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Artin's lemma in Galois Theory: when is [E:F] < |G|?

  1. May 14, 2014 #1
    I am trying to understand Galois Theory and reading through various theorems and lemmas, some of which are still confusing me.

    A lemma proved by Artin states that if F is the fixed field of a finite group G of automorphisms in a field E, then the degree [E:F] ≤ |G| = n. The proof relies on setting up a linear dependence relationship for any set of n+1 elements in E, using n equations in n+1 unknowns with coefficients obtained by every permutation in G.

    I am trying to get an intuitive understanding of the result. I already know that in the case where G is the Galois group G(E/F) and E is a splitting field of a separable polynomial over F, then F is the fixed field of G and [E:F] = |G|. But what properties of E and G could cause [E:F] < |G|?

    I can't find any example where strict inequality arises.
     
  2. jcsd
  3. May 14, 2014 #2
    You can't find a counterexample because I believe none exists. In fact, given any set [itex] S [/itex] of [itex] n[/itex] distinct isomorphisms of a field [itex] E_1 [/itex] into a field [itex] E_2[/itex] (here they don't even need to form a group just some set of isomorphisms), then [itex] [E:\mathrm{Fix}(S) ] \geq n[/itex].

    Whatever source you are using probably split the proof that if [itex] S[/itex] is a group then [itex] [E:\mathrm{Fix}(S) ] =n[/itex] into separate lemmas so I would imagine there is also a proof of the above fact nearby the proof of the other inequality in your book.
     
  4. May 14, 2014 #3
    Yes, I think you are right. I found out later in the book, they prove that if F is a fixed field of E under the finite group of automorphisms of E, then E is a finite, normal, separable extension of F (in fact the two properties are equivalent). This implies that [E:F] = |G(E/F)| ≥ |G|, which is the other inequality.

    I was scratching my head for a while before I realized this.

    An interesting side-effect of [E:F] = n is that it proves the set of all automorphisms of G is linearly independent over F, since from Artin's proof, the nxn system of equations must have only the trivial solution.
     
    Last edited: May 14, 2014
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