# Intermediate subfields of $\mathbb{Q}(\sqrt[4]{2}, i)/\mathbb{Q}$

A similar question was asked here but never received an answer: Find Galois group and all intermediate fields of the extension $L=\mathbb{Q}(\sqrt[4]{2},i)\supseteq \mathbb{Q}=K$. Forgive me if posting a new question is not the correct way to ask this question.

Set $G = \text{Gal}(\mathbb{Q}(\sqrt[4]{2}, i)/\mathbb{Q})$ and let $\sigma, \tau \in G$ such that $\sigma : \sqrt[4]{2} \mapsto i\sqrt[4]{2}, i \mapsto i$ and $\tau: \sqrt[4]{2} \mapsto \sqrt[4]{2}, i \mapsto -i$. It is clear to me that $G = \langle \sigma, \tau \mid \sigma^4 = \tau^2 = 1, \sigma\tau = \tau\sigma^{-1} \rangle \cong D_8$. There are eight non trivial subgroups of $G$, three of which have index $2$ in $G$ and five of which have index $4$ in $G$. I can easily compute the fixed fields corresponding to six of these subgroups but finding the fixed fields of the remaining two is giving me quite a bit of trouble.

The two subgroups which I am battling with are $\langle \tau\sigma \rangle$ and $\langle \tau\sigma^3 \rangle$ both of which have index four in $G$ so I am looking for fixed fields of degree $4$ over $\mathbb{Q}$. I also know that these two subgroups are contained in $\langle \tau\sigma, \sigma^2 \rangle$ so the fixed fields I am looking for must contain the degree $2$ extension $\mathbb{Q}(i\sqrt{2})$ being that this is the fixed field of $\langle \tau\sigma, \sigma^2 \rangle$.

Any hints of how I may proceed to find these fixed fields?

Here's a hint: Note that $\tau\sigma(\root4\of 2) = -i\root4\of 2$ and $\tau\sigma(i)=-i$, so $\tau\sigma(-i\root4\of 2) = \root4\of 2$. What is $\tau\sigma((1-i)\root4\of 2)$?
• How can one figure out that he needs precisely these elements $(1-i)\sqrt[4]{2},(1+i)\sqrt[4]{2}$? – Rodrigo Nov 8 '17 at 20:58
• Well, you obviously have to look at the action of the particular group element — here $\tau\sigma$ — on the basis vectors for the field extension. Then you either spot an invariant or you solve a system of linear equations to find it. – Ted Shifrin Nov 8 '17 at 21:08