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

[samkolb]

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.

 

[VKint]

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?)

 

[samkolb]

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.

 

[VKint]

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 $$n$$.