3

I'm trying to calculate the syzygies of a set of elements on the polinomial ring of 6 variables. But I'm trying to specify the number of generator in each degree the syzygies have. I know that Macaulay2 can give me the syzygies of the sistem very fast, but it is returning some columns with mixed degree, and because of this I don`t know how to determinate how many generator of each degree I have.

I thought in define the set of polynomials as a map, and find a resolution for the kernel of this map, but it seems that even with Macaulay2, if I go that way I will have aplenty of calculations to do... My question is if that is any other faster way to solve this.

Thanks in any advance.

Rodrigo de Azevedo
  • 23,223
User43029
  • 1,376
  • I'm having trouble understanding your question. What do you mean by "But I''m trying to specify the number of generator in each degree the syzygies have"? – Youngsu Nov 15 '15 at 03:50
  • Dear @Youngsu, I mean that the matrix that Macaulay return when one asks for the syzygies of a matrix (or an ideal) gives in the columns the generators of the syzygie module, but I want to count how many generators in each degree I have for my matrix, so I think I need to count how many columns in the syzygy matrix I have, the problem is that in some columns there are polynomials with degree one and two for instance. Thank you. – User43029 Nov 15 '15 at 13:23
  • I see. I skimmed through the answer posted below and your comments. You may want to check the grading first. Or the best is just posting the ideal (or module) you want to compute here. – Youngsu Nov 16 '15 at 04:57

1 Answers1

5

I think the problem here is understanding Macaulay2's notation. Let's take an example:

i1 : R = QQ[x,y,z]

o1 = R

o1 : PolynomialRing

i7 : I = ideal(random(2,R), random(3,R), random(2,R));

o7 : Ideal of R

i8 : betti res I

            0 1 2 3
o8 = total: 1 3 3 1
         0: 1 . . .
         1: . 2 . .
         2: . 1 1 .
         3: . . 2 .
         4: . . . 1

Here the last table gives us all the numerical information about the resolution of some ideal in $\mathbb Q[x,y,z]$. The way to read this is as follows: In the first syzygy $1$ there's two generators of degree $2$ and one of degree $3=1+2$. In the second syzygy there's one generator of degree $4=2+2$ and two of degree $5=3+2$.

In general, the number of generators of degree $i$ in the $j$th syzygy is found by reading the $(i,j-i)$ position in the Betti table.

Fredrik Meyer
  • 20,690
  • Dear @Fredrik Meyer thank you for your answer, the problem is exactly the problem that you pointed, I am trying to do a particular case, from a paper from V. Anconna and G. Ottaviani, ''On moduli of Instanton Bundles on $\mathbb{P}^{2n+1}$, example 2.1, when I ask for the syzygie matrix wich they suggest in the paper I can get the same number that they get to the degree 0 generators of the syzygies, but I'm obtaining the half of the generators of degree 1, I understand that this happen because some generators of degree 0 gives some generator of degree one, but I don't know how compute this. – User43029 Nov 15 '15 at 13:31
  • And asking for the betti table of the kernel of the map, does not give the numbers of generators of degree one as well, because of this I was wondering if the Betti table is what I am searching for. I will put the result of the Betti table below. Thank you so much for the help. – User43029 Nov 15 '15 at 13:35
  • when I ask for the Betti table of the resolution of the kernel of the matrix that V. Anconna and G. Ottaviani suggests in their paper I get the table: 0 1 2 3 4 o9 = total: 104 89 52 21 5|| 1: 14 . . . .|| 2: 90 78 25 . .|| 3: . 5 12 9 2|| 4: . 6 15 12 3|| In the paper they say that Syz_{0} = 14 and Syz_{1} = 174, but I don't see how they get the 174. Thank you for the help. – User43029 Nov 15 '15 at 14:01