5
$\begingroup$

I have polynomials {A1,B1} and then I find their Groebner basis to be {g1,g2,g3,g4}.

Is there a way for me to express each polynomial g1,g2,g3,g4 in terms of A1, B1? (ie have g1= h1 A1 + h2 B1 where h1,h2 are some polynomial)

I also have the following relation(if it helps), where {k1,k3,j1,j2,j3} are polynomials.

A1 = k1 g1 + k3 g3

B1 = j1 g1 + j2 g2 + j3 g3

Side note: The answer in the similar post was for a different question, and the article linked was too difficult for me to understand and use. Any help would be greatly appreciated.

Edit: Example Request

$A1=-\frac{(M - t)^2 (M + t)^2 (-1 + M t) (1 + M t) (M^2 + t^2)^2 (1 + M^2 t^2)}{M^6 t^6}$

$B1=-\frac{(-1 + M) (1 + M) (1 + M^2) (M - t) (M + t) (-1 + M t) (1 + M t) (M^2 + t^2) (1 + M^2 t^2) (-M^4 + t^4 - M^2 t^4 - M^6 t^4 + M^8 t^4 - M^4 t^8)}{M^{10} t^8}$

The solution given by Michael works wonderfully once I get rid of the denominator. (Though I will have to see whether this affects my problem or not.)

$\endgroup$
6
  • $\begingroup$ Duplicate of this? Also, an explicit example would be helpful here. $\endgroup$ Feb 3, 2018 at 22:22
  • $\begingroup$ @DanielLichtblau Yes it is a duplicate, but if you read the answer you will see that the original question was not answered. The article that was also linked, I could sadly not understand it. I just want want to express g1,g2,g3,g4 in terms of A1,B1. I am thinking this should be possible since g1,g2,g3,g4 are the groebner basis to A1,B1. $\endgroup$ Feb 3, 2018 at 23:12
  • 2
    $\begingroup$ Umm, what about an example? $\endgroup$ Feb 3, 2018 at 23:31
  • 1
    $\begingroup$ We not only like explicit examples, we really need them. In cut-and-pastable form. $\endgroup$ Feb 4, 2018 at 23:41
  • 1
    $\begingroup$ There is now ExtendedGroebnerBasis in the Wolfram Function Repository. $\endgroup$ Feb 2 at 17:15

1 Answer 1

7
$\begingroup$
bas = {x^2 + y^2 + z^2 - 1, x - 2 y^3 - 3};
{gb, mat} = GroebnerBasis`BasisAndConversionMatrix[bas, {x,y,z}, {}]
mat.bas == gb // Simplify

(* True *)
$\endgroup$
3
  • $\begingroup$ Is there a way for this to work if we are working with Laurent Polynomials? What you have put works wonderfully after I multiply out the denominator, but I'm not sure if it will affect my problem. Thank You so much for you help! $\endgroup$ Feb 6, 2018 at 18:11
  • $\begingroup$ @AndyNguyen I'm not sure. A copy-pastable example (code, not TeX), would make it easier to try out ideas. It seems like the monomial ordering of $1/t^{-1}, 1/t^{-2}$ is the opposite of $t,1$ in $(t+1)/t^2$ (for the purposes of a Gröbner basis). $\endgroup$
    – Michael E2
    Feb 7, 2018 at 0:44
  • $\begingroup$ It did affect our problem, but my professor found a way to circumvent it and still be able to use the solution you have provided. Thank you so much for helping. $\endgroup$ Feb 7, 2018 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.