Let be a ring, which may or may not have
We proved in here that if
for all
then
is commutative. A similar approach shows that if
for all
then
is commutative.
Problem. Prove that if for all
then
is commutative.
Solution. Clearly is reduced, i.e.
has no nonzero nilpotent element. Note that
for all
because
Hence
is an idempotent for every
because
Thus is central for all
by Remark 3 in this post. Therefore
is central for all
But
and hence is central. Thus
which gives
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?
See this post: https://ysharifi.wordpress.com/2018/03/14/rings-satisfying-xn-x-are-commutative/
The same for rings where x^2=x for every x in the ring.
Yes, but, as I’m sure you know, the case
is trivial.