## Jacobson’s theorem and further

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.

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Also note that $R$ is a comm. von Neumann regular ring (i.e. for each $r$ there is $s$ s.t. $r = r^2s$) and such rings are known to embedd into a product of fields. – Ralph Apr 5 2012 at 3:28

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

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Thanks, I got the idea. How about starting from the Jacobson radical of $R$? I think it is easier. Also, showing the Jacobson radical is zero, is easy. If $x\in J(R)$, then $x^n-1$ is unit for any $n\geq 1$. So, $x=0$. The Jacobson radical is intersection of maximal ideals, so we get $R/M$ is field for each maximal ideals $M$. – i707107 Apr 5 2012 at 2:20
Looks good. In particular, you get $R/M$ a field for free. In effect, since $R/P$ is a field, each $P$ is maximal and $J(R) = rad(R)$. But using $J(R)$ in the proof is certainly better. – Ralph Apr 5 2012 at 2:29

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.

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 It is not necessary to consider a fraction field $K$ of $R$ because you already know that "each non-zero element [of $R$] is a root of unity". Hence $R$ ist alreday a field. – Ralph Apr 5 2012 at 2:13 Ralph, you beat me by 6 seconds ! Mathoverflow is really becoming too competitive :-) – Joël Apr 5 2012 at 2:23 Thank you for you too Joel. Both answers are basically the same. So, I accepted the first one. – i707107 Apr 5 2012 at 2:26 Yes, your answer showed up directly after mine. Even more amazing is that we not just had the same idea of proof but used nearly the same words: "this is true", "is the intersection of all primes". In any case I wished it were possible in MO to accept more than one answer. – Ralph Apr 5 2012 at 2:33