Show that $|\operatorname{Aut}(\mathbb{Q}(\sqrt[10]2))|=2$

Question

Let $\mathbb{K}=\mathbb{Q}(\sqrt[10]2)$. Show that $|\operatorname{Aut}(\mathbb{Q}(\sqrt[10]2))|=2$.

Well, it is easy to see that the degree of this extension over $\mathbb{Q}$ is ten. Also, is easy to see that there are two $\mathbb{Q} $-automorphisms, namely the identity and the mapping $\psi:\mathbb{Q}(\sqrt[10]2) \longrightarrow \mathbb{Q}(\sqrt[10]2)$ determined by $$\sqrt[10]2 \mapsto -\sqrt[10]2.$$ But i can not ensure that there are no more.

Answer 1

Hint : an automorphism $\sigma$ is uniquely determined by its value at the generator $\sqrt[10]{2}$, which must be a root of $X^{10}-2$. Which of those roots lay in $\mathbb K$ ?