Ring question

  • Thread starter samkolb
  • Start date
  • #1
37
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.
 

Answers and Replies

  • #2
139
12
Let [tex] n = 2k [/tex]. What's [tex] (-x)^{2k} [/tex]?

(By the way, the proof you have for [tex] n = 2 [/tex] doesn't work for noncommutative rings. The above hint suggests a method that does. Can you see why?)
 
  • #3
37
0
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.
 
  • #4
139
12
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 [tex] (x + x)^2 = (2x)^2 = 2^2 x^2 = 4 x^2 [/tex]; my only point was that in a noncommutative ring, [tex] (ab)^k \neq a^k b^k [/tex] in general. However, I suppose it's true that [tex] [(n \cdot 1) b]^k = (n^k \cdot 1) b^k [/tex] for natural numbers [tex] n [/tex].
 

Related Threads on Ring question

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
907
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
888
Top