Let $f \in \mathbb Q[x]$ be an irreducible polynomial of degree $4$, such that two of its roots are in $\mathbb R$ and two are in $\mathbb C \setminus \mathbb R$. Let $E \subseteq \mathbb C$ be the splitting field of $f$ over $\mathbb Q$.
Prove that, if $L = E \cap \mathbb R$, then $|E:L| = 2$.
Here is my attempt:
Let $a_1, a_2 \in \mathbb R$ and $b_1, b_2 \in \mathbb C \setminus \mathbb R$ be, respectively, the two real roots and the two non real roots of $f$. We can write $E = \mathbb Q[a_1, a_2, b_1, b_2]$, since $E$ is the splitting field of $f$.
Since $f$ is a separable polynomial, being $f \in \mathbb Q[x]$, we have that $E|\mathbb Q$ is a Galois extension. Given that $\mathbb Q \subseteq L \subseteq E$, we have that $E|L$ is a Galois extension, too.
It must be then $|E:L|=|Gal(E|L)|$.
Beacuse of its definition, we can write $E=L[b_1, b_2]$, this means that the automorphisms of $Gal(E|L)$ depend on their effect on the set $\{b_1, b_2\}$, which generates $E$ over $L$.
In $E[x]$, we can write $f=c(x-a_1)(x-a_2)(x-b_1)(x-b_2)$, for some $c \in E$.
Let $\varphi \in Gal(E|L)$. We have:
- $\varphi(a_1) = a_1$ and $\varphi(a_2) = a_2$, since $a_1, a_2 \in L$
- $\varphi(b_1) = b_1$ or $\varphi(b_1) = b_2$, because it must be $f(\varphi(b_1))=0$
- $\varphi(b_2) = b_1$ or $\varphi(b_2) = b_2$, because it must be $f(\varphi(b_2))=0$
This implies that, if $\varphi_1 : E \to E$, and $\varphi_2 : E \to E$ are defined as:
- $\varphi_1(l) = l$ for $l \in L$ and $\varphi_1(b_1) = b_1, \varphi_1(b_2) = b_2$
- $\varphi_2(l) = l$ for $l \in L$ and $\varphi_2(b_1) = b_2, \varphi_2(b_2) = b_1$
We have that $\varphi_1, \varphi_2 \in Gal(E|L)$ and $Gal(E|L)=\{\varphi_1, \varphi_2 \}$.
This means that $2 = |Gal(E|L)| = |E:L|$.
I would like to know if my proof is correct, feel free to tell where I am wrong.