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Ring question

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Let R be a ring and suppose there exists a positive even integer n such that x^n = x for

    every x in R. Show that -x = x for every x in R.

    2. Relevant equations

    3. The attempt at a solution
    I solved the case where n = 2.

    Let x be in R.

    (x+x)^2= x+x = 2x,

    (x+x)^2 = 4x^2 = 4x.

    So 4x = 2x and 2x = 0. Done.

    I tried using this same method when n = 4 and got nowhere.
  2. jcsd
  3. May 27, 2009 #2
    Let [tex] n = 2k [/tex]. What's [tex] (-x)^{2k} [/tex]?

    (By the way, the proof you have for [tex] n = 2 [/tex] doesn't work for noncommutative rings. The above hint suggests a method that does. Can you see why?)
  4. May 28, 2009 #3
    Thanks for the hint. That works. But why does my proof for n=2 not work for noncommutative rings? Since the only terms in the expansion of (x+x)^2 are powers of x, I don't think I ever used commutativity.
  5. May 28, 2009 #4
    Yeah...you're right. I read your work as [tex] (x + x)^2 = (2x)^2 = 2^2 x^2 = 4 x^2 [/tex]; my only point was that in a noncommutative ring, [tex] (ab)^k \neq a^k b^k [/tex] in general. However, I suppose it's true that [tex] [(n \cdot 1) b]^k = (n^k \cdot 1) b^k [/tex] for natural numbers [tex] n [/tex].
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