Quadratic subfields of the cyclotomic field $\mathbb{Q}(\zeta_{14})$ 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?
 A: 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 $x^2 -x +2.$
A: It is not generally true that if you add up the values of a subgroup $H$ of the Galois group on a generating element for the whole field that you'll get an element generating the subfield fixed by $H$. For example,  ${\rm Gal}(\mathbf Q(\zeta_8)/\mathbf Q) \cong (\mathbf Z/8\mathbf Z)^\times$ and the subfield fixed by $H = \{1,5 \bmod 8\}$ is $\mathbf Q(i)$, but if you apply $H$ to $\zeta_8$ and add the results you get $\zeta_8 + \zeta_8^5$, which is $0$, so $\zeta_8 + \zeta_8^5$ is contained in $\mathbf Q(i)$ but that sum does not generate $\mathbf Q(i)$ over $\mathbf Q$. The fact that you did not run into such a problem using $\zeta_{14}$ depends on some accidental choices in your "random" example.
When $H$ is the subgroup of order $3$ in ${\rm Gal}(\mathbf Q(\zeta_{14})/\mathbf Q)$, summing the elements in the $H$-orbit of $\zeta_{14}$ gives you a generator of the subfield of $\mathbf Q(\zeta_{14})$ with codimension $3$ (equivalently, of dimension $6/3 = 2$ over the base field $\mathbf Q$) because (i) a sum over an $H$-orbit generates the subfield fixed by $H$ when you sum over the $H$-orbit of an element in a normal basis, as explained in Theorem 3.8 here, and (ii)
the primitive $n$th roots of unity form a normal basis for $\mathbf Q(\zeta_n)/\mathbf Q$ when $n$ is squarefree (e.g., $n = 14$), as explained in my answer on the MSE page here.
