id\`{e}leIdele2002-05-22 16:51:262005-01-14 06:44:41Definitionid\`{e}le groupgroup of id\`{e}les% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\ilim}{\,\underset{\longleftarrow}{\lim}\,}Let $K$ be a number field. For each finite prime $v$ of $K$, let $\o_v$ be the valuation ring of the completion $K_v$ of $K$ at $v$, and let $U_v$ be the group of units in $\o_v$. Then each group $U_v$ is a compact open subgroup of the group of units $K_v^*$ of $K_v$. The {\em id\`ele group} $\I_K$ 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 $\I_K$ 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 ad\`eles, the group $K^*$ is a discrete subgroup of the group of id\`eles $\I_K$, but unlike the case of ad\`eles, the quotient group $\I_K/K^*$ is not a compact group. It is, however, possible to define a certain subgroup of the id\`eles (the subgroup of norm 1 elements) which does have compact quotient under $K^*$.
{\bf Warning:} The group $\I_K$ is a multiplicative subgroup of the ring of ad\`eles $\A_K$, but the topology on $\I_K$ is different from the subspace topology that $\I_K$ would have as a subset of $\A_K$.