A friend suggested the following combinatorial game. At any time, the state of the game is a (commutative) Noetherian ring $\neq 0$. On a player's turn, that player chooses a nonzero non-unit element of the ring, and replaces the ring with its quotient by the ideal generated by that element. The player to make the last legal move wins, by passing the opponent a field.

So if the ring was $\mathbb C[x,y]/(x^2+y^2-1)$, and a player chooses $x$, the ring becomes $\mathbb C[y]/(y^2-1)$. This is a poor move, as his opponent can turn the ring into a field by choosing either $y+1$ or $y-1$, and win. A winning move would have been $x+iy+1$, which turns the ring into a field immediately and wins the game.

Problem: Given a ring, how can we tell if it is a winning position for the first player or for the second player?

(The game terminates since the original ring is Noetherian, and an unending game would be an infinite ascending chain in the original ring.)

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    $\begingroup$ Does it tell too much about me that I think this game would be kind of fun? $\endgroup$
    – Olivier
    Apr 6, 2012 at 8:06
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    $\begingroup$ Uhm..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... $\endgroup$ Apr 6, 2012 at 9:12
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    $\begingroup$ I 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? $\endgroup$
    – Malte
    Apr 6, 2012 at 11:40
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    $\begingroup$ Question: 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. $\endgroup$
    – Joël
    Apr 6, 2012 at 18:56
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    $\begingroup$ Does anyone have a idea if $k[x,y]$ wins or loses? $\endgroup$ May 4, 2012 at 14:25

4 Answers 4


I computed the nimbers of a few rings, for what it's worth. I don't see any sensible pattern so perhaps the general answer is hopelessly hard. This wouldn't be surprising, because even for very simple games like sprouts starting with $n$ dots no general pattern is known for the corresponding nimbers.

OK so the way it works is that the nimber of a ring $A$ is the smallest ordinal which is not in the set of nimbers of $A/(x)$ for $x$ non-zero and not a unit. The nimber of a ring is zero iff the corresponding game is a second player win -- this is a standard and easy result in combinatorial game theory. If the nimber is non-zero then the position is a first player win and his winning move is to reduce the ring to a ring with nimber zero.

Fields all have nimber zero, because zero is the smallest ordinal not in the empty set. An easy induction on $n$ shows that for $k$ a field and $n\geq1$, the nimber of $k[x]/(x^n)$ is $n-1$; the point is that the ideals of $k[x]/(x^n)$ are precisely the $(x^i)$. In general an Artin local ring of length $n$ will have nimber at most $n-1$ (again trivial induction), but strict inequality may hold. For example if $V$ is a finite-dimensional vector space over $k$ and we construct a ring $k\oplus \epsilon V$ with $\epsilon^2=0$, this has nimber zero if $V$ is even-dimensional and one if $V$ is odd-dimensional; again the proof is a simple induction on the dimension of $V$, using the fact that a non-zero non-unit element of $k\oplus\epsilon V$ is just a non-zero element of $V$, and quotienting out by this brings the dimension down by 1. In particular the ring $k[x,y]/(x^2,xy,y^2)$ has nimber zero, which means that the moment you start dealing with 2-dimensional varieties things are going to get messy. But perhaps this is not surprising -- an Artin local ring is much more complicated than a game of sprouts and even sprouts is a mystery.

Rings like $k[[x]]$ and $k[x]$ have nimber $\omega$, the first infinite ordinal, as they have quotients of nimber $n$ for all finite $n$. As has been implicitly noted in the comments, the answer for a general smooth connected affine curve (over the complexes, say) is slightly delicate. If there is a principal prime divisor then the nimber is non-zero and probably $\omega$ again; it's non-zero because P1 can just reduce to a field. But if the genus is high then there may not be a principal prime divisor, by Riemann-Roch, and now the nimber will be zero because any move will reduce the situation to a direct sum of rings of the form $k[x]/(x^n)$ and such a direct sum has positive nimber as it can be reduced to zero in one move. So there's something for curves. For surfaces I'm scared though because the Artin local rings that will arise when the situation becomes 0-dimensional can be much more complicated.

I don't see any discernible pattern really, but then again the moment you leave really trivial games, nimbers often follow no discernible pattern, so it might be hard to say anything interesting about what's going on.

