$\newcommand{\Q}{\mathbb{Q}}$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 $24$?

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    $\begingroup$ I suppose you mean "abelian extension". Otherwise, your statement is "obviously" false as Galois extensions with Galois group $S_4$ over $\mathbb Q$ have subfields of degree $4$ over $\mathbb Q$. For example, if $\alpha$ is a root of $P=X^4-X-1$, the splitting field of $P$ has Galois group $S_4$ over $\mathbb Q$. $\endgroup$ Mar 5, 2014 at 14:15
  • $\begingroup$ @EwanDelanoy No I don't... This is the question I have... So you think that this question asks to prove something that is false? Can you please explain more, I am not sure I understand your argument. You could give it as an answer if you want. $\endgroup$
    – Spenser
    Mar 5, 2014 at 14:27
  • $\begingroup$ You can write $K=Q(\alpha)$ for some $\alpha$ in $K$ because the extension $K/Q$ is finite and separable: this is the «primitive element theorem.» $\endgroup$ Jun 8, 2014 at 17:55

1 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 $[L:\mathbb{Q}]=8$.


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