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

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

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

    Now, I am having trouble showing that $\displaystyle \omega$ is a root of g(x) for $\displaystyle f(x)=(x^p - 1)g(x)$. 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 $\displaystyle \omega $ into $\displaystyle g(x) $ results in the sum of all $\displaystyle p^{th} $ roots of unity. This always equals $\displaystyle 0 $ for all $\displaystyle p $.

    Your original question though was to show $\displaystyle \omega $ is a root of $\displaystyle f(x) $. This is easy to see since $\displaystyle 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 $\displaystyle \omega $ into $\displaystyle g(x) $ results in the sum of all $\displaystyle p^{th} $ roots of unity. This always equals $\displaystyle 0 $ for all $\displaystyle 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 $\displaystyle g(\omega) = 0 $.

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

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

    Then we have $\displaystyle f(\omega) = 0 = a \cdot g(\omega) $. Since $\displaystyle a \neq 0 $, this implies $\displaystyle g(\omega) = 0 $. Now one must realize $\displaystyle g(\omega) $ is the sum of all $\displaystyle 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 $\displaystyle \omega = e^{(2\pi i)/p^2}$ and $\displaystyle f(x)= x^{p^2} - 1 $ where p is a prime > 2.

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

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

    Now, I am having trouble showing that $\displaystyle \omega$ is a root of g(x) for $\displaystyle f(x)=(x^p - 1)g(x)$. 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 $\displaystyle \omega$ is a primitive root of unity of order $\displaystyle p^2$ , and we have, using the sum of a geometric series :

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

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


    Tonio
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