Why is $\mathbb{Q}(\sqrt[8]{2},i)$ Galois over $\mathbb{Q}(i\sqrt{2})$?

Question

Question:

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 $Gal(K/F)$.

Answer:

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.

You can use the criterion here. Of course $K/\Bbb Q$ is Galois, since it is the splitting field of $x^8-2$. — Dietrich Burde, Jul 31 '23 at 12:45
Yeah you are right. I forgot that tower of extension argument. Thanks... — confused, Aug 01 '23 at 10:15

score 1 · Answer 1 · answered Jul 31 '23 at 12:41

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 $\mathbb{Q}$.

answered Jul 31 '23 at 12:41

ronno

11,390

Why is $\mathbb{Q}(\sqrt[8]{2},i)$ Galois over $\mathbb{Q}(i\sqrt{2})$?

1 Answers1