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 $\Bbb Q(\sqrt{-2}\sqrt3)$.

  • $\begingroup$ 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 $\eta'\notin \mathbb{Q}(\sqrt{-2}\sqrt{3})$? $\endgroup$ – The Coding Wombat Apr 7 at 17:59
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    $\begingroup$ @TheCodingWombat As $8/4=2$, one is seeking a quadratic extension... $\endgroup$ – Lord Shark the Unknown Apr 7 at 18:06
  • $\begingroup$ What are the 8 and 4 here? I guess $[L:\mathbb{Q}]=8$ and $4$ the size of the subgroup? $\endgroup$ – The Coding Wombat Apr 7 at 18:36
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    $\begingroup$ @TheCodingWombat Indeed $\endgroup$ – Lord Shark the Unknown Apr 7 at 19:02

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