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Help on finding Galois group

  1. Dec 11, 2007 #1
    I was asked to find the Galois group of [tex]x^8-1[/tex] over Q,

    I first find all the roots to it :
    [tex]\pm i[/tex] , [tex]\pm \sqrt{i}[/tex] , [tex]\pm i \cdot \sqrt{i}[/tex], [tex]\pm 1[/tex].

    Then since [tex]i \cdot \sqrt{i}[/tex] is just a multiple of i and sqrt(i)
    so I had Q(i, sqrt(i)) being the splitting field for the equation over Q.
    Next, [Q(i, sqrt(i)) :Q] = 4, so I conclude that the Galois group is a cyclic group of order 4.

    is the above correct? If not, can someone please tell me what's wrong? Thanks!
     
    Last edited: Dec 12, 2007
  2. jcsd
  3. Dec 12, 2007 #2

    morphism

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    It's not going to be the cyclic group of order 4.

    In general I would avoid writing [itex]\sqrt{i}[/itex]; wich square root of i are you taking?

    The roots of x^8 - 1 are going to be the 8th roots of unity, so the splitting field will be generated over Q by the primitive 8th root of unity [itex]\zeta_8 = e^{2 \pi i / 8}[/itex]. That is, the splitting field is going to be [itex]\mathbb{Q}(\zeta_8)[/itex]. Consequently [itex][\mathbb{Q}(\zeta_8) : \mathbb{Q}] = ???[/itex] and [itex]\text{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong ???[/itex].

    Try to take it from here.
     
  4. Dec 12, 2007 #3

    HallsofIvy

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    The square roots of i are [itex](\sqrt{2}/2)(1\pm i)[/itex].
     
  5. Dec 12, 2007 #4
    thanks~ pls tell me this is right

    for the 8 roots, eliminating the ones that can be expressed as multiples of others, i am left with [tex]e^{2\pi i/8}[/tex] and [tex]e^{\pi i/2}[/tex]=i where their orders are 4 and 2 respectively so i get a splitting field over Q of order 8 =>Gal( Q( [tex]e^{2\pi i/8}[/tex] , [tex]e^{\pi i/2})[/tex] / Q) [tex]\cong[/tex] cyclic group of order 8 (D4)??
     
  6. Dec 12, 2007 #5

    morphism

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    [tex](e^{2 \pi i /8})^2 = e^{\pi i /2}[/tex]

    The splitting field is Q(e^(2pi*i/8)) which is of order 4 over Q. So the Galois group is either C_2 x C_2 or C_4. It's not going to be C_4 (why?).

    [If you're comfortable with cyclotomic extensions, then this amounts to the fact that [itex]\text{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong (\mathbb{Z}/8\mathbb{Z})^* \cong C_2 \times C_2[/itex].]
     
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