The Idele Group

The invertible elements of any commutative topological ring $ R$ are a group $ R^*$ under multiplication. In general $ R^*$ is not a topological group if it is endowed with the subset topology because inversion need not be continuous (only multiplication and addition on $ R$ are required to be continuous). It is usual therefore to give $ R^*$ the following topology. There is an injection

$\displaystyle x\mapsto \left( x,    \frac{1}{x} \right)$ (21.1)

of $ R^*$ into the topological product $ R\times R$. We give $ R^*$ the corresponding subset topology. Then $ R^*$ with this topology is a topological group and the inclusion map $ R^*\hookrightarrow R$ is continous. To see continuity of inclusion, note that this topology is finer (has at least as many open sets) than the subset topology induced by $ R^*\subset R$, since the projection maps $ R\times R\to R$ are continuous.

Example 21.1.1   This is a ``non-example''. The inverse map on $ \mathbf{Z}_p^*$ is continuous with respect to the $ p$-adic topology. If $ a,b\in \mathbf{Z}_p^*$, then $ \left\vert a\right\vert=\left\vert b\right\vert=1$, so if $ \left\vert a-b\right\vert<\varepsilon $, then

$\displaystyle \left\vert\frac{1}{a} - \frac{1}{b}\right\vert
= \left\vert\frac{...
...a\right\vert}{\left\vert ab\right\vert} < \frac{\varepsilon }{1}=\varepsilon .

Definition 21.1.2 (Idele Group)   The $ \mathbb{I}_K$ of $ K$ is the group $ \AA _K^*$ of invertible elements of the adele ring $ \AA _K$.

We shall usually speak of $ \mathbb{I}_K$ as a subset of $ \AA _K$, and will have to distinguish between the $ \mathbb{I}_K$ and $ \AA _K$-topologies.

Example 21.1.3   For a rational prime $ p$, let $ \mathbf{x}_p\in \AA _\mathbf{Q}$ be the adele whose $ p$th component is $ p$ and whose $ v$th component, for $ v\neq p$, is $ 1$. Then $ \mathbf{x}_p \to 1$ as $ p\to\infty$ in $ \AA _\mathbf{Q}$, for the following reason. We must show that if $ U$ is a basic open set that contains the adele $ 1=\{1\}_v$, the $ \mathbf{x}_p$ for all sufficiently large $ p$ are contained in $ U$. Since $ U$ contains $ 1$ and is a basic open set, it is of the form

$\displaystyle \prod_{v\in S} U_v \times \prod_{v\not\in S} \mathbf{Z}_v,$

where $ S$ if a finite set, and the $ U_v$, for $ v\in
S$, are arbitrary open subsets of $ \mathbf{Q}_v$ that contain $ 1$. If $ q$ is a prime larger than any prime in $ S$, then $ \mathbf{x}_p$ for $ p\geq q$, is in $ U$. This proves convergence. If the inverse map were continuous on $ \mathbb{I}_K$, then the sequence of $ \mathbf{x}_p^{-1}$ would converge to $ 1^{-1}=1$. However, if $ U$ is an open set as above about $ 1$, then for sufficiently large $ p$, none of the adeles $ \mathbf{x}_p$ are contained in $ U$.

Lemma 21.1.4   The group of ideles $ \mathbb{I}_K$ is the restricted topological project of the $ K_v^*$ with respect to the units $ U_v=\O _v^*\subset K_v$, with the restricted product topology.

We omit the proof of Lemma 21.1.4, which is a matter of thinking carefully about the definitions. The main point is that inversion is continuous on $ \O _v^*$ for each $ v$. (See Example 21.1.1.)

We have seen that $ K$ is naturally embedded in $ \AA _K$, so $ K^*$ is naturally embedded in  $ \mathbb{I}_K$.

Definition 21.1.5 (Principal Ideles)   We call $ K^*$, considered as a subgroup of $ \mathbb{I}_K$, the .

Lemma 21.1.6   The principal ideles $ K^*$ are discrete as a subgroup of $ \mathbb{I}_K$.

Proof. For $ K$ is discrete in $ \AA _K$, so $ K^*$ is embedded in $ \AA _K\times
\AA _K$ by (21.1.1) as a discrete subset. (Alternatively, the subgroup topology on $ \mathbb{I}_K$ is finer than the topology coming from $ \mathbb{I}_K$ being a subset of $ \AA _K$, and $ K$ is already discrete in $ \AA _K$.) $ \qedsymbol$

Definition 21.1.7 (Content of an Idele)   The of $ \mathbf{x}=\{x_v\}_v \in \mathbb{I}_K$ is

$\displaystyle c(\mathbf{x}) = \prod_{\text{all }v} \left\vert x_v\right\vert _v \in \mathbf{R}_{>0}.

