How to find the fixed element for Galois group?

Question

This is extension of my question of the link How to find the fixed field for Galois group?

(I've already solved the question (1))

Question) Duplicated

Let be $K$ the finite splitting field of $f(x) (\in \Bbb Q[x])$ over the field, $\Bbb Q$(rational number set)

And say $E_H$ is a fixed field of $H\subset \operatorname{Gal}(K/Q) $.

Find the fixed field $E_H$

$f(x) = x^8 +1$, $H= \{ \sigma_1, \sigma_7 ,\sigma_9, \sigma_{15 } \}$

with$\sigma_n (\omega) = \omega \to \omega^n $ for $\omega = e^{{2\pi i} \over 16} $ and $gcd(n,16)=1$

I'm already knew the $\operatorname{Gal}(K/Q) \simeq Z_2 \times Z_4$

Also the $E_H \simeq Z_2 \times Z_2$

But I have a problem that finding the exact fixed element and field.

(I.E. All I have to do is the fixed element for finding the fixed field )

I can't figure out the how to take it. Cause the calculating process really complicated.

How to find the fixed element?

p.s.) Here is my process(This method is really complicated.)

First) Express the field's element, $x$ by the linear combination of the basis

Second) Find the $x$ $s.t. \sigma (x)=x$ by comparing coefficent for $\sigma \in \operatorname{Gal}(K/Q) $

Answer 1

A canonical basis for $K$ is $B = \{1, \omega, \omega^2,\ldots,\omega^7 \}$. Consider the map $\mathcal{S}: K \rightarrow E_H: u \mapsto \sum_{h \in H}{h(u)}$. $E_H$ is then generated by the vectors $\mathcal{S}(B)$. In this case they are $\{ 4, 0, 2\omega^2-2\omega^6, 0, 0, 0, -2\omega^2+2\omega^6, 0 \}$ and all of them are fixed by $H$. To extract an independent set of them, in order to obtain a basis for $E_H$, one can choose for example the set $\{1,\omega^2-\omega^6\}$.

Answer 2

For $K/\Bbb{Q}$ Galois, once you have a normal basis for $K/\Bbb{Q}$, that is some $a \in K$ such that $$K = \sum_{\sigma \in Gal(K/Q)} \sigma(a) \Bbb{Q}$$ then since the trace $Tr_{K/K^H}(b)= \sum_{h \in H} h(b)$ is surjective $K \to K^H$ $$K^H = \sum_{\sigma \in Gal(K/Q)} Tr_{K/K^H}(\sigma(a))\Bbb{Q}= \sum_{\sigma \in Gal(K/Q)/ H} Tr_{K/K^H}(\sigma(a))\Bbb{Q} = \sum_{\sigma \in Gal(K/Q)/ H} (\sum_{h \in H} h\sigma_j(a)) \Bbb{Q}$$ If instead you start from a primitive element $K = \Bbb{Q}(c)$ then $K = \sum_{j=1}^n c^j \Bbb{Q}$ so that $$K^H = \sum_{j=1}^n Tr_{K/K^H}(c^j) \Bbb{Q}=\Bbb{Q}(Tr_{K/K^H}(c),\ldots, Tr_{K/K^H}(c^n))$$