In an anwswer to a question on our sister site here I mentioned that a *reduced* commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r\in R $ the equation $r=r^2x$ has a solution $x\in R$.

A user asked in a comment whether this implies that an arbitrary product of zero-dimensional rings is zero dimensional.

I answered that indeed this is true and follows from von Neumann regularity **if** the rings are all reduced but I gave the following counterexample in the non reduced case:

Let $R$ be the product ring $R=\prod_{n=1}^\infty \mathbb Z/2^n\mathbb Z$.

Every $\mathbb Z/2^n\mathbb Z$ is zero dimensional but $R$ has $\gt 0$ dimension.

My argument was that its Jacobson radical $Jac(R)=\prod_{n=1}^\infty Jac (\mathbb Z/2^n\mathbb Z)=\prod_{n=1}^\infty2\mathbb Z/2^n\mathbb Z$ contains the **non-nilpotent** element $(2,2,\cdots,2,\cdots)$.

However in a zero dimensional ring the Jacobson radical and the nilpotent radical coincide and thus $R$ must have positive dimension.

My question is then simply: we know that $dim(R)\gt 0$, but what is the exact Krull dimension of $R$ ?

**Edit**

Many thanks To Fred and Francesco who simultaneously (half an hour after I posted the question!) referred to an article by Gilmer and Heinzer answering my question .

Here is a non-gated link to that paper.

Interestingly the authors, who wrote their article in 1992, explain that already in 1983 Hochster and Wiegand had outlined (but not published) a proof that $R$ was infinite dimensional.

Already after superficial browsing I can recommend this article, which contains many interesting results like for example infinite-dimensionality of $\mathbb Z^{\mathbb N}$.

**New Edit**

As I tried to read Hochster and Wieland's article, I realized that it refers to an article of Maroscia to which I have no access. Here is a more self-contained account of some of Hochster and Wieland's results.