A characterization of commutative rings with Krull dimension zero Recently, I saw the following characterization of zero dimensional commutative rings:

A commutative ring $R$ (with $1$) is $0$-dimensional if and only if $R/\sqrt 0$ is von Neumann regular. (Here $\sqrt 0$ is the nil radical of $R$, the set of all nilpotent elements of $R$.)

Where can I find a verification for it?
 A: If $R/\sqrt{0}$ is von Neumann regular, and $P$ is a prime ideal in $R$, then $R/P$ is an integral domain which is a quotient of $R/\sqrt{0}$ and so is von Neumann regular.  This implies easily that $R/P$ is a field, so $P$ is maximal.  Thus, $\dim(R)=0$.
Conversely, suppose that $\dim(R)=0$, and put $R'=R/\sqrt{0}$, so $R'$ is $0$-dimensional and reduced.  Given $a\in R'$, put $S=\{a^n(1+ab):n\in\mathbb{N},\;b\in R'\}$, and note that this is multiplicatively closed. If $0\notin S$ then we can find a prime ideal $P$ not meeting $S$.  Now $\dim(R')=0$ so $P$ is maximal, so either $a\in P$ or $1-ab\in P$ for some $b\in P$. Either possibility contradicts the fact that $P\cap S=\emptyset$. Thus, we must have $0\in S$ after all, so $a^n(1-ab)=0$ for some $n$ and $b$.  This means that $a(1-ab)$ is nilpotent, but $R'$ is reduced, so $a(1-ab)=0$.  It follows that $R'$ is von Neumann regular.
A: This follows from Theorem 1.16 in Goodearl's book "von Neumann regular rings" (Pitman, 1979).
A: This equivalence (among a couple others!) is worked out explicitly in Exercise 4.15 from T.Y. Lam's book "Exercises in Classical Ring Theory".
