$\newcommand{\Q}{\mathbb{Q}}$$\DeclareMathOperator{\Aut}{Aut}$Let $K/E$ and $E /F$ be Galois extensions. I would like to show that, if every $\sigma \in \Aut(E/F)$ is the restriction of an element of $\Aut(K/F)$, then $K \supset F$ is Galois.

I know that this is not necessarily true when only the information "Let $K /E$ and $E/F$ be Galois extensions." is given, for instance when $\Q \subset \Q(\sqrt2) \subset \Q(\sqrt[4]2)$, $K/F$ is not Galois.

I am not sure how to use the restriction aspect of the question, how to proceed?

Thank you.


$\DeclareMathOperator{\Aut}{Aut}$ Hint 1

You want to prove that if $a \in K \setminus F$, then there is $g \in \Aut(K/F)$ such that $a g \ne a$.

Hint 2

If $a \notin E$, then you even achieve that with an element of $g \in \Aut(K/E)$.

Hint 3

So you are left with $a \in E \setminus F$. You know that there is $h \in \Aut(E/F)$ such that $a h \ne a$.

Hint 4

And now use your hypothesis about the restriction...


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.