4
$\begingroup$

I was wondering what are the most efficient ways to find a cyclotomic field s.t. given $K/\mathbb{Q}$ we have $K\leq \mathbb{Q}(\zeta_n)$? For quadratic fields this is easy by just considering factors of $d$ for the adjoined square root $\sqrt{d}$. How about cubic extensions and higher?

I guess such a formula can't be solely based on the primes that ramify, because we could construct fields with discriminant of the form $p^aq^b$ for rather large $a$ and $b$, and this would require $n$ be large too.

  • 3
    The smallest such integer is the conductor, so you might google information about the conductors of abelian fields. For instance, you can find on the wikipedia page for conductors the fact that the conductor of your extension can be computed from the local conductors, and these are related to the unit groups in the completions at ramified primes. I don't know how practical that is for concrete computation, but for instance it shows that for a tamely ramified prime p, the power of p dividing the conductor will be 1.2011-05-03
  • 0
    @Barry: You can relate the conductor and discriminant of your extension using the conductor-discriminant formula. Regards,2011-05-03
  • 0
    I thought about mentioning that, but I think of that as a formula for computing the discriminant from conductors and not vice-versa. Can this viewpoint be reversed in a practical setting?2011-05-03
  • 0
    Since I can't edit the previous comment...to avoid confusion, I should have said for a tamely ramified p, the exact power of p dividing the conductor will be $p^1$2011-05-03
  • 0
    @Barry: Dear Barry, With regard to your question "Can this viewpoint be reversed ... ?", see my answer below for an indication as to how. Best wishes,2011-05-03
  • 0
    @Matt E: Ah, nice answer. Thanks.2011-05-03
  • 0
    @dstt Could you please either indicate where you think that the answers to your questions leave anything to be desired, or start accepting answers? This way, you will help building a useful resource of knowledge.2011-05-07

1 Answers 1