It is now time to prove the reciprocity law, the primary result in class field theory.  I know I haven’t posted on this topic in a little while, so new readers (if they don’t already know this material) may want to review the strategy of the proof and the meaning of the Artin lemma (which is useful in reducing this to the cyclotomic case).

1. The cyclic reciprocity law

Well, I’ve already stated it before multiple times, but here it is:

Theorem 1 (Reciprocity law, cyclic case) Let ${L/k}$ be a cyclic extension of number fields of degree ${n}$. Then the reciprocity law holds for ${L/k}$: there is an admissible cycle ${\mathfrak{c}}$ such that the kernel of the map ${I(\mathfrak{c}) \rightarrow G(L/k)}$ is ${P_{\mathfrak{c}} N(\mathfrak{c})}$, and the Artin map consequently induces an isomorphism$\displaystyle J_k/k^* NJ_L \simeq I(c)/P_{\mathfrak{c}} N(\mathfrak{c}) \simeq G(L/k).$

The proof of this theorem is a little sly and devious.

Recall that, for any admissible cycle ${\mathfrak{c}}$, we have

$\displaystyle (I(\mathfrak{c}): P_{\mathfrak{c}} N(\mathfrak{c})) = n$

by the conjunction of the first and second inequalities, and the Artin map ${I(\mathfrak{c}) \rightarrow G(L/k)}$ is surjective. If we prove that the kernel of the Artin map is contained in ${P_{\mathfrak{c}} N(\mathfrak{c})}$, then we’ll be done by the obvious count.

This is what we shall do. (more…)

This is the lemma we shall use in the proof of the reciprocity law, to reduce the cyclic case to the cyclotomic case:

Lemma 4 (Artin) Let ${L/k}$ be a cyclic extension of degree ${n}$ and ${\mathfrak{p}}$ a prime of ${k}$ unramified in ${L}$. Then we can find a field ${E}$, a subextension of ${L(\zeta_m)}$, with ${E \cap L = k}$ such that in the lattice of fields

we have:

1. ${\mathfrak{p}}$ splits completely in ${E/k}$

2. ${E(\zeta_m) = L(\zeta_m)}$, so that ${LE/E}$ is cyclotomic

3. ${\mathfrak{p}}$ is unramified in ${LE/k}$

Moreover, we can choose ${m}$ such that it is divisible only by arbitrarily large primes.

The proof of this will use the previous number-theory lemmas and the basic tools of Galois theory.

So, first of all, we know that ${E}$ is a subextension of some ${L(\zeta_m)}$. We don’t know what ${m}$ is, but pretend we do, and will start carrying out the proof. As we do so, we will learn more and more about what ${m}$ has to be like, and eventually choose it.
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So, the key to proving the Artin reciprocity law in general is to reduce it to the cyclotomic case (which has already been handled, cf. this). To carry out this reduction, start with a number field ${k}$ and a cyclic extension ${L}$. We will prove that there is an extension ${k'}$ of ${k}$ and ${L'}$ of ${L}$ such that we have a lattice of fields

(where lattice means ${L \cap k' = k}$, ${Lk' = L'}$) such that ${L'/k'}$ is cyclotomic and a given prime ${\mathfrak{p}}$ of ${k}$ splits completely in ${k'}$. This means that, in a sense, the Artin law for ${L/k}$ becomes reduced to that of ${L'/k'}$, at least for the prime ${\mathfrak{p}}$, in that ${( \mathfrak{p}, L'/k') = (\mathfrak{p}, L/k)}$. From this, we shall be able to deduce the Artin law for cyclic extensions, whence the general case will be a “mere” corollary.

1. A funny lemma

The thing is, though, finding the appropriate root of unity to use is not at all trivial. In fact, it requires some tricky number-theoretic reasoning, which we shall carry out in this post. The first step is a lemma in elementary number theory. Basically, we will need two large cyclic subgroups of multiplicative groups of residue classes modulo an integer, one of which is generated by a fixed integer known in advance. We describe now how to do this.

