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

samkolb
Messages
37
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
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...
Back
Top