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…)