Show that the ring $R$ is commutative with unity with $x^n=x$ for all $x\in R$ has finite characteristic.

Question

Suppose that $R$ commutative ring with unity and for all $x\in R$, $x^n=x$ and $n$ is fixed positive integer. Show that $R$ has finite characteristic.

How to prove that for unknown integer $n$. Also what is the possible numbers for characteristic . Is it the divisors of $n$.

I tried to start like this way :

$2x=(2x)^n=2^nx^n \Rightarrow (2^n-2)x=0 $. So this prove $R$ has finite characteristic, right ? but that will give me the possible values for characteristic for $R$ is the divisors of $2^n-2$. I am so confused . I appreciate any hint. Thanks

I know in this type of ring every prime ideal is maximal, reduced, and Von Neumman ring.

Answer 1

Consider $x=1+1$. Then $$0=(1+1)^n-(1+1)=(2^n-2) \cdot 1$$ so that the ring has a characteristic dividing $2^n-2$.