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?

  • $\begingroup$ 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

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.


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