How do I prove that there exists an automorphism of order 3? Let $\mathbb{Q}'$ be an algebraic closure of $\mathbb{Q}$. How do I prove that there exists an automorphism on $\mathbb{Q}'$ whose order is 3?
I first tried to find a subfield $F$ such that $[\mathbb{Q}',F]=3$, but then realized that it is much harder to find a field than to find an automorphism. So I tried to extend an $\mathbb{Q}$-isomorphism $\mathbb{Q}(e^{\frac{2\pi i }{3}})\rightarrow \mathbb{Q}(e^{\frac{2\pi i }{3}})$, which sends $e^{\frac{2\pi i}{3}}$ to $-e^{\frac{2\pi i}{3}}$, to an isomorphism between $\mathbb{Q}(e^{\frac{2\pi i }{3}})(S)$ and then to $\mathbb{C}$ and then restrct to $\mathbb{Q}'$ to get a desired automorphism, where $S$ is a transcendental basis, but it does not work well. (My plan was to construct an isomorphic copy of $\mathbb{Q}'$ from $\mathbb{Q}(e^{\frac{2\pi i }{3}})(S)$). How do I prove this?

An application of the Artin-Schreier theorem is that for any field $K$, with algebraic closure $\overline{K}$, the torsion elements of ${\rm Aut}(\overline{K}/K) \subset {\rm Aut}(\overline{K})$ have order $1$ or $2$, and a pair of non-commuting torsion elements have a product with infinite order. In particular, any nontrivial torsion element of the automorphism group of $\overline{Q}$ has order $2$. So order $3$ is not possible.
