If a.a=a prove R is commutative

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Homework Help Overview

The discussion revolves around proving that a ring R is commutative under the condition that for all elements a in R, the equation a.a = a holds. Participants explore the implications of this condition within the context of ring theory.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of R, questioning whether it is a group or a ring, and what properties can be derived from the given condition. There are attempts to manipulate expressions involving addition and multiplication to explore commutativity.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions about the structure of R. Some have suggested that R may be a Boolean ring, leading to further exploration of its properties. There is no explicit consensus yet, but various lines of reasoning are being examined.

Contextual Notes

Participants note the absence of a multiplicative identity in R and discuss the implications of this on the proof. The nature of R as a ring rather than a group is emphasized, which influences the direction of the discussion.

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Homework Statement


If a.a=a for all elements of R, prove R is commutative.

The Attempt at a Solution



(a+b)2=a+b=a2+ba+ab+b2=a+ba+ab+b

Then -ba=ab

Any suggestions of how to show that -b=b?
 
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Are we to assume, here, that R is a group? Then it is always true that ae= a (e is the identity) so if it is also true that aa= a, then we have aa= ae and, multiplying on the left by the inverse of a, a= e. So we have only the "trivial" group containing only the identity.

If R is not a group, what sort of algebraic entity is it?
 
Sorry I should have been more clear but R is ring. So we are trying to show that . is a commutative operation and therefore making (R,+,.) a commutative ring.(At least that is my understanding of a commutative ring)

If I understand correctly I don't think we can assume that R has a multiplicative identity(1).
 
The structure in your OP is called a Boolean ring. You can prove that any element a in your Boolean ring satisfies a+a=0.

Prove this by considering (a+a)(a+a)
 
(a+a)(a+a)=a+a
a2+a2+a2+a2=a+a
a+a+a+a=a+a
a+a=0
a=-a for all a in R

Thanks, that makes sense.
 

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