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What roots of unity can be contained in the abelian extensions of an imaginary quadratic number field $K = \mathbb{Q}(\sqrt{-d})$? In particular, I would like to know:

  1. Is $K(\zeta_n)/K$ an abelian extension for every $n$?

  2. What are the roots of unity in the ray class field of $K$ with conductor $\mathfrak{c}$?

  3. What are the roots of unity in the ring class field of the order $\mathcal{O} = \mathbb{Z} + f\mathcal{O}_K$ with conductor $f$?

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    $\begingroup$ Regarding your first question, since $Gal(K(\xi_n)/K)$ is a subgroup of $Gal(\mathbb Q(\xi_n)/\mathbb Q)$ (corresponding to the elements fixing $K$), it is always abelian (independently of whether $K$ is quadratic or not). About your second question, the conductor is giving you the posible ramification of the roots of unity, then I think you get the $c$-roots of unity, where $c$ is the generator of your conductor intersected with $\mathbb Z$ (or the norm of the ideal, you can check which one it is using class field theory). $\endgroup$
    – A. Pacetti
    Commented Jun 4, 2011 at 1:37
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    $\begingroup$ This is just elementary Galois theory. Cyclotomic extensions of number fields are always abelian and the proof is the same as over $\mathbb{Q}$ (or deduce it from the corresponding result over $\mathbb{Q}$). In your case, the ramification is also almost the same as in $\mathbb{Q}(\zeta_n)/\mathbb{Q}$, with slight differences if $K$ lies in this cyclotomic. So the standard theory of cyclotomic fields should immediately answer your last two questions. $\endgroup$
    – Alex B.
    Commented Jun 4, 2011 at 1:43
  • $\begingroup$ Thanks guys. So it is easier than I thought. I'm just learning this stuff on the fly, but I guess I should have thought it through a bit more before posting. $\endgroup$
    – Jon Yard
    Commented Jun 4, 2011 at 2:07

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Just as a minor warning: even if the conductor is $1$, there might be nontrivial roots of unity in the class field: take $K = {\mathbb q}(\sqrt{-5}\,)$ and ${\mathfrak c} = (1)$; then the ray class field is the Hilbert class field $K(\sqrt{-1})$, which contains the 4th roots of unity. The roots of unity in the Hilbert class field (i.e. for conductor $1$) lie in the genus class field and can be computed easily.

Any additional roots of unity must come from ramified extensions; a necessary condition for the $p$-th roots of unity to lie in the ray class field must be that the ry class number, which is easily computed, be divisible by $p-1$ (or $(p-1)/2$ if the genus class field contains the quadratic subfield of the $p$-th roots of unity).

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