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Math Help - Show (a + b) ^ p = a^p + b^p

  1. #1
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    Show (a + b) ^ p = a^p + b^p

    Hi I'm trying to show that for a commutative ring with characteristic p that
    (a + b)^p = a^p + b^p

    I have verified this with a few examples but can't seem to prove that the binomial coefficients C(p,r) = 0 (modp) where r /= 0 or p (i.e. the first or last term)

    Can anyone help me out? Cheers =)
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  2. #2
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    Quote Originally Posted by liedora View Post
    Hi I'm trying to show that for a commutative ring with characteristic p that
    (a + b)^p = a^p + b^p

    I have verified this with a few examples but can't seem to prove that the binomial coefficients C(p,r) = 0 (modp) where r /= 0 or p (i.e. the first or last term)

    Can anyone help me out? Cheers =)

    Hints:

    1) Since there's commutativity we can use Newton's binomial theorem: \displaystyle{(a+b)^p=\sum\imits_{k=0}^p\binom{p}{  k}a^{p-k}b^k

    2) Now prove that \displaystyle{\forall 1\leq k\leq p-1\,,\,\,p\mid\binom{p}{k} .

    Tonio
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  3. #3
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    Hi Tonio, the part 2) was where I was having the difficulty, I can't seem to prove that. Any advice of how I might get there?

    Thanks
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  4. #4
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    Quote Originally Posted by liedora View Post
    Hi Tonio, the part 2) was where I was having the difficulty, I can't seem to prove that. Any advice of how I might get there?

    Thanks

    \displaystyle{\binom{p}{k}=\frac{p!}{k!(p-k)!}=\frac{(k+1)(k+2)\cdot\ldots\cdot p}{(p-k+1)(p-k+2)\cdot\ldots\cdot (p-k)}} , and now

    check that no factor in the denominator divides p for <br />
1\leq k\leq p-1 , and since

    the binomial coefficient is always an integer then we're done.

    Tonio
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  5. #5
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    Cheers Tonio, got it now =)
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