# Find a symmetric polynomial with a projection divisible by a known polynomial

Consider the polynomial $Q$, a homogeneous quartic in seven variables:

$$Q(R, s_1, s_2, s_3, s_4, d_1, d_2) = \\ (d_1^2-(R+s_1+s_2-s_3-s_4)(R+s_1-s_2+s_3+s_4))\\(d_1^2-(R-s_1-s_2-s_3-s_4)(R-s_1+s_2+s_3+s_4))+\\ (d_2^2-(R-s_1-s_2-s_3-s_4)(R+s_1-s_2+s_3+s_4))\\(d_2^2-(R+s_1+s_2-s_3-s_4)(R-s_1+s_2+s_3+s_4))+\\ 2 d_1^2 d_2^2 - \\(R-s_1-s_2-s_3-s_4)(R+s_1+s_2-s_3-s_4)(R+s_1-s_2+s_3+s_4)(R-s_1+s_2+s_3+s_4)$$

I am trying to find a polynomial $P$ in nine variables, those of $Q$ along with $d_3$ and $d_4$, that has the following properties:

1. $P$ is invariant under the action of the symmetric group $S_4$ that simultaneously permutes the indices of the $s_i$ and the $d_i$. For example, $R^2 (s_1 d_1+s_2 d_2+s_3 d_3+s_4 d_4)$ would be invariant under this action.

2. $P(R, s_1, s_2, s_3, s_4, d_1, d_2, 0, 0)$ is a polynomial in the seven variables of $Q$ that is exactly divisible by $Q$.

3. $P$ is homogeneous, and of the lowest degree possible.

4. I don't expect $P$ to be unique, and if there is a family of polynomials meeting these conditions I would like to find a parameterised description of that family.

What I have been trying so far is, for successive degrees $n$:

1. Generating a list of all invariant polynomials that can be created by summing the elements of an orbit of the action of $S_4$ on suitable monomials in the nine variables.

2. Writing a candidate for $P$ as an arbitrary linear combination of those invariant polynomials with coefficients $g_i$.

3. Setting $d_3$ and $d_4$ to zero in the candidate polynomial.

4. Using the Mathematica function PolynomialReduce to compute the remainder of the candidate polynomial divided by $Q$.

5. Looking for a solution for the linear equations in the $g_i$ that would make the remainder polynomial identically zero.

For degrees 4 to 9 this has led to linear equations with no solutions.

Is there a better strategy?

The context for this question is a geometric problem that is discussed at length here. In brief, I believe $Q$ is a factor of a suitable projection of a larger, more symmetric polynomial $P$ because it describes a special case of the geometric problem for which I am seeking $P$.

Added 14 July 2014: I found a family of polynomials of degree 12, using the brute-force method described above.