Subfields of the splitting field of $x^4 - 2$ over $\mathbf Q$ I understand why the Galois group is isomorphic to $D_8$ let's say with 
\begin{align*}
r: 2^{\frac{1}{4}} &\mapsto 2^{\frac{1}{4}}i &&& s: 2^{\frac{1}{4}} &\mapsto 2^{\frac{1}{4}} \\
   i &\mapsto i &&&    i &\mapsto -i\\
\end{align*}
as generators. 
I don't understand how to find the fixed fields given a subgroup of the the Galois group. For example, I believe the fixed field of $\langle r^3s\rangle$ is $\mathbf Q((1 + i)2^{\frac{1}{4}}$) and the fixed field of $\langle rs \rangle$ is $\mathbf Q((1 - i)2^{\frac{1}{4}}$). However, is there an easy way I could see this without just plugging in many random elements to see if they are fixed?
 A: Let $L/\mathbb{Q}$ be a finite Galois extension and $K \subseteq L$ a subfield. Then the trace $$\mathrm{Tr}_{L/K} = \sum_{\sigma \in \mathrm{Gal}(L/K)} \sigma : L \rightarrow K$$ is surjective. Using this you can get a basis of $K/\mathbb{Q}$.
In your case there is a basis $$L = \mathrm{Span}\Big( 1,\sqrt[4]{2},\sqrt{2},\sqrt[4]{2}^3,i,\sqrt[4]{2}i,\sqrt{2}i,\sqrt[4]{2}^3 i \Big)$$ so the fixed field under $<rs> = \{\mathrm{id},rs\}$ is the image of $$\mathrm{Tr}_{L/K} = \mathrm{id} + rs : \begin{cases} 1 \mapsto 2; \\ \sqrt[4]{2} \mapsto \sqrt[4]{2} (1+i); \\ \sqrt{2} \mapsto 0; \\ \sqrt[4]{2}^3 \mapsto \sqrt[4]{2}^3 (1-i); \\ i \mapsto 0; \\ \sqrt[4]{2} i \mapsto \sqrt[4]{2} (1 +i); \\ \sqrt{2} i \mapsto 2 \sqrt{2} i; \\ \sqrt[4]{2}^3 i \mapsto \sqrt[4]{2}^3 (1-i). \end{cases}$$
and therefore $K = \mathbb{Q}(\sqrt[4]{2} (1+i)).$
A: let be $x\in K; K=\mathbb Q(\alpha,i); \alpha^4=2, i^2=-1$
$K$ splitting and separable field then $|Gal(K,\mathbb Q)|=[K,\mathbb Q]$
Galois group is $\{ id,\sigma,\sigma^2,\sigma^3,\tau, \sigma\tau,\sigma^2\tau,\sigma^3\tau\} ; \sigma: \sqrt[4] {2}\to i \sqrt[4]{2}; \tau: i\to -i $
let be $H=\{id,\sigma\tau  \}$
$id(x)=a_1+a_2\alpha+a_3\alpha^2+a_4\alpha^3+a_5i+a_6i\alpha+a_7i\alpha^2+a_8i\alpha^3$
$\sigma \tau(x)=a_1+a_6\alpha-a_3\alpha^2-a_8\alpha^3-a_5i+a_2i\alpha+a_7i\alpha^2-a_4i\alpha^3$
then $id(x)=\sigma\tau(x) \Leftrightarrow a_2=a_6, a_3=-a_3, a_4=-a_8, a_5=-a_5$
then $x\in K_H \Leftrightarrow x=a_1+a_2(1+i)+a_4(1-i)\alpha^3+a_7i\alpha^2$ then $K_H=\mathbb Q((1+i)\alpha)$
