Field extension of $\Bbb Q$ of degree $4$ has reals Suppose $L / \mathbb{Q}$ is a field extension with $[L : \mathbb{Q}]$ = 4, with $L \not\subset \mathbb{R}$. Is it true that $L \cap \mathbb{R} \neq \mathbb{Q}$? If not, is it true if $L / \mathbb{Q}$ is normal?
I have tried to suppose $a \in L \cap \mathbb{R} \implies a \in \mathbb{Q}$, then from $z = a + b i \in L$ we get, $z + \bar{z} = 2a$ ($\bar{z} \in L$ if $L$ is normal at least). So $a \in \mathbb{Q}$. But I don't know how to proceed from there. I guess one should be able to show that $L = \mathbb{Q}(i)$ which contradicts degree $4$.
Assume that $L/\Bbb Q$ is normal. Let $\sigma$ be the field automorphism given by complex conjugation (which is a field automorphism because the extension is normal). Then the subgroup $H$ of $\operatorname{Aut}(L)$ generated by $\sigma$ has order $2$, so $L$ has degree $2$ over the fixed field $L^H$. We get $[L^H:\Bbb Q]=4/2=2&gt;1$ and $L^H\subset \Bbb R$, i.e. $L\cap \Bbb R\ne \Bbb Q$.
Without the normal assumption it is not necessarily true, take a polynomial of degree $4$ with Galois group $S_4$ and a nonreal root $\alpha$. Then $L=\Bbb Q(\alpha)$ doesn't have any non-trivial intermediate fields at all, so since $L\nsubseteq\Bbb R$ we get $L\cap \Bbb R=\Bbb Q$.
Hint: As an example, consider a primitive 8th root of unity $\xi$. It obeys the equation $\xi^4=-1$.
Then $L = \{a+b\xi+c\xi^2+d\xi^3\mid a,b,c,d\in\Bbb Q\}$.