Rings satisfying x^4 = x are commutative

Posted: April 19, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,
3

Let R be a ring, which may or may not have 1. We proved in here that if x^3=x for all x \in R, then R is commutative.  A similar approach shows that if x^4=x for all x \in R, then R is commutative.

Problem. Prove that if x^4=x for all x \in R, then R is commutative.

Solution. Clearly R is reduced, i.e. R has no nonzero nilpotent element. Note that 2x=0 for all x \in R because x=x^4=(-x)^4=-x. Hence x^2+x is an idempotent for every x \in R because

(x^2+x)^2=x^4+2x^3+x^2=x^2+x.

Thus x^2+x is central for all x \in R, by Remark 3 in this post.  Therefore (x^2+y)^2+x^2+y is central for all x,y \in R. But

(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2

and hence x^2y+yx^2 is central. Therefore (x^2y+yx^2)x^2=x^2(x^2y+yx^2) which gives us xy=yx. \ \Box

Advertisements
Comments
  1. Esam Kamal says:
    December 5, 2013 at 1:02 pm

    But What about the following questin?
    If R is a ring such that a^5=a for every a in R. Is it commutative?
    Also What about the following question?
    If R is a ring such that a^n=a for every a in R where n is a natural number. Is it commutative?

    Reply
  2. Otman says:
    May 3, 2012 at 12:38 pm

    The same for rings where x^2=x for every x in the ring.

    Reply

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

w
Cancel

Connecting to %s