Here is something which came up in my algebraic geometry class, and I'm wondering if it has a deeper explanation. Let $F(w,x,y,z) = w^3+x^3+y^3+z^3$ and let $X$ be the cubic surface in $\mathbb{P}^3$ where $F$ vanishes. As is well known, there are $27$ lines on $X$ (in characteristic $\neq 3$).
I had my students verify that the obvious equations on the Grassmannian $G(2,4)$ define the $27$ points corresponding to these lines as a reduced scheme. This is easiest to do in an affine cover. For example, consider the affine chart of lines of the form $$\mathrm{RowSpan} \begin{pmatrix} 1 & 0 & p & q \\ 0 & 1 & r & s \end{pmatrix}.$$ This chart contains $18$ of the $27$ points in question. The obvious equations come from setting equal to $0$ the coefficients of $t^3$, $t^2 u$, $t u^2$ and $u^3$ in $$F(t,u,pt+ru,qt+su).$$ I.e. $$1 + r^3 + s^3 = 3 p r^2 + 3 q s^2= 3 p^2 r + 3 q^2 s= 1 + p^3 + q^3=0. \quad (\ast)$$
To verify that they define a radical ideal, one must check the Jacobian condition: I.e., that $$\det \begin{pmatrix} 0 & 0 & 3 r^2 & 3 s^2\\ 3 r^2 & 3 s^2 & 6pr & 6 qs \\ 6pr & 6 qs & 3 p^2 & 3 q^2 \\ 3 p^2 & 3 q^2 & 0 & 0 \\ \end{pmatrix} \neq 0$$ at each of the $18$ roots of $(\ast)$. (Actually, I only assigned them to do one root; then the large symmetry group of $X$ does the rest.)
So, here is the thing I can't explain. Out of curiosity, I computed the above determinant and factored it. It turns out that $$\det \begin{pmatrix} 0 & 0 & 3 r^2 & 3 s^2\\ 3 r^2 & 3 s^2 & 6pr & 6 qs \\ 6pr & 6 qs & 3 p^2 & 3 q^2 \\ 3 p^2 & 3 q^2 & 0 & 0 \\ \end{pmatrix} = - 81 \det\begin{pmatrix} p & q \\ r & s \end{pmatrix}^4 .$$
Is there any deep reason for this? Does that determinant even have any significance away from the $18$ points which describe lines on $X$?
