Finding ideal generators of the ideal of relations among a certain set of polynomials

Question

This is a rather calculational question. I am looking for the ideal of relations among a certain set of polynomials. In polynomial ring $k[X_0, X_1, X_2]$, where $k$ is an algebraic closed field, consider the elementary symmetric polynomials $e_1 = X_0 + X_1 + X_2$, $e_2 = X_0 X_1 + X_1 X_2 + X_0 X_2$, and $e_3 = X_0 X_1 X_2$. Then consider the seven polynomials $a = e_3^2$, $b=e_1 e_2 e_3$, $c=e_1^3 e_3$, $d=e_2^3$, $e=e_1^2 e_2^2$, $f = e_1^4 e_2$, and $g=e_1^6$. These are homogeneous polynomials of degree 6. Now I would like to know the ideal $I$ of relations among them, so I am looking for the kernel of the homomorphism of $k$-algebras $k[A, \dots, G] \rightarrow k[a, \dots, g], A \mapsto a, \dots, G \mapsto g$, where $k[A, \dots, G]$ is the polynomial ring in seven variables. So $I$ is homogeneous prime ideal.

I have found that the following eight Polynomials are in this kernel: $B^2 - AE, E^2 -DF, F^2 - EG, BE - CD, BC - AF, BG - CF, DG - EF, C^2 - AG$. Let $J$ be the ideal generated by these polynomials in $k[A, \dots, G]$. Now I also know that the zero sets of $I$ and $J$ in projective space $\mathbb{P}^6(k)$ coincide, e.g. $V(I) = V(J)$, which implies $\sqrt J = I$, by the projective Nullstellensatz of Hilbert. Also, $J$ is properly included in $I$, because for example the ninth polynomial $h = BF - CE$ doesn't lie in $J$, but it lies in $I$.

I was thinking, that $J + (h) = I$ holds. A way to prove this, is to show that $J + (h)$ is a radical ideal. That is where I am stuck, because I am not familiarized enough with techniques testing if an ideal is a radical ideal or not. I have looked for a Groebner basis of $J + (h)$, using reverse lexicographic order, and found that the nine polynomials $a, \dots, h$ already constitute a Groebner basis for this ideal. I don't know if this can help.

So I would like to know whether $J + (h)$ is a radical ideal or not, and if not, how I can find find additional polynomials from $I$ to $a, \dots, h$, so that they altogether generate the ideal of relations $I$ I was first interested in. I would be very thankful for any help!

Many thanks in advance.

Answer 1

In Macaulay2:

R=QQ[a..g]
S=QQ[e1,e2,e3,Degrees=>{1,2,3}]
phi=map(S,R,{e3^2,e1*e2*e3,e1^3*e3,e2^3,e1^2*e2^2,e1^4*e2,e1^6})
I=ker phi
--ideal(f^2-e*g,e*f-d*g,c*f-b*g,e^2-d*f,c*e-b*f,c*d-b*e,c^2-a*g,b*c-a*f,b^2-a*e)
radical I == I --true
J=ideal(f^2-e*g,e*f-d*g,c*f-b*g,e^2-d*f,c*d-b*e,c^2-a*g,b*c-a*f,b^2-a*e)
radical J == I --true

Answer 2

Here is the Magma code:

Q := RationalField();
R<X0,X1,X2> := PolynomialRing(Q,3);
e1 := X0+X1+X2;
e2 := X0*X1+X0*X2+X1*X2;
e3 := X0*X1*X2;
a := e3^2; 
b := e1*e2*e3; 
c := e1^3*e3; 
d := e2^2; 
e := e1^2*e2^2; 
f := e1^4*e2; 
g := e1^6;
L := [a,b,c,d,e,f,g];
S<A,B,C,D,E,F,G> := PolynomialRing(Q,7);
RelationIdeal(L,S); 

The output is:

Ideal of Polynomial ring of rank 7 over Rational Field
Order: Lexicographical
Variables: A, B, C, D, E, F, G
Inhomogeneous, Dimension >0
Basis:
[
    D^3*G - E^3,
    A*E - B^2,
    A*F - B*C,
    A*G - C^2,
    B*F - C*E,
    B*G - C*F,
    E*G - F^2
]

The 7 polynomials listed in the Basis give a full set of generators for the ideal of relations among your quantities $a,\ldots,g$.