17
$\begingroup$

One equivalent definition of a noetherian ring is that it satisfies the ascending chain condition. That is, for every ascending chain of ideals $$ I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots \subseteq I_{k} \subseteq I_{k+1} \dots $$ there is a j such that for all $k \ge j$, $I_{k} = I_{k+1}$.

What if this definition is extended into the ordinals? Could there be a ring which satisfies the $j = \omega^2 + \omega\cdot3 + 5$ noetherian condition, for example? That is, could there be an infinite $j$ such that there are no ascending chains of length greater than $j$. In this example, we would say that the ring is $\omega^2 +\omega\cdot3 + 5$ noetherian.

Most importantly, what kind of rings would be created? Same question with a variety of definitions of noetherian.

Edit: The way I am trying to define such rings seems to be problematic. I'd like to conceive a clearer definition, but I need to become a bit more familiar with noetherian rings.

$\endgroup$
  • 3
    $\begingroup$ The condition makes sense, but I'm afraid it's not as useful as in case $\kappa=\omega$. None of the special results satisfied by noetherian rings can be generalized. Notice that any ring satisfies such a chain condition if you take $\kappa$ bigger than the cardinal of the ring. $\endgroup$ – Fernando Muro Mar 12 '12 at 23:45
  • 4
    $\begingroup$ @Fernando: Just a nitpick: one needs to take $\kappa$ bigger than the cardinality of the powerset of the ring. It seems counterintuitive, but it is possible for a strictly increasing chain of subsets of a set to be of larger cardinality than the set itself. $\endgroup$ – Kevin Ventullo Mar 12 '12 at 23:59
  • 3
    $\begingroup$ Erin, perhaps the way ordinals arise naturally in the context of Noetherian rings is as follows: the ascending chain requirement essentially asserts that a certain order is well-founded (ideals ordered by reverse inclusion). Thus, this order admits an ordinal ranking function, so that the ordinal rank of an ideal is the supremum of the ranks plus one of the ideals in which it is included. Going to a larger ideal means reducing the rank. The rank of the ring would be the corresponding overall supremum. Among Noetherian rings, what are the interesting things to say about this rank? $\endgroup$ – Joel David Hamkins Mar 13 '12 at 0:13
  • 3
    $\begingroup$ About two weeks ago Hans Schoutens gave an interesting take at GC that I think is related to your question. The title of his talk was Ordinal-valued invariants. You can find an abstract of his talk here: websupport1.citytech.cuny.edu/faculty/lghezzi/Ghezzi/… If you see him around you can ask him, because I think this is related to what he was talking about in his talk. $\endgroup$ – Mahdi Majidi-Zolbanin Mar 13 '12 at 0:35
  • 9
    $\begingroup$ Kevin is correct in claiming that a set of size $\kappa$ can have chains of subsets of size $2^\kappa$, as for example is the case with the Dedekind cuts in the rationals, where we have a chain of $2^\omega$ many subsets of $\mathbb{Q}$, a countable set. But meanwhile, Fernando is correct that this situation cannot arise with well-ordered chains, as in the question, since between each set and the next one can pick a distinct element of the underlying set. $\endgroup$ – Joel David Hamkins Mar 13 '12 at 2:39
8
$\begingroup$

My own thoughts gravitate to valuation rings. These can be defined in a variety of ways; for example, as an integral domain $O$ such that for any nonzero element $x$ in its field of fractions $K$, either $x$ or $x^{-1}$ belongs to $O$. They are equivalently integral domains whose ideals are totally ordered by inclusion. The Noetherian valuation rings form a very special class called discrete valuation rings.

Associated with a valuation ring $O$ is its value group $K^\ast/O^\ast$, where $R^\ast$ denotes the group of units in a ring $R$. This is naturally a totally ordered abelian group. The positive elements in this group can be identified with principal ideals of the valuation ring, if I'm not mistaken.

One nice fact is that given any totally ordered abelian group $G$, there is a valuation ring having $G$ as its value group. The usual construction is given by taking formal Hahn series

$$\sum_{g \in G} a_g x^g$$

where the coefficients $a_g$ are elements of a field, and the set of them has well-ordered support.

