TITLE: Finite commutative ring with unity and without nilpotent elements
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.