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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.

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  • $\begingroup$ At least the way I've seen krull dimension constructed, we need to use Euler-Poincare maps, which don't make sense in higher cardinalities. Check out the chapter "Dimension Theory" in Atiyah-MacDonald, Hartshorne, and even Lang. They all define dimension in the same way, so there must be a reason for it. $\endgroup$ Jan 21, 2010 at 21:16
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    $\begingroup$ The problem is that the higher ordinals don't form a group. You may be able to rectify this by using surreals. You could also take the commutative monoid of ordinals less than a fixed inaccessible ordinal and perform the Grothendieck k construction on it, but this seems like it would be quite a challenge. $\endgroup$ Jan 21, 2010 at 21:19
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    $\begingroup$ Harry- I don't follow. Are you saying that this definition of Krull dimension is somehow ill-defined? or that it won't have some property one usually expects? What do Euler-Poincare maps have to do with anything? They certainly don't appear in the discussion on the Wikipedia page. $\endgroup$
    – Ben Webster
    Jan 21, 2010 at 21:25
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    $\begingroup$ As with Ben, I don't follow (and not for a lack of knowledge of the material in the list of books cited above). The Krull dimension is just a definition: what's the construction? $\endgroup$ Jan 21, 2010 at 21:40
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    $\begingroup$ So apparently I mixed up the hilbert-serre result with the definition of the length of a module. $\endgroup$ Jan 22, 2010 at 1:14

2 Answers 2

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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.

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    $\begingroup$ I find the definition interesting, especially since it's not obvious to me that it's equivalent to the usual definition (which uses prime ideals only)! [I am thinking here only about the commutative case.] Do they make any remark about the equivalence? Does anyone want to help me out? $\endgroup$ Jan 21, 2010 at 21:53
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    $\begingroup$ Pete, the proof that this coincides with the usual dimension in the nötherian commutative case is not as straightforward as one might wish---it is proved in detail in the book (which I emphatcally recommend!) $\endgroup$ Jan 21, 2010 at 22:01
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    $\begingroup$ I just have to ask: where did you get the habit of writing "nötherian"? I understand that this has the same German pronunciation, but "oe" is a normal feature of German names; the German word for Noetherian is "Noethersch." $\endgroup$
    – Ben Webster
    Jan 21, 2010 at 23:08
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    $\begingroup$ If you are having trouble locating the book of McConnell/Robson, then you could try searching for the book of K. Goodearl and R.B. Warfield, Jr. titled "An Introduction to Noncommutative Noetherian Rings." $\endgroup$ Jan 21, 2010 at 23:17
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    $\begingroup$ You're right, Ben. I have just checked some mathematics books in German (by Neukirch, Kunz, Fischer, Brieskorn,Schulze-Pillot,...) as well as the German Wikipedia: everybody writes Noether and Noethersch. $\endgroup$ Jan 21, 2010 at 23:27
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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

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    $\begingroup$ Certainly a paper on this topic should quote: H. Bass. Descending chains and the Krull ordinal of commutative Noetherian rings. J. Pure Appl. Algebra 1 (1971) 347-360, and maybe also H. Gulliksen. A theory of length for Noetherian modules. J. Pure Appl. Algebra 3 (1973), 159-170. $\endgroup$
    – YCor
    Sep 19, 2019 at 7:39

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