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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 $R$ ?

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.

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+1, However this question is more advanced of my understanding. :) – Babak S. Nov 7 '13 at 10:12

2 Answers 2

up vote 17 down vote accepted

The ring $R$ is infinite-dimensional. More generally, the product of a family of zero-dimensional rings has dimension $0$ if and only if it has finite dimension. This is proven as Theorem 3.4 in R. Gilmer, W. Heinzer, Products of commutative rings and zero-dimensionality, Trans. Amer. Math. Soc. 331 (1992), 663--680.

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You have beaten me for a few seconds :-) – Francesco Polizzi Mar 12 '12 at 13:09
Thanks a lot, Fred. – Georges Elencwajg Mar 13 '12 at 6:25

According to the paper Products of Commutative Rings and Zero-Dimensionality, your ring must be infinite-dimensional.

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Thanks a lot, Francesco. I'm sorry I can accept only one answer: I would have liked to accept yours too. – Georges Elencwajg Mar 13 '12 at 8:21
You are welcome Georges. Since Fred answered first, it is fair that you accept his answer – Francesco Polizzi Mar 13 '12 at 12:21

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