2
$\begingroup$

Trying to find generators of Galois group of K:Q where K is the splitting field for $x^4 - 6x^2 -2$. I have the four roots (plus minus $\alpha$ and plus minus $\beta$) and K is generated by $\alpha$ and the product $\alpha\beta=i\sqrt2 $. I have found all the elements and the order, however, I am stuck on trying to find the generators of the group? My next task is to list all the subgroups and find the subfields. Any help will be greatly appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

Hint: $Gal(K/\mathbb Q)$ must be the dihedral group of order 8 from Kaplansky's Theorem. Therefore, it must be generated by some $\rho$ and $\tau$ where $\tau^2 = 1$, $\rho^4 = 1$ and $\rho\tau = \tau\rho^{-1}$.

$\endgroup$
2
  • $\begingroup$ I have τ2 that fixes alfa and sends αβ to -αβ. However, the only generator of order 4 I can think of is such that sends α to β=(i(2)^(1/2)/α) and sends αβ to 1/αβ. Can't figure out how else to get rid of the i(2)^(1/2). However, 1/αβ is not a root. $\endgroup$
    – em4
    Nov 25, 2015 at 22:40
  • $\begingroup$ Also, the action of the group on the four zeros is faithful. So the group is isomorphic to a subgroup of $S_4$. Because the order is 8, it is a Sylow 2-subgroup, and hence isomorphic to $D_4$. Nothing wrong with your approach either :-) $\endgroup$ Nov 26, 2015 at 6:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .