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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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    See section 6 of this paper, by the way, to check your work for k[x,y] :) http://arxiv.org/pdf/1205.2884v1.pdf2012-06-16
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    @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 $k$.2013-03-09
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    @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.2013-03-09
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    Alex, what is a winning strategy for $\mathbb{Z}[x]$?2013-03-17
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    @MartinBrandenburg As I said, I was unable to determine one.2013-03-18
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    I've made some progress. I solved it for $\mathbb{Z}[x]$ (it will appear in the second version of my paper).2013-05-28
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    @MartinBrandenburg Nice. I look forward to reading it.2013-05-28
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    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: https://www.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.2014-12-15
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    It seems to me that the case $K[x, y]$ is still easy: the first player mods out $xy$, and the rest are clear.2015-07-23
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    ...and the second player mods out by $x-1$ and wins.2015-07-31
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    @MartinBrandenburg Your article is very interesting ! Where does $18x-8$ comes from in the last example 5.13 ?2016-06-09
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    @Xoff: It is a combination of all the proofs appearing before. We mod out $36=2^2 3^2$, so that we have a product of two rings by CRT, now apply the strategy for $\mathbb{Z}/p^2[X]/(X^2)$, etc. It is rather clumsy to write down all the details, and $18x-8$ is really just the end result after putting it all together. But you can also verify by hand that the quotient is, as stated, $\mathbb{Z}/4[X]/(X^2,2X)$. Why do we discuss this here, where $k[X,Y,Z]$ is asked? E-Mail would have been better and you would not have to wait 4 months.2016-10-21
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    The revised paper is now online: https://arxiv.org/abs/1205.2884 (also http://link.springer.com/article/10.1007/s00182-017-0577-7). Hence I have deleted the file game_3 linked above. I have made some progress concerning $R[X,Y]$ for PIDs $R$ which are no fields (in particular $K[X,Y,Z]$), I think I have found a winning move. But my proof is not complete yet. Actually it is just a bunch of conjectures which I could not refute so far. As in the case of $R[X]$, one has to work with several zero-dimensional rings first.2017-05-05

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