology. For instance, take ${K=\mathbb{Q}}$. Consider the sequence ${x^{(n)}}$ of ideles where ${x^{(n)}}$ is ${p_n}$ at ${v_{p_n}}$ (where ${p_n}$ is the ${n}$-th prime) and 1 everywhere else. Then ${x^{(n)} \rightarrow 0 \in \mathbf{A}_{\mathbb{Q}}}$ but not in ${J_{\mathbb{Q}}}$.

However, we still do have a canonical “diagonal” embedding ${K^* \rightarrow J_K}$, since any nonzero element of ${K}$ is a unit almost everywhere. This is analogous to the embedding ${K \rightarrow \mathbf{A}_K}$. (more…)