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Math Help - Commutative Ring

  1. #1
    pkr
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    Commutative Ring



    Help would be appreciated.
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  2. #2
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    We will first prove that if b^2 = 0 then 1+b is a unit. This is simple, (1+b)(1-b) = 1 - b^2 = 1. Thus, 1+b,1-b are inverses of eachother. Say that a is a unit. Since b^2 = 0 it means a^{-2}b^2 = 0 and so (a^{-1}b)^2 = 0 since the ring is commutative. Using the above result it means 1 \pm a^{-1}b is a unit. But then a(1\pm a^{-1}b) = a\pm b is a unit.
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  3. #3
    pkr
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    So for the 2X2 Matirx would you take B to be the matrix (0 1, 0 0) as that squared = (0 0, 0 0)

    This is confusing me slightly, I haven't seen the notation of a "unit" in this context before...What exactly would the conditions be I need to satisfy for A-B to not be a unit?
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  4. #4
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    Quote Originally Posted by pkr View Post
    So for the 2X2 Matirx would you take B to be the matrix (0 1, 0 0) as that squared = (0 0, 0 0)

    This is confusing me slightly, I haven't seen the notation of a "unit" in this context before...What exactly would the conditions be I need to satisfy for A-B to not be a unit?
    For matrices the identity is the matrix I = (1 0, 0 1).
    Therefore a matrix A is a unit if and only if there is B such that AB = BA = I.
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  5. #5
    pkr
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    Many thanks, all sorted now!
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