Lemma 21.1.8   The map $ \mathbf{x}\to c(\mathbf{x})$ is a continuous homomorphism of the topological group $ \mathbb{I}_K$ into $ \mathbf{R}_{>0}$, where we view $ \mathbf{R}_{>0}$ as a topological group under multiplication. If $ K$ is a number field, then $ c$ is surjective.

Proof. That the content map $ c$ satisfies the axioms of a homomorphisms follows from the multiplicative nature of the defining formula for $ c$. For continuity, suppose $ (a,b)$ is an open interval in $ \mathbf{R}_{>0}$. Suppose $ \mathbf{x}\in\mathbb{I}_K$ is such that $ c(\mathbf{x})\in (a,b)$. By considering small intervals about each non-unit component of $ \mathbf{x}$, we find an open neighborhood $ U\subset \mathbb{I}_K$ of $ \mathbf{x}$ such that $ c(U)\subset (a,b)$. It follows the $ c^{-1}((a,b))$ is open.

For surjectivity, use that each archimedean valuation is surjective, and choose an idele that is $ 1$ at all but one archimedean valuation. $ \qedsymbol$

Remark 21.1.9   Note also that the $ \mathbb{I}_K$-topology is that appropriate to a group of operators on $ \AA _K^+$: a basis of open sets is the $ S(C,U)$, where $ C,U\subset \AA _K^+$ are, respectively, $ \AA _K$-compact and $ \AA _K$-open, and $ S$ consists of the $ \mathbf{x}\in \mathbb{I}_J$ such that $ (1-\mathbf{x}) C\subset U$ and $ (1-\mathbf{x}^{-1}) C\subset U$.

Definition 21.1.10 ($ 1$-Ideles)   The subgroup $ \mathbb{I}_K^1$ of is the subgroup of ideles $ \mathbf{x}=\{x_v\}$ such that $ c(\mathbf{x})=1$. Thus $ \mathbb{I}_K^1$ is the kernel of $ c$, so we have an exact sequence

$\displaystyle 1 \to \mathbb{I}_K^1 \to \mathbb{I}_K \xrightarrow{c} \mathbf{R}_{>0}\to 1,

where the surjectivity on the right is only if $ K$ is a number field.

Lemma 21.1.11   The subset $ \mathbb{I}_K^1$ of $ \AA _K$ is closed as a subset, and the $ \AA _K$-subset topology on $ \mathbb{I}_K^1$ coincides with the $ \mathbb{I}_K$-subset topology on $ \mathbb{I}_K^1$.

Proof. Let $ \mathbf{x}\in \AA _K$ with $ \mathbf{x}\not\in\mathbb{I}_K^1$. To prove that $ \mathbb{I}_K^1$ is closed in $ \AA _K$, we find an $ \AA _K$-neighborhood $ W$ of $ \mathbf{x}$ that does not meet $ \mathbb{I}_K^1$.

1st Case. Suppose that $ \prod_v \left\vert x_v\right\vert _v<1$ (possibly $ =0$). Then there is a finite set $ S$ of $ v$ such that

  1. $ S$ contains all the $ v$ with $ \left\vert x_v\right\vert _v>1$, and
  2. $ \prod_{v\in S}\left\vert x_v\right\vert _v < 1$.
Then the set $ W$ can be defined by

$\displaystyle \left\vert w_v - x_v\right\vert _v$ $\displaystyle < \varepsilon \qquad v\in S$    
$\displaystyle \left\vert w_v\right\vert _v$ $\displaystyle \leq 1 \qquad v\not\in S$    

for sufficiently small $ \varepsilon $.

2nd Case. Suppose that $ C:=\prod_v \left\vert x_v\right\vert _v>1$. Then there is a finite set $ S$ of $ v$ such that

  1. $ S$ contains all the $ v$ with $ \left\vert x_v\right\vert _v>1$, and
  2. if $ v\not\in S$ an inequality $ \left\vert w_v\right\vert _v<1$ implies $ \left\vert w_v\right\vert _v < \frac{1}{2C}.$ (This is because for a non-archimedean valuation, the largest absolute value less than $ 1$ is $ 1/p$, where $ p$ is the residue characteristic. Also, the upper bound in Cassels's article is $ \frac{1}{2}C$ instead of $ \frac{1}{2C}$, but I think he got it wrong.)
We can choose $ \varepsilon $ so small that $ \left\vert w_v-x_v\right\vert _v<\varepsilon $ (for $ v\in
S$) implies $ 1<\prod_{v\in S} \left\vert w_v\right\vert _v < 2C.$ Then $ W$ may be defined by

