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 $4$.

Hint: As an example, consider a primitive 8th root of unity $\xi$. It obeys the equation $\xi^4=-1$.
Then $L = \{a+b\xi+c\xi^2+d\xi^3\mid a,b,c,d\in\Bbb Q\}$.
