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…)