# Subfields fixed by the Galois group of $x^4-4$

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.

• The polynomial is reducible. – i. m. soloveichik Nov 27 '12 at 17:33
• ok...and that means what? – 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.