The Automorphism Group $\Gamma(\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q})$ is trivial if $n$ is odd. The full question is: Given a group $F_n=\mathbb{Q}(\sqrt[n]{2})$ then prove that:

*

*(A) If $n$ is odd then $\Gamma(F_n:\mathbb{Q})$={$\rm{id}$} and

*(B) If $n$ is even then $\Gamma(F_n:\mathbb{Q})\cong\mathbb{Z_2}$
I am confused as to how to find either, I know that $F_n:\mathbb{Q}$ is a non-normal extension. We were given the hint:
"For any $\tau\in\rm{Aut}_\mathbb{Q}(F_n)$ we must have that $\tau(\sqrt[n]{2})\in F_n\subseteq\mathbb{R}$ is again a real $n$-th root of 2"
 A: The $\mathbb Q$-automorphisms of $F_n$ are completely determined by where they send $\sqrt[n]2$, and they must send it to another $n$-th root of $2$. But if $n$ is odd, then no other $n$-th root of $2$ is in $F_n$, and if $n$ is even, then only one other $n$-th root of $2$ is in $F_n$ (you might need to prove these facts beforehand). So there are only one/two possible $\mathbb Q$-automorphisms.
A: By Eisenstein's criterion, we know that $X^n-2$ is always irreducible over $\Bbb Q$. Hence $m(\sqrt[n]{2},\Bbb Q) = X^n-2$ and the roots of this polynomial are of the form $\xi_n^j\sqrt[n]{2}$ with $\xi_n$ an $n$-th primitive root and $1 \leq j \leq n$.
As you note, a morphism $\sigma \colon F_n/\Bbb Q \to \overline{\Bbb Q}$ will be determined by the image of $\sqrt[n]{2}$ in $\Bbb Q$ and, by the previous remark, this ought to be an element of the form $\xi_n^j \sqrt[n]{2}$. But if we want $\mathsf{im} \ \sigma$ to be in $\Bbb R$, can't we say something about $n$ and $j$?

 We need to have $\xi_n^j \in \Bbb R$. But a real root of unity is either $-1$ of $1$. Thus the options are $j = n, \xi_j^n = 1$ and $n \in 2\Bbb Z, j = n/2, \xi_n^j = -1$. When $n$ is even, both options can happen, hence the Galois group is cyclic generated by $\sigma$ such that $\sqrt[n]{2} \mapsto - \sqrt[n]{2}$. When $n$ is odd, the first case is the only one possible, thus every automorphism fixes $\sqrt[n]{2}$... but this means they were the identity to begin with.

