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Let $K$ be an extension of $F$. The Galois group of $K$ is defined to be the set of all $F$-automorphisms of $K$, i.e., all automorphisms of $K$ which fix $F$.

An extension $K$ over $F$ is Galois if $| \operatorname{Gal}(K/F)|=[K:F]$.

(I am aware some call the set of all $F$-automorphisms of $K$ the Galois group when $K$ is Galois over $F$.)

It is clear $\operatorname{Gal}(K/F) \subset \operatorname{Aut}(K)$.

When is $ \operatorname{Aut}(K)= \operatorname{Gal}(K/F)$?

user5826
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  • It's certainly correct if you have $F$ prime subfield of $K$; but you know that already I guess?? –  Apr 17 '18 at 21:03
  • Ohh I see there is an answer saying that, but just appeared to me I'm sorry. In fact your question is open 1 hour and for me was saying some minutes. weird. –  Apr 17 '18 at 21:06
  • Another "easy" example is if $Aut(K)$ is trivial; then $Aut(K) = Aut(K|F)$ for every subfield $F$ of $K$. Examples are $K=\Bbb R$ and $K= \Bbb Q_p$. On the other hand, for $K=\Bbb C$, what $Aut(K)$ is depends on whether you assume the axiom of choice or not. If yes, then $Aut(\Bbb C) = Aut(\Bbb C|\Bbb Q) \neq Aut(\Bbb C|F)$ for $F\neq \Bbb Q$. However, without choice it is possible that $Aut(\Bbb C) = Aut(\Bbb C|F)$ for every field $F\subseteq \Bbb R$. See https://math.stackexchange.com/a/412034/96384. So the problem seems wild in this generality, maybe you can narrow down the question. – Torsten Schoeneberg Apr 18 '18 at 03:42

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When $F$ is the prime field of $K$, i.e. the subfield of $K$, which does not contain any proper subfield, then we have equality. This is due to the fact that the prime field is fixed by any automorphism.

You can check this in two steps:

  • first prove that the prime field is always isomorphic to either $\mathbb{Q}$ (if $char(K)=0$) or $\mathbb{F}_p$ (if $char(K)=p$)
  • then show that any automorphism fixes the prime field (use that $\sigma(1)=1, \sigma(a+b)=\sigma(a)+\sigma(b)$ and $\sigma(a\cdot b)=\sigma(a)\sigma(b)$)
Notone
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