Timeline for Computing the bourbaki ideals
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2014 at 2:02 | comment | added | Youngsu | Aurora, please understand that I'm not trying to write a code which'd compute what you are interested. Instead, I'm trying write a code that would compute an example. So, it is better to try with examples which are of "small" size and you already know answers to. | |
| Nov 30, 2014 at 1:45 | comment | added | Aurora | @Youngsu: $n$ should be necessarily greater than or equal to $4$. I set $n=8$ to increase the dimension of $R/I$. Let me to try to find an example. | |
| Nov 30, 2014 at 1:40 | comment | added | Youngsu | Well, then I don't see why one needs to compute for $n = 8$. Can you give me an example you know the answer to? | |
| Nov 30, 2014 at 1:32 | comment | added | Aurora | @Youngsu: No, unfortunately. I don't know how can I compute the ideal $I$ even in that case. | |
| Nov 30, 2014 at 1:28 | comment | added | Youngsu | Do you have an answer to this case? For instance, say $i = 2$, and $j = 3$? | |
| Nov 30, 2014 at 1:20 | comment | added | Aurora | @Youngsu: Lِet, e.g., $R=K[X_1,\ldots,X_n]$ with $n=8$. You can, easily, assume that $M=L_1\bigoplus L_2$ where $L_1$ ($L_2$) is an $i$-th ($j$-th) (graded) syzygy of $R/(X_1,\ldots,X_n)$ with $i\neq j$ and $2\le i,j\le n-2$. | |
| Nov 30, 2014 at 1:00 | comment | added | Youngsu | Let me try. Would you mind posting an example you have in mind? | |
| Nov 30, 2014 at 0:44 | comment | added | Aurora | @Youngsu: Yes, in the Bourbaki exact sequence mentioned in my previous comment, as I know, $M$ should be torsion free and the free module $F$ satisfies $\text{rank}(F)=\text{rank}(M)-1$. Would you please kindly explain how can I obtain an ideal $I$ from a generic choice of minimal generating set of $M$ by means of Macaulay 2? | |
| Nov 29, 2014 at 22:08 | comment | added | Youngsu | I am not so sure if there is an easy way to convert a module into an ideal on Macaulay2. If the module $M$ has a rank, maybe even torsion free, then probably a generic choice of minimal generating set would produce a free submodule $F$ of rank one less than $M$. But here, your ideal $I$ may not be unique since it'd depend on the choice of the generators for $F$. | |
| Nov 29, 2014 at 7:53 | comment | added | Aurora | @abx: Let e.g. $R=K[X_1,...,X_n]$. Then the the Bourbaki ideal associated to an $R$-module $M$ is an ideal $I$ arising from an exact sequence $0\rightarrow F\rightarrow M\rightarrow I\rightarrow 0$ where $F$ is a finitely generated graded free $R$-module | |
| Nov 29, 2014 at 7:41 | comment | added | abx | Just out of curiosity, what is a Bourbaki ideal? | |
| Nov 29, 2014 at 7:28 | history | asked | Aurora | CC BY-SA 3.0 |