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?
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The second extension has degree $2$, not $3$. – Angina Seng Dec 09 '19 at 18:35
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$\mathbb{Q}(\zeta_{12})=\mathbb{Q}(\sqrt{3},i)$. – 19021605 Jun 30 '25 at 18:17
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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 $\Bbb Q(\sqrt3)$.
Angina Seng
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1If I have any such extension $Q(\zeta_{a})$, does every other automorphism beside conjugation have fixed field of the form $Q(\zeta_{a}^b)$? I guess what I'm asking is, in general, is there an easier way of finding fixed fields besides just trying every possibility? What if a is very large? – Victoria Dec 09 '19 at 18:39
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Nevermind - I ended up solving this myself. If $\sigma$ has order n, simply consider the following element $\zeta + \sigma(\zeta) + \sigma^2(\zeta) ... \sigma^n(\zeta)$, which will be fixed when we apply sigma one more time to it. – Victoria Dec 09 '19 at 21:30