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Thread: Primitive pth root of unity of a prime p

  1. #1
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    Primitive pth root of unity of a prime p

    Let $\displaystyle K= \mathbb {Q}[ \omega ] $, where $\displaystyle \omega $ is a primitive pth root of unity for an odd prime p, and write N for $\displaystyle N_{K}$.

    a) Let f be the minimum polynomial of $\displaystyle \omega $. Show that $\displaystyle f'( \omega ) = \frac {p}{ \omega ( \omega - 1)} $

    proof so far:

    Now, $\displaystyle x^p - 1 = (x-1)f(x) $
    Then $\displaystyle px^{p-1} = (x-1)f'(x) + f(x) $

    Plug in $\displaystyle \omega $, then $\displaystyle f( \omega ) =1 $, then we have:

    $\displaystyle p \omega ^{p-1} = ( \omega - 1) f'( \omega) + 1 $
    $\displaystyle \frac {p \omega ^{p-1} - 1} { \omega - 1 } = f' ( \omega ) $

    Am I doing this right so far?

    b) Show that $\displaystyle disc( \omega ) = -p^{p-2} $, in addition, $\displaystyle disc ( \omega ) = p^ {p-2} $ iff $\displaystyle p \equiv 1 \ (mod \ 4) $

    I just can't really get a hand on disc, any hint?

    Thank you!
    Last edited by tttcomrader; Apr 23rd 2008 at 06:46 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let $\displaystyle K= \mathbb {Q}[ \omega ] $, where $\displaystyle \omega $ is a primitive pth root of unity for an odd prime p, and write N for $\displaystyle N_{K}$.

    a) Let f be the minimum polynomial of $\displaystyle \omega $. Show that [tex] f'( \omega ) = \frac {p}{ \omega ( \omega - 1)}

    proof so far:

    Now, $\displaystyle x^p - 1 = (x-1)f(x) $
    Then $\displaystyle px^{p-1} = (x-1)f'(x) + f(x) $

    Plug in $\displaystyle \omega $, then $\displaystyle f( \omega ) =1 $, then we have:

    $\displaystyle p \omega ^{p-1} = ( \omega - 1) f'( \omega) + 1 $
    $\displaystyle \frac {p \omega ^{p-1} - 1} { \omega - 1 } = f' ( \omega ) $

    Am I doing this right so far?
    I would do it a little differently. The mimimal polynomial is $\displaystyle 1+x+...+x^{p-1}$. It happens to be irreducible so that is why it is minimal (you never said why you poynomial was minimal). And then differenciate term-by-term.

    b) Show that $\displaystyle disc( \omega ) = -p^{p-2} $, in addition, $\displaystyle disc ( \omega ) = p^ {p-2} $ iff $\displaystyle p \equiv 1 \ (mod \ 4) $

    I just can't really get a hand on disc, any hint?
    I never seen "disk" before.
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  3. #3
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    Disc is the discriminant, so is the determinant of $\displaystyle \{ \omega ^{1}, \omega ^{2}, . . . , \omega ^{p-1} \} $
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