I need to give a counter-example against the following theorem:

Suppose $H \subset \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is a subgroup. Then we have $\mathbb{Q}(\zeta_n)^H = \mathbb{Q}(\eta_H)$, with $\eta_H = \sum_{\sigma \in H} \sigma(\zeta_n)$, the Gaussian-period.

This theorem is true for $n = p$ prime, but not for general $n$. So far, my attempt is the following. We take $n = 8$. Then $\operatorname{Gal}(\mathbb{Q}(\zeta_8)/\mathbb{Q}) \cong (\mathbb{Z}/8\mathbb{Z})^\ast = \{1,3,5,7\} \cong \mathbb{Z}/4\mathbb{Z}$.

Now we have the subfield $\mathbb{Q}(\zeta_4) = \mathbb{Q}(i)$, since $4 \mid 8$. We also have the subfield $\mathbb{Q}(\sqrt{2})$ since $\zeta_8 + \zeta_8^{-1} = 2 \cos(2\pi/8) = \sqrt{2}$. Hence, we have also the quadratic subfield $\mathbb{Q}(\sqrt{-2})$. Notice that $1,5$ keep $i$ fixed, so according to the above theorem, we have $$ \mathbb{Q}(i) = \mathbb{Q}(\zeta_8)^{\{1,5\}} = \mathbb{Q}(\zeta_8 + \zeta_8^5). $$ I am stuck at this point, how to derive a contradiction from this?

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    $\begingroup$ $(\mathbb{Z}/8\mathbb{Z})^*$ is not isomorphic to $\mathbb{Z}/4\mathbb{Z}$. That wouldn't make sense because $\mathbb{Z}/4\mathbb{Z}$ has three subgroups, but we've already seen five subfields, $\mathbb{Q},\mathbb{Q}(i),\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{-2}),$ and $\mathbb{Q}(\zeta_8)$. The Galois group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ – A. Kriegman Jun 25 at 14:22
  • $\begingroup$ Yes, you are right, I made a mistake there. Apart from that, do you think the proof is correct, adding the extra line of Angina? $\endgroup$ – Sigurd Jun 25 at 14:40
  • $\begingroup$ IMO, the point of proof is that you don't need someone else to tell you if it's correct. $\endgroup$ – A. Kriegman Jun 25 at 15:07
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    $\begingroup$ Somewhat related. The primitive roots of unity of order $n$ form a basis (necessarily then a normal basis) of $\Bbb{Q}(\zeta_n)/\Bbb{Q}$ if and only if $n$ is square-free. I think in that case the Gaussian period always generates the fixed field. So, just like you did, you need to choose $n$ such that it is divisible by some square to get your counterexample. $\endgroup$ – Jyrki Lahtonen Jun 25 at 17:25
  • $\begingroup$ @JyrkiLahtonen That's a nice addition, thanks! I didn't realize that, it was a coincidence, but that will narrow down the searching. $\endgroup$ – Sigurd Jun 25 at 18:18

This is because $\zeta_8^5=-\zeta_8$, so that $\zeta_8+\zeta_8^5=0$.

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  • $\begingroup$ Indeed! Adding this line, is the rest of my proof correct do you think? $\endgroup$ – Sigurd Jun 25 at 14:09

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