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…)