TITLE: A Game on Noetherian Rings
QUESTION [216 upvotes]: 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.)

REPLY [33 votes]: 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.