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If a.a=a prove R is commutative

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    If a.a=a for all elements of R, prove R is commutative.

    3. The attempt at a solution

    (a+b)2=a+b=a2+ba+ab+b2=a+ba+ab+b

    Then -ba=ab

    Any suggestions of how to show that -b=b?
     
  2. jcsd
  3. Nov 11, 2012 #2

    HallsofIvy

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    Are we to assume, here, that R is a group? Then it is always true that ae= a (e is the identity) so if it is also true that aa= a, then we have aa= ae and, multiplying on the left by the inverse of a, a= e. So we have only the "trivial" group containing only the identity.

    If R is not a group, what sort of algebraic entity is it?
     
  4. Nov 11, 2012 #3
    Sorry I should have been more clear but R is ring. So we are trying to show that . is a commutative operation and therefore making (R,+,.) a commutative ring.(At least that is my understanding of a commutative ring)

    If I understand correctly I don't think we can assume that R has a multiplicative identity(1).
     
  5. Nov 11, 2012 #4

    micromass

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    The structure in your OP is called a Boolean ring. You can prove that any element a in your Boolean ring satisfies a+a=0.

    Prove this by considering (a+a)(a+a)
     
  6. Nov 11, 2012 #5
    (a+a)(a+a)=a+a
    a2+a2+a2+a2=a+a
    a+a+a+a=a+a
    a+a=0
    a=-a for all a in R

    Thanks, that makes sense.
     
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