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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.

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.

We could also play this 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 mod out the smallest normal subgroup containing $a \neq 1$. The ending condition holds iff the ascending chain condition with respect to 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: Given a variety in the sense of universal algebra, start with with an algebra $A$. A move consists in replacing $A$ by $A/a \sim b$, where $a,b \in A$ with $a \neq b$. The game proposed by Will Sawin is the game on rings, including the zero ring, under the misère play rule, i.e. the last one moving loses (see Tom Goodwillie's comment).

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the article Algebraic games. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ (i.e. are losing positions):

Let $A$ be a finitely generated abelian group.

• Under the normal play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some finitely generated abelian group $B$.

• Under the misère play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.

Now for something on-topic, some results about the game on rings:

If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Lemma 6.2). Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$. Otherwise, $R$ is $\mathcal{P}$ (Prop. 6.3). It follows, for example, that $k[x,y]/(y^2-x^3+x-1)$ is $\mathcal{P}$. Hence, $k[x,y]$ is $\mathcal{N}$. Section 6.2 is devoted to zero-dimensional rings (whose complexity was already mentioned by Tom Goodwillie), finally showing that the cusp $k[x,y]/(y^2-x^3)$ is $\mathcal{P}$.

Many problems remain open, for example if $k[x_1,\dotsc,x_n]$ is $\mathcal{N}$ for all $n \geq 1$.

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.

We could also play this 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 mod out the smallest normal subgroup containing $a \neq 1$. The ending condition holds iff the ascending chain condition with respect to 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: Given a variety in the sense of universal algebra, start with with an algebra $A$. A move consists in replacing $A$ by $A/a \sim b$, where $a,b \in A$ with $a \neq b$. The game proposed by Will Sawin is the game on rings, including the zero ring, under the misère play rule, i.e. the last one moving loses (see Tom Goodwillie's comment).

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the article Algebraic games. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ (i.e. are losing positions):

Let $A$ be a finitely generated abelian group.

• Under the normal play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some finitely generated abelian group $B$.

• Under the misère play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.

Now for something on-topic, some results about the game on rings:

Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$. Otherwise, $R$ is $\mathcal{P}$ (Prop. 6.3). It follows, for example, that $k[x,y]/(y^2-x^3+x-1)$ is $\mathcal{P}$. Hence, $k[x,y]$ is $\mathcal{N}$. Section 6.2 is devoted to zero-dimensional rings (whose complexity was already mentioned by Tom Goodwillie), finally showing that the cusp $k[x,y]/(y^2-x^3)$ is $\mathcal{P}$.

Many problems remain open, for example if $k[x_1,\dotsc,x_n]$ is $\mathcal{N}$ for all $n \geq 1$.

We could also play this 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 mod out the smallest normal subgroup containing $a \neq 1$. The ending condition holds iff the ascending chain condition with respect to 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: Given a variety in the sense of universal algebra, start with with an algebra $A$. A move consists in replacing $A$ by $A/a \sim b$, where $a,b \in A$ with $a \neq b$. The game proposed by Will Sawin is the game on rings, including the zero ring, under the misère play rule, i.e. the last one moving loses (see Tom Goodwillie's comment).

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the article Algebraic games. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ (i.e. are losing positions):

Let $A$ be a finitely generated abelian group.

• Under the normal play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some finitely generated abelian group $B$.

• Under the misère play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.

Now for something on-topic, some results about the game on rings:

If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Lemma 6.2). Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$. Otherwise, $R$ is $\mathcal{P}$ (Prop. 6.3). It follows, for example, that $k[x,y]/(y^2-x^3+x-1)$ is $\mathcal{P}$. Hence, $k[x,y]$ is $\mathcal{N}$. Section 6.2 is devoted to zero-dimensional rings (whose complexity was already mentioned by Tom Goodwillie), finally showing that the cusp $k[x,y]/(y^2-x^3)$ is $\mathcal{P}$.

Many problems remain open, for example if $k[x_1,\dotsc,x_n]$ is $\mathcal{N}$ for all $n \geq 1$.