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.