Let $\alpha = \sqrt{2 + \sqrt{5}}, \beta = \imath\sqrt{\sqrt{5} - 2}$. The Galois group of $\mathbb{Q}(\alpha, \beta):\mathbb{Q}$ is isomorphic to $D_4$. Find subfield corresponding to subgroup (cyclic subgroup of order $4$) given by permutations $\{id, (1324), (12)(34), (1423)\}$, where $1$ represents $\alpha, 2$ represents $-\alpha, 3$ repr. $\beta, 4$ repr. $-\beta$. I was able to find subfields corresponding for all the other subgroups, but I'm stuck here.
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$\begingroup$ if it helps, I could list all subfields of the other subgroups $\endgroup$– Pan MiroslavMay 21, 2017 at 12:35
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1 Answer
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The automorphism of order $4$ maps $\alpha\to\beta\to-\alpha$. It maps $\sqrt5=\alpha^2-2$ to $\beta^2-2=-\sqrt5$ and $i=\alpha\beta$ to $-\alpha\beta=-i$. It fixes $\sqrt{-5}$. The fixed field is $\Bbb Q(\sqrt{-5})$.
answered May 21, 2017 at 12:10
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$\begingroup$ But if we consider automorphism which takes $\alpha \mapsto \beta$ and $\beta \mapsto -\alpha$, then $\alpha(-\alpha) = -(2 + \sqrt{5}) \mapsto \beta(-\beta) = \sqrt{5} - 2$, hence it is not fixed by $(1324)$, is it? $\endgroup$ May 21, 2017 at 12:18
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