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Question about ideals.

  1. Oct 27, 2008 #1
    Lets say that we know that for [tex]a \in A[/tex] and [tex]b \in B[/tex] where A and B are
    ideals of R, [tex]a - b = 1[/tex]. Do we then know that [tex](a-b)(a-b) = a^2 -2ab+b^2
    = 1[/tex]? In that case, what is the meaning of the number 2 in the term 2ab? We
    don't know that 2 is in R, R could be anything.
     
  2. jcsd
  3. Oct 30, 2008 #2

    morphism

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    First of all, if R really is "anything" (in particular, if it's noncommutative), then (a-b)(a-b) is not equal to a^2-2ab+b^2 - it's equal to a^2-ab-ba+b^2.

    Second, if R contains 1, then R contains 1+1, and 1+1+1, and so on. The positive number n used used to denote these. The negative number -n is used to denote their additive inverses, i.e. -(1+1), -(1+1+1), and so on. 0 of course denotes 0. In this way every unital ring contains a homomorphic (warning: not isomorphic) copy of the integers.
     
  4. Oct 30, 2008 #3
    I should have said that R is a commutative ring with identity in this context.
     
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