MDD

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These are questions asked by MDD

Hi Dears

I need some random zero-dimensional binomial ideals (20 ideals or more) with two, three, or four ... generators with 4 variables atmost. Then I want to regenerate each of them such that some of their generators are not binomial and the obtained ideals are equal to the first corresponding original binomial ideals. How can do I this automatically?

As a simple example let I be an ideal generated by {x-1, y-1, z-1} which is zero-dim. We can obtain J=<x-z, x+z-2, y+z-2> that is equal to I.

Thank you in advance.

Hi

Is there any command to compute the Gauss-Jordan form and the Inverse of a Matrix simultaneously?

Thank you in advance

Sincerely

Hi every one

Let F= [x^2-z^2, x*y+y*z, x*z-y*z, y^2+y*z,w^2] be a special list of monomials and binomials. How to convert F into the following list:

FF=[x^2=z^2, x*y=-y*z, x*z=y*z, y^2=-y*z,w^2=0]. Please note that the left side of each equation of FF is greater than the right side w.r.t. lex ordering plex(x,y,z,w). 

Thank you in advance.

Hi every one

How do I automatically change the polynomial ring R=K[x1,...,xn] into the quotient ring R/I when I is a homogeneous ideal of degree d in R?

For example, let I be a polynomial ideal generated by [x-y, x^2+y^2+z^2] in R=K[x,y,z]. In the middle of the computations, the ideal B=[x-y] is computed and now we have to continue the calculation at the quotient ring R/B. How does change R into R/B automatically in Maple and then [x^2+y^2+z^2] changes into [2y^2+z^2] and K[x,y,z]/[x-y] ----> K[y,z]?

Thank you in advance.

Hi every one

Let F=[f1,...,f10] be a  list of homogeneous polynomials with different degrees. I want to create a list of lists s.t. any list satisfies the following conditions:

1. all elements of a list have the same degree.

2. the lists contained in the main list sort increasingly.

For example if F=[x-y, y+z, x^3-xyz+z^3, y^3-xz^2, y^6-x^4y^2, x^5y-z^6+x^2y^2z^2] then 

L=[[x-y, y+z],[x^3-xyz+z^3, y^3-xz^2],[y^6-x^4y^2, x^5y-z^6+x^2y^2z^2]] is the output.

Thanks for your answers.

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