$Q(\alpha)$ is the extension $E_0=R\cap Q(\alpha)=Q(\alpha+\frac{1}{\alpha})$ This is part of Problem 3.8(b) from Falko Lorenz Algebra:Fields and Galois Theory.
$Q(\alpha)$ is the extension $E_0=R\cap Q(\alpha)=Q(\alpha+\frac{1}{\alpha})$ where $Q$ is rational number, $R$ is real number and $\alpha$ is the $n-th$ root of $x^n=1$ for $n\geq 3$. I want to show $R\cap Q(\alpha)=Q(\alpha+\frac{1}{\alpha})$.
It is clear for $Q(\alpha+\frac{1}{\alpha})\subset R\cap Q(\alpha)$. I would like to show the other direction.
The following is my approach to the problem. Since I do not see any instantaneous trick to solve the problem. I see that for $x\in R\cap Q(\alpha)$ with $x=\bar{x}$, $x=\sum_ia_i(\alpha^i+\frac{1}{\alpha^i})$,$a_i\in Q$. I tried to solve the equation $x=\sum_jb_j(\alpha+\frac{1}{\alpha})^j$ for coefficient $b_j\in Q$. It can be done to show that there is solution to $b_j$'s by inducting from top degree in $\alpha$ due to symmetry of $\alpha^i$ and $\frac{1}{\alpha}^i$ and symmetry of expansion of binomials. This is basically change of coordinates.
The other possible way to do this is by considering $[R\cap Q(\alpha):Q(\alpha+\frac{1}{\alpha})]\leq 2$ as $[Q(\alpha):Q(\alpha+\frac{1}{\alpha})]\leq 2$ where latter equality will be enforced once this problem is done. So one has to show the degree of extension is strictly less than 2. I do not know how to proceed this way.
Is there any trick to solve this problem? 
 A: As far as I can see, both of your methods are fine. I'll fill in the logic for your second argument.


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*Showing that $[\mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1})] \leq 2$:
We may use the standard fact:
If $K \subseteq L$ is a field extension and if $L = K(\alpha)$ for some element $\alpha \in L$, then the degree of the extension $K \subseteq L$ is equal to the degree of the minimal polynomial for $\alpha$. Taking $K = \mathbb Q (\alpha + \alpha^{-1})$ and $L = K(\alpha) = \mathbb Q(\alpha)$, we observe that $\alpha$ is a root of the quadratic polynomial $$X^2 - (\alpha + \alpha^{-1})X + 1 \in K[X].$$ Therefore, the minimal polynomial of $\alpha$ over $K$ must have degree at most two, and hence, $$[K:L] = [\mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1})] \leq 2.$$

*Showing that $[\mathbb R \cap \mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1} )] < 2$, with a strict inequality:
Consider the following tower of field extensions:
$$ \mathbb Q(\alpha + \alpha^{-1} ) \subseteq \mathbb R \cap \mathbb Q(\alpha) \subseteq \mathbb Q(\alpha).$$
By the tower law, we know that
$$ [\mathbb Q(\alpha) : \mathbb R \cap \mathbb Q(\alpha)] \times [\mathbb R \cap \mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1})]= [\mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1})]  \leq 2.$$
Clearly, $\mathbb Q(\alpha) \neq \mathbb R \cap \mathbb Q(\alpha)  $ when $\alpha$ is the $n$th primitive root of unity with $n \geq 3$ (for example, $\alpha$ itself is not real), so  $$[\mathbb Q(\alpha) : \mathbb R \cap \mathbb Q(\alpha)]  \geq 2.$$ Therefore, the only possibility is that $$[\mathbb R \cap \mathbb Q(\alpha) : \mathbb Q(\alpha + \alpha^{-1})] = 1,$$
i.e. $\mathbb R \cap \mathbb Q(\alpha) = \mathbb Q(\alpha + \alpha^{-1})$.
