The galois group of x^4 + 1 over Q

  • Thread starter Daveyboy
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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.
 

Answers and Replies

  • #2
matt grime
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Then I want to permute these roots so the Galois group is just S4.
...but what I can not find what is wrong with the logic.
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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