I'm trying to find the Galois group of $x^4-x^2-6$. I think there are 4 roots, thus I guess the Galois group is $A_4$? But I don't know in general how to solve this.


By transforming $x^2\mapsto y$ the equation becomes $y^2-y-6$. Using quadratic formula, $y=-2,y=3$ are the roots. So, $x^4-x^2-6=(x^2+2)(x^2-3)$. Thus, the roots are $\sqrt3,-\sqrt3,\sqrt{-2},-\sqrt{-2}$. Galois map should take the roots of each irreducible polynomial to itself. Thus, there are four Galois maps taking $\sqrt3,\sqrt{-2}$ to 4 different values. Square of any Galois map should be the identity as we have two irreducible quadratics. So, the Galois group is the Klein-4 group

  • $\begingroup$ You can also finish by noticing that the splitting field is a square root tower. $\endgroup$ – Emre May 20 '16 at 6:23
  • $\begingroup$ So, why we can deduce is the Klein-4 group? I see Klein 4 group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}.$ And is this Galois group solvable? $\endgroup$ – Winterfell May 20 '16 at 6:29
  • $\begingroup$ First question: there are two groups of order 4, the cyclic group and Klein-4 and the cyclic group has elements of order 4, where our does not as I proved. Second question: Yes, as it is Abelian. $\endgroup$ – Emre May 20 '16 at 6:30
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    $\begingroup$ Also the fact that every root has been written down is a pretty good indication that the Galois group is solvable. $\endgroup$ – pjs36 May 20 '16 at 6:33
  • $\begingroup$ How about $x^5+1.$ There are 5 roots. Is the Galois group $\mathbb{Z}_5$? $\endgroup$ – Winterfell May 20 '16 at 6:43

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