Let n be a positive number, $\alpha=\frac{1+\sqrt{5}}{2}$ be a root of $x^2-x-1$. Then the splitting field K of $x^{2n}-x^{n}-1$ is generated by $\alpha^{\frac{1}{n}}$ and $\zeta_{2n}$, so we may compute the order of Galois group by $[K:\Bbb Q]$= $ \frac{[\Bbb Q(\alpha^{\frac{1}{n}}) :\Bbb Q][\Bbb Q(\zeta_{2n}):\Bbb Q]}{[\Bbb Q(\alpha^{\frac{1}{n}}) \cap \Bbb Q(\zeta_{2n}) :\Bbb Q]}$. But what is the Galois group? Can we describe $Gal(\Bbb Q(\alpha^{\frac{1}{n}},\zeta_{2n})/\Bbb Q)$ using some more familiar groups like dihedral group?

  • $\begingroup$ Related to this question. $\endgroup$ – Dietrich Burde Jul 25 '17 at 9:08
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    $\begingroup$ One can decide whether or not the Galois group is the dihedral group using the criteria given in this article. For $n=2$ it is true (which is easy to see, with the polynomial $x^ 4-x^ 2-1$). $\endgroup$ – Dietrich Burde Jul 25 '17 at 9:43
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    $\begingroup$ Am I missing something or is there not the following problem with even $n$? The roots of the quadratic are $\alpha$ and $\beta=-1/\alpha$. So for even $n$ adjoining $\alpha^{1/n}$ does not automatically adjoin the $n$ the root $\beta^{1/n}=(-1)^{1/n}/\alpha^{1/n}$. For example, $\Bbb{Q}(\sqrt{\alpha},\zeta_2)$ is real, but $x^4-x^2-1$ has only two real roots. When $n$ is odd you get the $n$th roots of $-1$ as the negatives of the $n$th roots of unity. $\endgroup$ – Jyrki Lahtonen Jul 26 '17 at 10:18
  • $\begingroup$ @JyrkiLahtonen You're right, I will recorrect the mistake. $\endgroup$ – sawdada Jul 26 '17 at 12:22

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