4
$\begingroup$

Let $F$ be a square-free (as a polynomial) ternary cubic form over $\mathbb{C}$, and let $H_F$ be its Hessian determinant... which is also a ternary cbic form. If $F$ splits over $\mathbb{C}$, so that it can be written as the product of three linearly independent linear forms (i.e., the $3 \times 3$ matrix formed by the coefficients of the three linear forms is non-singular), then one has that $F, H_F$ are proportional. Indeed, since $H_F$ is a covariant and any such $F$ is $\text{GL}_3(\mathbb{C})$-equivalent to $xyz$, it suffices to check this assertion for this form, and this is trivial.

The converse is not true. For example, for $F = x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz)$ we also have $F, H_F$ are proportional. Moreover, since the quadratic factor is equal to $((x-y)^2 + (x-z)^2 + (y-z)^2)/2$ which is positive definite and non-singular, it cannot be a product of linear forms over $\mathbb{C}$.

Thus, a necessary condition for $F$ to split into linear factors over $\mathbb{C}$ is for $F, H_F$ to be proportional, but this is not sufficient by the example above. What additional condition is needed to ensure that $F$ is in fact a product of linear forms?

Edit: it seems I was too quick to claim that $(x-z)^2 + (x-y)^2 + (y-z)^2$ is positive definite. In fact it has the non-trivial solution $x = y = z = 1$. Moreover, the matrix $\left(\begin{smallmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{smallmatrix}\right)$ is singular. As Will Jagy points out below, it factors as $(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$ for $\omega$ a primitive third root of unity.

$\endgroup$
  • 6
    $\begingroup$ The quadratic form you list is semidefinite and equals $(x+y \omega + z \omega^2)(x+y\omega^2 + z \omega),$ where $\omega \neq 1$ but $\omega^3 = 1.$ The condition is also sufficient $\endgroup$ – Will Jagy Oct 13 '18 at 2:08
  • 4
    $\begingroup$ Recall that the intersection of $H_F=0$ and $F=0$ consists of all inflection points of the curve $F=0$. If $H_F$ is a multiple of $F$ then every point of $F=0$ is an inflection point. At least in characteristic zero this means that every component of $F=0$ is a line. $\endgroup$ – Noam D. Elkies Oct 13 '18 at 2:19
  • $\begingroup$ Stanley, I added a difficult example. Well, hard enough. $\endgroup$ – Will Jagy Oct 13 '18 at 2:29
  • 1
    $\begingroup$ Note that "proportional" includes the possibility that $H_F$ is identically zero, if the three lines are coincident (then after a change of coordinates $F$ is a cubic form in $x,y$, so the $z$ row and column of the Hessian matrix vanish, so the Hessian determinant vanishes too). $\endgroup$ – Noam D. Elkies Oct 13 '18 at 2:38
  • $\begingroup$ Just a quick link to mathoverflow.net/questions/164641/… and mathoverflow.net/questions/109334/… with references to Brill's equations which answer the OP's question in much greater generality. I suppose a good exercise (which I didn't do) is to specialize the Brill-Gordan solution to the ternary cubic case. $\endgroup$ – Abdelmalek Abdesselam Oct 15 '18 at 21:37
5
$\begingroup$

a more difficult complete factorization, in this case reals are enough: $$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) $$ which is zero when $x=y=z$

The roots of $\eta^3 - 3 \eta - 1 = 0$ are $$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$

$$ \color{red}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} $$

The condition that the Hessian determinant is a constant multiple of the original ternary cubic is necessary and sufficient for it to factor completely over the complex numbers. This was first published by S. Aronhold in 1849. There is a complete proof of both directions in a 2016 paper by G. Brookfield:

enter image description here

Sorry. The special fact you want is in a book by Schinzel called Polynomials with Special Regard to Reducibility

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But that's a criterion for reducibility (factorization as linear x quadratic), not complete reducibility (factorization as three linears). $\endgroup$ – Noam D. Elkies Oct 13 '18 at 2:14
  • $\begingroup$ @NoamD.Elkies I had not realized that the original question incorrectly reports on complete factorization. It is if and only if, complete factorization over the complexes if and only if the Hessian determinant is a constant mutliple of the original ternary cubic. Note that $x^2 + y^2 + z^2 - yz-zx-xy$ is semidefinite, actually zero when $x=y=z$ $\endgroup$ – Will Jagy Oct 13 '18 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.