Question

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.

Answer 1

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 $x = x^n = x^k x^{n-k} = 0$.

Answer 2

This is true. Let $R$ be a ring satisfying your property. Then $R$ has no nilpotent element since if $x^a=0$, then for $b$ an integer such that $n(x)^b > a$ we have $x=x^{n(x)^b}=0$. So the radical of $R$ is $0$, and since the radical is the intersection of all prime ideals of $R$, we see that the natural map $R \mapsto \prod_{P} R/P$ is an injection. It therefore suffices to prove the result for $R/P$, that is for a domain.

Assuming no that $R$ is a domain, the equation $x^{n(x)}-x=0$ factors as $x=0$ or $x^{n(x)-1}=0$. So, in $R$, every non-zero element is a root of unity. This is also true of the fraction field $K$ of $R$. No $K$ is of characteristic $p>0$, since otherwise it would contain $\mathbb{Q}$ which contains many non-roots-of-unity such as $2$. And clearly, $K$ is algebraic over the prime field. So we're done.