# Group of finite ideles

Let $K$ be a number field. Let $\mathbb{I}_f$ denote the group of finite ideles, and let $\phi: K^{\times} \rightarrow \mathbb{I}_f$ be the diagonal embedding. On page 167 of his notes on Class Field Theory, Milne states the following result without proof:

"The induced topology on $K^{\times}$ has the following description: $U_K = \mathcal{O}_K^{\times}$ is open, and a fundamental system of neighborhoods of $1$ is formed by the subgroups of $U_K$ of finite index."

Can someone point me to a source where this is proved, or outline an argument for a proof of this theorem if it is not too much trouble.

-
 In case you didn't know about it, this profinite topology on ${\mathcal O}_K^\times$ can be described by congruence conditions. That is, any subgroup of finite index in the unit group contains a subgroup of the form $\{u : u \equiv 1 \bmod \mathfrak a\}$ for some (nonzero) ideal $\mathfrak a$ in ${\mathcal O}_K$, so your subgroup can be described by congruence conditions (whichever cosets mod $\mathfrak a$ happen to fill up your subgroup). This is due to Chevalley. See www.math.uconn.edu/~kconrad/blurbs/gradnumthy/chevalleyunit.pdf. – KCd May 14 '12 at 9:02

For the first claim: The subset of the finite ideles given by $\prod_v \mathcal{O}_{K_v}^\times$ is open by the definition of the topology on the restricted direct product, and an element in $K^\times$ is in this subset iff it is a unit in each completion, which is true iff it's in $\mathcal{O}^\times$.
The second claim is just a generalization of this first argument. A basis of open neighborhoods of 1 in the finite ideles is given by picking any open subgroup you choose of $K_v^\times$ in finitely many of the $v$-spots and picking $\mathcal{O}_v^\times$ in all the rest of them. If we just want a basis of open subsets, we may as well only look at really small ones, so we can assume that in the finitely many $v$-spots where we didn't pick $\mathcal{O}_v^\times$, we picked some small open subset of $\mathcal{O}_v^\times$. Note that subgroups of $\mathcal{O}_v^\times$ are all finite index and generated by a power of the uniformizer. Note also that all these open subsets of the ideles are subsets of the one we discussed in checking the first claim.
What happens when we intersect such a neighborhood with the diagonal image of $K^\times$? We get the subgroup of elements in $\mathcal{O}_K^\times$ which are in all the open subgroups of $\mathcal{O}_v^\times$ that we picked at the special places $v$. The index is just the product of the indices of those subgroups, so that's a finite index subgroup. Conversely, any finite index subgroup of $\mathcal{O}_K^\times$ is determined by the index of its image in each completion.