Show that $\mathbb{Q}(\sqrt[3]{2})$ does not have non trivial automorphisms Since I started the abstract algebra course, I haven't had exercises associated with the existence or non-existence of automorphisms, and I haven’t found exercises similar to this either so I don't know how to proceed. I would greatly appreciate your help.

The field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal (and therefore not galois) as the minimal polynom $X^3-2$ of $\sqrt[3]{2}$ has other solutions ($\zeta_3\sqrt[3]{2}$ and $\zeta_3^2\sqrt[3]{2}$ with $\zeta_3=e^{\frac{2\pi i}{3}}$) that are not in $\mathbb{Q}(\sqrt[3]{2})$. Let $H=\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2}))=\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$, then according to Artin's theorem, we have:
$$|H|
=[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})^H]
=\frac{[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{2})^H:\mathbb{Q}]}
\Rightarrow
\frac{3}{|H|}
=[\mathbb{Q}(\sqrt[3]{2})^H:\mathbb{Q}]\in\mathbb{N},$$
which implies $|H|=1$ or $|H|=3$. The latter case is not possible as $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not galois.
