Uncountable chain of prime ideals in an arbitrary direct product of rings

I am only considering commutative rings with $1$. Dimension refers to Krull dimension.

In the paper "Products of commutative rings and zero-dimensionality", Gilmer and Heinzer give necessary and sufficient conditions for an arbitrary direct product of rings to be at least countably infinite-dimensional. As there are uncountably many prime ideals in an arbitrary product of rings, I wonder whether there is an uncountable chain of prime ideals in the infinite-dimensional case.

An analogous problem has been studied for power series rings. See "How to construct huge chains of prime ideals in power series rings" by Kang and Toan.

Yes, whenever the product is not zero-dimensional there is an uncountable chain of primes (in fact, a chain of cardinality $\mathfrak{c}$ and uncountable cofinality). Let $(A_\alpha)_{\alpha\in I}$ be an infinite collection of rings such that $\prod A_\alpha$ is not zero-dimensional. It follows from Gilmer and Heinzer's Theorem 3.4 that we can partition $I$ into infinitely many sets $I_n$ such that $\prod_{I_n} A_\alpha$ is infinite-dimensional for each $n$. It thus suffices to prove the following: Let $A_n$ be an infinite sequence of rings such that $\dim A_n\geq n$ for each $n$. Then there is an uncountable chain of primes in $A=\prod A_n$.
First, for each $n$ choose a chain of primes $P^n_0\subset P^n_1\subset\dots\subset P^n_n\subset A_n$ for each $n$ and fix a nonprincipal ultrafilter $U$ on $\mathbb{N}$. For any function $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)\leq n$ for all $n$, let $P_f$ be the ideal of sequences $(y_n)\in A$ such that $\{n:y_n\in P^n_{f(n)}\}\in U$. Since $U$ is an ultrafilter, $P_f$ is prime. Write $f\sim g$ if $\{n:f(n)=g(n)\}\in U$ and $f<g$ if $\{n:f(n)<g(n)\}\in U$. Since $U$ is an ultrafilter, $<$ is a total ordering on the set of $\sim$-equivalence classes of functions. Furthermore, whenever $f<g$, $P_f\subset P_g$. The collection of all primes of the form $P_f$ is thus a chain, and is in bijection with the set of $\sim$-equivalence classes of functions $f$ such that $f(n)\leq n$. This latter set, also known as the ultraproduct $\prod_U [n]$ of the sets $[n]=\{0,1\dots,n\}$, is well-known to be countably saturated and have cardinality $\mathfrak{c}$.
• Assume the continuum hypothesis is true. Given a answer to my previous comment, I hope that the Krull dimension of an arbitrary non-zero dimensional product of rings exists. At least for a countable index set, would the dimension equal $\mathfrak{c}$? – Andrew Chiriac Mar 24 '15 at 13:31
• For your first question, there is always a surjection onto the set of ultrafilters of the index set, given by looking at elements which are $1$ or $0$ on every coordinate. This is a bijection iff each factor is local and the product is zero-dimensional (iff there is an integer $n$ such that every non-unit $x$ in a factor satisfies $x^n=0$). – Eric Wofsey Mar 24 '15 at 14:57
• For your second question, there are multiple ways you might define an infinite Krull dimension (do you just care about cardinality of chains, or do you want the length of well-ordered chains, or something else?). I don't know of an upper bound for the dimension of a countable product of finite-dimensional rings under any of these definitions, though. For instance, suppose $A_n$ is of the form $k[x_i]/(x_i^n)$, where $\{x_i\}$ is a very large infinite set of variables. It seems plausible to me that the product might have at least as many prime ideals as there are variables. – Eric Wofsey Mar 24 '15 at 15:13