# idèle

Let $K$ be a number field. For each finite prime $v$ of $K$, let $\mathfrak{o}_{v}$ be the valuation ring of the completion $K_{v}$ of $K$ at $v$, and let $U_{v}$ be the group of units in $\mathfrak{o}_{v}$. Then each group $U_{v}$ is a compact open subgroup of the group of units $K_{v}^{*}$ of $K_{v}$. The idèle group $\mathbb{I}_{K}$ of $K$ is defined to be the restricted direct product of the multiplicative groups $\{K_{v}^{*}\}$ with respect to the compact open subgroups $\{U_{v}\}$, taken over all finite primes and infinite primes $v$ of $K$.

The units $K^{*}$ in $K$ embed into $\mathbb{I}_{K}$ via the diagonal embedding

 $x\mapsto\prod_{v}x_{v},$

where $x_{v}$ is the image of $x$ under the embedding $K\hookrightarrow K_{v}$ of $K$ into its completion $K_{v}$. As in the case of adèles, the group $K^{*}$ is a discrete subgroup of the group of idèles $\mathbb{I}_{K}$, but unlike the case of adèles, the quotient group $\mathbb{I}_{K}/K^{*}$ is not a compact group. It is, however, possible to define a certain subgroup of the idèles (the subgroup of norm 1 elements) which does have compact quotient under $K^{*}$.

Warning: The group $\mathbb{I}_{K}$ is a multiplicative subgroup of the ring of adèles $\mathbb{A}_{K}$, but the topology on $\mathbb{I}_{K}$ is different from the subspace topology that $\mathbb{I}_{K}$ would have as a subset of $\mathbb{A}_{K}$.

Title idèle Idele 2013-03-22 12:39:28 2013-03-22 12:39:28 djao (24) djao (24) 7 djao (24) Definition msc 11R56 Adele idèle group group of idèles