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Consider the following code :

R.<x,y,z> = QQ[];

p=(x-y)(y-z)(x-z);

J = matrix(R,[[x-y],[y]]);

J.rank();

The answer displayed is 1. Is this the rank over the polynomial ring ? I would like to compute rank over the field of rationals. What is the way to do it ?

Consider the following code :

R.<x,y,z> = QQ[];

p=(x-y)(y-z)(x-z);

J = matrix(R,[[x-y],[y]]);

J.rank();

The answer displayed is 1. Is this This is the rank over the polynomial polynomia ring ? I suppose. I would like to compute rank over the field of rationals. That is if there are rational numbers a and b such that a diff(p,x) + b diff(p,y) =0 but a,b non zero then rank should be 2.

What is the way to do it ?

Consider the following code :

R.<x,y,z> = QQ[];

p=(x-y)(y-z)(x-z);

J = matrix(R,[[x-y],[y]]);

J.rank();

The answer displayed is 1. This is the rank over the polynomia ring I suppose. I would like to compute rank over the field of rationals. That is if there are rational numbers a a,b and b c such that a diff(p,x) + b diff(p,y) + c diff(p,z) =0 but a,b a,b,c non zero then rank should be 2. 3.

What is the way to do it ?

Consider the following code :

R.<x,y,z> = QQ[];

p=(x-y)(y-z)(x-z);

J = matrix(R,[[x-y],[y]]);

J.rank();

The answer displayed is 1. This is the rank over the polynomia ring I suppose. I would like to compute rank over the field of rationals. That is if there are rational numbers a,b and c such that a diff(p,x) + b diff(p,y) + c diff(p,z) =0 but a,b,c non zero then rank should be 3.

What is the way to do it ?