  • $\begingroup$ 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? $\endgroup$ Apr 8, 2012 at 3:14
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    $\begingroup$ 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. $\endgroup$ Apr 8, 2012 at 7:57
  • $\begingroup$ @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. $\endgroup$
    – YCor
    Apr 8, 2012 at 18:28
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    $\begingroup$ @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 $\endgroup$ Apr 9, 2012 at 3:13
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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Apr 27, 2012 at 2:59

Let us call a Noetherian commutative ring $R$ an $\mathcal{N}$-position if the next player has a winning strategy; otherwise the previous player has a winning strategy and we will call it a $\mathcal{P}$-position. The zero ring is allowed and hence we will choose the misère play rule, i.e. the player with the last move loses (see Tom Goodwillie's comment). For example, it is clear that the zero ring is $\mathcal{N}$, that fields are $\mathcal{P}$, and that PIDs which are no fields are $\mathcal{N}$.

In general, $R$ is $\mathcal{P}$ iff $R/\langle x \rangle$ is $\mathcal{N}$ for all $0 \neq x \in R$, and $ R$ is $\mathcal{N}$ iff either $R=0$ or $R/\langle x \rangle$ is $\mathcal{P}$ for some $0 \neq x \in R$. As a general theme, $\mathcal{P}$-positions are quite rare and hard to find, and every $\mathcal{P}$-position is responsible for many $\mathcal{N}$-positions which are then more easy to find.

The following results are proven here:

  • If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Rem. 5.1).
  • Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$; otherwise, $R$ is $\mathcal{P}$ (Prop. 5.6).
  • It follows that $K[X,Y]/\langle f \rangle $ is $\mathcal{P}$ when $f$ is a Weierstrass equation and $K$ is an algebraically closed field (Prop. 5.7).
  • One can also show that the coordinate ring of the cusp $K[X,Y]/\langle Y^2-X^3 \rangle$ is $\mathcal{P}$ (Prop. 5.13).
  • Hence, $K[X,Y]$ is $\mathcal{N}$ if $K$ is algebraically closed.
  • This actually holds for every field $K$, because $K[X,Y]/\langle X^2 \rangle$ is $\mathcal{P}$ (Prop. 5.16).
  • For this one needs that $K[X,Y]/\langle X^2,f^{n+1},f^n X \rangle$ is $\mathcal{P}$ for every irreducible $f \in K[Y]$ (special case of Prop. 5.15).

I conjecture that $K[X_1,\dotsc,X_n]$ is $\mathcal{N}$ for all $n \geq 1$.

We can also play such a game with groups. One starts with a group $G$. A move consists in replacing $G$ by $G/\langle\langle a \rangle\rangle$, i.e. we quotient out the smallest normal subgroup containing some element $a \neq 1$. The ending condition holds iff the ascending chain condition on normal subgroups holds (do these groups have a name?). When $G$ is abelian, this means that $G$ is finitely generated.

Actually we can play this game for every algebraic structure: We start with an algebra $A$ of a given signature. A move consists in replacing $A$ by $A/(a \sim b)$, where $a,b \in A$ with $a \neq b$.

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the mentioned article. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ under both play rules:

If $A$ is a finitely generated abelian group, then

  • $A$ is a normal $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some abelian group $B$.
  • $A$ is a misère $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.
  • $\begingroup$ Interesting. If we restrict to abelian $p$-groups, then the game with $\mathbf{Z}/p^{a_1} \mathbf{Z} \oplus \ldots \oplus \mathbf{Z}/p^{a_n} \mathbf{Z}$ seems to be equivalent to Nim with $(a_1,\ldots,a_n)$. $\endgroup$ Apr 22, 2012 at 9:24
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    $\begingroup$ @Francois: No since we can also mod out, say, $(1,1,\dotsc)$. $\endgroup$ Apr 22, 2012 at 9:38

This exact game was studied and published in 1970: Henson's "Winning Strategies for the Ideal Game".

In particular, if $R$ is a single-variable polynomial over any PID then the first player wins. Generally, if $R$ is an integral domain with an element $x$ such that $R/(x)$ is a PID but not a field then the first player wins, and the second player wins the corresponding $R/(x^2)$ game.


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

  • $\begingroup$ 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. $\endgroup$
    – Pat Devlin
    Apr 6, 2012 at 16:50
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    $\begingroup$ I doubt it can be extended to a UFD, since UFDs can have quotients that are not UFDs. $\endgroup$
    – Will Sawin
    Apr 6, 2012 at 18:08
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    $\begingroup$ 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. $\endgroup$ Apr 7, 2012 at 10:52
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    $\begingroup$ 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? $\endgroup$ Apr 7, 2012 at 13:33

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