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Math Help - Solution of x^p = 1 in a finite field of char p

  1. #1
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    Solution of x^p = 1 in a finite field of char p

    I was wondering if x^p = 1 implies that x =1 in a finite field of characteristic p, or in other words
    If x^p = 1 (mod p), does that imply that x = 1 (mod p) ?

    If so why?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by petersmith View Post
    I was wondering if x^p = 1 implies that x =1 in a finite field of characteristic p, or in other words
    If x^p = 1 (mod p), does that imply that x = 1 (mod p) ?

    If so why?
    x cannot be =0, otherwise, it wouldn't be possible that x^p=1
    Considering that x is an element of the field, \neq 0, we have by Fermat's little theorem :
    x^p \equiv x (\bmod{p})
    So if x^p \equiv 1 (\bmod{p}), then ...

    Hmmm I think it killed the question... or I've made a mistake ^^'
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  3. #3
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    If \text{char}(F)=p then (x+y)^p = x^p+y^p.
    Thus if x^p = 1 \implies x^p -1 = 0 \implies (x-1)^p = \implies x=1.

    Quote Originally Posted by Moo
    Hmmm I think it killed the question... or I've made a mistake ^^'
    What you did is correct if F=\mathbb{F}_p.
    However, a finite field need not be a the integers modulo p.
    Thus, it does not cover all the cases.

    This is Mine 14th Post!!!
    Last edited by ThePerfectHacker; August 31st 2008 at 08:38 AM.
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