# Finding a counter-example for Gaussian-periods for non-primes

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?

• $(\mathbb{Z}/8\mathbb{Z})^*$ 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 $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. – A. Kriegman Jun 25 at 14:22
• 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? – Sigurd Jun 25 at 14:40
• IMO, the point of proof is that you don't need someone else to tell you if it's correct. – A. Kriegman Jun 25 at 15:07
• 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. – Jyrki Lahtonen Jun 25 at 17:25
• @JyrkiLahtonen That's a nice addition, thanks! I didn't realize that, it was a coincidence, but that will narrow down the searching. – Sigurd Jun 25 at 18:18

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