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.

  • $\begingroup$ The polynomial is reducible. $\endgroup$ – i. m. soloveichik Nov 27 '12 at 17:33
  • $\begingroup$ ok...and that means what? $\endgroup$ – Frank White Nov 27 '12 at 17:58

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.

  • $\begingroup$ Oh, crap. I totally screwed this one up. Thank you. $\endgroup$ – Frank White Nov 27 '12 at 18:23

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