How to find generators for the subfields of $\mathbb{Q}(\zeta_{12})$

Question

This is somewhat of a follow-up to this question: A complete picture of the lattice of subfields for a cyclotomic extension over $\mathbb{Q}$.

After reading this, I am still confused on how to find generators for the fixed fields. I know that in the case of $\zeta_p$ where $p$ is prime and $\zeta_p$ is a primitive $p^\text{th}$ root of unity, we can find generators as follows: let $H$ be a subgroup of $\text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$. Then a generator for the fixed field $\text{Fix}(H)$ is given as $$ \alpha_H=\sum_{\sigma\in H}\sigma\zeta_p. $$ In other words, we just sum over the Galois conjugates of $\zeta_p$ by the elements of $H$. This is all fine and dandy, since for example, in the case of $\mathbb{Q}(\zeta_{13})$, the Galois group is cyclic on $12$ elements with generator $\sigma:\zeta_{13}\mapsto\zeta_{13}^2$, and a generator for the fixed field corresponding to the subgroup of order $3$ is $$ \zeta_{13}+\sigma^4\zeta_{13}+\sigma^8\zeta_{13}=\zeta+\zeta^{2^4}+\zeta^{2^8}=\fbox{$\zeta+\zeta^3+\zeta^9$.} $$ My question: is there anything like this that we can do for $\mathbb{Q}(\zeta_{12})$? I have found that the Galois group of $\mathbb{Q}(\zeta_{12})$ over $\mathbb{Q}$ is $C_2\times C_2$ where $C_2$ is the cyclic group on $2$ elements. For the fixed fields, I can find the obvious generator $\zeta_{12}+\zeta_{12}^{-1}$ corresponding to a subgroup of order $2$ (complex conjugation action), but I am stuck otherwise. Can anyone give me some pointers, or explain to me what I am missing about the question I linked to above? Any feedback is much appreciated.

Is it not the case that $\text{Gal}(\mathbb{Q}(\zeta_{13})/\mathbb{Q})\cong(\mathbb{Z}/13\mathbb{Z})^\times\cong C_{12}$? — wormram, May 20 '20 at 20:40
In the present case you can use the observations that $\zeta_4$ and $\zeta_3$ are elements of $\Bbb{Q}(\zeta_{12})$. Surely you know the quadratic fields those two generate, and can then guess what the third quadratic intermediate field must be. — Jyrki Lahtonen, May 20 '20 at 20:42
I was using the case of $\zeta_{13}$ as an example of how this is easier to figure out when $p$ is prime. — wormram, May 20 '20 at 20:43
You can follow the recipe you explained for the case of $\zeta_p$ if and only if the conjugates of $\zeta_n$ form a so called normal basis. Unfortunately that happens if and only if $n$ is square-free. — Jyrki Lahtonen, May 20 '20 at 20:46
For this particular field, see also this old question or this. — Jyrki Lahtonen, May 20 '20 at 21:18

score 4 · Answer 1 · answered May 20 '20 at 20:42

Actually, $\Bbb Q(\zeta_{12})=\Bbb Q(i,\sqrt3)$ so the quadratic subfields are $\Bbb Q(i)$, $\Bbb Q(\sqrt3)$ and $\Bbb Q(i\sqrt3)$.

But suppose one didn't know that. The Galois group is $G=\{\text{id},\sigma_5,\sigma_7,\sigma_{11}\}$ where $\sigma_a(\zeta)=\zeta^a$.

The fixed field of the subgroup $H_a=\{\text{id},\sigma_a\}$ contains $\zeta+\zeta^a$ and more generally also $\zeta^k+\zeta^{ak}$ for any $k$.

The fixed field of $H_5$ contains $\zeta+\zeta^5=2i\sin(\pi/6)=i$. So this field is $\Bbb Q(i)$.

The fixed field of $H_7$ contains $\zeta+\zeta^7=0$ which is a bit useless. But it also contains $\zeta^2+\zeta^{14}=2\zeta^2=2\cos(\pi/3)+2i\sin(\pi/3)=1+i\sqrt3$. So this field is $\Bbb Q(i\sqrt3)$.

The fixed field of $H_{11}$ contains $\zeta+\zeta^{11}=2\cos(\pi/6)=\sqrt3$. So this field is $\Bbb Q(\sqrt3)$.

Okay, this makes sense, thanks. One question, I am not the best at readily determining the explicit form of expressions like $2\zeta^2$ or $\zeta+\zeta^5$. I see geometrically why $\zeta+\zeta^5=i$ and $\zeta+\zeta^7=0$, but in general I am not sure. How do you recommend I practice this? Is there a method you prefer to use? Do you convert to polar coordinates and go from there? — wormram, May 20 '20 at 21:21

How to find generators for the subfields of $\mathbb{Q}(\zeta_{12})$

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