Generalization of Krull dimension for commutative rings

Question

In the paper How to construct huge chains of prime ideals in power series rings by B. Kang and P. Toan the Krull dimension of a commutative ring with $1$ is defined as follows:

Let $R$ be a commutative ring and $\mathfrak{C}$ be an arbitrary chain of prime ideals in $R$. Then length of $\mathfrak{C}$ is defined by $\left\vert{\mathfrak{C}}\right\vert-1$. The Krull dimension of $R$ is the largest cardinal number $\alpha$ (if any) such that there exists a chain of prime ideals in $R$ whose length is equal to $\alpha$. We write $\dim(R)\geq \alpha$ if there is a chain of prime ideals in $R$ whose length is $\geq \alpha$.

Note that this agrees with the usual definition of dimension when $\dim(R)$ is finite. However, there is no guarantee that the dimension of a ring exists in general. In the above paper, there are classes of rings with uncountable chains of prime ideals. Assuming that the Continuum hypothesis is true, there are a few examples where the Krull dimension is determined. This is done by noting that $\dim(R) \leq \left\vert{2^R}\right\vert$.

Does the dimension of a commutative ring exist in general? Is the existence of dimension for all commutative rings equivalent to the generalized continuum hypothesis?

Answer 1

Here is one stupid obstruction.

Take any limit cardinal $\alpha = \bigvee_{i \in I} \alpha_i$, $\alpha_i < \alpha$, such that there is a ring $R_i$ with chains of any length $< \alpha_i$ but not of length $\alpha_i$. Then the ring $R$ defined as the unitalizatioon of $\bigoplus_i R_i$ doesn't have a dimension.

Indeed, any prime ideal is necessarily supported on some $i$, i.e. is pulled back from one of $R_i$, so every chain of prime ideals of $R$ is pulled back from a chain in some $R_i$. These don't have a maximal cardinality.