MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f$?

share|cite|improve this question
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). – A. Pacetti Jun 4 '11 at 1:37
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. – Alex B. Jun 4 '11 at 1:43
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. – Jon Yard Jun 4 '11 at 2:07
up vote 3 down vote accepted

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.