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.
