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With respect to the definition of Galois field, for $E$ an extension of $F$ ($E$ and $F$ are finite fields) $\mathrm{Gal}\,(E/F)$ is the set of automorphisms of $E$ which fix $F$ pointwise. So I think that we can distinguish between $\mathrm{Gal}\,(E/F)$ and $\mathrm{Aut}\,(E)$, right?

Brian Fitzpatrick
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Nil
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1 Answers1

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You don't need to assume that $E$ and $F$ are finite fields. Every automorphism of a field fixes its prime field, by which I mean $\mathbb Q$ or $\mathbb F_p$, the field generated by $1\in E$. However it doesn't need to fix any other subfield, so that's what makes the Galois group different.

Ian Coley
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  • threfore the order of Aut(E) > Gal(E/F) ? E is considered as a vector space over F. – Nil Feb 08 '14 at 23:40
  • Certainly $\geq$, since every element of the Galois group is an automorphism of $E$. – Ian Coley Feb 08 '14 at 23:47
  • Okay, then what about the difference between $Gal(E/F)$ and $Aut(E/F)$? ...seems to sometimes depend on the author; you need to say something about normality. – j0equ1nn Oct 22 '14 at 04:44