# Should Krull dimension be a cardinal?

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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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. – Harry Gindi Jan 21 '10 at 21:16
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. – Harry Gindi Jan 21 '10 at 21:19
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. – Ben Webster Jan 21 '10 at 21:25
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? – Pete L. Clark Jan 21 '10 at 21:40
So apparently I mixed up the hilbert-serre result with the definition of the length of a module. – Harry Gindi Jan 22 '10 at 1:14

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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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? – Pete L. Clark Jan 21 '10 at 21:53
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." – Ben Webster Jan 21 '10 at 23:08
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." – Manny Reyes Jan 21 '10 at 23:17
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. – Georges Elencwajg Jan 21 '10 at 23:27
@Ben: I dunno, I like umlaute I guess :P I've seen it in various places, though. www-history.mcs.st-andrews.ac.uk/Biographies/Noether_Max.html tells us that Max apparently preferred to write oe... – Mariano Suárez-Alvarez Jan 21 '10 at 23:42