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I've read that if $L/K$ is a field extension and $|Aut(L/K)|=|L/K|$, then L/K is a galois extension. I was wondering whether the converse is true, i.e if $|Aut(L/K)|\neq |L/K|$, then can we just assume that L/K is not a galois extension or can someone give me a counter-example for this statement.

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  • $\begingroup$ What is your definition of a Galois extension? An extension that's normal and separable? $\endgroup$ – Kaj Hansen Nov 11 '14 at 23:42
  • $\begingroup$ The field extension $L/K$ is a galois extension if $L^{Aut(L/K)}=K$, where $Aut(L/K)$ is the set of all automorphisms on L to K. Also, $L^S:=${$\alpha\in L|\sigma(\alpha)=\alpha\space\forall\sigma\in S$} $\endgroup$ – Andrew Brick Nov 12 '14 at 0:00
  • $\begingroup$ What do you mean by a normal extension, also what does being separable have to do with the extension being a galois extension? $\endgroup$ – Andrew Brick Nov 12 '14 at 0:01
  • $\begingroup$ There are lots of equivalent definitions @Nabeel. See here: en.wikipedia.org/wiki/Galois_extension $\endgroup$ – Kaj Hansen Nov 12 '14 at 0:08
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I assume your extension is finite. Your claim is true and can be found in any decent book on Galois theory (say Morandi's Field and Galois theory for concreteness; which I usually consult for elementary Galois theory); the equality $|Aut(L/K)|=|L/K|$ it is one of E.Artin's characterizations of Galois extensions. If your definition of Galois extension is 'normal and separable' then you need some work to give the equivalence. For instance, Morandi's definition of Galois extension is that the fixed field of all $K$-automorphisms of $L$ is actually $K$ and nothing bigger. Then he uses the linear independence of automorphisms to get the important inequality $|Aut(L/K)|\leq |L/K|$ and uses that to show that equality holds if and only if the extension is Galois. He later proves that Galois extensions by his definition are precisely the normal and separable ones, finishing the characterization.

Edit: It seems your definition of Galois is that the fixed field of automorphisms is $K$. In that case, use the linear independence of automorphisms to get $|Aut(L/K)|\leq |L/K|$ and then if the inequality were strict, take $K'$ to be the actual fixed field of automorphims so $|Aut(L/K)|=|L/K'|$ because now THIS is Galois, and use the multiplicativity of degree $|L/K'||K'/K|=|L/K|$ to conclude $K\neq K'$.

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  • $\begingroup$ Thank you, I'll take a look at the proof you mentioned in the book. Just to confirm, you're saying that $|Aut(L/K)|=|L/K|\iff L/K$ is a galois extension. $\endgroup$ – Andrew Brick Nov 12 '14 at 0:05
  • $\begingroup$ Yes, this is correct. $\endgroup$ – guest Nov 12 '14 at 0:08

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