Let $\alpha_i,\; 1 \leq i \leq 4$ be the four roots of the polynomial. I know that the Galois group is isomorphic to $D_4$. To find the intermediate fields of the splitting field of the polynomial, I find the fixed fields of the corresponding subgroups. However I was not able to find the intermediate field corresponding to $H = \langle r_{180} \rangle$ (the rotation by $180^\circ$). Any hints or solutions would be appreciated.
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3$\begingroup$ That subgroup is the intersection of all the index two subgroups of $D_4$. Therefore the fixed field is the compositum of the quadratic intermediate fields. Have you identified those? $\endgroup$– Jyrki LahtonenMar 18, 2021 at 21:40
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$\begingroup$ @JyrkiLahtonen thank you very much,I understand your reasoning, it seems I just need to research dihedral groups a bit more. $\endgroup$– MathRMar 18, 2021 at 22:31
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$\begingroup$ I assume you have a list of zeros of the quartic. Their form yields an obvious quadratic subfield, and using their products you find another. It is difficult to give a more helpful hint because you did not share everything you know about this splitting field. Taken somewhat literally, you gave the impression that you have a description of the intermediate fields corresponding to other subgroups. Is that correct? $\endgroup$– Jyrki LahtonenMar 19, 2021 at 4:57
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$\begingroup$ @JyrkiLahtonen yes I have a list of all intermediate fields and subgroups, I just could not figure out, how to use galois correspondence to arrive at the fixed field for the rotation by 180 degrees. Your initial comment has already explained how to. $\endgroup$– MathRMar 19, 2021 at 14:54
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1 Answer
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Hint: $x^4 + 2 x^2 -2 = (x^2+1)^2 - 3.$
answered Mar 18, 2021 at 21:46
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$\begingroup$ I have a few ideas about the rotation acting invariantly on the sqaured roots of the polynomial over the algebraic closure however I cant come up with concrete solution to my problem from this idea yet. $\endgroup$– MathRMar 18, 2021 at 22:52
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1$\begingroup$ @MathR from the hint, we have the polynomial splitting over the $\sqrt3$ extension as $(x^2+1+\sqrt3)(x^2+1-\sqrt3)$. $\endgroup$ Apr 18, 2021 at 18:15