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If $K$ is an imaginary quadratic field, then one can show that the orders in $K$ are precisely the rings $\mathbb{Z}+f\mathscr{O}_K$ for $f\geq 1$, where $\mathscr{O}_K$ is the ring of integers of $K$. Fix such an order $\mathscr{O}$. Extension of ideals from $\mathscr{O}$ to $\mathscr{O}_K$ gives an isomorphism between $\mathrm{Pic}(\mathscr{O})$ and the quotient $I_K(f\mathscr{O}_K)/P_{K,\mathbb{Z}}(f\mathscr{O}_K)$, where the top group is the subgroup of fractional ideals of $\mathscr{O}_K$ generated by the primes not dividing $f\mathscr{O}_K$, and the bottom group consists the principal ideals which admit a generator $\alpha$ congruent to some integer $m$ relatively prime to $f$ mod $f\mathscr{O}_K$, i.e., $\alpha\equiv m\pmod{f\mathscr{O}_K}$. Because $P_{K,\mathbb{Z}}(f\mathscr{O}_K)$ contains the group $P_K(f\mathscr{O}_K)$ of the principal ideals with a generator $\alpha\equiv 1\pmod{f\mathscr{O}_K}$, it is a "congruence subgroup mod $f\mathscr{O}_K$" (in the language of the ideal-theoretic development of global class field theory for $K$). Thus we can associate to $P_{K,\mathbb{Z}}(f\mathscr{O}_K)$ a subfield $L$ of the ray class field $K_f$ mod $f$, such the the ideal-theoretic Artin map gives an isomorphism $I_K(f\mathscr{O}_K)/P_{K,\mathbb{Z}}(f\mathscr{O}_K)\cong\mathrm{Gal}(L/K)$.

With all that exposition out of the way, my question is: is it possible to explicitly write down an open subgroup of the idele class group $C_K$ which cuts out the subfield $L$ (the so called ring class field of conductor $f$)? I think it must be, but I've never seen it actually done. The reason I ask is because, in the literature, people frequently assert that the Galois group of the union of all $p$-power conductor ring class fields has a certain structure, with a proof consisting of the phrase "by class field theory," and I'm trying to make this explicit (for my own sanity). I imagine one could do this using the ideal language, but I'm hoping it can also be done using ideles.

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Yes, absolutely. Hint: looking only at the finite places, the group you want is the profinite completion of the unit group $\mathcal{O}^{\times}$. –  Pete L. Clark Mar 31 '12 at 2:30
    
@Pete I'm not sure I understand. The group of units of the order is finite, right? So won't it be its own profinite completion? –  Keenan Kidwell Mar 31 '12 at 14:07
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Maybe I want the unit group of the completion of $\mathcal{O}$? –  Keenan Kidwell Mar 31 '12 at 20:19
    
$@$Keenan: yes, sorry for the inaccuracy. –  Pete L. Clark Apr 2 '12 at 1:11
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