How to show that any field extension $K/\mathbb{Q}$ of degree 4 that is not Galois has a quadratic extension $L$ that is Galois over $\mathbb{Q}$.

Question

$\newcommand{\Q}{\mathbb{Q}}$Let $K/\Q$ be a field extension of degree $4$ that is not Galois. How to show that there exists an extension $L\supseteq K$ such that $[L:K]=2$ and $L/\Q$ is Galois?

I know the example of $\Q(\sqrt[4]{2})$ which is not Galois but is contained in the splitting field of $x^4-2$ which is Galois and of degree $8$, and I am trying to generalize this. But I am not even sure if we can write $K=\Q(\alpha)$ for some $\alpha$. Anyway, if this is the case, then the splitting field $L$ of the minimal polynomial of $\alpha$ would be Galois and of degree $8$, $12$ or $24$ since $\mathrm{Gal}(L/\Q)$ would be a subgroup of $S_4$. But how to rule out $12$ and $24$?

Answer 1

As Ewan Delanoy remarked: pick any $\alpha \in \overline{\mathbb{Q}}$ root of $P=x^4-x-1$. Then, for $K:=\mathbb{Q}(\alpha)$, you have $[K:\mathbb{Q}]=4$. Now if $[L:K]=2$ and $L/\mathbb{Q}$ is Galois, then $L$ would contain the splitting field of $P=x^4-x-1$. This is not possible because the splitting field of $P=x^4-x-1$ over ${\mathbb{Q}}$ has degree $4!=24$ which is bigger than $[L:\mathbb{Q}]=8$.