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Math Help - Boolean Ring

  1. #1
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    Boolean Ring

    For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R

    What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4?
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  2. #2
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    Quote Originally Posted by knguyen2005 View Post
    For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R

    What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4?
    well, if a^n = a, for all a \in R, then (2^n -2)a=0, \ \forall a \in R. because for any a \in R: 2^n a = (2a)^n = 2a. that's jsut trivial! a non-trivial (even hard probably) question is this:

    suppose R \neq (0) is a ring with identity element and n \geq 3 is the smallest integer for which a^n = a, \ \forall a \in R. what is the smallest integer m for which ma=0, \ \forall a \in R?

    for example if n=3, would it be possible to have m=2 or 3?
    Last edited by NonCommAlg; October 17th 2009 at 07:06 PM.
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    Thanks for a quick reply

    If we use the formula
    Then for the case n = 2, we have : 2a = 0 implies that a + a = 0. So the formula works
    when n = 3, 6a = 0 hence 3a+3a = 0

    But when I do the calculation:
    Let a in R, a^3 = a ad so (-a)^3 = (-a) by hypothesis
    But (-a)^3 = (-a)(-a)(-a) = (a^2)(-a)
    Thus, (-a) = (a^2)(-a) hence a^2 = 1.

    What am I doing wrong here?
    Why did I not get the result 6a = 0 when n = 3?

    Thanks again
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  4. #4
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    Quote Originally Posted by knguyen2005 View Post

    Thus, (-a) = (a^2)(-a) hence a^2 = 1.
    how did you get that? also we can't cancel -a out because -a is not necessarily a unit in R.
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    I knew my calculation was wrong, I just want to do another way to get to the answer.

    If I dont want to use the formula when n = 3. How am I supposed to get to the answer 6a = 0 ? I mean can we do it another way instead of the formula u gave?

    Thank you so much for your valuable time
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  6. #6
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    Quote Originally Posted by knguyen2005 View Post
    I knew my calculation was wrong, I just want to do another way to get to the answer.

    If I dont want to use the formula when n = 3. How am I supposed to get to the answer 6a = 0 ? I mean can we do it another way instead of the formula u gave?

    Thank you so much for your valuable time
    a+a=(a+a)^3 = (2a)^3=8a^3=8a. so 8a=2a and thus 6a=0.
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