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For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

The adele ring is the restricted productAddition and multiplication are componentwise. If $x_v,y_v\in {\mathcal{O}}_v$ outside finite sets, then $x_v+y_v,x_v{y}_v\in {\mathcal{O}}_v$ outside their union, so these operations are closed on ${\mathbb{A}}_K$. The componentwise ring axioms make it a ring. Its restricted product topology has basic open sets ${\prod }_v{U}_v$ with each $U_v\subseteq K_v$ open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

The diagonal mapis an injective homomorphism because every completion map $K\to K_v$ is injective. The diagonal copy of $K$ is discrete in the adele ring. Choose an adelic neighbourhood $W$ of $1$ with $W\cap K=\{1\}$. By the subspace description of the idele topology, $W\cap {\mathbb{I}}_K$ is an idele neighbourhood of $1$ meeting diagonal $K^{\times }$ only at $1$. Translation proves that $K^{\times }$ is a discrete subgroup of ${\mathbb{I}}_K$.