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Let $E$ be the set of cubic ternary forms $$E=\{ P=aX^3+bY^3+cZ^3+3dX^2Y+3eY^2Z+3fZ^2X+3gXY^2+3hYZ^2+3iZX^2+6jXYZ : (a,b,c,d,e,f,g,h,i,j) \in \mathbb{C}^{10} \}.$$

Consider the subset $F$ of forms in $E$ which are the product of three linear forms. By definition, $F$ is the image of $$\phi:(m_{ij})_{1\leq i,j\leq 3} \mapsto (m_{11}X+m_{12}Y+m_{13}Z)(m_{21}X+m_{22}Y+m_{23}Z)(m_{31}X+m_{32}Y+m_{33}Z).$$

By a Theorem of Chevalley, $F$ is a constructible set. Let $\bar{F}$ be its Zariski closure.

Question : Can we find explicit polynomials defining $\bar{F}$, and how ?

Unfortunately (for me), I know little of algebraic geometry. Here some remarks:

1) An open subset of the image is a homogeneous space $GL(3,\mathbb{C})/Stab(XYZ)$, the stabilizer being (up to finite index) the set of diagonal matrices of determinant one. Thus the expected dimension of $\bar{F}$ is 9-2=7.

2) I am aware that cubic ternary forms have two invariants $S,T$, called the Aronhold invariants, homogeneous of degree $4$ and $6$, that generates the algebra of invariant polynomial under the action of SL(3) by substitution. (For a generic element $P$ of $E$, these two polynomials seem to be related to the $g_2$ and $g_3$ coefficients of the underlying elliptic curve described by $P(X,Y,1)=0$, although the exact relationship isn't clear to me). Namely, using the coordinates above $$S = a*g*e*c - a*g*h^2 - a*j*b*c + a*j*e*h + a*f*b*h - a*f*e^2 - d^2*e*c + d^2*h^2 + d*i*b*c - d*i*e*h + d*g*j*c - d*g*f*h - 2*d*j^2*h + 3*d*j*f*e - d*f^2*b - i^2*b*h + i^2*e^2 - i*g^2*c + 3*i*g*j*h - i*g*f*e - 2*i*j^2*e + i*j*f*b + g^2*f^2 - 2*g*j^2*f + j^4.$$ and an awfully long formula for $T$. I checked (or rather, I asked Sagemath to check) that we have on $F$ the relationship

$$64*S^3=T^2.$$ I guessed this because of the discriminant of elliptic curves : the ones in $F$ are fairly degenerate (union of three lines), so have discriminant zero. That's one equation, but i miss (a least) two others to describe a dimension 7 algebraic variety. Moreover, it is of degree 12, so I suspect it is consequence of smaller degree relationships.

3) The question here seems related :

How to characterize the image of $PGL(V)$ in $\mathbb{P}(W)$ for an irreducible $GL(V)$-representation $W$

user120527
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  • I did a rather artificial one last week https://math.stackexchange.com/questions/2678486/how-to-show-that-if-x-y-z-are-rational-numbers-satisfying-x-y-z3-9/2678675#2678675 – Will Jagy Mar 14 '18 at 18:04
  • @WillJagy Yes, that's a good example : how did you guess that the ternary cubic form (x+y+z)^3-9(x^2y+y^2z+z^2x) is a product of three linear forms ? – user120527 Mar 14 '18 at 18:29
  • I think it began with If we have an irrational root $p$ of $p^3 + 3 p^2 - 9p-3=0$ we have some other irrational real $q$ such that $$ 2z-x-y = q(y-x) ; , $$ $$ (q-1)x + (-q-1) y + 2 z = 0 ; ; , $$ which is evidently a plane containing the line $x=y=z.$ – Will Jagy Mar 14 '18 at 18:39

1 Answers1

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This seems worth an answer: by Aronhold(1849). Given a form, the Hessian refers to the determinant of the matrix of second partial derivatives. Given a ternary cubic form $f,$ the Hessian $ \mathcal H$ is also a ternary cubic form.

Theorem. $f$ is completely reducible if and only if there is a scalar $\lambda \in \mathbb C$ such that $$ \mathcal H = \lambda f \; \; . $$ Here $\lambda$ turns out to just be double the determinant of the 3 by 3 matrix of linear coefficients of the three linear factors.

Source Gary Brookfield pdf

There is an 1852 book by Salmon that seems to discuss this or generalizations, I have requested a library copy (of one of the reprint editions)

Look at that, there is a 2000 book by A. Schinzel called Polynomials with Special Regard to Reducibility. I requested that also, should be easier to read. Apparently, on page 213, there is a discussion of when the form is not completely reducible but is reducible, which would be a linear times an irreducible quadratic form.

For example, the form I did at How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$ has Hessian determinant $0.$ 

parisize = 4000000, primelimit = 500000
? f = (x+y+z)^3 - 9 *(x^2 * y + y^2 * z + z^2 * x)
%1 = x^3 + (-6*y + 3*z)*x^2 + (3*y^2 + 6*z*y - 6*z^2)*x + (y^3 - 6*z*y^2 + 3*z^2*y + z^3)
? fx = deriv(f,x)
%2 = 3*x^2 + (-12*y + 6*z)*x + (3*y^2 + 6*z*y - 6*z^2)
? fxx = deriv(fx,x)
%3 = 6*x + (-12*y + 6*z)
? fxy = deriv(fx,y)
%4 = -12*x + (6*y + 6*z)
? fxz = deriv(fx,z)
%5 = 6*x + (6*y - 12*z)
? fy = deriv(f,y)
%6 = -6*x^2 + (6*y + 6*z)*x + (3*y^2 - 12*z*y + 3*z^2)
? fyy = deriv(fy,y)
%7 = 6*x + (6*y - 12*z)
? fyz = deriv(fy,z)
%8 = 6*x + (-12*y + 6*z)
? fz = deriv(f,z)
%9 = 3*x^2 + (6*y - 12*z)*x + (-6*y^2 + 6*z*y + 3*z^2)
? fzz = deriv(fz,z)
%10 = -12*x + (6*y + 6*z)
? 
? m = [ fxx, fxy, fxz; fxy,fyy,fyz; fxz,fyz,fzz]
%11 = 
[6*x + (-12*y + 6*z) -12*x + (6*y + 6*z)  6*x + (6*y - 12*z)]

[-12*x + (6*y + 6*z)  6*x + (6*y - 12*z) 6*x + (-12*y + 6*z)]

[ 6*x + (6*y - 12*z) 6*x + (-12*y + 6*z) -12*x + (6*y + 6*z)]

? h = matdet(m)
%12 = 0
Will Jagy
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