Timeline for A game on Noetherian rings
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2012 at 13:33 | comment | added | Emil Jeřábek | If we are interested in the game starting from a particular ring $R$, the relevant part of $G$ can be represented as follows: vertices are the ideals of $R$, and there is an edge from $I$ to $J$ if $J=I+aR$ for some $a\notin I$. In any case, it seems to me that this is just a trivial restatement of the problem (any game can be represented as a graph). Is there anything the upvoters see and I am missing? | |
| Apr 7, 2012 at 10:52 | comment | added | Philip van Reeuwijk | It might be nice to consider a generalization of the ideal class group that measures how far a ring is from being a PID. Then you could hope this group is finite, and that the parity of its cardinality solves the problem... just a guess, though. | |
| Apr 6, 2012 at 18:08 | comment | added | Will Sawin | I doubt it can be extended to a UFD, since UFDs can have quotients that are not UFDs. | |
| Apr 6, 2012 at 16:50 | comment | added | Pat Devlin | If $R$ is a PID then the problem has a complete solution. Case I: If $R$ is a field, player 1 loses. Case II: If $R$ is not a field, let $I \subseteq R$ be a maximal proper ideal. Then since $R$ is a PID, $I=(r)$ for some $0 \neq r \in R$. But then $R/I = R/(r)$ is a field [because $I$ was a maximal ideal]. Therefore, with this move for player 1, he can make it so that player 2 loses. I do not know if this can be extended to $R$ a UFD or not. | |
| Apr 6, 2012 at 12:56 | history | answered | Pat Devlin | CC BY-SA 3.0 |