Collecting examples of all rings R such that $x^3=x \forall x \in R$ 
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*$\mathbb{Z_{3}}$ or its Cartesian products..

*The Boolean ring R which is defined as a ring in which every element is idempotent, that is $ \forall x \in R$ we have $x^2=x$ and hence $x^3=x^2x=xx=x$ in fact here $x^n=x$ ..


Any more???
Thanks for any help!!
 A: Let me start out by saying we'll assume you believe Jacobson's famous theorem which proves such a ring is commutative.
The condition that $x^3=x$ implies three things:


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*There are non nonzero nilpotent elements, so the intersection of all prime ideals is zero; and

*Every prime ideal is maximal, and moreover given a prime ideal $P$, $R/P$ is isomorphic to the field of two or three elements; and

*The Jacobson radical is zero (this is just a combination of the last two points.)


The first one I hope is obvious, and the second one is clear when you look at $R/P$ for a prime ideal $P$: the equation $0=x^3-x=x(x-1)(x+1)$ in the field of fractions of $R/P$ has exactly three solutions $\{0, 1, -1\}$, and all of those happen to be in $R/P$ already. So the quotient is either the field of two elements $F_2$ or the field of three elements $F_3$, and $P$ is maximal.
Now we can form the product ring $S=\prod R/M$ where $M$ ranges over the complete set of maximal ideals of $R$ (or just a subset of the maximals whose intersection is zero, if you like), and we see it is a (potentially infinite) product ring of copies of $F_2$ and/or $F_3$. Now, there is the diagonal mapping $r\mapsto (\ldots,r+M ,\ldots)$. The Jacobson radical is the kernel of this mapping, so we have actually shown that $R$ injects into a ring like $S$.
It also should be obvious that any subring of a ring like $S$ satisfies $x^3=x$.
This proves that the set of rings (with identity) satisfying $x^3=x$ is exactly the class of subrings of rings of the form $\prod_{i\in I} F_i$ where $F_i\in \{F_2, F_3\}$ and $I$ is an index set. 
One interesting case occurs when the set of maximal ideals is finite: the diagonal map mentioned above is onto by the Chinese Remainder Theorem, and therefore you ring is isomorphic to a finite product of copies of $F_2$ and/or $F_3$.
