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

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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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