What does the 2 mean in (a-b)² = a² - 2ab + b² when a - b = 1 in an ideal?

[futurebird]

Lets say that we know that for $$a \in A$$ and $$b \in B$$ where A and B are
ideals of R, $$a - b = 1$$. Do we then know that $$(a-b)(a-b) = a^2 -2ab+b^2<br /> = 1$$? 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.

 

[morphism]

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.

 

[futurebird]

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.

I should have said that R is a commutative ring with identity in this context.