Let $R$ be a commutative ring with 1. An element $a$ in $R$ is said $\pi$-regular if $a^n = a^nra^n$ for some $r \in R$ and a natural number $n$. A ring $R$ is called $\pi$-regular if every element is $\pi$-regular.
Is any zero dimensional ring a $\pi$-regular ring?
Note: Zero dimensional= every prime ideal is maximal.
On
I have found the following results, so I think it is an answer for the question.
[1, Corollary 11]: Every zero dimensional ring is clean.
[2, Corollary 2.8]: The following statements are equivalent for a commutative ring $R$:
(a) $R$ is a clean ring with Krull dimension zero;
(b) $R$ is a $\pi$-regular ring.
By combining above results we have: a commutative ring is zero dimensional if and only if it is $\pi$-regular!!!
[1] D. D. Anderson and V. P. Camillo, Commutative rings whose elements are a sum of a unit and an idempotent, Comm. Algebra 30 (2002), 3327-3336.
[2] D. Lu, W. Yu, On prime spectrums of commutative rings, Comm. Algebra, 34, (2006), 2667-2672.
On
This is in some sense "obvious" from the basic algebro-geometric theory of reduced $0$-dimensional (i.e., von Neumann regular) rings. In particular, suppose $R$ is a $0$-dimensional ring and let $N$ denote its nilradical. Then $R/N$ is von Neumann regular, and so any principal ideal is generated by an idempotent. In geometric terms, if $a\in R/N$ generates the same ideal as the idempotent $e$, then $e$ gives a partition of $\operatorname{Spec} R/N$ into two clopen sets such that $a$ is $0$ on one of them and a unit on the other. (Explicitly, if $e=ab$ and $a=ea$, then $a$ is a unit with inverse $b$ on the clopen set obtained by inverting $e$, but is $0$ on its complement where $1-e$ is inverted.)
Now let's pull things back to $R$. Given an element $a\in R$, its image in $R/N$ gives a partition of $\operatorname{Spec} R/N$ into clopen sets $U$ and $V$ such that $a$ is $0$ on $U$ and a unit on $V$. But the natural map $\operatorname{Spec} R/N\to \operatorname{Spec} R$ is a homeomorphism, so we can consider $U$ and $V$ as subsets of $\operatorname{Spec} R$ and the statements above will be true modulo nilpotent elements. Specifically, $a$ will be nilpotent on $U$ and a unit on $V$ (since a unit plus a nilpotent is still a unit). But now it is easy to prove $a$ is $\pi$-regular: if $n$ is large enough then $a^n$ is $0$ on $U$ and still a unit on $V$, and so we can take $r$ to be any element of $R$ which is the inverse of $a^n$ on $V$.
Alternatively, here's a more direct geometric proof that for any $a\in R$, there is a partition of $\operatorname{Spec} R$ into clopen sets $U$ and $V$ such that $a$ is nilpotent on $U$ and a unit on $V$. Note that for any prime $P\subset R$, every element of $R_P$ is either a unit or nilpotent (since it has only one prime ideal, which thus is the nilradical). Note moreover that if $a$ is nilpotent (or a unit) in $R_P$, there is an entire open neighborhood of $P$ on which $a$ is nilpotent (or a unit), since we can invert a single element of $R\setminus P$ to witness this. So, if $U$ is the set of $P\in\operatorname{Spec} R$ such that $a$ is nilpotent in $R_P$ and $V$ is the set of $P\in\operatorname{Spec} R$ such that $a$ is a unit in $R_P$, then $U$ and $V$ are both open. But since they are complements of each other, they are both clopen.
Now we just have to verify that $a$ is globally nilpotent on $U$ and globally a unit on $V$, rather than just locally. For $V$, this is just the fact that a ring element that is not in any prime ideal is a unit (or more geometrically, since inverses are unique, we can patch together local inverses to $a$ on $V$ to get a global inverse). For $U$, we use the fact that $U$ is covered by open sets on which $a$ is nilpotent (since as above, if $a$ is nilpotent in $R_P$ it is nilpotent on an open neighborhood of $P$). But $U$ is closed in $\operatorname{Spec} R$ and thus quasicompact, so it is covered by finitely many open sets on which $a$ is nilpotent. We can thus find $n$ such that $a^n=0$ on all of these finitely many open sets, so $a$ is nilpotent on $U$.
