I recently read about the Ring Game on MathOverflow, and have been trying to determine winning strategies for each player on various rings. The game has two players and begins with a commutative Noetherian ring $R$. Player one mods out a nonzero non-unit, and gives the resulting ring to player 2, who repeats the process. The last person to make a legal move (i.e. whoever produces a field) wins. For PIDs the game is trivial: player 1 wins by modding out a single prime element. However, the game becomes far more complicated even in the case of 2-dimensional UFDs. After several days I was unable to determine a winning strategy for either player for $\mathbb Z[x]$.

I believe the most tractable class of rings are finitely generated commutative algebras over an algebraically closed field $K$, as for these we can take advantage of the nullstellensatz. So far, I've been able to deal with the cases of $K[x]$ and $K[x,y]$. Player 1 has a trivial winning strategy for $K[x]$, as it is a PID. For $K[x,y]$, player 1 has a winning strategy as follows:

$1.$ Player 1 plays $x(x+1)$.

$2.$ Player 2 plays $f(x,y)$ which vanishes somewhere on $V(x(x+1))$ but not everywhere (this describes all legal moves). Note that $V(f(x,y),x(x+1))$ is a finite collection of points possibly union $V(x)$ or $V(x+1)$ but not both. Furthermore, $V(f(x,y),x(x+1))$ cannot be a single point as the projection of $V(f(x,y))$ onto the $x$-axis yields an algebraic set, which must either be a finite collection of points or the entire line and so intersects $V(x(x+1))$ in at least two points, thus $R/(x(x+1),f(x,y))$ is not a field.

$3.$ Player 1 plays $ax+by+c$ which vanishes at exactly one point in $V(f(x,y),x(x+1))$. This is always possible, since any finite collection of points can be avoided and the lines $V(x)$ or $V(x+1)$ will intersect $V(ax+by+c)$ at most once. The resulting ring is a field, as the ideal $I$ generated by the three plays contains either $(ax+by+c,x)$ or $(ax+by+c,x+1)$, which are maximal, hence is equal to one of these and $R/I$ is a field.

However, I've no idea where to go for $K[x,y,z]$ and beyond. I suspect that $K[x,y,z]$ is a win for player 2, since anything player 1 plays makes the result look vaguely like $K[x,y]$, but on the other hand player 1 could make some pretty nasty plays which might trip up any strategy of player 2's.

So my question is: which player has a winning strategy for $K[x,y,z]$, and what is one such strategy?

Edit: As pointed out in the comments, this winning strategy is wrong.

Simon Fraser
  • 2,428
  • 10
  • 27
Alex Becker
  • 58,347
  • 7
  • 124
  • 180
  • 24
    See section 6 of this paper, by the way, to check your work for k[x,y] :) http://arxiv.org/pdf/1205.2884v1.pdf – Dylan Wilson Jun 16 '12 at 07:23
  • 21
    @Alex: I haven't checked your proof for $k[x,y]$ in full detail, but it seems to be wrong. With the notation of my paper mentioned by Dylan, you claim that $k[x,y]/(x(x+1)) \cong k[y] \times k[y]$ is $\mathcal{P}$, but this contradicts Proposition 6.2. In fact, we can mod out $(y,1)$ and get the field $k$. – Martin Brandenburg Mar 09 '13 at 03:10
  • 9
    @MartinBrandenburg Ah, I see. Mentally for some reason I must have been picturing $V(f(x,y))$ as the graph of some polynomial $g(x)$, which is of course ridiculous. I'll edit my post soon. – Alex Becker Mar 09 '13 at 10:24
  • 2
    Alex, what is a winning strategy for $\mathbb{Z}[x]$? – Martin Brandenburg Mar 17 '13 at 15:26
  • 3
    @MartinBrandenburg As I said, I was unable to determine one. – Alex Becker Mar 18 '13 at 01:01
  • 33
    I've made some progress. I solved it for $\mathbb{Z}[x]$ (it will appear in the second version of my paper). – Martin Brandenburg May 28 '13 at 10:11
  • 3
    @MartinBrandenburg Nice. I look forward to reading it. – Alex Becker May 28 '13 at 12:08
  • 16
    Just seen the bounty. Perhaps I should make my proof for $R[x]$ available where $R$ is a PID. I don't want to change the arXiv file, so for now: https://www.dropbox.com/s/hrhx0qewpsah3cu/game_3.pdf?dl=0 (Section 5.3). The next case would be $R[x,y]$ where $R$ is a PID (this includes $k[x,y,z]$ as a special case), and I've tried this a lot some time ago, but didn't succeed. There are many more lower-dimensional cases which one has to consider first. – Martin Brandenburg Dec 15 '14 at 19:15
  • 2
    It seems to me that the case $K[x, y]$ is still easy: the first player mods out $xy$, and the rest are clear. – WhatsUp Jul 23 '15 at 12:03
  • 4
    ...and the second player mods out by $x-1$ and wins. – Jeroen Jul 31 '15 at 02:45
  • 2
    @MartinBrandenburg Your article is very interesting ! Where does $18x-8$ comes from in the last example 5.13 ? – Xoff Jun 09 '16 at 07:20
  • 3
    @Xoff: It is a combination of all the proofs appearing before. We mod out $36=2^2 3^2$, so that we have a product of two rings by CRT, now apply the strategy for $\mathbb{Z}/p^2[X]/(X^2)$, etc. It is rather clumsy to write down all the details, and $18x-8$ is really just the end result after putting it all together. But you can also verify by hand that the quotient is, as stated, $\mathbb{Z}/4[X]/(X^2,2X)$. Why do we discuss this here, where $k[X,Y,Z]$ is asked? E-Mail would have been better and you would not have to wait 4 months. – Martin Brandenburg Oct 21 '16 at 14:14
  • 3
    The revised paper is now online: https://arxiv.org/abs/1205.2884 (also http://link.springer.com/article/10.1007/s00182-017-0577-7). Hence I have deleted the file game_3 linked above. I have made some progress concerning $R[X,Y]$ for PIDs $R$ which are no fields (in particular $K[X,Y,Z]$), I think I have found a winning move. But my proof is not complete yet. Actually it is just a bunch of conjectures which I could not refute so far. As in the case of $R[X]$, one has to work with several zero-dimensional rings first. – Martin Brandenburg May 05 '17 at 07:19

1 Answers1


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.