I'm having trouble identifying the subfields fixed by the Galois group of $x^4-4$ over $\mathbb{Q}$. I have the conjugation automorphism as $\phi$ and the automorphism $\psi$ defined by $\psi(\sqrt[4]{4})=i\sqrt[4]{4}$, with the Galois group calculated as $D_8$. I'm trying to identify the subfields fixed by the subgroups of the Galois group, but I'm stuck.
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$\begingroup$ The polynomial is reducible. $\endgroup$– i. m. soloveichikNov 27, 2012 at 17:33
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$\begingroup$ ok...and that means what? $\endgroup$– Frank WhiteNov 27, 2012 at 17:58
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1 Answer
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Notice that $x^4-4=(x^2-2)(x^2+2)$ so the splitting field is actually given by $\mathbb Q(\sqrt{2},i)$. In particular the Galois group is $\mathbb Z_2 \times \mathbb Z_2$ or klein-4 not $D_8$. So the non-trivial subfields are all degree $2$ extensions. I think you can finish it from here.
answered Nov 27, 2012 at 17:58
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$\begingroup$ Oh, crap. I totally screwed this one up. Thank you. $\endgroup$ Nov 27, 2012 at 18:23