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Math Help - Galois Theory Question

  1. #1
    Junior Member Mathhead200's Avatar
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    Galois Theory Question

    Let E = Q(α,ω), α = 21/3, ω = e2πi/3, (Q is the field of rationals)
    Is E a simple extension of Q?

    My attempt:
    Let β = α + ω, find a monic irreducible polynomial, p in Q[x] with root β.
    p should have degree 6, implying E = Q(β)

    β3 = (α + ω)3
    = α3 + 3α2ω + 3αω2 + ω3
    = 2 + 3α2ω + 3αω2 + 1
    = 3αω(α + ω) + 3
    = 3αωβ + 3
    β3 – 3 = 3αωβ
    3 – 3)3 = (3αωβ)3
    3)3 + 3(β3) 2(-3) + 3(β3) (-3)2 + (-3)3 = 33α3ω3β3
    β9 - 9β6 + 27β3 – 27 = 9(2)(1)β3
    = 18 β3
    β9 - 9β6 + 9β3 – 27 = 0

    However, if I define p(x) = x9 – 9x6 + 9x3 – 27 = 0, my calculator tells me p(21/3 + e2πi/3) ≠ 0.
    Also I can’t see how to reduce p such that p = fg with f,g in Q[x], deg(f) = 6, and f(β) = 0.
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  2. #2
    Member Sylvia104's Avatar
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    Re: Galois theory question

    Try \beta=\alpha\omega. Then \beta is a root of x^6-4.
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  3. #3
    Junior Member Mathhead200's Avatar
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    Re: Galois theory question

    x^6 - 4 = (x^3 - 2)(x^3 + 2)
    In fact, E is a splitting field for x^3 - 2 = (x - \alpha)(x - \alpha\omega)(x - \alpha\omega^2)

    I need an irreducible polynomial in Q
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  4. #4
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    Re: Galois Theory Question

    i have a different idea, prove Q(α+ω) = Q(α,ω). obviously Q(α,ω) contains Q(α+ω). so if we can show α,ω are in Q(α+ω), we're done. in fact all we NEED to do is show ω is in Q(α+ω), because then: α = α+ω - ω is also in Q(α+ω).

    note that: 0 = α3 - 2 = (α+ω - ω)3 - 2 = (α+ω)3 - 3(α+ω)2ω + 3(α+ω)ω2 - ω3 - 2

    = (α+ω)3 - 3(α+ω)2ω + 3(α+ω)(-ω-1) - 3 (since ω2+ω+1 = 0)

    = (α+ω)3 - 3(α+ω)2ω - 3(α+ω)ω - 3(α+ω) - 3, so that:

    ω(3(α+ω)2+3(α+ω)) = (α+ω)3 - 3(α+ω) - 3, thus:

    ω = [(α+ω)3 - 3(α+ω) - 3]/(3(α+ω)2+3(α+ω)), which is in Q(α+ω).

    by the way, there is an error in your p(x), it should be:

    p(x) = x9-9x6-27x3-27 (unless i've also made an error. could be.)
    Thanks from Mathhead200
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