Finite commutative ring with unity and without nilpotent elements Let $R$ be a commutative ring with unity such that for each $x \in R$
there exists a $n \in \mathbb{N}$, $n>1$, such that $x^n = x$. Then show that
$$
R\simeq F_{1}\times F_{2}\times \cdots\times F_{n}
$$ 
where $F_k$ are fields.
I am facing difficulty in proving the above without using Artin–Wedderburn theorem (or its proof). There exists an elementary proof?
 A: The same logic from this solution to a special case applies here, except that you disregard the comments about $F_2$ and $F_3$ and just settle for all the quotients by prime ideals being fields.
The battle plan is, briefly:


*

*The intersection of all prime ideals is the zero ideal

*The quotient by any prime ideal is in fact a field.

*The ring embeds into a product of quotient rings of the form $R/P$ where $P$ is a prime ideal. (And of course, that is a product of fields.)


So such a ring is a subring of a product of fields, and is not necessarily the whole product, or a finite product.
If the ring is finite, or better yet, only has finitely many maximal ideals, then the injection above is into a product of only finitely many fields. A collection of maximal ideals is always comaximal, so the Chinese Remainder Theorem says the map is surjective, and so it is actually an isomorphism.
A: If the ring is finite, then it is an artinian reduced ring and therefore splits as the product of artinian local rings. Every term of the product must be reduced and an artinian local reduced ring is a field, proving your claim.
