Galois correspondence and invariant fields of L of $f(x)=x^4-2x^2-2$

Let $$L\subset \mathbb{C}$$ be a splitting field of $$f(x)=x^4-2x^2-2\in\mathbb{Q}[x]$$ and let $$\alpha\in L$$ be a real zero of $$f(x)$$.

I want to find the invariant fields corresponding to the subgroups of $$\text{Gal}(L/\mathbb{Q})$$ of order $$4$$.

I have set $$\alpha=\sqrt{1+\sqrt{3}}$$ and have determined that the other zeros are $$-\alpha$$, $$\sqrt{-2}/\alpha$$, $$-\sqrt{-2}/\alpha$$. So $$\pm\sqrt{1\pm\sqrt{3}}$$. We also have $$|\text{Gal}(L/\mathbb{Q}|=8$$.

We create the maps

$$\sigma: \alpha \mapsto \sqrt{-2}/\alpha, \quad \sigma: \sqrt{-2}\mapsto -\sqrt{-2}$$

$$\tau: \alpha \mapsto \alpha, \quad \tau: \sqrt{-2}\mapsto -\sqrt{-2}$$.

We have that the order of $$\sigma$$ is 4, and the order of $$\tau$$ is 2, and $$\sigma\tau=\tau\sigma^3$$. Therefore it is isomorphic to the dihedral group of order 8 ($$D_4$$).

The subgroups of order 4 are $$a=\{id, \sigma, \sigma^2, \sigma^3\}, b= \{id, \tau, \sigma^2, \tau\sigma^2\}, c=\{id, \tau\sigma, \sigma^2, \tau\sigma^3\}$$.

I have found that the invariant field corresponding to $$b$$ is $$\mathbb{Q}(\alpha^2)=\mathbb{Q}(\sqrt{3})$$, and that the invariant field corresponding to $$c$$ is $$\mathbb{Q}(\sqrt{-2})$$.

I don't know how to find the invariant field of $$a$$. I have tried the following things:

$$\sigma(\alpha)=\sqrt{-2}/\alpha$$, so no

$$\sigma(\sqrt{-2})=-\sqrt{-2}$$, so no

$$\sigma(\alpha^2)=-2/\alpha^2$$, so no

Is there a better way of finding these invariant fields? I feel like trying all possibilities isn't the best way. And which elements do I have to try?

As $$a=\{1,\newcommand{\si}{\sigma}\si,\si^2,\si^3\}$$ then $$\eta=\xi+\si(\xi)+\si^2(\xi)+\si^3(\xi)$$ is an element of the fixed field of $$a$$. So just put in various possible $$\xi$$ until one gets some $$\eta\notin\Bbb Q$$.
I'll try $$\xi=\sqrt{-2}\alpha^2=\sqrt{-2}(1+\sqrt3)$$. Then $$\si(\xi)=(-\sqrt{-2})(1-\sqrt3)$$, $$\si^2(\xi)=\xi$$ etc, so that $$\eta=4\sqrt{-2}\sqrt3$$. Then the fixed field is $$\Bbb Q(\sqrt{-2}\sqrt3)$$.
• Does this confirm that $\mathbb{Q}(\sqrt{-2}\sqrt{3})$ is the whole of the invariant field? Or does one have to try again for $\eta'\notin \mathbb{Q}(\sqrt{-2}\sqrt{3})$? – The Coding Wombat Apr 7 at 17:59
• @TheCodingWombat As $8/4=2$, one is seeking a quadratic extension... – Lord Shark the Unknown Apr 7 at 18:06
• What are the 8 and 4 here? I guess $[L:\mathbb{Q}]=8$ and $4$ the size of the subgroup? – The Coding Wombat Apr 7 at 18:36