If $E = KL$, then $\operatorname{Aut}(E/L)\simeq\operatorname{Aut}(K/F)$

Asked Sep 24 '21 at 14:35

Active Dec 10 '21 at 05:04

Viewed 63 times

$\newcommand{\Gal}{\operatorname{Gal}}$ My question arose from my previous post. I want to say $\Gal(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q(\sqrt[n]{p}))\simeq \Gal(\Bbb Q(\eta)/\Bbb Q)$ so that the sequence splits. So in general situation, what is a sufficient condition to be:

If $E =KL$ is a composite field and field extension of $F$, then $\operatorname{Aut}(E/K)\simeq\operatorname{Aut}(L/F)$.

Any help will be appreciated.

Answer by user8268 : $[E:K] = [L:F]$.

edited Dec 10 '21 at 05:04

asked Sep 24 '21 at 14:35

one potato two potato

4,981

that would be true under the condition that $K\otimes_FL$ is a field (which is then $E$), or equivalently if $[E:K]=[L:F]$ (if the extensions are finite) – user8268 Sep 24 '21 at 14:59
@user8268 Why is that? – one potato two potato Sep 24 '21 at 15:04
A small remark to the title. $F$ comes from nowhere, if we only know $E=KL$. – Dietrich Burde Sep 24 '21 at 16:02
@DietrichBurde but in the main post, I indicated $F$ is a base field of $E$ – one potato two potato Sep 24 '21 at 16:14
For this reason it was a remark concerning the title (which one reads first - and sometimes only... :) ) – Dietrich Burde Sep 24 '21 at 16:31
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the same Galois groups certainly imply the same degrees; in the other direction, every $F$-automorphism of $L$ lifts canonically to a $K$-automorphism of $K\otimes_FL$ – user8268 Sep 24 '21 at 17:24
@user8268 Thanks. Is that an if and only if statement? – one potato two potato Sep 24 '21 at 23:36
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yes - it's enough to think about the degrees of the extensions, orders of groups, and that lifting off automorphisms – user8268 Sep 25 '21 at 07:45

If $E = KL$, then $\operatorname{Aut}(E/L)\simeq\operatorname{Aut}(K/F)$

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