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.


1 Answer 1


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 $$D_L = \frac{(8k^2)^{\phi(k)}}{\displaystyle \prod_{p \mid \phi(k)} p^{2\phi(k)/(p-1)}}. $$

  • $\begingroup$ 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

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