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Conjugacy of Galois subgroups

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Let K be a Galois extension of F. Two intermediate fields E and L of field F are said to be conjugate if there exists
    [tex]\sigma\in\text Gal_F K [/tex] such that [tex]\sigma (E) = L[/tex].
    Prove that E and L are conjugates of F if and only if [tex]\text Gal_E K[/tex] and [tex]\text Gal_L K[/tex] are conjugate subgroups of [tex]\text Gal_F K[/tex].

    3. The attempt at a solution
    From left to right, I have it already. I can't figure out how to get anywhere going from the right to left part of the proof.
    I want to show that [tex]\sigma (E)=L [/tex] for some [tex]\sigma\in Gal_F K[/tex].
    Let [tex]\alpha\in Gal_L K , \beta\in Gal_E K[/tex].
    Since [tex]Gal_L K, Gal_E K [/tex]are conjugates, then [tex] Gal_E K=\{\sigma\alpha\sigma^{-1} | \alpha\in Gal_L K\}[/tex] or [tex] Gal_L K=\{\sigma^{-1}\beta\sigma | \beta\in Gal_E K\}[/tex].

    [tex]\alpha=\sigma\beta\sigma^{-1}[/tex]
    [tex]\beta=\sigma^{-1}\alpha\sigma[/tex]

    I know that [tex]\beta[/tex] fixes E and that [tex]\alpha[/tex] fixes L.

    [tex]\alpha (L)=\sigma\beta\sigma^{-1} (L)[/tex]
    [tex]\beta (E)=\sigma^{-1}\alpha\sigma (E)[/tex]

    [tex]L=\sigma\beta\sigma^{-1} (L)[/tex]
    [tex]E=\sigma^{-1}\alpha\sigma (E)[/tex]

    [tex]\sigma^{-1} (L)=\beta (\sigma^{-1} (L))[/tex]
    [tex]\sigma (E)=\alpha (\sigma (E))[/tex]

    [tex]\beta [/tex] fixes [tex]\sigma^{-1} (L)[/tex] and [tex]\alpha[/tex] fixes [tex]\sigma (E)[/tex]

    [tex]\beta\in Gal_{\sigma^{-1} (L)} K[/tex] and [tex]\alpha\in Gal_{\sigma (E)} K[/tex]

    From earlier: [tex]\beta\in Gal_E K[/tex] and [tex]\alpha\in Gal_L K[/tex]

    So the only conclusion that I have is that the [tex]Gal_{\sigma (E)} K \cap Gal_L K [/tex] and [tex] Gal_{\sigma^{-1} (L)} K \cap Gal_E K[/tex] are nontrivial.

    I feel like I should be using the Galois extension, i.e. Galois correspondence, to my advantage here, but I just don't see how it is applicable. It gives me that there exists a isomorphism from the intermediate fields to their respective Galois subgroup, i.e. [tex]\tau (E)=Gal_E K[/tex]. And since isomorphisms are order preserving, I get [tex]\left| {Gal_E K} \right|=\left| {E} \right|[/tex] and [tex]\left| {Gal_L K} \right|=\left| {L} \right|[/tex].

    Any direction on the problem is appreciated.

    Thanks in advance.
     
    Last edited: May 7, 2009
  2. jcsd
  3. May 7, 2009 #2
    Shouldn't that argument be carried out on elements of E (and elements of L) rather than on the set E itself (resp. L)? Still get the same conclusion though.

    No, better than that, didn't you show

    [tex]Gal_L K\subset Gal_{\sigma (E)} K [/tex] and [tex] Gal_E K\subset Gal_{\sigma^{-1} (L)} K[/tex] ?


    Would it help now to use [tex]G_1 \subset G_2[/tex] implies [tex]K_{G_1} \supset K_{G_2}[/tex] where [tex]K_G[/tex] denotes the subfield of K fixed by G?
     
  4. May 7, 2009 #3
    Let [tex] G = \textrm{Gal}_L K [/tex] and [tex] H = \textrm{Gal}_E K [/tex]. You have [tex] G = \sigma H \sigma^{-1} [/tex]. Why is it true that if [tex] a \in E [/tex] and [tex] \tau \in G [/tex], then [tex] \tau \circ \sigma (a) = \sigma(a) [/tex]? Why does this tell you that [tex] \sigma(E) \subseteq L [/tex] (hint: fundamental theorem of Galois theory)? (The reverse inclusion is basically the same.)
     
  5. May 8, 2009 #4
    Thank you both for your kind help. I see how [tex]
    \sigma(E) \subseteq L
    [/tex] , but i dont see how to get [tex]
    L \subseteq \sigma(E)
    [/tex]

    Thanks again!
     
    Last edited: May 8, 2009
  6. May 8, 2009 #5
    You can get [tex] \sigma^{-1}(L) \subseteq E[/tex] so then you have it.
     
  7. May 10, 2009 #6
    Wow, yeah, it's so obvious now. Thank you all for your time. I get it now.
     
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