algebraic number theory

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…)

We continue our quest to climb Mount Takagi-Artin.

In class field theory, it will be important to compute and keep track of the orders of groups such as {(K^*:NL^*)}, where {L/K} is a Galois extension of local fields. A convenient piece of machinery for doing this is the Herbrand quotient, which we discuss today. I only sketch the proofs though, and a little familiarity with the Tate cohomology groups will be useful (but is not strictly necessary if one accepts the essentially combinatorial results without proof or proves them directly).

1. Definition

Let {G} be a cyclic group generated by {\sigma} and {A} a {G}-module. It is well-known that the Tate cohomology groups {H^i_T(G, A)} are periodic with period two and thus determined by {H^0} and {H^{-1}}. By definition,

\displaystyle H^0(G,A) = A^G/ NA ,

where {A^G} consists of the elements of {A} fixed by {G}, and {N: A \rightarrow A} is the norm map, {a \rightarrow \sum_g ga}. Moreover,

\displaystyle H^{-1}(G, A) = \mathrm{ker} N/ (\sigma -1) A.

(Normally, for {G} only assumed finite, we would quotient by the sum of {(\sigma - 1)A} for {\sigma \in G} arbitrary, but here it is enough to do it for a generator—easy exercise.)

If both cohomology groups are finite, define the Herbrand quotient {Q(A)} as

\displaystyle Q(A) = \frac{ |H_T^0(G,A)|}{|H_T^{-1}(G,A)|}. (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…)

Adele – c’est un nom si belle. (Oops, that’s bad French, isn’t it?)

I actually will not be able to finish the proof of the unit theorem here, because I don’t get to the ideles in this post. That will come next time (there are some of the same themes as here).

Let {K} be a global field, i.e. a finite extension of either {\mathbb{Q}} or {\mathbb{F}_p(t)}. Then we can consider the set absolute values on {K}. In the number field case, these are extensions (up to a power) of the archimedean absolute value on {\mathbb{Q}} or the {p}-adic absolute values by a theorem of Ostrowski classifying absolute values on {\mathbb{Q}}. In the function field case, we need another result.

Here’s how we define the adele ring. It is the restricted direct product

\displaystyle \mathbf{A}_K := \prod'_v K_v

where restricted means that any vector {(x_v)_{v \in V} \in \mathbf{A}_K} is required to satisfy {|x_v|_v \leq 1} for almost all {v}. This becomes a topological ring if we take a basis of the form

\displaystyle \prod_{v \in S} T_v \times \prod_{v \notin S} \mathcal{O}_v

where {T_v \subset k_v} are open and {\mathcal{O}_v} is the ring of integers, and {S} is a finite set containing the archimedean places. It is clear that addition and multiplication are continuous, and that {A_K} is locally compact. For {S} finite and containing the archimedean absolute values {S_\infty}, there is a subring {\mathbf{A}_K^S = \prod_{v \in S} K_v \times \prod_{v \notin S} \mathcal{O}_v}, and {\mathbf{A}_K} is the union of these subrings.

Since any {x \in K} is contained in {\mathcal{O}_v} for almost all {v} (this is analogous to a rational function on a curve having only finitely many poles), there is an injective homomorphism {K \rightarrow \mathbf{A}_K}.

Next, we may define a Haar measure on {\mathbf{A}_K^S} by taking the product of the Haar measures {\mu_v} on {K_v}, normalized such that {\mu_v(O_v )=1} for {v \notin S_\infty}. Thus one gets a (i.e., the) Haar measure on {\mathbf{A}_K} itself. (more…)

There is another major result in algebraic number theory that we need to get to!  I have this no longer secret goal of getting to class field theory, and if it happens, this will be a key result.  The hard part of the actual proof (namely, the determination of the rank of a certain lattice) will be deferred until next time; it’s possible to do it with the tools we already have, but it is cleaner (I think) to do it once ideles have been introduced.

Following the philosophy of examples first, let us motivate things with an example. Consider the ring {\mathbb{Z}[i]} of Gaussian; as is well-known, this is the ring of integers in the quadratic field {\mathbb{Q}(i)}. To see this, suppose {a+ib, a, b \in \mathbb{Q}} is integral; then so is {a-ib}, and thus {2a,2b} are integers. Also the fact {(a+ib)(a-ib) = a^2+b^2 } must be an integer now means that neither {a,b} can be of the form {k/2} for {k} odd.

What are the units in {\mathbb{Z}[i]}? If {x} is a unit, so is {\bar{x}}, so the norm {N(x)} must be a unit in {\mathbb{Z}[i]} (and hence in {\mathbb{Z}}). So if {x=a+ib}, then {a^2+b^2=1} and {x = \pm 1} or {\pm i}. So, the units are just the roots of unity.

In general, however, the situation is more complicated. Consider {\mathbb{Z}[\sqrt{2}]}, which is again integrally closed. Then {x=a+b\sqrt{2}} is a unit if and only if its norm to {\mathbb{Q}}, i.e. {a^2 - 2b^2} is equal to {\pm 1}. Indeed, the norm {N(x)} of a unit {x} is still a unit, and since {\mathbb{Z}} is integrally closed, we find that {N(x)} is a {\mathbb{Z}}-unit. In particular, the units correspond to the solutions to the Pell equation. There are infinitely many of them.

But the situation is not hopeless. We will show that in any number field, the unit group is a direct sum of copies of {\mathbb{Z}} and the roots of unity. We will also determine the rank. (more…)

This post won’t be as cool as the title sounds. But I will prove something neat, and it will lead to neater things as time goes on (assuming I keep posting on this topic).

We will now return to algebraic number theory, following Lang’s textbook, and study the distribution of points in parallelotopes.

The setup is as follows. {K} will be a number field, and {v} an absolute value (which, by abuse of terminology, we will use interchangeable with “valuation” and “place”) extending one of the absolute values on {\mathbb{Q}} (which are always normalized in the standard way); we will write {|x|_v} for the output at {x \in K}.

Suppose {v_0} is a valuation of {\mathbb{Q}}; we write {v | v_0} if {v} extends {v_0}. Recall the following important formula from the theory of absolute values an extension fields:

\displaystyle |N^K_{\mathbb{Q}}(x)|_{v_0} = \prod_{v | v_0} |x|_v^{ [K_v: \mathbb{Q}_{v_0} ] }.

Write {N_v := [K_v: \mathbb{Q}_{v_0} ] }; these are the local degrees. From elementary algebraic number theory, we have {\sum_{v | v_0} N_v = N := [K:\mathbb{Q}].} This is essentially a version of the {\sum ef = N} formula.

The above formalism allows us to deduce an important global relation between the absolute values {|x|_v} for {x \in K}.

Theorem 1 (Product formula) If {x \neq 0},\displaystyle \prod_v |x|_v^{N_v} = 1.


The proof of this theorem starts with the case {K=\mathbb{Q}}, in which case it is an immediate consequence of unique factorization. For instance, one can argue as follows. (more…)

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