1
$\begingroup$

Let $S=k[x_1,...,x_9]$ be a polynomial ring over field $k$. Set $q_1=(x_1,x_2,x_5,x_6)$, $q_2=(x_1,x_2,x_6,x_7)$, $q_3=(x_2,x_3,x_7,x_8)$, $q_4=(x_1,x_5,x_6,x_7)$, $q_5=(x_1,x_6,x_7,x_8)$, $q_6=(x_2,x_6,x_7,x_8)$, $I=(x_1,..,x_4)$. How to compute $\operatorname{Depth}_I(S/\cap_{i=1}^{6}q_i)$? in general, Is there a way to compute $\operatorname{Depth}_I(S/\cap_{i=1}^{t}q_i)$ when $S=k[x_1,...,x_n]$, $q_i=(x_{i_1},...,x_{i_r})$? Is there any algorithm to compute it (using Macaulay2)?

background:

the common length of the maximal $M$-sequences in $I$ is called the grade of $I$ on $S$-module $M$ denoted by $\operatorname{Depth}_I(M)$.

Improve this question
$\endgroup$
8
  • 3
    $\begingroup$ Use Macaulay 2. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Oct 6, 2013 at 21:06
  • $\begingroup$ @MahdiMajidi-Zolbanin I wanna comput it manually. $\endgroup$
    – ann
    Commented Oct 7, 2013 at 5:56
  • 1
    $\begingroup$ In order to get people motivated you may want to add answers to the questions like: What is your expectation of the depth? Why is this question interesting to you, or where is this question coming from? If it is just your own curiosity, then I guess mathstackexchange might be a better place for this question. $\endgroup$
    – Youngsu
    Commented Oct 7, 2013 at 7:09
  • 1
    $\begingroup$ I just used Macaulay2 to compute it. Macaulay asserts it has Depth = 2. I can post the Macaulay2 code here as an answer if you'd like. (Make sure to load the "Depth" package). $\endgroup$
    – Karl Schwede
    Commented Oct 9, 2013 at 15:23
  • 1
    $\begingroup$ math.uiuc.edu/Macaulay2/doc/Macaulay2-1.6/share/doc/Macaulay2/… $\endgroup$
    – Karl Schwede
    Commented Oct 16, 2013 at 21:52

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .