12
$\begingroup$

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.)

  • 1
    Check Zev's answer, I was gonna ask you calculate $\sigma(\zeta \sqrt[8]{2})$ .2011-04-22

2 Answers 2

7

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 $\sigma^4(\sqrt[8]{2})$.

  • 1
    No problem, glad to help :)2011-04-22
4

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 $\sqrt{-2}$.