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…)