Jacobson's theorem states that If $R$ is a ring, and for every $x\in R$, there exists $n(x)\geq 2$ such that $x^{n(x)}=x$. Then $R$ is commutative.

I wonder if the following stronger assertion(in case $R$ has unity) is true.

Let $R$ be a ring with unity. For every $x$ in $R$, there exists $n(x)\geq 2$ such that $x^{n(x)} = x$. Then, $R$ is embedded in a product(possibly infinite) of fields $F_i$, where each $F_i$ is an algebraic extension of $F_{p_i}$ (prime field of $p_i$ elements).

If this is not true, then I am also interested in counterexample.

Thanks in advance.