Let $K$ be a Galois extension of $\mathbb{Q}$. View $K$ as a subfield of $\mathbb{C}$. If $\sigma$ is complex conjugation, show that $\sigma(K) = K$, so $\sigma|_K\in \text{Gal}(K/\mathbb{Q})$. Show that $F(\sigma|_K) = K\cap \mathbb{R}$, and conclude that $[K:K\cap \mathbb{R}]\leq 2$. Give examples to show that both $[K:K\cap \mathbb{R}]= 1$ and $[K:K\cap \mathbb{R}]= 2$ can occur.
I am very confused solving this problem. How many subfields of $\mathbb{C}$ are there that are an extension of Gaois of $\mathbb{Q}$? If $a\in K$ then I can say that $\text{Im}(a)=0$ and thus conclude that $\sigma(a)=a$? To show $F(\sigma|_K) = K\cap \mathbb{R}$, I think I can try the inclusion $"\supseteq" $, but the other inclusion I do not know how to prove it, could you please help me? I can say that $|\text{Gal}(K/\mathbb{Q})|\leq 2$ for being $K$ a subfield of $\mathbb{C}$ and so $[K:K\cap \mathbb{R}]\leq 2$ ?. How can I find such examples? Thank you very much.