$\displaystyle \left\vert w_v - x_v\right\vert _v$ $\displaystyle < \varepsilon \qquad v\in S$    
$\displaystyle \left\vert w_v\right\vert _v$ $\displaystyle \leq 1 \qquad v\not\in S.$    

This works because if $ \mathbf{w}\in W$, then either $ \left\vert w_v\right\vert _v=1$ for all $ v\not\in S$, in which case $ 1 < c(\mathbf{w}) < 2c$, so $ \mathbf{w}\not\in \mathbb{I}_K^1$, or $ \left\vert w_{v_0}\right\vert _{v_0}<1$ for some $ v_0\not\in S$, in which case

$\displaystyle c(\mathbf{w}) = \left(\prod_{v\in S} \left\vert w_v\right\vert _v...
\left\vert w_{v_0}\right\vert \cdots < 2C \cdot \frac{1}{2C} \cdots < 1,$

so again $ \mathbf{w}\not\in \mathbb{I}_K^1$.

We next show that the $ \mathbb{I}_K$- and $ \AA _K$-topologies on $ \mathbb{I}_K^1$ are the same. If $ \mathbf{x}\in \mathbb{I}_K^1$, we must show that every $ \AA _K$-neighborhood of $ \mathbf{x}$ contains an $ \AA _K$-neighborhood and vice-versa.

Let $ W\subset \mathbb{I}_K^1$ be an $ \AA _K$-neighborhood of $ \mathbf{x}$. Then it contains an $ \AA _K$-neighborhood of the type

$\displaystyle \left\vert w_v - x_v\right\vert _v$ $\displaystyle < \varepsilon \qquad v\in S$ (21.2)
$\displaystyle \left\vert w_v\right\vert _v$ $\displaystyle \leq 1 \qquad v\not\in S$ (21.3)

where $ S$ is a finite set of valuations $ v$. This contains the $ \mathbb{I}_K$-neighborhood in which $ \leq$ in (21.1.2) is replaced by $ =$.

Next let $ H\subset \mathbb{I}_K^1$ be an $ \mathbb{I}_K$-neighborhood. Then it contains an $ \mathbb{I}_K$-neighborhood of the form

$\displaystyle \left\vert w_v - x_v\right\vert _v$ $\displaystyle < \varepsilon \qquad v\in S$ (21.4)
$\displaystyle \left\vert w_v\right\vert _v$ $\displaystyle = 1 \qquad v\not\in S,$ (21.5)

where the finite set $ S$ contains at least all archimedean valuations $ v$ and all valuations $ v$ with $ \left\vert x_v\right\vert _v\neq 1$. Since $ \prod\left\vert x_v\right\vert _v=1$, we may also suppose that $ \varepsilon $ is so small that (21.1.4) implies

$\displaystyle \prod_v\left\vert w_v\right\vert _v < 2.

Then the intersection of (21.1.4) with $ \mathbb{I}_K^1$ is the same as that of (21.1.2) with $ \mathbb{I}_K^1$, i.e., (21.1.4) defines an $ \AA _K$-neighborhood. $ \qedsymbol$

By the product formula we have that $ K^*\subset \mathbb{I}_K^1$. The following result is of vital importance in class field theory.

Theorem 21.1.12   The quotient $ \mathbb{I}_K^1/K^*$ with the quotient topology is compact.

Proof. After the preceeding lemma, it is enough to find an $ \AA _K$-compact set $ W\subset \AA _K$ such that the map

$\displaystyle W\cap \mathbb{I}_K^1 \to \mathbb{I}_K^1/K^*

is surjective. We take for $ W$ the set of $ \mathbf{w}=\{w_v\}_v$ with

$\displaystyle \left\vert w_v\right\vert _v\leq \left\vert x_v\right\vert _v,

where $ \mathbf{x}=\{x_v\}_v$ is any idele of content greater than the $ C$ of Lemma 20.4.1.

Let $ \mathbf{y}=\{y_v\}_v\in \mathbb{I}_K^1$. Then the content of $ \mathbf{x}/\mathbf{y}$ equals the content of $ \mathbf{x}$, so by Lemma 20.4.1 there is an $ a\in K^*$ such that

$\displaystyle \left\vert a\right\vert _v \leq \left\vert\frac{x_v}{y_v}\right\vert _v$   all $v$$\displaystyle .$

Then $ a\mathbf{y}\in W$, as required. $ \qedsymbol$

Remark 21.1.13   The quotient $ \mathbb{I}_K^1/K^*$ is totally disconnected in the function field case. For the structure of its connected component in the number field case, see papers of Artin and Weil in the ``Proceedings of the Tokyo Symposium on Algebraic Number Theory, 1955'' (Science Council of Japan) or [AT90]. The determination of the character group of $ \mathbb{I}_K/K^*$ is global class field theory.

William Stein 2004-05-06