The idea of a hierarchy on Noetherian rings as suggested earlier is good, but it doesn't reflect the structure of the problem. For example, a move need not reduce the "type" of a ring (it is clear from definition however, that a move cannot reduce the type by more than 1). For instance, the ring $\mathbb{C}[x,y] / (x^2 + y^2-1)$ is of "type 1" (as noted in the original post). If we mod out by $x$, we obtain the ring $\mathbb{C}[y]/(y^2 - 1)$, which is also of "type 1".

It is also not clear (at all!) that an optimal move "ought" to decrease the "type" by 1 if possible.

Here is a different idea that resolves this issue and creates a hierarchy of Noetherian rings preserving a "win/lose" structure:

First, slightly change the original game so that you *are* allowed to mod out by a unit element and the win conditions become "the first player to pass his opponent the zero ring loses" [this game is the same as the original one].

Then we construct the following rooted directed acyclic graph, $G$:

The vertices of $G$ are the "set" of Noetherian rings, where two rings are the same vertex iff they are isomorphic [to even discuss this collection of isomorphism classes requires the axiom of choice, and it's probably too big to be called a set, but I don't think that's too important].

The directed edges of $G$ are you draw an edge from $R$ to $S$ iff there is some $0 \neq r \in R$ such that $S \cong R/(r)$ [i.e., iff a player can get to $S$ from $R$ in one move].

Then the graph $G$ is acyclic and rooted (with root $0$) since any path in this graph must have finite length terminating in 0 (since all the rings are Noetherian).

Finally, to analyze the game, we just need to mark each vertex as "win for player 1" or "win for player 2" in the usual way.

Things to consider

I believe the "type" of a ring, $R$, as defined earlier is just the length of the shortest path from $R$ to a field.

To what extent can we algorithmically determine (pieces of) the graph $G$?

If we were magically given the graph $G$, to what extent can we use it to algorithmically determine who wins in each case?

-Pat Devlin

60Does it tell too much about me that I think this game would be kind of fun? – Olivier – 2012-04-06T08:06:44.923

3Uhm..nice! I hoped to reduce to regular rings, so the winning position would coincide with the value $\pmod{2}$ of the Krull dimension. But since quotients of regular rings may fail to be regular again (see $\mathbb{Z}/p^2$), it leads nowhere... – Filippo Alberto Edoardo – 2012-04-06T09:12:34.833

1@FAE: I don't believe that Krull dimension is the right thing to consider. In a geometric situation like that of the OP, the minimal dimension of an irreducible component should play a role (you can win in one move if you have the union of a line and a plane for instance). And there are reduction issues too (start with a double line). – Jérôme Poineau – 2012-04-06T10:37:32.973

3I am absolutely shocked no one has pointed out the similarities to Choquet's game. Baire's theorem gives a characterization of the existence of a winning strategy for one of the players: www.math.auckland.ac.nz/~moors/game.pdf

Question: Why not use limites to extend the game to any type of ring? – Malte – 2012-04-06T11:40:16.073

2@Jerôme: I absolutely agree with you, I was reporting on a bad idea... – Filippo Alberto Edoardo – 2012-04-06T12:00:21.680

4Question: are the algebra softwares strong enough so that it would possible to actually implement this game ? If so, we could organized a tournament. That would be really fun. – Joël – 2012-04-06T18:56:19.623

1Does anyone have a idea if $k[x,y]$ wins or loses? – Martin Brandenburg – 2012-05-04T14:25:02.023

If $k$ is algebraically closed, the player who plays on it wins. Any Weierstrauss equation will do, since the opponent must pass you a ring of finite $\geq 2$ dimension over $k$, with which you can always pass them a field. If it's not algebraically closed, this might not work, and I'm not sure what to do. My guess is that if $k$ is a number field then the first player to play loses. – Will Sawin – 2012-05-04T19:56:16.353

In fact, Hilbert irreducibility implies that this is true. – Will Sawin – 2012-05-04T19:58:03.017

@Will: Thanks; but I don't understand the argument. I don't see why every ring of finite vector space dimension $\geq 2$ over $k$ can be moved to some field immediately, i.e. that it has some principal maximal ideal. – Martin Brandenburg – 2012-05-09T19:36:04.123

The ring is a quotient of a Dedekind domain, thus, a product of quotients of DVRs. Choose the maximal ideal from one DVR and the unit ideal from the others.

Also: In my Hilbert irreducibility argument, I neglected to mention the obviously critical assumption that the original polynomial is irreducible. Playing something reducible, like x(x-1), forces them to pass you a finite-dimension non-field (with a principal maximal ideal.) – Will Sawin – 2012-05-09T19:47:36.683

1This is also similar to Sylver Coinage. – Gabriel C. Drummond-Cole – 2014-08-20T16:39:49.490

1@Joël: I have written a GAP-program which helps to analyse the game for finite rings. One has to input the initial ring using the method RingByStructureConstants. – Martin Brandenburg – 2017-05-05T08:15:15.367