Timeline for Krull dimension and elimination theory over the integers
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2020 at 14:07 | comment | added | Stein Chen | Ok, thank you very much. | |
| Dec 28, 2020 at 1:15 | comment | added | R. van Dobben de Bruyn | The ring $R_{\mathbf Q} = R \otimes_{\mathbf Z} \mathbf Q$ is a localisation of $R$, so $\dim R_{\mathbf Q} \leq \dim R$. But in fact it is true that $\dim R_{\mathbf Q} \leq \dim R-1$, since $\dim \mathbf Z = 1$ and components of $R$ that do not dominate $\operatorname{Spec} \mathbf Z$ disappear in $R_{\mathbf Q}$. In particular my comment applies when $\dim R \leq 1$. | |
| Dec 27, 2020 at 22:55 | vote | accept | Stein Chen | ||
| Dec 27, 2020 at 22:22 | comment | added | Stein Chen | Last question: If the Krull dimension of $R$ in the $K=\mathbb{Z}$ case is zero, does this imply that the Krull dimension of $R$ in the $K=\mathbb{Q}$ case is zero (via Groebner bases)? Sorry for these trivial questions...I am not familiar with this area of mathematics. | |
| Dec 27, 2020 at 22:09 | history | undeleted | RumDiary | ||
| Dec 27, 2020 at 22:09 | history | deleted | RumDiary | via Vote | |
| Dec 27, 2020 at 22:05 | comment | added | Stein Chen | Ok, thank you very much. | |
| Dec 27, 2020 at 22:00 | comment | added | R. van Dobben de Bruyn | There is provably no algorithm that determines whether a system of polynomials over $\mathbf Z$ has a solution (over $\mathbf Z$), so in particular no algorithm to find all solutions. However, in the baby case of my comment you may be able to do it (find the components of $R\otimes_{\mathbf Z}\mathbf Q$, determine which ones have length $1$ over $\mathbf Q$, and then determine which of these solutions have no denominators). Beyond that, already for smooth curves it is a very hard problem that seems out of reach with current technology. | |
| Dec 27, 2020 at 21:56 | comment | added | Stein Chen | Ok, thanks. If we know that there are only finitely many solutions, is there an algorithm to find all of them? | |
| Dec 27, 2020 at 21:52 | comment | added | R. van Dobben de Bruyn | Of course one sufficient criterion is that $R \otimes_{\mathbf Z} \mathbf Q$ has Krull dimension $0$, for then there are only finitely many rational solutions. | |
| Dec 27, 2020 at 21:49 | comment | added | Stein Chen | Ok, thank you very much. Do you know a sufficient criterion in the $K=\mathbb{Z}$ case which guarantees that there are only finitely many solutions, if the criterion is fulfilled? | |
| Dec 27, 2020 at 21:46 | comment | added | RumDiary | No, I'm sorry. But I think that you shouldn't expect a "nice" criterion. | |
| Dec 27, 2020 at 21:44 | comment | added | Stein Chen | Ok, thanks. Do you happen to know, if there is a criterion which says sth. like There are finitely many solutions in the $K=\mathbb{Z}$ case $\Leftrightarrow \dots$ ? | |
| Dec 27, 2020 at 21:40 | history | answered | RumDiary | CC BY-SA 4.0 |