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. 



Jacobson's theorem and further
Yes, this is true. 


By Jacobson's theorem, $R$ is commutative. Now the radical of $R$ is the intersection of all primes $P$ of $R$. Hence we have an embedding 
$$\phi: R/rad(R) \to \prod\_P R/P.$$
For each $x \in R/P, x \neq 0$ there is $n \ge 1$ such that $x(x^n-1)=0$ and since $R/P$ is a domain, $x^n =1$, i.e. $x$ is a unit. Thus $R/P$ is a field. Since elements $\neq \pm 1$ of $\mathbb Q$ aren't roots of unity, $R/P$ has prime characteristic and each element of $R/P$ is algebraic. Therefore $R/P$ is an algebraic extension of some $\mathbb{F}\_p$. 


To finish the proof, we have to show $rad(R) = 0$. Let $x \in rad(R)$. Let $k>0$ be minimal with $x^k=0$ and let $n \ge 2$ with $x^n=x$. Suppose $k > n$. Write $k=qn+r$ with $q > 0$, $0 \le r < n$. Then $0 = x^k = (x^n)^q x^r = x^{q+r}$. Minimality of $k$ implies $k = q+r$, i.e. $n=1$, in contradiction to the assertion $n \ge 2$. Hence $k\le n$ and $x = x^n = x^k x^{n-k} = 0$.


