Timeline for Is every ordinal the nimber of a ring?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 5, 2017 at 7:29 | history | bounty ended | CommunityBot | ||
| S Nov 5, 2017 at 7:29 | history | notice removed | CommunityBot | ||
| S Oct 28, 2017 at 6:19 | history | bounty started | Christopher King | ||
| S Oct 28, 2017 at 6:19 | history | notice added | Christopher King | Improve details | |
| May 5, 2017 at 15:49 | answer | added | Will Sawin | timeline score: 3 | |
| May 5, 2017 at 15:33 | comment | added | Will Sawin | @Gro-Tsen An easier special case of the question is "Does every ordinal $< \omega^\omega$ arise as the Grundy value of a Noetherian ring of finite Krull dimension?" If I understand correctly, Theorem 2.3 of the first paper you link implies that every Noetherian ring of finite Krull dimension has length $< \omega^\omega$ and that arbitrarily high such ordinals appear already in the rings $k[x_1,\dots,x_n]$. This question is already nontrivial. | |
| May 5, 2017 at 12:42 | comment | added | Gro-Tsen | I'm not saying Gulliksen's work would give the answer, but I'm saying the question you ask is probably at least as complicated as the one he answers, because of the trivial fact that the Gulliksen length gives an upper bound on the nimber (which I prefer to call Grundy value, btw). So if you want to construct rings with arbitrarily high nimbers, you must construct rings with arbitrarily high Gulliksen lengths, which is already not so trivial. | |
| May 5, 2017 at 11:59 | comment | added | Joel David Hamkins | Ah, I see now. This isn't at all like the count-down game on the ordinals, since one can go immediately to a maximal ideal, in a PID, for example, skipping over many smaller ideals. | |
| May 5, 2017 at 11:24 | comment | added | Martin Brandenburg | No, for many reasons (for example, as I've said, the nimber is not just about the partial order of ideals, and also, $\mathrm{mex}$ differs from $\mathrm{sup}^+$). To get a feeling for this game, please have a look at my paper or the linked MO discussion. $K[X,Y]/\langle X^2,XY,Y^2\rangle$ has nimber $1$ (because it is $\mathcal{P}$), but its options have nimbers $0,2$. Not every ring is "Nim-like", its options may have larger nimbers. | |
| May 5, 2017 at 11:21 | comment | added | Joel David Hamkins | The rank of an ideal $I$ is $\sup\{\text{rank}(J)+1\mid I\subset J\}$. Isn't this the same recursion as what you wrote? | |
| May 5, 2017 at 11:16 | comment | added | Martin Brandenburg | @JoelDavidHamkins: No, the nimber does not only depend on the partial order of ideals. Therefore (@Gro-Tsen) I also doubt that Gulliksen's work is related to this. | |
| May 5, 2017 at 11:07 | comment | added | Joel David Hamkins | It seems to me that what you call the nimber of the ring is the same as the ordinal rank of the collection of ideals under inclusion (larger ideals are lower), which is a well-founded relation exactly in a Noetherian ring and hence has an ordinal rank there. | |
| May 5, 2017 at 8:38 | comment | added | Gro-Tsen | (contd.) Since already this "one player case" is fairly non-trivial, I suspect the two player nimber won't get any easier, and at any rate, probably requires a deep understanding of Gulliksen's techniques (which I certainly don't have). | |
| May 5, 2017 at 8:34 | comment | added | Gro-Tsen | I suppose you're aware, but the case where the game has one player (playing against a ticking ordinal clock), i.e., if you replace $\mathrm{mex}$ by $\mathrm{sup}^+$ (=smallest ordinal greater than) in your definition, has been studied (under the name of "length") by Tor Gulliksen in two papers in 1973–1974: here where he studies it for Noetherian modules in general, and there where he constructs rings of arbitrary length. (contd.) | |
| May 5, 2017 at 7:10 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |