Set $f(X) = X^4 − 6X^2 − 2$ and denote by $K$ the splitting field of $f$ over $\mathbb{Q}$. I am interested in finding the subgroups of $\operatorname{Gal}(K : Q)$ and hence finding all subfields of $K$.

I have already found the roots of $f$ to be $\pm\alpha$ and $\pm\beta$, where $\alpha=\sqrt{3+\sqrt{11}}$ and $\beta=\sqrt{3-\sqrt{11}}$, and $K=\mathbb{Q}(\alpha,\alpha\beta)$. The extension $K:\mathbb{Q}$ is then Galois, non abelian with order 8. I just don't know how to do the last step to find these subgroups and subfields.

  • 1
    $\begingroup$ Computing the discriminant should help narrow down the Galois group. $\endgroup$
    – take008
    Nov 22, 2019 at 14:11

1 Answer 1


The Galois group contains the following three maps: $$\begin{align} \alpha\mapsto - \alpha&, \beta\mapsto \beta\\ \alpha\mapsto \alpha&, \beta\mapsto - \beta\\ \alpha\mapsto\beta&, \beta\mapsto\alpha \end{align} $$ Call these three maps $n_\alpha, n_\beta$ and $s$ ("Negate $\alpha$", "Negate $\beta$", and "Swap") respectively. They all have oder $2$, but $sn_\alpha$ and $sn_\beta$ have order $4$. Also, $n_\alpha$ and $n_\beta$ commute, but neither of them commute with $s$ (we have $n_\alpha s = sn_\beta$), although $n_\alpha n_\beta$ does commute with $s$.

This gives us the 8 elements $$ e, n_\alpha, n_\beta, n_\alpha n_\beta,\\ s, sn_\alpha, sn_\beta, sn_\alpha n_\beta $$ It doesn't take much work from here to prove that this is the dihedral group $D_4$.

And with that, we can start to look at what subgroups we have and what fixed subfields they give rise to. First the order $2$ subgroups, giving rise to degree $4$ extensions:

  • $\langle n_\alpha\rangle$ clearly fixes $\beta$, so $\Bbb Q(\beta)$ is the corresponding fixed field
  • $\langle n_\beta\rangle$ clearly fixes $\alpha$, so $\Bbb Q(\alpha)$ is the corresponding fixed field
  • $\langle n_\alpha n_\beta\rangle$ fixes $\alpha\beta = i\sqrt2$ as well as $\alpha^2 = 3 + \sqrt{11}$, so the corresponding fixed field is $\Bbb Q(\alpha\beta, \alpha^2) = \Bbb Q(i\sqrt2, \sqrt{11})$
  • $\langle s\rangle$ fixes $\alpha + \beta$, so the fixed field is $\Bbb Q(\alpha + \beta)$
  • $\langle sn_\alpha n_\beta)$ fixes $\alpha - \beta$, so the fixed field is $\Bbb Q(\alpha - \beta)$

Then the order $4$ subgroups:

  • $\langle sn_\alpha\rangle$ fixes $\alpha\beta(\alpha^2-\beta^2) = 2i\sqrt{22}$, so the fixed field is $\Bbb Q(i\sqrt{22})$
  • $\langle n_\alpha, n_\beta\rangle$ fixes $\alpha^2-\beta^2 = 2\sqrt{11}$, so the fixed field is $\Bbb Q(\sqrt{11})$
  • $\langle s, sn_\alpha n_\beta\rangle$ fixes $\alpha\beta=i\sqrt2$, so the fixed field is $\Bbb Q(i\sqrt2)$

And finally, of course, the trivial subgroup fixes all of $\Bbb Q(\alpha, \beta)$, and the whole group fixes only $\Bbb Q$.


You must log in to answer this question.