Let $K$ be the fixed field of the complex-conjugation automorphism of the algebraic closure of $\mathbb{Q}$. Prove that $K$ is not a simple extension of $\mathbb{Q}$.
I think, if we take $K$ to be $\mathbb{Q}(\sqrt{2})$ for example, then $K$ would be fixed by the conjugation automorphism of the algebraic closure of $\mathbb{Q}$. What is wrong with this example?
Any help would be appreciated!
$K$ is way larger than $\mathbb Q(\sqrt{2})$. Actually $K = \overline{\mathbb Q} \cap \mathbb R$.
But it is not relevant how $K$ looks like. Important is that $K/\mathbb Q$ is infinite and algebraic. But simple algebraic extensions are finite.