Computing subfields of $Q(\zeta_{12})$

Question

I computed that Gal($Q(\zeta_{12})$/Q) $\cong Z_2 \times Z_2$, and the elements are of the form $ \sigma_a(\zeta_{12}) = \zeta_{12}^{a} $, where a is 1,5,7 or 11. We know that the Klein four group has 3 non-trivial subgroups, so now by Galois correspondence I just have to find the fixed fields of the group generated by each automorphism. By brute forcing I managed to find that two fixed fields are $Q(\zeta_{12}^{3})$ and $Q(\zeta_{12}^4)$ How do I find the last fixed field, since brute force doesn't appear to work in this case? Additionally, shouldn't all the fixed fields be degree 2 extensions by Galois correspondence? Why is the second extension above degree 3?

Answer 1

As $\zeta^3=i$, and $\zeta^4=\frac12(-1+\sqrt{-3})$, you have two quadratic extensions. One of the automorphisms of $\Bbb Q(\zeta)$ is complex conjugation, taking $\zeta$ to $\zeta^{-1}$. Its fixed field contains $\zeta+\zeta^{-1}=\sqrt3$ and so equals $\Bbb Q(\sqrt3)$.