Find $ [\mathbb{Q(\alpha)}: \mathbb{Q}] $ where $ \alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9 $ Suppose $   \zeta$ is a primitive  $   11$-th root of unity and  $ \alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9  $

Find $ [\mathbb{Q(\alpha)}: \mathbb{Q}]  $

Could someone please give me a hint how too do that?
 A: You may notice that
$$ \alpha = \sum_{n=1}^{5}\zeta^{n^2} $$
from which it follows that (just square the previous sum):
$$ \alpha^2 = -\frac{5+i\sqrt{11}}{2} $$
and $\alpha=\pm \frac{1}{2}(1-i\sqrt{11})$. Gauss sums hence prove that $[\mathbb{Q}(\alpha):\mathbb{Q}]=\color{red}{2}$.
You may also use the general identity:
$$ \sum_{k=1}^{p-1}\left(\frac{k}{p}\right)\exp\left(\frac{2\pi i k}{p}\right) = \sqrt{\pm p}$$
where the sign in the RHS just depends on $p\pmod{4}$.
A: Here’s another method: Consider the Galois group of $\Bbb Q(\zeta_{11})$ over $\Bbb Q$. It’s cyclic, isomorphic to $(\Bbb Z/11\Bbb Z)^\times$, which is generated by $2$, that is, $\sigma:\zeta\mapsto\zeta^2$ generates the group. There’s only one subgroup of order five, generated by $\sigma^2:\zeta\mapsto\zeta^4$. Since the powers of $4$ in $(\Bbb Z/11\Bbb Z)^\times$ are $1$, $4$, $5$, $9$, and $3$, your number is the trace from the big field down the fixed field of that group of order five.
That should be enough for you to finish the argument.
