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:

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

  2. 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.

  3. 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.

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    $\begingroup$ See section 6 of this paper, by the way, to check your work for k[x,y] :) arxiv.org/pdf/1205.2884v1.pdf $\endgroup$ – Dylan Wilson Jun 16 '12 at 7:23
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    $\begingroup$ @Alex: I haven't checked your proof for $k[x,y]$ in full detail, but it seems to be wrong. With the notation of my paper mentioned by Dylan, you claim that $k[x,y]/(x(x+1)) \cong k[y] \times k[y]$ is $\mathcal{P}$, but this contradicts Proposition 6.2. In fact, we can mod out $(y,1)$ and get the field $k$. $\endgroup$ – Martin Brandenburg Mar 9 '13 at 3:10
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    $\begingroup$ @MartinBrandenburg Ah, I see. Mentally for some reason I must have been picturing $V(f(x,y))$ as the graph of some polynomial $g(x)$, which is of course ridiculous. I'll edit my post soon. $\endgroup$ – Alex Becker Mar 9 '13 at 10:24
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    $\begingroup$ I've made some progress. I solved it for $\mathbb{Z}[x]$ (it will appear in the second version of my paper). $\endgroup$ – Martin Brandenburg May 28 '13 at 10:11
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    $\begingroup$ Just seen the bounty. Perhaps I should make my proof for $R[x]$ available where $R$ is a PID. I don't want to change the arXiv file, so for now: dropbox.com/s/hrhx0qewpsah3cu/game_3.pdf?dl=0 (Section 5.3). The next case would be $R[x,y]$ where $R$ is a PID (this includes $k[x,y,z]$ as a special case), and I've tried this a lot some time ago, but didn't succeed. There are many more lower-dimensional cases which one has to consider first. $\endgroup$ – Martin Brandenburg Dec 15 '14 at 19:15

protected by user26857 May 6 '15 at 20:18

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