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Boolean rings with identity can only take 2 elements?

  1. Mar 27, 2007 #1
    Using the theorem that in any boolean ring a+a=0 for all a in boolean ring R.

    Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.
     
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  3. Mar 27, 2007 #2

    matt grime

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    What? That makes no sense. Mainly because you do not ask a question at all.
     
  4. Mar 27, 2007 #3

    Dick

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    Ok, so '2' isn't in the ring. Why does that mean there is nothing else in the ring? There's no rule that says you can generate everything in a ring by adding 1's.
     
  5. Mar 27, 2007 #4
    I was looking for a confirmation that my claim is correct.

    So far we know 0 and 1 is in R. Take all 4 combinations of these two elements with respect to the 2 operations. So 8 additions and multiplications all together, all of which gives 0 or 1 so R contains only two elements 0 and 1. The only non trivial one is 1+1 which has to be 0 as by the theorem in the OP.
     
    Last edited: Mar 27, 2007
  6. Mar 27, 2007 #5

    Dick

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    Your 'claim' is not only incorrect. It's ridiculous. So all rings are generated by 0 and 1? You had better add that to the axiom list, because its not there yet, last I heard.
     
  7. Mar 28, 2007 #6

    matt grime

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    Here is a meta-answer that shows the claim to be silly. If there is a unique boolean ring, why did you ask about boolean ringS?
     
  8. Mar 28, 2007 #7
    Remember I am talking about a boolean wring with multiplicative identity.

    I am claiming there is only one unique boolean ring with multiplicative identity.

    Or with even this consideration taken into account, it is not correct? If so why?

    If the boolean ring dosen't have multiplicative identity than other combinations may be possible.
     
    Last edited: Mar 28, 2007
  9. Mar 28, 2007 #8

    HallsofIvy

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    I don't see why those prevent other elements. What about a set {0, 1, a, b} with operation tables:
    addition:
    0 1 a b
    0 0 1 a b
    1 1 0 b a
    a a b 0 1
    b b a 1 0

    multiplication
    0 1 a b
    0 0 0 0 0
    1 0 1 a b
    a 0 a b 1
    b 0 b 1 a
     
  10. Mar 28, 2007 #9

    matt grime

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    and?

    but that is patently silly.

    Just write down something to produce a counter example. It is trivial to produce such example. Hell, diagonal matrices with 0s and 1s on the diagonals gives infinitely many counter examples without having to think at all.
     
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