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.



Uncountable chain of prime ideals in an arbitrary direct product of rings
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 $\mathfrak{c}$.


