Proving -x = x with Positive Even Integer n in R

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SUMMARY

The discussion centers on proving that if there exists a positive even integer n such that x^n = x for every x in a ring R, then -x = x for every x in R. The initial proof for n = 2 demonstrates that (x+x)^2 = 4x^2 leads to the conclusion that 2x = 0, confirming -x = x. However, the proof does not extend to noncommutative rings due to the property that (ab)^k ≠ a^k b^k in general, which affects the validity of the expansion used in the proof.

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Homework Statement


Let R be a ring and suppose there exists a positive even integer n such that x^n = x for

every x in R. Show that -x = x for every x in R.


Homework Equations





The Attempt at a Solution


I solved the case where n = 2.

Let x be in R.

(x+x)^2= x+x = 2x,

(x+x)^2 = 4x^2 = 4x.

So 4x = 2x and 2x = 0. Done.


I tried using this same method when n = 4 and got nowhere.
 
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Let n = 2k. What's (-x)^{2k}?

(By the way, the proof you have for n = 2 doesn't work for noncommutative rings. The above hint suggests a method that does. Can you see why?)
 
Thanks for the hint. That works. But why does my proof for n=2 not work for noncommutative rings? Since the only terms in the expansion of (x+x)^2 are powers of x, I don't think I ever used commutativity.
 
Since the only terms in the expansion of (x+x)^2 are powers of x, I don't think I ever used commutativity.

Yeah...you're right. I read your work as (x + x)^2 = (2x)^2 = 2^2 x^2 = 4 x^2; my only point was that in a noncommutative ring, (ab)^k \neq a^k b^k in general. However, I suppose it's true that [(n \cdot 1) b]^k = (n^k \cdot 1) b^k for natural numbers n.
 

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