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Roots of unity proof

  1. Jan 8, 2007 #1
    I'm having trouble following one step in a proof I'm studying. I'm sure I'm missing something obvious, but I just can't get it to work out (it supposed to be "obvious" which is why they left out the details).

    Anyway, it's part of a proof showing that if you have a monic polynomial with all integer coefficients and all the zeros have absolute value 1 then all the zeros are roots of unity.

    So say we have our polynomial:


    A simple argument is then used to show that you can create new monic integer polynomials from this by raising the alpha's to powers. For example:


    Then it's easy to put a finite limit on the the absolute value of the coefficients, from which you get a finite set of inequalities which have a finite set of solutions, and then it is an easy leap to see that there must be two different of these "power polynomials" that are equal, for different m.

    So far so good (rushing through the proof here). We then create a permutation function [itex]\pi[/itex] so we end up with for m not equal to g:

    [itex]\alpha^m_j = \alpha^g_{\pi (j)}[/itex]

    And then comes the next step that I don't really see. I think you only need the above, and not all the background. But my book claims that a simple induction argument gives you from the above that:

    [itex]\alpha^{m^r}_j = \alpha^{g^r}_{\pi^r (j)}[/itex]

    And I just can't see how to get this. The rest of the proof is also easy. It's just this one step that is bugging me. So I would really appreciate an explanation. Thanks.
  2. jcsd
  3. Jan 8, 2007 #2
    nevermind, it was trivial. I just had a blind spot.
  4. Jan 8, 2007 #3

    Gib Z

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    Homework Helper

    Lol i've had those before, It laster much longer and it was much more obvious though. It was proving the identity for cos(x-y) and I couldnt see why the angle between the y angle and x angle was x-y...took me 6 months to realize :D
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