Proposition 1 Let ${a, n \in \mathbb{N} - \{1\}}$. Then there exists ${m \in \mathbb{N}}$, prime to ${a}$, such that ${a \in (\mathbb{Z}/m\mathbb{Z})^*}$ has order divisible by ${n}$. In addition, there exists ${b \in (\mathbb{Z}/m\mathbb{Z})^*}$ of order divisible by ${n}$ such that the cyclic subgroups generated by ${a,b}$ have trivial intersection.

Finally if ${N \in \mathbb{N}}$, we can assume that ${m}$ is divisible only by primes ${>N}$.
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It is now time to begin the final descent towards the Artin reciprocity law, which states that for an abelian extension ${L/k}$, there is an isomorphism

$\displaystyle J_k/k^* NJ_L \simeq G(L/k).$

We will actually prove the Artin reciprocity law in the idealic form, because we have only defined the Artin map on idelas. In particular, we will show that if ${\mathfrak{c}}$ is a suitable cycle in ${k}$, then the Artin map induces an isomorphism

$\displaystyle I(\mathfrak{c}) / P_{\mathfrak{c}} N(\mathfrak{c}) \rightarrow G(L/k).$

The proof is a bit strange; as some have said, the theorems of class field theory are true because they could not be otherwise. In fact, the approach I will take (which follows Lang’s Algebraic Number Theory, in turn following Emil Artin himself).

So, first of all, we know that there is a map ${I(\mathfrak{c}) \rightarrow G(L/k)}$ via the Artin symbol, and we know that it vanishes on ${N(\mathfrak{c})}$. It is also necessarily surjective (a consequence of the first inequality). We don’t know that it factors through ${P_{\mathfrak{c}}}$, however.

Once we prove that ${P_{\mathfrak{c}}}$ (for a suitable ${\mathfrak{c}}$) is in the kernel, then we see that the Artin map actually factors through this norm class group. By the second inequality, the norm class group has order at most that of ${G(L/k)}$, which implies that the map must be an isomorphism, since it is surjective.

In particular, we will prove that there is a conductor for the Artin symbol. If ${x}$ is sufficieintly close to 1 at a large set of primes, then the ideal ${(x)}$ has trivial Artin symbol. This is what we need to prove.

Our strategy will be as follows. We will first analyze the situation for cyclotomic fields, which is much simpler. Then we will use some number theory to reduce the general abelian case to the cyclotomic case (in a kind of similar manner as we reduced the second inequality to the Kummer case). Putting all this together will lead to the reciprocity law.
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We continue (and finish) the proof started in the previous post of the second inequality.

3. Construction of ${E}$

So, here’s the situation. We have a cyclic extension ${L/k}$ of degree ${n}$, a prime number, and ${k}$ contains the ${n}$-th roots of unity. In particular, we can write (by Kummer theory) ${L = k(D^{1/n})}$ for ${D}$ a subgroup of ${k^*}$ such that ${(k^{*n} D: k^{*n}) = n}$, in particular ${D}$ can be taken to be generated by one element ${a}$.

We are going to prove that the norm index of the ideles is at most ${n}$. Then, by the reductions made earlier, we will have proved the second inequality.

3.1. Setting the stage

Now take a huge but finite set ${S}$ of primes such that:

1. ${J_k = k^* J_S}$

2. ${a}$ is an ${S}$-unit

3. ${S}$ contains all the primes dividing ${n}$

4. ${S}$ contains the ramified primes We will now find a bigger extension of ${L}$ whose degree is a prime power. We consider the tower ${k \subset L \subset M = k(U_S^{1/n})}$ for ${U_S}$ the ${S}$-units. We have the extension ${[M:k]}$ whose degree we can easily compute; it is

$\displaystyle (k^{*n} U_S: k^{*n}) = (U_S: U_S^n) = n^{|S|}$

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…)

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…)

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…)