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