### algebraic number theory

Today, we will prove the second inequality: the norm index of the ideles is at most the degree of the field extension. We will prove this using ideles (cf. the discussion of how ideles and ideals connect to each other), and some analysis.

1. A Big Theorem

We shall use one key fact from the theory of L-series. Namely, it is that:

Theorem 1 If ${k}$ is a number field, we have

$\displaystyle \sum_{\mathfrak{p}} \mathbf{N} \mathfrak{p}^{-s} \sim \log \frac{1}{s-1} (*)$

as ${s \rightarrow 1^+}$. Here ${\mathfrak{p}}$ ranges over the primes of ${k}$. The notation ${\sim}$ means that the two differ by a bounded quantity as ${s \rightarrow 1^+}$.

This gives a qualititative expression for what the distribution of primes must kinda look like—with the aid of some Tauberian theorems, one can deduce that the number of primes of norm at most ${N}$ is asymptotically ${N/\log N}$ for ${N \rightarrow \infty}$, i.e. an analog of the standard prime number theorem. In number fields. We actually need a slight refinement thereof.

Theorem 2 More generally, if ${\chi}$ is a character of the group ${I(\mathfrak{c})/P_{\mathfrak{c}}}$, we have$\displaystyle \sum_{ \mathfrak{p} \not\mid \mathfrak{c}} \chi(\mathfrak{p}) \mathbf{N} \mathfrak{p}^{-s} \sim \log \frac{1}{s-1}$

if ${\chi \equiv 1}$, and otherwise it tends either to a finite limit or ${-\infty}$.

Instead of just stating this as a random, isolated fact, I’d like to give some sort of context.  Recall that the Riemann-zeta function was defined as ${\zeta(s)=\sum_n n^{-s}}$. There is a generalization of this to number fields, called the Dedekind zeta function. The Dedekind-zeta function is not defined by summing over ${\sum |N(\alpha)|}$ for ${\alpha}$ in the ring of integers (minus 0). Why not? Because the ring of integers is not a unique factorization domain in general, and therefore we don’t get a nice product formula. (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…)

We shall now consider a number field ${k}$ and an abelian extension ${L}$. Let ${S}$ be a finite set of primes (nonarchimedean valuations) of ${k}$ containing the ramified primes, and consider the group ${I(S)}$ of fractional ideals prime to the elements of ${S}$. This is a free abelian group on the primes not in ${S}$. We shall define a map, called the Artin map from ${I(S) \rightarrow G(L/k)}$.

1. How does this work?

Specifically, let ${\mathfrak{p} \notin S}$ be a prime in ${k}$. There is a prime ${\mathfrak{P}}$ of ${L}$ lying above it. If ${A,B}$ are the rings of integers in ${k,L}$, respectively, then we have a field extension ${A/\mathfrak{p} \rightarrow B/\mathfrak{P}}$. As is well-known, there is a surjective homomoprhism of the decomposition group ${G_{\mathfrak{P}}}$ of ${\mathfrak{P}}$ onto ${G(B/\mathfrak{P} / A/\mathfrak{p})}$ whose kernel, called the inertia group, is of degree ${e(\mathfrak{P}|\mathfrak{p})}$.

But, we know that the extension ${B/\mathfrak{P} / A/\mathfrak{p}}$ is cyclic, because these are finite fields. The Galois group is generated by a canonically determined Frobenius element which sends ${a \rightarrow a^{|A/\mathfrak{p}|}}$. We can lift this to an element ${\sigma_{\mathfrak{p}}}$ of ${G_{\mathfrak{P}}}$, still called the Frobenius element. (more…)

We are now (finally) ready to start handling the cohomology of the idele classes.  The previous few posts in this series contain important background computations of the Herbrand quotients of local fields and units, which should be read before this.  Let ${L/k}$ be a finite cyclic extension of global fields of degree ${n}$. In the following, the Herbrand quotient will always be respect to the Galois group ${G=G(L/k)}$.

Theorem 1 We have ${Q(J_L/L^*) = n}$. In particular,$\displaystyle (J_k: k^* NJ_L) \geq n.$

1. Some remarks

The point of class field theory, of course, is that there is exactly an equality in the above statement, which is induced by an isomorphism between the two groups, and which holds for an arbitrary abelian extension of number fields.

Before we prove this theorem, let’s review a little. We know that ${G}$ acts on the ideles ${J_L}$, and also on ${L^*}$ (clearly). As a result, we get an action on the idele classes ${C_L=J_L/L^*}$. There is a map ${C_k = J_k/k^* \rightarrow C_L}$; I claim that it is an injection, and the fixed points of ${G}$ in ${C_L}$ are precisely the points of ${C_k}$. This can be proved using group cohomology. We have an exact sequence ${0 \rightarrow L^* \rightarrow J_L \rightarrow C_L \rightarrow 0}$, and consequently one has a long exact sequence

$\displaystyle 0 \rightarrow H^0(G, L^*) \rightarrow H^0(G, J_L) \rightarrow H^0(G, C_L) \rightarrow H^1(G, L^8) = 0$

by Hilbert’s Theorem 90, and where ${H^0}$ is the ordinary (non-Tate) cohomology groups, that is to say just the ${G}$-stable points. Since we know that ${H^0(G, L^*)=k^*}$ and ${H^0(G, J_L)=J_k}$, we find that ${(C_L)^G = C_k}$, q.e.d.

So, anyhow, this Big Theorem today computes the Herbrand quotient ${Q(C_L)}$. It in particular implies that ${H_T^0(G, C_L) \geq n}$, and since this group is none other than

$\displaystyle C_k/NC_L = J_k/k^* NC_L$

we get the other claim of the theorem. We are reduced to computing this messy Herbrand quotient, and it will use all the tools that we have developed up to now.

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

$\displaystyle \mathbb{R} \otimes_{{\mathbb Q}} \hom_{\mathbb{Q}[G]}( L_{{\mathbb Q}}, L_{{\mathbb Q}}) \simeq \hom_{{\mathbb R}[G]}(L_{{\mathbb R}}, L'_{{\mathbb R}}).$ (more…)

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. (more…)

The following situation—namely, the cohomology of induced objects—occurs very frequently, and we will devote a post to its analysis. Let ${G}$ be a cyclic group acting on an abelian group ${A}$. Suppose we have a decomposition ${A = \bigoplus_{i \in I} A_i}$ such that any two ${A_i}$ are isomorphic and ${G}$ permutes the ${A_i}$ with each other. It turns out that the computation of the cohomology of ${A}$ can often be simplified.

Then let ${G_0}$ be the stabilizer of ${A_{i_0}}$ for some fixed ${i_0 \in I}$, i.e. ${G_0 = \{g: gA_{i_0} = A_{i_0} \}.}$ Then, we have ${A = \mathrm{Ind}_{G_0}^G A_{i_0}}$. This is what I meant about ${A}$ being induced.

I claim that

$\displaystyle \boxed{ H_T^i(G, A) \simeq H_T^i(G_0, A_{i_0}) , \quad i = -1, 0. }$

In particular, we get an equality of the Herbrand quotients ${Q(A), Q(A_{i_0})}$. (more…)