Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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

2 Answers 2

up vote 15 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.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.