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This question is about the game of Noetherian rings, see MO/93276. Here I will include the zero ring in order to get better formulas.

The nimber of a Noetherian ring is an ordinal number. It is defined recursively by

$$\alpha(R) = \mathrm{mex} \bigl\{\alpha\bigl(R/\langle x \rangle\bigr) : 0 \neq x \in R\bigr\}.$$

where $\mathrm{mex} \, S$ denotes the smallest ordinal not contained in $S$. I am only interested in commutative rings here. You can find some basic computations in Section 5.4 here. But I do not know yet the nimber of $K[X,Y]$ for instance (this is work in progress). What I would like to know:

Does every ordinal number arise as the nimber of a Noetherian commutative ring?

For example, every ordinal number $<\omega^2$ is realizable: For all $k,n<\omega$ we have $$\alpha(R_1 \times \dotsc \times R_k \times S/p^n) = \omega \cdot k + n,$$ where $R_1,\dotsc,R_k,S$ are principal ideal domains which are no fields, and $p \in S$ is a prime element.

Examples and partial results as answers are very much appreciated!

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    $\begingroup$ I suppose you're aware, but the case where the game has one player (playing against a ticking ordinal clock), i.e., if you replace $\mathrm{mex}$ by $\mathrm{sup}^+$ (=smallest ordinal greater than) in your definition, has been studied (under the name of "length") by Tor Gulliksen in two papers in 1973–1974: here where he studies it for Noetherian modules in general, and there where he constructs rings of arbitrary length. (contd.) $\endgroup$ – Gro-Tsen May 5 '17 at 8:34
  • $\begingroup$ (contd.) Since already this "one player case" is fairly non-trivial, I suspect the two player nimber won't get any easier, and at any rate, probably requires a deep understanding of Gulliksen's techniques (which I certainly don't have). $\endgroup$ – Gro-Tsen May 5 '17 at 8:38
  • $\begingroup$ It seems to me that what you call the nimber of the ring is the same as the ordinal rank of the collection of ideals under inclusion (larger ideals are lower), which is a well-founded relation exactly in a Noetherian ring and hence has an ordinal rank there. $\endgroup$ – Joel David Hamkins May 5 '17 at 11:07
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    $\begingroup$ Ah, I see now. This isn't at all like the count-down game on the ordinals, since one can go immediately to a maximal ideal, in a PID, for example, skipping over many smaller ideals. $\endgroup$ – Joel David Hamkins May 5 '17 at 11:59
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    $\begingroup$ I'm not saying Gulliksen's work would give the answer, but I'm saying the question you ask is probably at least as complicated as the one he answers, because of the trivial fact that the Gulliksen length gives an upper bound on the nimber (which I prefer to call Grundy value, btw). So if you want to construct rings with arbitrarily high nimbers, you must construct rings with arbitrarily high Gulliksen lengths, which is already not so trivial. $\endgroup$ – Gro-Tsen May 5 '17 at 12:42
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I know how to construct all ordinals less than $\omega^3$.

Let $k$ be an infinite field.

Lemma 1: For any Noetherian ring $R$, $$\alpha( R \times k ) \geq \alpha(R)+1.$$

Proof: $R \times k$ can move to $R$ and it can also move to every element $R$ can move to, so $\alpha(R \times k) \geq \alpha(R)$.

Lemma 2: For any Noetherian ring $R$, $$\alpha\left(R \times k[x]\right) \geq \alpha(R) + \omega.$$

Proof: By inductively applying Lemma 1, every ordinal $< \alpha(R) + \omega$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k$. Hence every ordinal $< \alpha(R) + \omega$ is equal to the ordinal of a quotient of such a ring by a single element, which is necessarily a quotient of $R$ by zero or one elements times a finite product of copies of $k$, which is manifestly a quotient of $R \times k[x] $ by one element.

Lemma 3: For any Noetherian ring $R$, $$\alpha\left( R \times k[x,y] \times k[x]\right) \geq \alpha(R) + \omega^2$$

Proof: By inductively applying Lemma 2, every ordinal $< \alpha(R) + \omega^2$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k[x]$. So it is equal to the ordinal of a quotient of $R$ by $0$ or $1$ elements, times a finite product of copies of $k[x]$, times a finite product of quotients of $k[x]$ by one element. The first is a quotient of $R$ by $0$ or $1$ elements, the second is a quotient of $k[x,y]$ by one element, and the third, crucially, is a quotient of $k[x]$ by one element, because we can fit any finite number of zero-dimensional schemes that individually embed into the line together in the same line so that they do not intersect. Hence every ordinal less than $\alpha(R) + \omega^2$ is the ordinal of a quotient of $R \times k[x,y] \times k[x]$ by one element.

This last step prevents us from extending the argument one dimension higher, because there is no Noetherian affine scheme such that a finite union of hypersurfaces in $\mathbb A^2$ is itself a hypersurface in that scheme - at least I don't think there is.

However, by iterating Lemma 3, we can construct rings with nimbers at least $ n \omega^2$ for all natural numbers $n$, and hence verify the existence of rings with any given nimber $< \omega^3$.

Note that we have done this inductive argument without at any point upper bounding the nimber of any ring. I am guessing that at some point upper bounds for certain rings will be needed to lower bound the nimbers of other rings.

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  • $\begingroup$ Short remark: In Lemma 1, inequality can hold: When $R=k[X,Y]/\langle X^2,XY,Y^2\rangle$, then $\alpha(R)=1$, but $\alpha(R \times K)=4$. $\endgroup$ – Martin Brandenburg May 5 '17 at 16:11
  • $\begingroup$ @MartinBrandenburg Yes, and presumably the same sort of example exists for Lemma 2 and Lemma 3 even if our current nimber-computing abilities are not strong enough to verify it. $\endgroup$ – Will Sawin May 5 '17 at 16:13
  • $\begingroup$ Thank you for your answer; I really like your proof method. I have a question about Lemma 3. You have said (geometrically) that for every two non-zero $f,g \in k[x]$ there is some non-zero $h \in k[x]$ with $k[x]/\langle f \rangle \times k[x]/\langle g \rangle \cong k[x]/\langle h \rangle$. If $k$ is algebraically closed (or at least, if $f,g$ factor over $k$), I can prove this because there is some $\lambda \in k$ such that $f(x)$ and $g(x+\lambda)$ are coprime, and $h(x) = f(x) g(x+\lambda)$ does the job. How does one argue in the general case? $\endgroup$ – Martin Brandenburg May 5 '17 at 16:39
  • $\begingroup$ Ah, I see: $\{b-a : a \in \overline{k},\, b \in \overline{k},\, f(a)=0,\, g(b)=0\}$ is a finite subset of $\overline{k}$, and since $k \subseteq \overline{k}$ is infinite, we may choose some element $\lambda \in k$ outside of this set. $\endgroup$ – Martin Brandenburg May 5 '17 at 16:59
  • $\begingroup$ @MartinBrandenburg In fact that's a little more elegant than the argument I was thinking of. $\endgroup$ – Will Sawin May 5 '17 at 17:05

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