Let $K$ be a number field, and write $\mathbb{A}_{K,f}^\times$ for the group of finite ideles of $K$. That is $$ \mathbb{A}_{K,f}^\times = \{(u_v)_v \in \prod_{v \nmid \infty} K_v^\times : v(u_v) = 0 \text{ for all but finitely many $v$}\}, $$ where the product is over all finite places of $K$. For each ideal $I$ of $K$, define $$ U_{K,I} = \{u \in \mathbb{A}_{K,f}^\times : v(u_v) = 0 \text{ and } v(u_v - 1) \geq v(I)\text{ for all finite places $v$}\}. $$ Then I think that the ray class group of $K$ modulo $I$ is isomorphic to the double quotient $$ K^\times \backslash \mathbb{A}_{K,f}^\times / U_{K,I}, $$ but I have not been able to find a reference. Of course, there are lots of references for the analogous statement without the finiteness (indeed many authors define ray class groups that way). It's not too hard to deduce the statement I want using the definitions, but ideally I'd like a quick reference without having to write out a proof myself.
1 Answer
The ray class group is a more general object than the one considered in the original post. It is defined as $$K^\times \backslash \mathbb{A}_{K}^\times /(U_\infty U_{K,I}),$$ where $U_\infty$ is an open subgroup of $$K_\infty^\times=\prod_{v\mid\infty}K_v^\times.$$ That is, $$U_\infty=\prod_{v\mid\infty}U_v,$$ where $U_v=\mathbb{C}^\times$ for each complex place $v$ of $K$, and $U_v$ is either $\mathbb{R}^\times$ or $\mathbb{R}_{>0}$ for each real place $v$ of $K$.
The case considered in the original post is when $U_\infty=K_\infty^\times$, that is, when there are no archimedan conditions. Indeed, in this case the group in question equals $$K^\times \backslash \mathbb{A}_{K}^\times /(K_\infty^\times U_{K,I}),$$ which is isomorphic to $$K^\times \backslash \mathbb{A}_{K,f}^\times / U_{K,I}$$ by the second group isomorphism theorem. This isomorphism theorem (and its application) is so standard (in my opinion) that one does not even need to mention it.