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Algebraic and normal field extensions.

  1. Jul 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex]K[/itex] be an algebraic field extension of a field [itex]F[/itex], and let [itex]L[/itex] be a subfield of [itex]K[/itex] such that [itex] F \subseteq L[/itex] and [itex]L[/itex] is normal in F. Show that if [itex]\sigma[/itex] is an automorphism of [itex]K[/itex] over [itex]F[/itex], then [itex]\sigma(L)=L[/itex].




    2. Relevant equations



    3. The attempt at a solution

    I've been thinking about this for a while, but I couldn't really prove anything. So I'm just looking for a hint that will help me get started...
     
  2. jcsd
  3. Jul 29, 2013 #2

    micromass

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    What is your definition of a normal extension?
     
  4. Jul 29, 2013 #3
    If [itex]f \in F[x][/itex] has one root in [itex]L[/itex], then it splits over [itex]L[/itex].
     
  5. Jul 29, 2013 #4

    micromass

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    Take an element ##a\in L##. Use that ##K## is algebraic to find a polynomial ##f\in F[x]## such that ##f(a)=0##. What can you tell about ##f(\sigma(a))##?
     
  6. Jul 29, 2013 #5
    I'm guessing that I can assume that ##\sigma## fixes ##F## since it says "##\sigma## is an automorphism of ##K## over ##F##". If that's true, then

    ##f(a)=b_0 + b_1a + b_2a^2 + ... + b_na^n = 0##

    So,

    ##f(\sigma(a))= b_0 + b_1\sigma(a) + ... + b_n\sigma(a) = \sigma(b_0+b_1a+...+b_na^n)= \sigma(0)=0.##

    So ##\sigma(a)## must be a root of ##f##. But since #L# is normal, it must contain all the other roots of ##f##. So, basically, ##\sigma## just permutes the roots of a given polynomial over ##F## with roots in ##L##, meaning that ##\sigma(L)=L##. Is that correct?
     
  7. Jul 29, 2013 #6

    micromass

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    Yes, that's what I had in mind.
     
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