6
$\begingroup$

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

$\endgroup$
  • 4
    $\begingroup$ 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 $\mathbb Q$. $\endgroup$ – Ewan Delanoy Mar 5 '14 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 '14 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.» $\endgroup$ – Mariano Suárez-Álvarez Jun 8 '14 at 17:55
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.