Timeline for Is every ordinal the nimber of a ring?
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| May 5, 2017 at 23:15 | comment | added | Martin Brandenburg | Lemma 3: $\alpha(R \times S \times T) \geq \alpha(R) + \omega^2$ holds when there is a field $k$ such that (a) as before, and (b) every option of a product $S^n$ is an option of $S \times T$. (This holds for instance when (b1) every product $S^n$ is an option of $T$, and (b2) every product of options of $S$ is an option of $S$.) Lemma 4: $\alpha(R \times S \times T \times U) \geq \alpha(R) + \omega^3$ holds when there is a field $k$ such that (a),(b) as before, and (c) every option of a product $(S \times T)^n$ is an option of $S \times T \times U$. The question is: Do $S,T,U$ exist? | |
| May 5, 2017 at 23:14 | comment | added | Martin Brandenburg | Because, as you say, the analogue of Lemma 3 probably does not work the same in the next dimension, perhaps it is useful to generalize the Lemmas, keeping the dimensions, but replacing polynomial rings by more general rings. Lemma 1: $\alpha(R \times S) \geq \alpha(R)+1$ holds for every non-zero ring $S$. Lemma 2: $\alpha(R \times S) \geq \alpha(R)+\omega$ holds when there is a non-zero ring $k$ such that every option of a product $k^n$ is an option of $S$. If $k$ is a field, we only have to require that every product $k^n$ is an option of $S$. | |
| May 5, 2017 at 20:52 | comment | added | Will Sawin | @MartinBrandenburg I don't know. I guess the first step would be to calculate the nimber of $k[x,y]$ or, even better, find all nimbers that can be written as nimbers of quotients of $k[x,y]$. | |
| May 5, 2017 at 19:41 | comment | added | Martin Brandenburg | Is there any explicit ring with nimber $=\omega^2$ which we can extract from your arguments? Since $k[x,y] \times k[x]$ has nimber $\geq \omega^2$ by Lemma 3, either this ring or some option of it has nimber $=\omega^2$ (this uses LEM!). If the latter holds, which option exactly has nimber $\omega^2$, is it perhaps $k[x,y]$? I know that this question has no simple answer, but I think it is interesting to have explicit witnesses for given ordinals. | |
| May 5, 2017 at 17:05 | comment | added | Will Sawin | @MartinBrandenburg In fact that's a little more elegant than the argument I was thinking of. | |
| May 5, 2017 at 16:59 | comment | added | Martin Brandenburg | 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. | |
| May 5, 2017 at 16:39 | comment | added | Martin Brandenburg | 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? | |
| May 5, 2017 at 16:25 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
added 15 characters in body
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| May 5, 2017 at 16:13 | comment | added | Will Sawin | @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. | |
| May 5, 2017 at 16:11 | comment | added | Martin Brandenburg | 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 $\alpha(R \times K)=4$. | |
| May 5, 2017 at 15:49 | history | answered | Will Sawin | CC BY-SA 3.0 |