Should Krull dimension be a cardinal?

Question

A totally ordered finite set $\quad \mathcal P_0 \varsubsetneq \mathcal P_1\varsubsetneq \dots \mathcal \varsubsetneq \mathcal P_n \quad$ of prime ideals of a ring $A$ is said to be a chain of length $n$. As is well known, the supremum of the lengths of such chains is called the Krull dimension $\dim(A)$ of the ring $A$.

If the lengths of these chains are not bounded, the ring is said to be infinite dimensional: $\dim(A)=\infty$.This can happen, surprisingly, even for a Noetherian ring $A$.

But in the infinite dimensional case we could consider arbitrary totally ordered subsets $\Pi \subset Spec(A)$ of prime ideals, their cardinality $card(\Pi)$ and then take the sup of all those cardinals. Let us call this sup the cardinal Krull dimension of the ring $A$.

An equality $\dim(A)=\aleph$ would then be a more quantitative measure of the infinite dimensionality of $A$ than just $\dim(A)=\infty$

My question is whether results are known related to that cardinal Krull dimension. For example: for X a topological space, has the cardinal Krull dimension of $\mathcal C(X)$ (the ring of continuous functions on $X$) been calculated? I don't find this trivial, even for $X=\mathbb R$. There are obvious variants of this question concerning rings of differentiable functions on manifolds, etc. Thanks in advance for any information on this topic.

Answer 1

The Krull dimension, as defined by Gabriel and Rentschler, of not-necessarily commutative rings is an ordinal. See, for example, [John C. McConnell, James Christopher Robson, Lance W. Small, Noncommutative Noetherian rings].

More generally, they define the deviation of a poset $A$ as follows. If $A$ does not have comparable elements, $\mathrm{dev}\;A=-\infty$; if $A$ is has comparable elements but satisfies the d.c.c., then $\mathrm{dev}\;A=0$. In general, if $\alpha$ is an ordinal, we say that $\mathrm{dev}\;A=\alpha$ if (i) the deviation of $A$ is not an ordinal strictly less that $\alpha$, and (ii) in any descending sequence of elements in $A$ all but finitely many factors (ie, the intervals of $A$ determined by the successive elements in the sequence) have deviation less that $\alpha$.

Then the Gabriel-Rentschler left Krull dimension $\mathcal K(R)$ of a ring $R$ is the deviation of the poset of left ideals of $R$. A poset does not necessarily have a deviation, but if $R$ is left nötherian, then $\mathcal K(R)$ is defined.

A few examples: if a ring is nötherian commutative (or more generally satisfies a polynomial identity), then its G-R Krull dimension coincides with the combinatorial dimension of its prime spectrum, so in this definition extends classical one when these dimensions are finite. A non commutative example is the Weyl algebra $A_{n}(k)$: if $k$ has characteristic zero, then $\mathcal K(A_n(k))=n$, and if $k$ has positive characteristic, $\mathcal K(A_n(k))=2n$. The book by McConnel and Robson has lots of information and references.

Answer 2

I realize this question was asked some years ago, but some coauthors and I recently published a paper on this topic in J. of Algebra: "An infinite cardinal-valued Krull dimension for rings".

https://arxiv.org/abs/1711.08554