Say we have a an irreducible polynomial $f \in \mathbb{Q}[x]$ whose roots are not all real, and $K$ a splitting field for $f$ over $\mathbb{Q}$. Why is the degree of the field extension $[K : \mathbb{Q}]$ necessarily even?
I'm unsure of the right approach towards this. I'm probably missing something obvious, but can't figure it out!
$K$ embeds in the complex numbers, but not in the real numbers. Applying complex conjugation to this embedding thus gives a nontrivial automorphism with order $2$, so $2$ divides the order of the Galois group.