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 ${K \rightarrow \mathbf{A}_K}$.

2. Idele classes

In analogy to the whole business of adele classes modulo the ground field, we will consider the idele class group ${C_K = J_K/K^*}$. This will be super-important in class field theory. In fact, class field theory exhibits a bijection between the open finite-index subgroups of ${C_K}$ and the abelian extensions of ${K}$.

First, though, let’s prove that ${K^*}$ is discrete as a subset of ${J_K}$. Cassels-Frohlich suggest that, rather than an actual proof, a metamathematical argument is more convincing: it is impossible to imagine any natural topology on the global field ${K}$ other than the discrete one!

But, let’s give a legitimate proof. Consider the unit idele ${\mathbf{1}=(1)_v}$. We can find a small neighborhood of it of the form

$\displaystyle N = \prod_{v \ \mathrm{arch}} D_{1/2}(1) \times \prod_{v \ \mathrm{nonarch}} U_v$

and if ${x \in K^* - \{1 \}}$ belonged to ${N}$, then ${x-1}$ would not satisfy the product formula, contradiction. This completes the proof of the claim.

3. Compactness

In the adele case, we had ${\mathbf{A}_K/K}$ compact. This doesn’t work in the idele case. Indeed, first define the idele norm

$\displaystyle \left \lVert (x_v) \right \rVert = \prod_v | x_v |_v^{N_v}$

so that by the product formula, ${\left \lVert x \right \rVert = 1}$ for ${x \in K^*}$. This is evidently continuous since it is continuous on each ${J_K^S}$, and it factors to become a continuous map ${C_K \rightarrow \mathbb{R}^+}$ whose image is easily seen to be unbounded.

However, we do have:

Theorem 1 The group ${C_K^0}$ of idele classes of norm 1 is compact in the induced topology. More generally, ${J_K^{S,0}/K_S}$ of ${S}$-ideles of norm 1 modulo ${S}$-units is compact whenever ${S}$ contains the archimedean places.

We have an embedding ${J_K^{S,0}/K_S \rightarrow C_K}$ in an obvious manner, so it suffices to prove the first claim. However, it even suffices to prove that the topological group ${C_K^{t}}$ of ideles of norm ${t}$ (where ${t \in \mathbb{R}^+}$) is compact for some ${t}$. We can take ${t}$ very large if we want. This is because the groups ${C_K^0}$ and ${C_K^t}$ are topologically isomorphic, the isomorphism being given by multiplication by an idele which is ${t}$ at one archimedean place and 1 everywhere else.

Lemma 2 There is ${t}$ so large such that if ${i \in J_K}$ has norm ${t}$, then there is ${x \in K^*}$ with ${1 \leq |xi_v|_v \leq t}$ for all ${v}$.

This will be an application of the business of counting lattice points in parallelotopes, i.e. the “Riemann-Roch problem” in number fields.

Note that the family ${C_v = |i_v|_v}$ for any idele ${i}$ is a divisor ${\mathfrak{c}}$ in the sense of that post, and ${\left \lVert i \right \rVert}$ is just the norm of ${\mathfrak{c}}$. So if ${\left \lVert i \right \rVert}$ is large enough (depending only on ${K}$), say at least ${t}$, then there exists ${y \in K^*}$ with

$\displaystyle |y|_v \leq |i_v|_v, \quad \forall v$

which means that ${|y^{-1} i_v|_v \geq 1}$ for each ${v}$. In addition, since each ${|y^{-1} i_w|_w \geq 1}$,

$\displaystyle |y^{-1} i_v|_v \leq \frac{ \prod_w |y^{-1} i_w|_w }{ \prod_{w \neq v} |y^{-1} i_w|_w } \leq \left \lVert i \right \rVert.$

So if ${\left \lVert i \right \rVert = t}$, we get the lemma with ${x = y^{-1}}$.

We can finish the proof of the compactness theorem. Now consider the group

$\displaystyle G = \prod_v \{ x\in K_v: 1 \leq |x|_v \leq t \}$

with the product topology. It is evidently compact, by Tychonoff’s theorem. (Incidentally, does this mean algebraic number theory depends on the axiom of choice?) I claim that it is actually a topological subgroup of the idele group ${J_K}$. But this is because, for almost all ${v}$, the inequality ${1 \leq |x|_v \leq t}$ implies ${|x|_v = 1}$ by the way these absolute values are normalized. Then ${G}$ surjects on ${C_K^t}$, so ${C_K^t}$ is compact, proving the theorem.

4. Application I: The unit theorem

We can now finish the proof of the unit theorem (started here). Consider the continuous map ${h: J_K^{S,0} \rightarrow \mathbb{R}^S}$ sending

$\displaystyle i \rightarrow \{ N_w \log |i_v|_{v \in S} \}.$

The image ${V}$ is contained in the subspace ${W \subset \mathbb{R}}$ of vectors such that the sum of the coordinates is zero. Also, it is easy to see that ${V}$ contains a lattice ${L}$ of maximal rank in ${W}$ (and much more, actually). The image ${V_K}$ of the ${S}$-units ${K_S}$ is a lattice in this subspace, as we discussed already. Now ${V/V_K}$ is compact (as the continuous image of the compact set ${J_K^{S,0}/V_K}$. This means that ${(L+V_K)/V_K}$ is compact, but it is also discrete (${L}$ being a lattice), which means that ${V_K}$ must be of the same rank as ${L}$. In particular, ${V_K=h(K_S)}$ has maximal rank in ${W}$. Since all that was left in the prooof of the unit theorem was determining the rank of ${h(K_S)}$, we have completed the proof.

5. Application II: Finiteness of the class number

This is a big theorem:

Theorem 3 If ${K}$ is a number field, then the ideal class group of ${K}$ is finite.

Indeed, we have a map ${g}$ from ${J_K}$ (subgroup of ideles of norm 1) to the group ${F}$ of fractional ideals of ${K}$. This sends ${(i_v)_v}$ to ${\prod_{v \ \mathrm{nonarch}} \mathfrak{p}_v^{\mathrm{ord}_v i_v}}$ If we endow ${F}$ with the discrete topology, then ${g}$ is easily seen to be continuous. Moreover, ${g}$ becomes a continuous map ${J_K/K^* \rightarrow F/P}$, where ${P}$ is the subgroup of principal ideals.

But ${g}$ is surjective. In fact, since the values at archimedean places don’t affect ${g}$, it is even surjective as a map ${C_K^0 \rightarrow F/P}$. Since ${C_K^0}$ is compact, its image (i.e. the ideal class group) is both compact and discrete, hence finite.