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The galois group of x^4 + 1 over Q

  1. May 19, 2009 #1
    This thing splits if we adjoin e^ipi/4.
    Let [tex]\zeta[/tex]=e^ipi/4 =[tex]\frac{\sqrt{2}}{2}[/tex]+[tex]\frac{i\sqrt{2}}{2}[/tex]
    so x4+1=

    (x-[tex]\zeta[/tex])(x-[tex]\zeta[/tex]2)(x-[tex]\zeta[/tex]3)(x-[tex]\zeta[/tex]4).

    Then I want to permute these roots so the Galois group is just S4.

    But, Q([tex]\zeta[/tex])=Q(i,[tex]\sqrt{2}[/tex]) and [Q(i,[tex]\sqrt{2}[/tex]):Q]=4 (degree)

    I have the theorem that Galois group [tex]\leq[/tex] degree of splitting field over base field.

    Since |S4|=24 something is wrong, but what I can not find what is wrong with the logic.
     
  2. jcsd
  3. May 20, 2009 #2

    matt grime

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    What's wrong is that you've forgotten the definition of a Galois group - it is the group of field automorphisms, not just the arbitrary permutation of roots.
     
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