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.

1. Investigation of {\mathbf{A}_{\mathbb{Q}}}

We want to prove compactness results about the adele ring, and it is easiest to start with the case {K=\mathbb{Q}}.

Theorem 1 {\mathbb{Q}} is a discrete subset of {\mathbf{A}_\mathbb{Q}}, and {\mathbf{A}_{\mathbb{Q}}/\mathbb{Q}}, with the quotient topology, is compact.

 

First, recall that {\mathbf{A}_{\mathbb{Q}}} is a restricted direct product

\displaystyle \mathbf{A}_\mathbb{Q} = \mathbb{R} \times \prod'_p \mathbb{Q}_p .

We will find a closed neighborhood of zero containing no other element of {\mathbb{Q}}. For this, consider

\displaystyle N = [-1/2, 1/2] \times \prod'_p \mathbb{Z}_p.

Since any nonzero {x \in \mathbb{Q}} satisfies the product formula {\prod_{v } |x|_v = 1} (where {v} ranges over all normalized absolute values of {\mathbb{Q}}), we cannot have {x \in N} or else {\prod_v |x|_v \leq \frac{1}{2}}. This proves discreteness. The proof generalizes, since the product formula, and shows that {K} is discrete in {\mathbf{A}_K} for any number field {K}.

Now, we show that any {(x_v) \in \mathbf{A}_\mathbb{Q}} can be represented as {r + c} where {r \in \mathbb{Q}} and {c \in N} where {N} is the compact set as above. Indeed, let {S} be the finite set of nonarchimedean absolute values where {x_v \notin O_v}. Then set {r_1 = \sum_{p \in S } p^{\mathrm{ord}_p(x_p)} \in \mathbb{Q}}. Then clearly

\displaystyle x - r_1 \in \mathbb{R} \times \prod'_p \mathbb{Z}_p

since {1/p \in O_q = \mathbb{Z}_q} for {q \neq p}. Also, by subtracting an integer {r_2}, we can arrange it so that {x - r_1 - r_2 \in N} (since {r_2 \in \mathbb{Z}_p, \forall p}). Then take {r = r_1+r_2}. Since {N} is compact (by Tychonoff), it follows that {\mathbf{A}_\mathbb{Q}/\mathbb{Q}} is compact as well.

2. Changing fields

We now want to prove an extension of the previous result, namely:

Theorem 2 If {K} is a number field, then {K \subset \mathbf{A}_K} is discrete and {\mathbf{A}_K/K} is compact.

 

The fact about discreteness follows from the product formula in the same way. Compactness is tricker. We will need a discussion about tensor products and completions first with which we can reduce to the case {K = \mathbb{Q}}.

Let {k} be a field with a valuation {v} (possibly archimedean) and {L} a finite separable extension.

Proposition 3 We have an isomorphism of {L}-algebras and topological rings,\displaystyle L \otimes_k k_v \simeq \bigoplus_{w|v} L_w,

where {k_v, L_w} denote completions and {w} ranges over all extensions of {v} to {L}. 

 

First, {L \otimes_k k_v} is a direct product of fields. Indeed, we can write {L= k(\alpha) = k[X]/(P(X))} for suitable {\alpha \in L, P(X) \in k[X]} by the primitive element theorem. Let {P} split in {k_v[X]} as {P=P_1\dots P_r}; there are no repeated factors by separability. Then

\displaystyle L \otimes_k k_v \simeq k_v[X]/(P(X)) \simeq \bigoplus_{i} k_v[X] /(P_i).

So {L \otimes_k k_v} is isomorphic to a direct product (i.e. direct sum) of fields { k_v[X] /(P_i)}.

I claim now that the fields {L_i= k_v[X] /(P_i)} are precisely the {L_w} for {w} prolonging {v}. Indeed, first note that {k_v[X]/(P_i)}, as a finite extension of the complete field {k_v}, has a unique extension valuation with respect to which it is complete. The composite map {L \rightarrow L \otimes_k k_v \rightarrow k_v[X]/(P_i)} induces a map of {L \rightarrow L_i} and a valuation on {L} extending {v}. {L} is dense in {L_i}, since {\alpha} gets mapped to the image of {X} in {k_v[X]/(P_i)}. So {L_i} is the completion of {L} with respect to the valuation of {L} induced on it.

What we now need to show is that the valuations induced on {L} are distinct and that every {L}-valuation prolonging {v} can be realized in this way. Start with the first task.

{L \otimes_k k_v} is a finite-dimensional space over {k_v}, so it has a canonical topology induced by a (non-unique) norm, and the topologies on the {L_i} are the subspace topologies. Also, {L} is dense in it. If the morphisms {L \rightarrow L_i, L \rightarrow L_j} induced the same valuation on {L} for different {i,j}, then we would not be able to approximate a vector in {\bigoplus L_r} whose {i}-th coordinate was 1 and whose {j}-th coordinate was zero arbitrarily closely by elements of {L}.

Suppose we have a valuation {w} prolonging {v} on {L}. Then {w} extends to a multiplicative function on {L \otimes_k k_v} by continuity. This must be the norm on one of the {L_i}, since it is not identically zero. So we can get any {w} by pulling back from some {L_i}.

Corollary 4 For a finite separable extension {L/k},\displaystyle \sum_{w|v} [L_w:k_v] = [L:k]

In the case of {v} a discrete valuation, this is just the {\sum ef = n} equality.

Corollary 5 Let {L/k} be an extension of number fields. As topological rings and {L}-algebras,\displaystyle \mathbf{A}_k \otimes_k L \simeq \mathbf{A}_L.

Here {\mathbf{A}_k \otimes L} is as a topological group {\mathbf{A}_k \oplus \dots \oplus \mathbf{A}_k}. This is the restricted direct product

\displaystyle \prod'_v k_v \otimes_k L = \prod'_v \prod_{w|v} L_w = \mathbf{A}_L.

Now, we shall prove that {\mathbf{A}_K/K} is compact for any number field {K}. As a group, this is topologically isomorphic to {\bigoplus_{[K:\mathbb{Q}]} \mathbf{A}_\mathbb{Q}/\mathbb{Q}} under the identification {K = \bigoplus \mathbb{Q}} though, and this last group is compact.