- #1
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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.
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.