Calculating fixed fields I have to show that $\mathbb{Q}(\zeta)^{<\sigma>}$ $=$ $\mathbb{Q}(\zeta + \frac{1}{\zeta})$ 
$\sigma \colon L \to L$ is defined by $\sigma(\alpha) = \overline{\alpha}$, where $\overline{\alpha}$ is the complex conjugate of $\alpha$, and $L$ is a subfield of $\mathbb{C}$.
Also note that $\zeta = exp(\frac{2\pi i}{p})$ where $p$ is an odd prime. 
Now i know that $\beta \in \mathbb{Q}(\zeta)^{<\sigma>}$ $\iff$ $\sigma(\beta) = \beta$
So if we let $\beta = a_0 +a_1\zeta + \dots + a_{p-1}\zeta^{p-1}$, then apply $\sigma$ to $\beta$ and use the fact that $\zeta = exp(\frac{2\pi i}{p})$ and even Euler's formula, but I seem to get that all the coefficients apart from $a_0$ are equal to $0$, giving $\mathbb{Q}(\zeta)^{<\sigma>} = \mathbb{Q}$. Any hints?
 A: Since $\;\zeta\;$ is an element on the unit circle, we have that $\;\zeta^{-1}=\overline\zeta\;$ , so that
$$\sigma\left(\zeta+\zeta^{-1}\right)=\sigma\zeta+\sigma\overline\zeta=\overline\zeta+\overline{\overline\zeta}=\zeta+\overline\zeta=\zeta+\zeta^{-1}\implies$$
$$\Bbb Q\left(\zeta+\zeta^{-1}\right)\subset\Bbb Q(\zeta)^{\langle\sigma\rangle}$$
Now suppose 
$$\;\beta=\sum_{k=0}^{p-1}a_k\zeta^k\in\Bbb Q(\zeta)^{\langle\sigma\rangle}\implies \beta=\sigma\beta=\sum_{k=0}^{p-1}a_k\sigma\beta^k=\sum_{k=0}^{p-1}a_k\overline{\zeta^k}=\sum_{k=0}^{p-1}a_k\zeta^{-k}$$ 
Now, when $\;0\le k,m\le p-1\;$ , we have that 
$$\zeta^k=\zeta^{-m}\iff\zeta^{k+m}=1\iff (k+m)=0\pmod p\iff k=-m\pmod p$$
so that
$$\sum_{k=0}^{p-1}a_k\zeta^k=\sum_{k=0}^{p-1}a_k\zeta^{-k}\iff a_k+a_{p-k}=0\;,\;\;1\le k\le p-1\;,\;\;\text{since}\;\;\zeta^k=\zeta^{-(p-k)}$$
and we thus get that
$$\beta=a_0+a_1\left(\zeta+\zeta^{-1}\right)+a_2\left(\zeta^2+\zeta^{-2}\right)+\ldots\in\Bbb Q\left(\zeta+\zeta^{-1}\right)\;\;\text{(why?)}$$
