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Let $K = \mathbb{Q}(\sqrt[4]{2},i)$. Am I correct to say that $K$ has a 8-th primitive root: $\zeta_8 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$?

The 8-th cyclotomic polynomial is $\Omega_8 = X^4+1$ (is irreducible over $\mathbb{Q}$). The number of primitive 8-th roots of unity is given by the Euler $\varphi$ function: $\varphi(8) = \varphi(2^3) = 2^{3-1}(2-1) = 4$.

I found that $\text{Gal}(K/\mathbb{Q}) = \text{D}_4$, and $D_4$ has 5 subgroups of order 2 (by the Galois correspondence these give intermediate fields of order 4).

Two generators for the intermediate fields of order 4 are: $\sqrt[4]{2}$ and $\zeta_8$ but I'm having trouble to find the other 3.

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    $\begingroup$ How about $\Bbb Q(\zeta_8\sqrt[4]2)\supset\Bbb Q$ ? And how about some compositums of quadratic extensions of $\Bbb Q$ ? $\endgroup$ – Lubin May 30 '18 at 21:29
  • $\begingroup$ My answer here might be helpful $\endgroup$ – Stefan4024 May 30 '18 at 22:55
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I think the most systematic way to approach this is to write down the automorphisms generating each of these order-five subgroups, and find the intermediate fields fixed by each of these subgroups. That way, you can be absolutely certain that the intermediate fields you find will be distinct, and cover all the possibilities.

So let's establish some notation. ${\rm Gal}(K/ \mathbb Q) \cong D_4$ is generated by the automorphisms: $$ \sigma: \sqrt[4]{2}\mapsto i\sqrt[4]{2}, \ i\mapsto i$$

$$ \iota : \sqrt[4]{2}\mapsto \sqrt[4]{2}, \ i\mapsto -i.$$

It's also helpful to work out how $\zeta_8$ transforms under $\sigma$ and $\iota$, and I'll leave that to you.

The order two subgroups are $\left< e, \sigma^2 \right> $, $\left< e, \iota\right>$, $\left< e, \iota \sigma \right>$, $\left< e, \iota \sigma^2 \right>$ and $\left< e, \iota \sigma^3 \right>$, and you need to find the subfields fixed by each.

Can you see which order-two subgroup fixes $\mathbb Q(\zeta_8)$? What about $\mathbb Q(\sqrt[4]{2})$? And what about $\mathbb Q(\zeta_8 \sqrt[4]{2}$) (Lubin's example)? Can you identify and extend the pattern in the last two examples to come up with generators for the fixed fields for the remaining two subgroups?

Finally, it's good to appreciate that a subfield can be written in more than one way, and this can be confusing. For example, $\mathbb Q(i, \sqrt{2})$ is an extension of order four over $\mathbb Q$, so it must correspond to one of the five items in the list. Which one is it?

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