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(Relatively) Trivial question about extension field degrees (pertains to Galois theor

  1. May 7, 2012 #1
    So if we have an extension of E of F, then we can consider E as a vector space over F.
    The dimension of this space is the degree of the field extension, I think most people use [E:F].
    This is correct in most people's books, right?

    Defining [itex] \omega = cos (2\pi /7) + i sin (2\pi / 7) [/itex]

    Why is it that:
    "Since [itex] x^7 -1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1) [/itex],
    |Gal(Q(ω)/Q)|=[Q(ω):Q]≤ 6"
    ?
    With another piece of information he gets that the degree is 6. Since the vector space of Q(ω) over Q has degree 6, we should be able to represent it with 6 basis vectors...but this just feels weird since ω is a seventh root of unity and it feels like we should have something like:
    a+bω+cω2+...gω6 a,b,...,g in Q, right?

    What am I missing? I have no reason to doubt Gallian, I even looked up the general method for calculating degrees of Q(n-th roots of unity)/Q and I know that 6 is correct. . .I just don't see why and I can't construct 6 basis vectors.

    [This is from Gallian's Abstract Algebra book, 7th edition, in Ch. 32 on page 550]
     
  2. jcsd
  3. May 7, 2012 #2
    Re: (Relatively) Trivial question about extension field degrees (pertains to Galois t

    ....
     
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