5

I have a set of polynomials $$ a a_1, bb_1, c, c_1, ab, da_1 + bd_1, fa_1+cf_1+d_1e, eb_1+ce_1, de-bf, f_1, e_1, d_1. $$

Is there some software which can determine that they are algebraically independent or not. If dependent, how to find the relations they satisfy. Thank you very much.

LJR
  • 14,520

2 Answers2

6

Yes, what you're looking for is the relation ideal. Given a sequence $p_1(T), \dots, p_n(T) \in k[T_1,\dots,T_k]$ of polynomials, their relation ideal is the ideal $I \subseteq k[X_1,\dots,X_n]$ given by $I := \{ f(X) \in k[X_1,\dots,X_n] \mid f(p_1,\dots,p_n) = 0 \}$. If this is the zero-ideal, the $p_i$ are algebraically independent.

The ideal $I$ can be computed, for instance, using Gröbner bases. Define the ideal $J \subseteq k[X_1,\dots,X_n,T_1,\dots,T_k]$ by $J := (X_1 - p_1(T), \dots, X_n - p_n(T))$. Then $I$ is equal to the elimination ideal $J \cap k[X_1,\dots,X_n]$. This can be computed by taking an admissible monomial ordering in which all the $X_i$ are smaller than all the $T_j$. Compute a Gröbner basis $G$ of $J$ with respect to such an ordering. Then the polynomials of $G$ that are purely polynomials in the $X_j$ form a Gröbner basis of $I$. If there are none, $I = (0)$.

Most computer algebra systems will implement such a method. For instance, in Magma you could say

Q := RationalField();
R<t1,t2> := PolynomialRing(Q,2);
f1 := t1*t2;
f2 := t1^2;
f3 := t2^2;
L := [f1,f2,f3];
S<x1,x2,x3> := PolynomialRing(Q,3);
RelationIdeal(L,S);

and it will reply

Ideal of Polynomial ring of rank 3 over Rational Field
Order: Lexicographical
Variables: x1, x2, x3
Homogeneous, Dimension >0
Basis:
[
    x1^2 - x2*x3
]

A good reference for this at the undergraduate level is Ideals, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra by Cox, Little, and O'Shea.

Magdiragdag
  • 15,049
3

Here's a sketch of how you can check this on a computer, e.g. in Macaulay2. I'll do it in the case $f_1=x^2$, $f_2=xy$ and $f_3=y^2$.

R = QQ[t_1..t_3]
S = QQ[x,y]
f1 = x^2;
f2 = x*y;
f3 = y^2;
f = map(S,R,{f1,f2,f3});
ker f

The output is then

            2
o8 = ideal(t  - t t )
            2    1 3

o8 : Ideal of R

So in this case, the polynomials were algebraically dependent, and satisfies the relation showed above.

In fact, what we have just proved, is that $\mathbb P^1$ can be embedded via the $2$-uple embedding into $\mathbb P^2$ as the zero set of $t_2^2-t_1t_3=0$.

Fredrik Meyer
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