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Integral Domain, r^2 = r proof that r = 0 or 1

  1. Feb 14, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let r be an element of an integral domain R such that r^2 = r. Show that either r = 0_R or 1_R

    2. Relevant equations
    integral domain means no zero divisors.

    3. The attempt at a solution

    This is fundamental as 0 and 1 solve r^2 = r and are the only solutions.

    However, I'm not really sure what I can play with to show this fact.

    We have r*r = r
    There are no zero divisors so no such thing as r*s = 0.

    If we start this proof off by assuming r is not 1_R or 0_R then maybe we could get somewhere, but It doesn't feel promising.

    There has to be a way to use the fact that r^2 = r.

    (r^2)*r = r^2 ?
    Then we are left with r*r = r, that's no good.

    Is there any trick that I am neglecting? I feel this is just a simple multiplication of two things and boom we have r(r^2) = 0 or 1 or something.

    How about r^2 - r = r-r
    then we have
    r(r-1)=0
    r is either 0 or 1.

    BOOM?
     
  2. jcsd
  3. Feb 14, 2016 #2

    RJLiberator

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    When the problem talks about such things as 0_R and 1_R is it safe to assume that they just mean 0_R = 0 and 1_R =1 ?

    For instance, my proof has the following:

    r^2 = r
    r^2 - r = r - r
    r(1-r) = 0
    Therefore r = 0 or r = 1

    But is that a complete proof in consideration of 0_R and 1_R ?
     
  4. Feb 14, 2016 #3

    Samy_A

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    The last one is correct. BOOM! :oldsmile:
     
  5. Feb 14, 2016 #4

    RJLiberator

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    Samy_A do you have a moment to check out my reply? I am wondering on the terminology of this problem. :)
     
  6. Feb 14, 2016 #5

    Samy_A

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    Yes. 0_R is the identity element for addition, 1_R is the identity element for multiplication. It is common to simply write them as 0 and 1, as long as it is clear what ring we are talking about.
     
  7. Feb 14, 2016 #6

    HallsofIvy

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    "[itex]0_R[/itex]" and "[itex]1_R[/itex]" specifically mean "the additive identity" and the "multiplicative identity" of the ring, R, respectively. I believe SamyA is saying that it is alright to drop the "R" if it is clear what ring you mean. You cannot assume that they are the numbers 0 and 1 if that was what you meant.
     
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