Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
7  
See section 6 of this paper, by the way, to check your work for k[x,y] :) arxiv.org/pdf/1205.2884v1.pdf –  Dylan Wilson Jun 16 '12 at 7:23
8  
@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 $k$. –  Martin Brandenburg Mar 9 '13 at 3:10
    
@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. –  Alex Becker Mar 9 '13 at 10:24
12  
I've made some progress. I solved it for $\mathbb{Z}[x]$ (it will appear in the second version of my paper). –  Martin Brandenburg May 28 '13 at 10:11
3  
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. –  Martin Brandenburg Dec 15 '14 at 19:15

protected by user26857 May 6 at 20:18

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Browse other questions tagged or ask your own question.