1
$\begingroup$

Let $F \subseteq E $ be extension of fields. If $Gal(E/F)$ is a cyclic group, does it imply that the extension $E/F$ is a Galois extension? If not, any example?

$\endgroup$
2
  • $\begingroup$ A trivial group is a cyclic group of order $1$. $\Bbb Q(\sqrt[3]{2})/\Bbb Q$ is not Galois. $\endgroup$ – Balarka Sen Aug 22 '15 at 17:51
  • $\begingroup$ @ Balarka Sen Thank you. So the group is cyclic but $Q(\sqrt[3]{2})/Q$ is not Galois extension, so it does not imply, right? $\endgroup$ – MATH Aug 22 '15 at 17:59
6
$\begingroup$

It depends. First one should note that usually writing $\text{Gal}(E/F)$ is reserved for when the extension is already known to be Galois, and we write $\text{Aut}(E/F)$ otherwise, but this is not that important. Now let $\mathbb{F}_p(t^p)[x]/(x^p - t^p) = E$, and $\mathbb{F}_p(t^p) = F$, this has trivial automorphism group and is not Galois as it is not separable. In particular now one can take the compositum $E = \mathbb{F}_q(t^p)[x]/(x^p - t^p)$, $F = \mathbb{F}_p(t^p)$ which should have galois group $\mathbb{Z}/n\mathbb{Z}$ where $q = p^n$ but not be a Galois extension.

In general there are not any conditions I am aware of that one can put on the automorphism group of a field extension that is not Galois that ensure that it is Galois.

$\endgroup$

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.