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Math Help - Min polynomial

  1. #1
    Ant
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    Min polynomial

    Hi,

    I'm trying to find the minimal polynomial of the 12th primitive root of unity over \mathbb{Q}

    Is there a best way of going about this?

    Thanks,
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  2. #2
    MHF Contributor

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    Re: Min polynomial

    I would start by finding the primitive roots then constructing the minimal polynomial from that.

    The twelfth roots of unity, of course, satisfy the equation x^{12}= 1. They are of the form cos(n\pi/6)+ i sin(n\pi/6) with n from 0 to 11: 1, \sqrt{3}/2+ i/2, 1/2+ i\sqrt{3}/2, i, -1/2+ i\sqrt{3}{2}, -\sqrt{3}/2+ i/2, -1, -\sqrt{3}/2- i/2, -0.5- i\sqrt{3}/2, -i, 1/2- i\sqrt{3}/2, and \sqrt{3}/2- i/2.
    A primitive root is one that does NOT satisfy x^n- 1= 0 for any smaller value of n which removes 1, -1, i, and -i. What about cube roots?

    It might help to note that x^{12}= 1 is equivalent to
    x^{12}- 1= (x^3)^2- 1= (x^3- 1)(x^3+ 1)= (x- 1)(x^2+ x+ 1)(x+ 1)(x^2- x+ 1)= 0.
    Last edited by HallsofIvy; May 18th 2013 at 03:39 PM.
    Thanks from Ant
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  3. #3
    Ant
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    Re: Min polynomial

    Perhaps my terminology isn't quite right, by the "12th primitive root of unity" I meant:

     e^{i\pi/6} =: \omega

    So I am trying to find [\mathbb{Q}(\omega):\mathbb{Q}]

    As \omega is a root of x^{12}-1 we have that the min poly divides x^{12}-1.

    I think perhaps the factorisation is off by power 2?

    x^{12} -1 = (x^6)^2-1 = (x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1)

    Hence, the min polynomial is one of the above bracketed terms:

     \omega^4 = \dfrac{1}{2}(-1+\sqrt3 i) \  \omega^2=\dfrac{1}{2}(1+ \sqrt3 i)

    Hence x^4-x^2+1 is the min poly.

    Thanks!
    Last edited by Ant; May 18th 2013 at 03:58 PM.
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  4. #4
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    Re: Min polynomial

    Quote Originally Posted by HallsofIvy View Post
    It might help to note that x^{12}= 1 is equivalent to
    x^{12}- 1= (x^3)^2- 1
    Little mistake here but you can still go through the whole difference of 2 squares process with the higher index.
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