0
$\begingroup$

If $K/F$ is splitting field, $M,L$ are intermediate fields, how to prove $\operatorname{Gal}(K/LM)=\operatorname{Gal}(K/L) \cap \operatorname{Gal}(K/M)?$

$\endgroup$
0
$\begingroup$

An automorphism is in $\newcommand{\Gal}{\text{Gal}}\Gal(K/LM)$ iff it fixes all elements of $LM$.

An automorphism is in $\Gal(K/L)\cap\Gal(K/M)$ iff it fixes all elements of $L$ and all elements of $M$.

Now why must an automorphism fixing all elements of $M$ and of $L$ also fix all elements of $LM$ and vice versa?

$\endgroup$

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.