5
$\begingroup$

Let $k$ be a positive even square-free integer and let $$L:=\mathbb{Q}(\zeta_k,\sqrt{2}).$$

This is the maximal abelian extension of $L$ contained in $\mathbb{Q}(\zeta_k,2^{\frac{1}{k}})$. I would like to ask whether there is an explicit formula for the discriminant of $L$ only in terms of $k$, as the one that is given here for example.

$\endgroup$

1 Answer 1

2
$\begingroup$

If $k$ is squarefree and even then $L = \mathbf{Q}(\zeta_{k/2}, \sqrt{2})$, hence $L$ is the compositum of the linearly disjoint (over $\mathbf{Q}$) fields $K_1 = \mathbf{Q}(\zeta_{k/2}) = \mathbf{Q}(\zeta_k)$ and $K_2 = \mathbf{Q}(\sqrt{2})$. Since the discriminants are relatively prime (only odd primes are ramified in $K_1$ and $K_2$ is ramified at 2), there is an explicit formula for the discriminant of the compositum, namely $D_L = (D_{K_1})^{[K_2:\mathbf{Q}]} (D_{K_2})^{[K_1:\mathbf{Q}]}$, cf. Proposition I.2.11 in Neukirch's Algebraic Number Theory.

To be more explicit, the discriminant of $L$ is $$D_L = \frac{(8k^2)^{\phi(k)}}{\displaystyle \prod_{p \mid \phi(k)} p^{2\phi(k)/(p-1)}}. $$

$\endgroup$
3
  • $\begingroup$ That is extremely great, thanks! $\endgroup$
    – Mr. No
    Mar 3, 2017 at 22:59
  • 1
    $\begingroup$ You will also need that the discriminants are coprime. $\endgroup$ Mar 4, 2017 at 11:02
  • $\begingroup$ @franz Of course, thanks. It's updated now. $\endgroup$ Mar 4, 2017 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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