I'm facing this question:

Let $E/F$ be a finite field extension such that its Galois group $\text{Gal}(E/F)$ is abelian. Is it necessary that $E/F$ is a Galois extension?


My first guess is no, but apparently I can't come up with any counter example. Maybe choosing $F$ not to be a perfect field would help, but I'm not sure in what direction to move to get an abelian Galois group.

My attempt to prove the statement was also unfruitful; I tried showing that the field of fixed elements of $E$ through $\text{Gal}(E/F)$ is precisely $F$, namely $E^{\text{Gal}(E/F)}=F$, which I know is equivalent to $E/F$ being a Galois extension. The inclusion $F\subset E^{\text{Gal}(E/F)}$ is always true; for the other inclusion: $[E:F]=[E:E^{\text{Gal}(E/F)}]\cdot[E^{\text{Gal}(E/F)}:F]$, but since $E/F$ is finite, $[E:E^{\text{Gal}(E/F)}]=|\text{Gal}(E/F)|$. I suppose that, if this is the right way, the fact that the Galois group is abelian comes in here. But I can't prove that $[E:F]=|\text{Gal}(E/F)|$ with only this assumption.

Any counterexample or hint is greatly appreciated.


Hint: The automorphism group of a nontrivial non-Galois field extension can be trivial! (hence abelian)

Take $\mathbb Q(\sqrt[3]2)/ \mathbb Q$: any automorphism fixes $\sqrt[3]2$, hence $\operatorname{Gal}(\mathbb Q(\sqrt[3]2)/ \mathbb Q) = 1$!

  • $\begingroup$ Huh, it was that simple. I'm not very comfortable handling those things; Thank you, I appreciate the simplicity. $\endgroup$ May 15 '18 at 15:37
  • 2
    $\begingroup$ You're welcome. I'm sure you've got a better understanding of Galois theory while attempting to prove it, so that's a good thing! $\endgroup$ May 15 '18 at 15:52

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