1
$\begingroup$

Given a finite extension $E/k$ of a field $k$, how do I prove the following? $$|\operatorname{Gal}(E/k)| \text{ divides } [E:k].$$ Thanks.

  • 0
    You mean, given a finite *Galois* extension ...2012-03-19
  • 1
    @ÁlvaroLozano-Robledo Hm, have I blundered? If $E/k$ is Galois then it seems like (by definition, for some) we should have $|\operatorname{Gal}(E/k)| = [E : k]$. (Personally I would not write $\operatorname{Gal}$ for something that wasn't Galois, but it's not so hard to tell what is meant here.)2012-03-19
  • 0
    If $E/k$ is not Galois, then how can you talk about $\operatorname{Gal}(E/k)$?2012-03-19
  • 0
    I assumed that it was what I would write as $\operatorname{Aut}(E/k)$.2012-03-19
  • 0
    @Álvaro: Many authors use $\mathrm{Gal}(E/k)$ to denote the group of automorphisms, whether or not the extension is assumed to be Galois, and *define* Galois extension as one in which the fixed field of $\mathrm{Gal}(E/k)$ is $k$ (or in the finite case, where $|\mathrm{Gal}(E/k)| = [E:k]$).2012-03-19
  • 0
    @ÁlvaroLozano-Robledo, I am sorry. I was mistaken. Gal just means Aut.2012-03-19

1 Answers 1

4

The first thing that comes to mind is the following. It's a theorem of Artin that if $G$ is a finite group of automorphisms of $E$ then $E$ is Galois over the fixed field $E^G$ with Galois group $G$. See Corollary 3.5 and the surrounding paragraphs in Milne's notes. Since \[ [E : k] = [E : E^G][E^G : k] = |G|[E^G : k] \] this gives the result.

  • 0
    @ Dylan, that was nice. Thank you.2012-03-19