Show that $\mathbb{C}(t)$ is a Galois extension of $\mathbb{C}(t^{n} + t^{−n})$, where $t$ denotes an indeterminate. Find the Galois group of this extension.
Let $F = \mathbb{C}(t^{n} + t^{−n})$ and $E = \mathbb{C}(t)$. Since $t \in E$, we have $t^n \, , 1/t^n \, \in E$ and thus $F \subset E$. Now we need to show that $E/F$ is an algebraic extension. For $t^{n} + t^{−n} \in F$, $(t^{n} + t^{−n})^2 = t^{2n}+t^{-2n}+2$, therefore $t^{2n}+t^{-2n} \in F$, using strong induction one can show that $t^{kn}+t^{-kn} \in F$ for all $k \in \mathbb{N}$. Using this we can see that $Irr(t^m,F) = (t^{mn}+t^{-mn})^{-1}(x^{2n}+1)-x^n$ for $1\le m \le n-1$. I am not sure how to find Irr polynomials for a general $p(t) \in E$ ($t$ is algebraic).
This field extension is also seperable since $Char(F) = 0$. To show that this is a normal extension, it is enough to show that every $F$-embedding $\sigma: E \to \bar{F}$ is an automorphism of $E$. But i am unable to show this.
Edit 1: For $m=1$, let $f(x)= (t^{n}/(t^{2n}+1))(x^{2n}+1)-x^n$, roots of $f(x)$ are $x = t\omega^{i}$ and $t^{-1}\omega^{i}$, where $\omega$ is an nth root of unity. Therefore $E$ is a splitting field of a seperable polynomial in $F$, thus it is Galois.
Edit 2: Define maps $\sigma(t)= \omega t$ and $\tau(t) = \omega t^{-1}$, then $\sigma^{n} =Id$, $\tau^{2} = Id$ and $\sigma \circ \tau = \tau \circ \sigma^{n-1}$. The galois group must be $D_{2n}$ :)
$$(X^n-t^n)(X^n-\sigma(t^{n}))\in \mathbb{C}(t^{n} + t^{−n})[X]$$ where $\sigma(t^n)$ is the $\mathbb{C}(t^{n} + t^{−n})$-conjugate of $t^n$.
So $$\sigma(t^n) =\ ?$$