1

I'm looking for a (easy) procedure of some sort. I also know a little bit of Singular and CoCoA, and was wondering if you can do that in there?

Mark
  • 2,063
  • 1
  • 14
  • 25
  • 2
    There is no 'easy' procedure. You can do this with Groebner bases. – MooS Jun 21 '16 at 07:22
  • @MooS Okay, but I want a linear combination in terms of the given generators of $I$, not a combination in the polynomials of the Groebner basis. – Mark Jun 21 '16 at 07:34
  • When you produce a Gröbner basis of $I$ you can keep track of the progress, and will then know how to write the elements of the resulting Gröbner basis in terms of the original set of generators. – Jyrki Lahtonen Jun 21 '16 at 07:41
  • @JyrkiLahtonen It's sometimes not very easy to compute a basis by hand, so if I'm using a computer to find a Groebner basis, how do I keep track of everything then? – Mark Jun 21 '16 at 07:46
  • As I said: There is no easy procedure. – MooS Jun 21 '16 at 07:55
  • You could of course let the computer only do the division algorithms and keep track of the S-polynomials by hand (or by a self-written script). – MooS Jun 21 '16 at 07:57

1 Answers1

2

Here's how in Macaulay2.

Let's say your ideal is $I=(x^2+2y^2-3,y^2-yx)$.

Then I claim that $y^3-y$ is in the ideal. Then we can write

i59 : f

       2     2
o59 = x  + 2y  - 3

o59 : R

i60 : g

               2
o60 = - x*y + y

o60 : R

i61 : f = x^2+2*y^2-3

       2     2
o61 = x  + 2y  - 3

o61 : R

i62 : g = y^2-x*y

               2
o62 = - x*y + y

o62 : R

i63 : I = ideal(f,g)

              2     2               2
o63 = ideal (x  + 2y  - 3, - x*y + y )

o63 : Ideal of R

i64 : (y^3-y) // gens I

o64 = {2} | 1/3y      |
      {2} | 1/3x+1/3y |

              2       1
o64 : Matrix R  <--- R

Hence we see that $y^3-y = \frac 13 f +\frac 13(x+y)g$.

Fredrik Meyer
  • 20,228
  • in $k[x_1,x_2,\ldots,x_n]$ when $k =\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$ it reduces to solving a system of linear equations ? – reuns Jun 22 '16 at 09:47
  • 1
    @user1952009 Well, yes, but you don't a priori know the degree of $a,b$ in $af+bg$, so that you don't know a priori how big a system you must solve. – Fredrik Meyer Jun 22 '16 at 09:54
  • I didn't notice that problem, can the degree of $a_1,a_2,\ldots$ really be arbitrary big with respect to the degree of $P \in I$ ? – reuns Jun 22 '16 at 09:58