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

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

    proof so far:

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

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

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

    Am I doing this right so far?

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

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

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

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

    proof so far:

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

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

     p \omega ^{p-1} = ( \omega - 1) f'( \omega) + 1
     \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 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 disc( \omega ) = -p^{p-2} , in addition,  disc ( \omega ) = p^ {p-2} iff  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  \{ \omega ^{1}, \omega ^{2}, . . . , \omega ^{p-1} \}
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