TITLE: Generalization of Krull dimension for commutative rings
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.