2
$\begingroup$

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?

Attempt:

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.

$\endgroup$
4
$\begingroup$

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 $\operatorname{Gal}(\mathbb Q(\sqrt[3]2)/ \mathbb Q) = 1$!

$\endgroup$
  • $\begingroup$ Huh, it was that simple. I'm not very comfortable handling those things; Thank you, I appreciate the simplicity. $\endgroup$ – JustDroppedIn 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! $\endgroup$ – punctured dusk May 15 '18 at 15:52

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.