TITLE: Is $\mathbb Q(\sqrt2)$ the fixed field of some automorphism of $\overline{\mathbb Q}$?
QUESTION [8 upvotes]: Is is possible that $\mathbb{Q}(\sqrt{2})$ is the fixed field of an automorphism $\sigma$ of $\overline{\mathbb{Q}}$?
Thanks in advance for your help.

REPLY [15 votes]: First note that $\sqrt{3}, \sqrt{5}, \sqrt{15} \not\in \mathbf{Q}(\sqrt{2})$.
Assume, if possible, that the fixed field of $\sigma$ is $\mathbf{Q}(\sqrt{2})$. Then $\sigma(\sqrt{3}) \ne \sqrt{3}$, so we must have $\sigma(\sqrt{3}) = -\sqrt{3}$. Similarly, $\sigma(\sqrt{5}) = -\sqrt{5}$. Therefore $\sigma(\sqrt{15}) = \sqrt{15}$, contradicting the assumption.