Proving the Commutativity of a Ring with R satisfying a^2 = a

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SUMMARY

The discussion centers on proving that a ring R, where every element satisfies the equation a^2 = a, is commutative. A participant initially attempted to show that (ab - ba)^2 = (ba - ab)^2 leads to ab = ba, but was corrected on the flawed application of algebraic properties. The key takeaway is that the closure of R under multiplication and addition negates the need to prove ab - ba is an element of R, and participants suggested exploring the implications of (a + b)^2 = a + b for further insights.

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  • Understanding of ring theory and its properties
  • Familiarity with algebraic structures and operations
  • Knowledge of the concept of commutativity in mathematics
  • Basic proficiency in manipulating algebraic expressions
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  • Study the implications of the equation (a + b)^2 = a + b in ring theory
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This discussion is beneficial for mathematics students, particularly those studying abstract algebra, as well as educators and researchers interested in ring theory and its applications.

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Homework Statement


Let R be a ring that satisfies a^2 = a for all a in R. Prove that R is a commutative ring

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The Attempt at a Solution


My attempt at this solution is (ab-ba)^2 = (ba-ab)^2 is true for any ring R => (ab-ba) = (ba - ab) => 2ab = 2ba => ab = ba. The problem here is I have no method to prove that ab-ba is indeed an element of R; I'm needing help with that or a totally alternate approach to this problem is welcomed so I can perhaps gain insight
 
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You don't need to prove ab-ba is an element of R. It's a ring. It's closed under multiplication and addition. But your method is flawed from the start. Stating that (ab-ba)^2=(ba-ab)^2 uses the property that (-1)^2=1. But (assuming the ring has a unit 1) your assumption that a^2=a for a in R would mean (-1)^2=(-1). It's a big danger in working with rings to apply algebra rules that apply to reals, but not to a general ring. Why don't you start by seeing what conclusions you can draw from (a+b)^2=a+b?
 

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