The point I am trying to make is that just as you can manufacture totally ordered abelian groups with arbitrarily long increasing ordinal chains of elements, so you can construct valuation rings with arbitrarily long ordinal chains of ideals. One way of getting lots of examples of totally ordered abelian groups is to take any (set-sized) subgroup of Conway's class of surreal numbers (as a group under addition).

The ordinal $\kappa = \omega^2 + 3 \omega + 5$ is kind of a fun and kooky example (it's the 5 that throws me off a little), but it could be that if you consider products of rings where one is a finite local ring like $\mathbb{Z} mod 64$, you might be able to get a $\kappa$-Noetherian ring of that type. But I'm spitballing a little here.

$\endgroup$
6
$\begingroup$

Interesting. Here are a few thoughts.

  1. It seems that current formulation allows bad examples like this: Let $R=\mathbb{Z}$, which is of course noetherian. Pick any ordinal $\kappa$ and let $I_i=(0)$ for $i\le \kappa$ and $I_{\kappa+1}=(2)$. But I don't think this is what you intended. So let me reformulate this a bit. Say that a commutative ring $R$ is $\alpha$-noetherian, with $\alpha$ an ordinal, if there a no strictly increasing chains of ideals of length $\alpha$. Ditto for modules, where ideals are replaced by submodules. Clearly noetherian=$\omega$-noetherian.

  2. Suppose that all ideals (or submodules) are generated by less that $\kappa$ elements where $\kappa$ is a cardinal. Then the ring (or module) is $\kappa$-noetherian. To see this, take the union of the chain and use the hypothesis. Thus a vector space $V$ is $\kappa$-noetherian for any cardinal $\kappa>\dim V$. As for a ring example, presumably, there is some version of the Hilbert basis theorem, which would imply that if $R$ is noetherian, and $\lbrace x_i\mid i\in I\rbrace$ a set of variables, then the polynomial ring $R[x_i\mid i\in I]$ is $\kappa$-noetherian for any $\kappa> card(I)$.

Although I have to agree with Fernando's comment, that it's not going to be as useful as the usual noetherian condition, there is clearly something here. It may be fun to work out more of the story.

$\endgroup$
6
$\begingroup$

Possibly this is already too far away to qualify as 'answer', but it is long for a comment and might be of interest.

There is a classical construction of descending (not ascending), so rather matching artinian than noetherian, subgroups of abelian groups (or, to artificial stay a bit closer let's say submodules of Z-modules, but then this got generalized to modules over other rings so it is not totally artificial), where indexing by ordinals is done and relevant. They are called Ulm subgroups.

Let $G$ be an abelian group and $p$ some (fixed) prime. Than the (classical) p-height of an element $g$ is the supremum of all naturals such that the equation $p^nx = g$ has a solution $x$ in $G$.

One can express this differently by defining $p^nG$ to be the set, in fact it is a subgroup, of all $h\in G$ of the form $p^ng$ and define the p-height as the supremum of all $n$ such that $g \in p^nG$.

Now, one can continue and say $p^{\omega}G$ is the inersection of all the $p^nG$ and so on.

Or formally, $p^0G = G$, $p^{\alpha+1}G= p(p^{\alpha}G)$ and $p^{\beta}G =\bigcap_{\alpha \lt \beta}$ for $\beta$ a limit ordinal.

This forms a descending chain of subgroups that (thus) eventually stabilzes; the ordinal where it stabilzes is called Ulm length. One can now also generalize the notion of p-height of g by defnining it as the ordinal $\sigma$ such that $g \in p^{\sigma}G$ yet not in $g \in p^{\sigma + 1}G$ and $\infty$ if such an ordinal does not exist.

And this notion is quite useful. For example, a classical result of Ulm (1930s) then says that two countable abelian $p$-groups (ie, order of each element a prime power) $G$ and $H$ are isomophic if and only if $p^{\alpha}/p^{\alpha+1}G$ and $p^{\alpha}H/p^{\alpha+1}H$ are isomorphic for each $\alpha$ (these are called the Ulm factors) and the stable parts are isomorphic.

And conversely for any ordinal one can construct a group of that Ulm length (and in addition one can prescribe to a considerable extent the Ulm factors at each point until there).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy