# Prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$

I seem to have reached a contradiction. I am trying to prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$.

I could not think of a clever way to do this, so I decided to just list all the automorphisms of $\mathbb{Q}(\sqrt[8]{2}, i)$ that fix $\mathbb{Q}$ and hand-pick the ones that fix $i\sqrt{2}$. By the Fundamental Theorem of Galois Theory, those automorphisms should be a subgroup of the ones that fix $\mathbb{Q}$. I proved earlier that those automorphisms are given by $\sigma: \sqrt[8]{2} \mapsto \zeta^n\sqrt[8]{2}, i \mapsto \pm i$, where $n \in [0, 7]$ and $\zeta = e^\frac{2\pi i}{8}$.

However, I am getting too many automorphisms. One automorphism that fixes $i\sqrt{2}$ is $\sigma: \sqrt[8]{2} \mapsto \zeta\sqrt[8]{2}, i \mapsto -i$. However, this means all powers of $\sigma$ fix $i\sqrt{2}$, and I know $Q_8$ does not contain a cyclic subgroup of order $8$. What am I doing wrong?

(Please do not give me the answer. I have classmates for that.)

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$i \rightarrow i^2=-1$ cannot be an automorphism. Also, what is $\zeta^2$? – N. S. Apr 21 '11 at 23:58
My bad. I meant $i \mapsto \pm i$. I'll change that in the problem. (And $\zeta^2 = i$, but that still makes it an automorphism that fixes $i\sqrt{2}$!) – badatmath Apr 22 '11 at 0:05
Check Zev's answer, I was gonna ask you calculate $\sigma(\zeta \sqrt[8]{2})$ . – N. S. Apr 22 '11 at 0:13

Hint: The order of $\sigma$ is not 8.

Note that $\sigma(\sqrt{2})=\sigma((\sqrt[8]{2})^4)=(\sigma(\sqrt[8]{2}))^4=\zeta_8^4(\sqrt[8]{2})^4=-\sqrt{2}$.

Note that $\zeta_8=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$.

Now compute $\sigma(\zeta_8)$.

Now compute $\sigma^{2}(\sqrt[8]{2})=\sigma(\zeta_8\sqrt[8]{2})=\sigma(\zeta_8)\sigma(\sqrt[8]{2})$, and then $\sigma^4(\sqrt[8]{2})$.

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So I got $\sigma^2(\sqrt[8]{2}) = i\sqrt[8]{2}$, and $\sigma^4(\sqrt[8]{2}) = -\sqrt[8]{2}$, and $\sigma^8(\sqrt[8]{2}) = \sqrt[8]{2}$. Why isn't the order $8$? – badatmath Apr 22 '11 at 1:20
@badatmath - no, $$\sigma(\zeta_8)=\frac{1}{(-\sqrt{2})}+\frac{(-i)}{(-\sqrt{2})}=\frac{1}{\sqrt{‌​2}}+\frac{i}{\sqrt{2}}=\zeta_8^3,$$ so that $\sigma^2(\sqrt[8]{2})=\zeta_8^3\zeta_8\sqrt[8]{2}=-\sqrt[8]{2}$, and thus $\sigma^4(\sqrt[8]{2})=\sqrt[8]{2}$. The order of $\sigma$ is 4. – Zev Chonoles Apr 22 '11 at 1:24
Huh. It seems I get a different answer every time I calculate this. I must be bad at computation today. Thank you :) – badatmath Apr 22 '11 at 1:34
No problem, glad to help :) – Zev Chonoles Apr 22 '11 at 1:37

Would it be easier to notice that extension $\mathbb{Q}(\sqrt[8]{2},i)$ is equal to $\mathbb{Q}(\sqrt[8]{2},\zeta)$ which is a cyclotomic extension followed by Kummer extension? You can then work out which elements of its Galois group fix $\sqrt{-2}$.

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