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 $\dim(R) \leq \left\vert{2^R}\right\vert$.



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> 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?
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Generalization of Krull dimension for commutative rings
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.


