A friend suggested the following combinatorial game. At any time, the state of the game is a (commutative) Noetherian ring ≠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 C[x,y]/(x2+y2−1), and a player chooses x, the ring becomes C[y]/(y2−1). 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.)

**Answer**

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.

**Attribution***Source : Link , Question Author : Will Sawin , Answer Author : Kevin Buzzard*