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Math Help - Factor and root of f(x), dealing with e^(2 pi i/p^2)

  1. #1
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    Factor and root of f(x), dealing with e^(2 pi i/p^2)

    Let \omega = e^{(2\pi i)/p^2} and  f(x)= x^{p^2} - 1 where p is a prime > 2.

    Show that \omega is a root of f(x). done...
    Show that (x^p - 1) is a factor of f(x). done...

    I used geometric series to extrapolate f(x)
    = x^{p^2} - 1
    = (x^p)^p - 1
     = u^p - 1
     = (u-1)(1 + u + u^2 + u^3 + ... + u^{p - 1})
     = (x^p - 1)(1 + x^p + x^2p + x^3p + ... + x^{p^2 - p})

    Now, I am having trouble showing that \omega is a root of g(x) for f(x)=(x^p - 1)g(x). When I just try to plug and play in g(x), I don't get anything that heads towards 0. Step by step or a skeleton is fine with me, I just need a bit of a push.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Plugging  \omega into  g(x) results in the sum of all  p^{th} roots of unity. This always equals  0 for all  p .

    Your original question though was to show  \omega is a root of  f(x) . This is easy to see since  f(\omega) = e^{2\pi i}-1 = 0 .
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
    Plugging  \omega into  g(x) results in the sum of all  p^{th} roots of unity. This always equals  0 for all  p .
    Are you able to give me more details as to why this is? This really is the question and the place I get stuck since, as you noted, the other part is easy. Thank you!
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Well, I'd say this question could be used to prove why  g(\omega) = 0 .

    We know  f(x) = (x^p-1)g(x) and  f(\omega) = 0 .

    All you need to do is to verify that  \omega^p-1 = a \neq 0 , which is easy to see.

    Then we have  f(\omega) = 0 = a \cdot g(\omega) . Since  a \neq 0 , this implies  g(\omega) = 0 . Now one must realize  g(\omega) is the sum of all  p^{th} roots of unity.
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  5. #5
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    Thank you for having patience and pointing that out. I should have taken a break I guess and come back to it.
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  6. #6
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    Quote Originally Posted by tcRom View Post
    Let \omega = e^{(2\pi i)/p^2} and  f(x)= x^{p^2} - 1 where p is a prime > 2.

    Show that \omega is a root of f(x). done...
    Show that (x^p - 1) is a factor of f(x). done...

    I used geometric series to extrapolate f(x)
    = x^{p^2} - 1
    = (x^p)^p - 1
     = u^p - 1
     = (u-1)(1 + u + u^2 + u^3 + ... + u^{p - 1})
     = (x^p - 1)(1 + x^p + x^2p + x^3p + ... + x^{p^2 - p})

    Now, I am having trouble showing that \omega is a root of g(x) for f(x)=(x^p - 1)g(x). When I just try to plug and play in g(x), I don't get anything that heads towards 0. Step by step or a skeleton is fine with me, I just need a bit of a push.

    Your \omega is a primitive root of unity of order p^2 , and we have, using the sum of a geometric series :

    g(\omega)=a+\omega^p+\omega^{2p}+\cdots +\omega^{p^2-p}=1+\omega^p+\left(\omega^p\right)^2+\cdot +\left(\omega^p\right)^{p-1}=\frac{\omega^{p^2}-1}{\omega^p-1}=0

    Of course, it's way easier if we note that \omega is a root of f(x)=(x^p-1)g(x) , and since \omega^p\neq 1 then it must be a root of g(x)...


    Tonio
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