Finally, let me mention that the converse (every $\pi$-regular ring is $0$-dimensional) is true and comparatively trivial. Indeed, suppose $R$ is $\pi$-regular and let $P\subset R$ be a prime ideal. Replacing $R$ with $R/P$, we may assume in fact $R$ is a $\pi$-regular domain and we wish to show it is a field. But in a domain, $a^n=a^nra^n$ implies either $a^n=0$ (and thus $a=0$) or $1=ra^n$. So for any $a$, either $a=0$ or $a$ is a unit, which means $R$ is a field.
Results like this have definitely been known since the $80's$, when there was some interest in embeddability of rings in $0$-dimensional rings (like the work of Arapovic in the early 80's). Robert Gilmer wrote and coauthored a number of papers about $0$-dimensional rings in the late 80's and early 90's. A number of proofs of this fact might start with an element-wise observation like the following.
Proof:
Given $(1)$, take $e = a^n r$ and $r' = ra^{n-1}$ and the check that $a[(1-e)]^n = 0$ and that $e(ra - 1) = 0$ with $r = r'$ is straightforward, which gives $(3)$
Given $(3)$, since $(1-e)$ is idempotent we have $a^n(1-e) = 0$. Then substituting $e = era$ we have $a^n(1 - era) = 0$. So $a^n = a^{n+1}er$ and $(2)$ holds.
Given $(2)$, take $r' = r^n$. Indeed, we have $a^n = a^{n+1}r = ar a^n$, so iterating we get $a^n = a^n r^n a^n$.
Given $(3)$, note that the relation $a^n(1-e) = 0$ implies $e$ divides $a^n$, and the relation $e(ra-1) = 0$ implies that $e = e^n = er^na^n$ and hence $a^n$ divides $e$. Thus $eR = a^nR$, and the idempotent $e$ is unique because $eR = fR$ always implies $e = f$ (indeed, if $es = f$ and $ft = e$ then $ef = e(es) = es = f$ and $ef = ft(f) = ft = e$ so $f = e$).
Given $(4)$, it's clear that $(1-e)$ annihilates $eR = a^nR$, so $[a^n(1-e)]^n = 0$. Also there is an $r'$ such that $r'a^n = e$, and hence $r'a^ne = e$ and $e(r'a^n - 1) = 0$, so taking $r = r'a^{n-1}$ we are done. Since the idempotents in $(3)$ and $(4)$ are the same by proof, uniqueness carries over from $(4)$ to $(3)$. $\square$
As a bonus, condition $(3)$ makes transparent that
We also have
This follows immediately from $(4)$ and the fact that in a local ring the only idempotents are $1$ and $0$.
Standard, but a quick sketch is.... ($\implies$). Consider $aR/a^2R$ for any $a$. Using the facts that localization commutes with quotients and that $R_P$ is a field for any minimal prime of a reduced ring, and that being $0$ is a local property of modules, it's clear that $aR/a^2R$ is the zero module, hence $aR = a^2R$. Conversely, if $R$ is a VNR, then (M) says $aR = eR$ for some idempotent $e$, and so $a$ is either a unit or $0$ in any localization. Thus every localization is a field and $R$ is zero-dimensional.
We prove this by reducing to the reduced case, which was taken care of above.
Note that $R' = R/Rad(R)$ (where $Rad(R)$ is the nilradical of $R$) has the same Krull dimension as $R$, and that the homomorphic image of a $\pi$-regular ring is obviously $\pi$-regular. That $R/Rad(R)$ being $\pi$-regular implies $R$ being $\pi$-regular follows because if $a^2R' = R'$ then by definition $(a-a^2r) \in Rad(R)$ for some $r \in R$. Thus there exists an $n$ such that $(a - a^2r)^n = 0$, and binomial expansion immediately shows that $a^{n+1}$ divides $a^n$ in $R$.