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Math Help - Binomial Theorem and Commutative Rings

  1. #1
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    Binomial Theorem and Commutative Rings

    If R is a ring of characteristic p, prove (x+y)^p = x^p + y^p for all x,y in R. We need to use the rule that for each n>= 1, (x+y)^n = (sigma, k=0 to infinity)(n choose k) x^k y^(n-k).

    I truly don't know where to start other that maybe setting the summation = x^p + y^p
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  2. #2
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    Quote Originally Posted by KrazEcat01 View Post
    If R is a ring of characteristic p, prove (x+y)^p = x^p + y^p for all x,y in R. We need to use the rule that for each n>= 1, (x+y)^n = (sigma, k=0 to infinity)(n choose k) x^k y^(n-k).

    I truly don't know where to start other that maybe setting the summation = x^p + y^p
    Indeed we have to use the binomial theorem. By the way, notice that for the binomial theorem to hold in a ring R, it is very important that the ring is commutative.

    So (notice you wrote \infty instead of p at the top of the sum):

    (x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k}.

    Notice that in this sum, we get for k=0 the term y^p, and for k=p, we get the term x^p. So if all other terms vanish, then we are done.

    The ring has characteristic p, so any term, which p divides, will vanish. The following result holds true:

    Lemma: If 0<k<p, then p divides the binomial coefficient \binom{p}{k}.

    Notice how this lemma implies that every term of the sum above vanishes, except for x^p and y^p.

    You might know this lemma from your book, otherwise let me know.
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