Intermediate field of a cyclotomic extension

Question

Let $L:=\mathbb{Q}(\xi)$, where $\xi$ is a primitive $n^{\text{th}}$-root of unity with $n>2$ and let $K:=\mathbb{Q}(\xi+\xi^{-1})$. Prove that $K$ is the fixed field of the subgroup $\langle\sigma\rangle <\text{Aut}(L|K)$, where $\sigma$ is determined by $\sigma(\xi)=\xi^{-1}$, and determine for which values of $n$ we have that $K$ is the only subfield with $[L:K]=2$.

It is obvious that $K$ is contained in the fixed field, but I'm really struggling to prove the opposite inclusion, i.e., if $\frac{p(\xi)}{q(\xi)}=\frac{p(\xi^{-1})}{q(\xi^{-1})}$ for $p,q\in \mathbb{Q}[X]$, then $\frac{p(\xi)}{q(\xi)}=\frac{r(\xi+\xi^{-1})}{s(\xi+\xi^{-1})}$, for some $r,s\in \mathbb{Q}[X]$. I tried brute force, but it didn't work out. For finding the values of $n$, I really have no clue.

Any tips? Thank you!

Answer 1

Since cyclotomic extensions are algebraic, you only have to consider polynomials in $\xi$ as elements of the fixed field of $\langle \sigma\rangle$. Next you should convince yourself that any polynomial in the fixed field of sigma is of the form $f(\xi) = a_{n-1}\xi^{n-1} + \cdots+ a_1\xi^1 + a_0 + a_1 \xi^{-1} + \cdots + a_{n-1}\xi^{-(n-1)}$.

From here you can show that $f(\xi)$ is actually a polynomial $g(\xi + \xi^{-1})$ (to do this, induct by degree). For instance,

$$\xi^{2} + \xi^{-2} = (\xi + \xi^{-1})^2 - 2.$$

As far as the last part, that is equivalent to asking for which $n$ does the Galois group have a unique subgroup of order $2$. The Galois group of $\mathbb Q(\xi_n)$ is $\left(\mathbb{Z}_n\right)^*$. You can find more informations about such groups here.