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Boolean Ring

  1. Dec 5, 2016 #1
    1. The problem statement, all variables and given/known data
    If ##a^2 = a## for all ##a \in R##, then ##R## is commutative.

    2. Relevant equations


    3. The attempt at a solution
    I have been working on this problem for a few hours without any success; I literally have nothing but a's and b's scrawled over a bunch of papers. I could use a hint.
     
    Last edited: Dec 5, 2016
  2. jcsd
  3. Dec 5, 2016 #2

    fresh_42

    Staff: Mentor

    Have you calculated ##(a-b)^2## on your papers? And ##(1+1)^2##.
     
    Last edited: Dec 5, 2016
  4. Dec 5, 2016 #3
    I just did now, but I don't see how this is helpful. Here are some of my calculations:

    ##a-b = (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2## or ##a-b = a - ab -ba + b## or ##2b = ab + ba##...

    In fact, here is something strange I encountered: ##(-1)^2 = -1## becomes ##1 = -1##. Isn't this a contradiction? If so, then Boolean rings cannot exist.
     
  5. Dec 5, 2016 #4

    fresh_42

    Staff: Mentor

    No, it's no contradiction. Therefore you should compute ##(1+1)^2=1^2+ 1 \cdot 1+ 1 \cdot 1 + 1^2=1+1##, hence ##1+1=0##, i.e. the characteristic of this ring is two. This means, as you've noticed: ##+1=-1##.
    Now you know what ##b+b## and ##-ab## are.
     
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