In a nutshell, my question is: what is degree of the field extension $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) $ over $\mathbb{Q}$?
As to why I'm asking this, I was trying to find the subfields of $\mathbb{Q}\, (\zeta_{14})$, the cyclotomic field of order $14$, and I found that the subfield $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) \subseteq \mathbb{Q}\, (\zeta_{14})$ should have degree $2$ over $\mathbb{Q}$.
This is because it is the fixed field of the following subgroup of automorphisms
$$\{1,9,11 \} \subset (\mathbb{Z}/14\mathbb{Z})^{\times} \cong \text{Gal} \, \mathbb{Q}\, (\zeta_{14}) / \mathbb{Q}, $$
and since this subgroup has index $2$, the corresponding fixed field must have degree $2$ over $\mathbb{Q}$ by Galois Theory.
However, when I put the sum $\zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}$ in Wolfram alpha, I get the following expansion:
$$ \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11} = \sqrt[-7]{-1} + (-1)^{2/7} – (-1)^{4/7},$$
which doesn't look like the root of a quadratic polynomial! Futhermore, I also get that this number is a root of the sixth degree polynomial $x^6 – x^5 + x^4 – 15x^3 + 22x^2 – 8x + 8$. This suggests that this number isn't the root of a quadratic polynomial, which goes against the fact that the extension
$\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) $
should be quadratic by the Galois theory reasoning I gave above.
My question is: am I right to assume that the extension above is a degree $2$ extension of $\mathbb{Q}$, by Galois theory? And if I am right, why does this element of a quadratic extension seem to have degree $>2$ over the rationals?
Best Answer
I'm not sure what went wrong when you plugged it into Wolfram Alpha, but you could try and compute the minimal polynomial (which is indeed quadratic by your own reasoning) of $\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11}$ directly. Upon squaring, you should find that
$$(\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11} )^2 = (\zeta_{14}^2+ \zeta_{14}^4+\zeta_{14}^{6} + \zeta_{14}^{8} + \zeta_{14}^{10}+\zeta_{14}^{12}) + (\zeta_{14}^{6} +\zeta_{14}^{10}+\zeta_{14}^{12}) $$
Here I grouped the terms that are the $7^{th}$ roots of unity. I can now exploit the fact that the sum of the $n^{th}$ roots of unity is $0$ and obtain that the above sum is equal to $(-1) + \zeta_{14}^{6} + \zeta_{14}^{10} + \zeta_{14}^{12}$. You can draw these roots of unity on the unit circle to convince yourself that
$$\zeta_{14}^{6} + \zeta_{14}^{10} + \zeta_{14}^{12} = -(\zeta_{14}^{13} + \zeta_{14}^{3} + \zeta_{14}^{5} ) . $$
Notice that the $3$ primitive $14^{th}$ roots of unity that are NOT the ones you've given appear. We can use this to exploit the fact that the sum of $14^{th}$ roots of unity is $0$. If we set $\alpha = \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}$, we obtain that
$$ \underbrace{(-1) - (\zeta_{14}^{13} + \zeta_{14}^{3} + \zeta_{14}^{5})}_{\alpha^2} -\underbrace{(\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11})}_{\alpha} = (\zeta_{14}^2+ \zeta_{14}^4+\zeta_{14}^{6} + \zeta_{14}^{8} + \zeta_{14}^{10}+\zeta_{14}^{12}) + \zeta_{14}^7 = -2$$
We've concluded that $\alpha^2 - \alpha = -2 \ $ so we have the minimal polynomial being $x^2 -x +2.$