Let $K=\mathbb{Q}(\sqrt[8]{2},i)$ and $F=\mathbb{Q}(i\sqrt{2})$. Show that $K$ is Galois over $F$ and determine the Galois group $Gal(K/F)$.


By assuming the first part of the question, I managed to show that $Gal(K/F)\cong Q_8$. However, I couldn't do the first part. I know that it suffices to find an irreduicble polynomials in $F[x]$ that splits in $K[x]$. Any help/hint would be appreciated. Thanks in advance...

Related to this and this question.

  • 2
    $\begingroup$ You can use the criterion here. Of course $K/\Bbb Q$ is Galois, since it is the splitting field of $x^8-2$. $\endgroup$ Jul 31, 2023 at 12:45
  • $\begingroup$ Yeah you are right. I forgot that tower of extension argument. Thanks... $\endgroup$
    – confused
    Aug 1, 2023 at 10:15

1 Answer 1


Hint: $K = \mathbb{Q}(\sqrt[8]{2}, i) = \mathbb{Q}(\sqrt[8]{2}, \frac{1 + i}{\sqrt{2}})$ is the splitting field of $x^8 - 2$ over $\mathbb{Q}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .