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…)

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…)