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Let $K=\mathbb{Q}(\sqrt[8]{2},i)$ and let $F=\mathbb{Q}(i)$. I need to show that $\operatorname{Gal}(K/F)\cong Z_8$ where $Z_8$ is the cyclic group of order 8.

I have already shown that $[K:F]=8$, i.e. the dimension of $K$ over $F$ is 8. I have also already shown that $K/F$ is Galois so that $|\operatorname{Gal}(K/F)|=[K:F]=8$.

I am not sure how to go about picking the automorphisms I want. I think I want to pick all the automorphisms that fix $\mathbb{Q}(i)$. Thus I want to choose $\sigma$ so that $$\sigma(\sqrt[8]{2})=\zeta_8\sqrt[8]{2},$$ where $\zeta_8$ is a primitive $8^{\text{th}}$ root of unity. Thus $\langle \sigma\rangle=\operatorname{Gal}(K/F)\,\,$ is cyclic of order 8 and we conclude that $\operatorname{Gal}(K/F)\cong Z_8$.

Am I on the right track?

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Your general approach seems reasonable, but you need to be careful. (You might add that such a $\sigma$ is forced to exist since the Galois group acts transitively on the roots of $x^8-2$.)

Note that $\zeta_8 \not\in \Bbb Q(i)$, so that $\sigma^2(\sqrt[8]{2}) = \sigma(\zeta_8)\sigma(\sqrt[8]{2}) = \sigma(\zeta_8)\zeta_8\sqrt[8]{2}$. To actually check that $\sigma$ is order $8$, you're going to need to compute $\sigma(\zeta_8)$ in order to continue this computation. This is not too onerous, though, since $\zeta_8\in \Bbb Q(i, \sqrt{2})$.

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  • $\begingroup$ Thanks for the confirmation and advice. Your second paragraph is especially insightful for me. There was an example in my text that calculated all the elements of $\operatorname{Gal}(K/\mathbb{Q})$ and the maps they generated tracked what happened to $\zeta_8$, but I couldn't understand why that was important. Now I get it! $\endgroup$ – Laars Helenius Apr 15 '15 at 2:35

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