Is every ordinal the nimber of a ring? 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!
 A: 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.
