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Nilpotent element

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data

    An element a of a ring is nilpotent if a^n = 0 for some positive integer n.
    Prove that R has no nonzero nilpotent elements if and only if 0 is the unique
    solution of the equation x^2 = 0

    2. Relevant equations

    I think nilpotent means that not only that a^n = 0 for some positive integer
    n but that n is the least positive integer such that a^n = 0. ?


    3. The attempt at a solution

    Proving ->
    if R has no nonzero nilpotent elements then there is no nonzero element a such that a^n = 0, and specifically in the case of n = 2, there is no such a.
    Hence 0 is the unique solution of x^2 = 0


    Proving <-(via contradiction)
    if 0 is the unique solution of x^2 = 0 then
    if R does have some nonzero nilpotent element a, then a^n = 0 for
    some positive inetger n.

    then either
    n is even where a^n = a^(n/2) * a^(n/2) = 0
    but since 0 is the unique solution of x^2 = 0, then
    a^(n/2) = 0. Which contradicts the fact that n is the least
    positive integer for which a^n = 0.

    Or n is odd, where
    a^n = 0 ---> (a^n)*(a) = 0 * (a) (multiplying both sides by a)
    ---> a^(n+1) = 0, now n+1 is an even integer,
    which can then be factored into a square
    so a^(n+1/2) * a^(n+1/2) = 0, but since 0 is the unique
    solution of x^2 = 0, then a^(n+1/2) = 0, which contradicts the
    fact that n is the least positive integer for which a^n = 0.

    I would very much appreciate any evaluation of my attempt. Especially because I feel unsure of the n is odd case, since I just multiplied both sides by a so i could get the square.
     
  2. jcsd
  3. Feb 22, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What could be wrong with that? I think you are just lacking confidence. You may want to say explicitly that a^n=0 and a^m!=0 for m<n. As you say, make sure n is the least nilpotent power of a.
     
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