Timeline for A game on Noetherian rings
Current License: CC BY-SA 4.0
22 events
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| Oct 26 at 19:08 | comment | added | WhatsUp | @TimothyChow I don't quite agree with the sentence "Nimbers are inevitable". Of course it depends on how to interpret it; but e.g. for Wythoff's game, it's easy to describe the winning/losing positions, yet the nimbers are chaotic. So, it could happen that determining whether the nimber is zero is a much easier problem than finding the pattern in the nimbers. | |
| May 20, 2023 at 9:30 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| May 5, 2017 at 7:37 | comment | added | Martin Brandenburg | One gets better formulas if one includes the zero ring; for example then $k[x]/(x^n)$ has nimber $n$, and this is even better if we have products of such rings. See also Section 5.4 of my revised paper arxiv.org/abs/1205.2884 | |
| Mar 22, 2016 at 19:17 | review | Suggested edits | |||
| Mar 22, 2016 at 19:41 | |||||
| May 5, 2012 at 8:25 | comment | added | Kevin Buzzard | The $k$ in my assertion is not the ground field $k$, it can be bigger. Sorry for the confusion. I'm just using that any ideal is the product of prime ideals and the chinese remainder theorem. It doesn't matter for the purposes of this question anyway -- it's a far stronger assertion than what is necessary. | |
| May 3, 2012 at 8:07 | comment | added | Martin Brandenburg | @Kevin: Could you explain your statement "because any move will reduce the situation to a direct sum of rings of the form $k[x]/(x^n)$"? | |
| Apr 27, 2012 at 2:59 | comment | added | Will Sawin | This seems problematic because one should then be able to divide a higher edge by a lower edge, and get a non-invertible number that does not clearly correspond to any edge. | |
| Apr 27, 2012 at 2:58 | comment | added | Will Sawin | @Bruce: So you want to construct a ring on a rooted graph where there is a bijection between ideals and sets of edges closed upwards, with principal ideals being the ones generated by a particular edge? | |
| Apr 25, 2012 at 19:33 | comment | added | Bruce Westbury | I can try. Let me formulate it differently. There is a variation of this question with Noetherian categories. A move consists of nominating a morphism and taking the quotient by the ideal it generates. This seems like playing Hackenbush with directed graphs. You could now allow a move to specify an arbitrary ideal and get a different game. You could then take the algebra of the category and play the game in the question. A fourth variation is to start with a Noetherian ring and a move consists in passing to the quotient by a principal ideal. | |
| Apr 25, 2012 at 6:44 | comment | added | Kevin Buzzard | OK. Can you be more explicit about the "by taking path algebras with relations" bit? | |
| Apr 24, 2012 at 16:04 | comment | added | Bruce Westbury | I am not starting from an arbitrary game nor am I looking at a game tree. I am starting from an impartial Hackenbush position. | |
| Apr 21, 2012 at 18:15 | comment | added | Kevin Buzzard | It sounds to me like a hard problem to produce a Noetherian ring corresponding to an arbitrary game tree such that all paths through it are finite. Do you want to be more explicit about the "by taking path algebras with relations" bit?? | |
| Apr 21, 2012 at 17:18 | comment | added | Bruce Westbury | Am I right in thinking that by taking path algebras with relations that say two paths are equal that you can realise any impartial Hackenbush game as the game of a Noetherian ring (which is Artinian iff the Hackenbush graph is finite)? | |
| Apr 20, 2012 at 6:16 | comment | added | Martin Brandenburg | Oh, I understand: If $R$ is a finite-dim vector space over $k$, then its nimber is $<\omega$. | |
| Apr 19, 2012 at 18:51 | comment | added | Kevin Buzzard | Right -- I was implicitly invoking the fact that a ring for which the game only has a finite number of plays, has finite nimber. | |
| Apr 19, 2012 at 16:18 | comment | added | Martin Brandenburg | @Kevin: Your reasoning for $k[x]$ only shows that the nimber is $\geq \omega$, right? For equality, we have to show that the proper quotients $k[x]/(f)$ have nimber $<\omega$. For this, one shows by induction on $\mathrm{deg}(f)$ that the nimber is $\leq \mathrm{deg}(f)-1$. | |
| Apr 9, 2012 at 3:13 | comment | added | Timothy Chow | @Kevin: Thanks for the additional comment. By the way, it's true in surprisingly great generality that determining whether or not the nimber is zero is no easier than computing the nimber itself. See "Nimbers are inevitable," arxiv.org/abs/1011.5841 | |
| Apr 8, 2012 at 18:28 | comment | added | YCor | @Kevin: By your remarks, the problem is also interesting for one-dimensional domains, for which you can sometimes give an answer (with examples with nimber zero or not) without fully solving the Artinian case. | |
| Apr 8, 2012 at 7:57 | comment | added | Kevin Buzzard | The OP is asking "is the nimber zero or not" -- but the only algorithm I know for figuring out whether the nimber is zero or not is easily modified to one which computes the nimbers anyway, and is of the form "compute what's going on with all the quotients". The algorithm for computing a nimber is recursive -- you compute nimbers of all quotients by principal ideals and then take the mex. I just ran this algorithm for some simple Artin rings. Even for Artin rings the answer will be very complicated I should think. | |
| Apr 8, 2012 at 3:14 | comment | added | Timothy Chow | Nice answer. The OP seems to be asking a simpler question, though: Is there an algorithm for computing the nimber of a given Noetherian ring? Obviously this will depend on how the ring is specified, but I assume that you managed to figure out an algorithm for certain kinds of specifications? | |
| Apr 7, 2012 at 11:31 | history | edited | Kevin Buzzard | CC BY-SA 3.0 |
added a bit about smooth curves
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| Apr 7, 2012 at 11:15 | history | answered | Kevin Buzzard | CC BY-SA 3.0 |