-1

Possible Duplicate:
Ring such that $x^4=x$ for all $x$

Let $R$ be a ring such that $a^4=a$ $ ,\forall a \in R$. How do I show that $R$ is commutative?

Community
  • 1
Mohan
  • 15,494
  • 1
    Related: http://math.stackexchange.com/questions/16535/is-such-ring-commutative – Hans Lundmark Nov 16 '11 at 09:27
  • 3
    I answered this here: http://math.stackexchange.com/questions/76792/ring-such-that-x4-x-for-all-x/77016#77016 –  Nov 16 '11 at 09:43

1 Answers1

1

If 2 is invertible, it may work this way:

For arbitrary elements $a,b$ calculate $ab-ba = (ab-ba)^4 = (ba-ab)^4 = ba-ab$. So $2ab=2ba$ and hence $ab=ba$.

MaoK
  • 117
  • 4
    2 cannot be invertible, right? Otherwise, for any $a\in R$, $a=a^4=(-a)^4=-a$. So $2a=0$, which implies $a=0$. – Paul Nov 16 '11 at 09:19
  • 1
    you are totally right, thanks. So the above is useless. – MaoK Nov 16 '11 at 09:22
  • 1
    no, it's a good try. I think it's not easy to prove it, maybe I am not good at algebra. – Paul Nov 16 '11 at 09:28