small side question (field theory, galois groups)

Question

why does $\mathbb{Q}(3^{1/8},\zeta)=\mathbb{Q}(3^{1/8},2^{1/2}, i)$?

I know that $\mathbb{Q}(3^{1/8},\zeta)=\mathbb{Q}(3^{1/8},3^{1/8}e^{2\pi/8},3^{1/8}e^{4\pi/8},3^{1/8}e^{6\pi/8},3^{1/8}e^{8\pi/8},3^{1/8}e^{10\pi/8},3^{1/8}e^{12\pi/8},3^{1/8}e^{14\pi/8})$ $=\mathbb{Q}(3^{1/8},3^{1/8}e^{\pi/4},3^{1/8}e^{\pi/2},3^{1/8}e^{3\pi/4},-3^{1/8},3^{1/8}e^{5\pi/4},3^{1/8}e^{3\pi/2},3^{1/8}e^{7\pi/4})$, but that's all.

Answer 1

$=\mathbb{Q}(3^{1/8},3^{1/8}e^{\pi/4},3^{1/8}e^{\pi/2},3^{1/8}e^{3\pi/4},-3^{1/8},3^{1/8}e^{5\pi/4},3^{1/8}e^{3\pi/2},3^{1/8}e^{7\pi/4})$ = $=\mathbb{Q}(3^{1/8},3^{1/8}*[\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}],3^{1/8}i,3^{1/8}*[\frac{-\sqrt2}{2}+i\frac{\sqrt2}{2}],-3^{1/8},3^{1/8}*[-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}],3^{1/8}[-i],3^{1/8}*[\frac{\sqrt2}{2}+-\frac{\sqrt2}{2}])$ $=\mathbb{Q}(3^{1/8},2^{1/2},i)$

Answer 2

It's enough to show that $\Bbb Q(\zeta) = \Bbb Q(\sqrt 2, i)$ and since $\displaystyle \zeta = \frac{\sqrt 2}{2}+ \frac {\sqrt 2}{2} i$, it's obvious that $\zeta \in \Bbb Q(\sqrt 2, i)$, so $\Bbb Q(\zeta) \subseteq \Bbb Q( \sqrt 2, i)$.

To prove the other containment, just note that $\displaystyle \zeta^{-1}=\frac{\sqrt 2}{2}- \frac {\sqrt 2}{2} i$, so $\zeta+ \zeta^{-1}= \sqrt 2 \in \Bbb Q( \zeta)$, and $\displaystyle \frac{1}{\sqrt 2} \left ( \zeta - \zeta^{-1} \right ) = i \in \Bbb Q( \zeta)$.