All rings considered are commutative with $1$. In my study of arbitrary chains of prime ideals of a ring $R$, I have noticed two possiblities for the cardinality of a chain when $\dim(R)=\infty$:


1. All chains of primes are finite, but their lengths are unbounded.
2. R contains an uncountable chain of prime ideals.


For instance, if $T$ is a finite-dimensional non-SFT ring, then $\dim(T[[X]])\geq \aleph\_1$(Kang et al 2013). If $T$ is an almost Dedekind ring, we have $\dim(T[[X]])\geq 2^{\aleph\_1}$ (Chang et al 2015). Assuming the continiuum, this bound cannot be improved.
Let $S$ be an arbitrary direct product of rings. Then $\dim(S)=0$ or $\dim(S)\geq \aleph\_1$ ([Uncountable chain of prime ideals in an arbitrary direct product of rings](https://mathoverflow.net/questions/200803/)). An arbitrary direct product of copies of $\mathbb{Z}$ has dimension at least $2^{\aleph\_1}$. Once again, assuming the continuum hypothesis, this bound cannot be improved. The ring of entire functions also has dimension $2^{\aleph\_1}$ (assuming the continuum hypothesis). The proofs of all these results are quite difficult and technical, and I wonder whether there is a more general argument that would work for all cases.


The direct sum of rings $\{R\_i\}$ can be embedded in a ring $U$. It can be shown that each chain of prime ideals in $U$ corresponds to a chain of prime ideals in some $R\_i$. Thus if $\dim(R\_i)=i$ for each $i$, then $U$ is infinite-dimensional, but each chain is finite.


I know of no example of a ring that contains a countably infinite chain of prime ideals but not an uncountable chain. Does such a ring exist?



Question regarding arbitrary chains of prime ideals
See [Example Tag 01IY](http://stacks.math.columbia.edu/tag/01IY) of the Stacks project.


