# Discriminant of homogeneous polynomials

Let $$f$$ be a homogeneous polynomial in variables $$x,y,z$$.

Suppose that the sum of coefficients of $$\frac{\partial^i f}{\partial x^i}$$ is $$0$$ for each $$0 \leq i \leq r$$.

I believe that, in this situation, $$(y-z)^r$$ must divide $$\textrm{Disc}_x(f)$$.

Let us give you a simple example. Let $$f=x^2-2xy+z^2$$. Then $$\frac{\partial f}{\partial x}=2x-2y$$ and

$$\textrm{Disc}_x(f)=4y^2-4z^2=4(y-z)(y+z).$$

There are many examples which I computed using program.

Why this happens? Is there any reference which mention on this situation?

• The equations $f = 0$ and $\frac {\partial^i f} {\partial x^i} = 0$ respectively define a curve $\mathcal C$ and its $i$-th polar $\mathcal C^{(i)}$ with respect to $(1 : 0 : 0)$. The sum of the coefficients of a polynomial $g$ is $0$ iff $P = (1 : 1 : 1)$ belongs to the corresponding curve. Finally, the degree of $(y - z)$ in $\mathrm {Disc}_x (f)$ is the intersection multiplicity $m_P (\mathcal C, \mathcal C')$ at $P$. So, if it helps, you can equivalently state your claim as: if $P$ belongs to $\mathcal C, \mathcal C', \dotsc, \mathcal C^{(r)}$, then $m_P (\mathcal C, \mathcal C') \ge r$. – Luca Bressan Jul 24 at 17:20
• Notice that the case $r = 1$ is trivial: if the sums of the coefficients of $f$ and of $\frac {\partial f} {\partial x}$ are $0$, then $P$ belongs to both $\mathcal C$ and $\mathcal C'$, therefore $m_P (\mathcal C, \mathcal C') \ge 1$, which implies that $(y - z)$ divides $\mathrm {Disc}_x (f)$. – Luca Bressan Jul 24 at 17:25
• Beessan // Thank you. I understand what you say vaguely (with helps of wikipidea), but to solve the problem, It might be helpful that I learn about basic intersection theory and algebraic geometry. Can you recommand any referrence? – LWW Jul 25 at 2:32
• Bressan // And why the degree of $y-z$ in $\mathrm {Disc}_x (f)$ is the same as intersection multiplicity? – LWW Jul 25 at 2:50
• A classic reference is Plane Algebraic Curves by Brieskorn and Knörrer. If $f = 0$ and $g = 0$ are equations of $\mathcal C$ and $\mathcal D$, we can write the resultant $\mathrm{Res}_x (f, g)$ as $\prod_{k=1}^n (b_k y - a_k z)^{r_k}$. Then, for any $k$ the point $(1 : a_k : b_k)$ belongs to both $\mathcal C$ and $\mathcal D$ and by definition its intersection multiplicity is $r_k$. Since the factors of $\mathrm{Disc}_x (f)$ are the same as those of $\mathrm{Res}_x (f, f')$, it follows that the degree of $(y - z)$ in $\mathrm{Disc}_x (f)$ is the intersection multiplicity at $(1 : 1 : 1)$. – Luca Bressan Jul 25 at 7:23

Assume $$f$$ is of degree $$n$$. We can write the polynomial as: $$f(x,y,z) = (x^n)P_{n} + (x^{n-1})P_{n-1} + ... (x^{0})P_{0}$$ where $$P_k$$ denotes a homogeneous polynomial in $$z,y$$, of degree $$n-k$$. We can substitute $$z$$ with $$1$$ in $$f$$ at all places (let $$f_{1}(x,y) = f(x,y,1)$$), and simplify the problem thusly: $$P_{k}'s$$ become heterogeneous polynomials in $$y$$, the condition on the coefficients of $$f$$ and its $$x$$ derivatives remains the same, and $$(y-1)^r$$ must divide $$\textrm{Disc}_x(f_1)$$.

We see that the sum of coefficients of $$f_{1}(x,y) = (x^n)P_{n}(y) + (x^{n-1})P_{n-1}(y) + ... (x^{0})P_{0}(y)$$ can be computed as the sum (for $$k = 1 .. n$$) of the sums of the coefficients of the individual $$P_{k}$$'s. We can also observe that the sum of coefficients of $$P_{k}(y)$$ is equal to the 0'th degree coefficient of $$P_{k}(y+1)$$. So it would be best to change the variables again such that our old $$y$$ gets offset as $$y+1$$. We can study $$f_{2}(x,y) = f_{1}(x,y+1)$$, and group those 0'th degree coefficients together. By writing $$f_{2}(x,y) = y Q(x,y) + K(x)$$ where $$K(x)$$ is heterogeneous in $$x$$ and $$Q(x,y)$$ is heterogeneous in $$x,y$$ we see that the sum of $$f_{1}$$'s coefficients is equal to the sum of $$K(x)$$'s coefficients, where K is the above component of $$f_2$$. Our problem becomes to prove that $$\textrm{Disc}_x(f_2)$$ is divisible by $$(y+1 - 1)^r = y^r$$ (we've offset $$y$$ by 1), where the sum of the coefficients of $$\frac {\partial^i K} {\partial x^i}$$ is $$0$$, $$\forall 0 \leq i \leq r$$. However we can again exploit the fact that the sum of coefficients of $$K(x)$$ is equal to the 0'th degree coefficient of $$K(x+1)$$. This means (applying to derivatives) that $$K(x+1) = x^{r+1}J(x)$$. Having $$f_{3}(x,y) = f_{2}(x+1,y)$$ (and knowing that the value of the discriminant is unaffected by translation in its variable) our final restatement of the problem is:

Let $$f_{3}(x,y) = y M(x,y) + x^{r+1}J(x)$$. Prove $$y^r$$ divides $$\mathrm{Disc}_x (f_3)$$.

This is a much simpler reformulation of our problem. Unfortunately, I'm out of "clever" tricks as to how to solve this. A direct proof can be made by writing the discriminant as the determinant of the Sylvester matrix of $$f_{3}$$ and $$\frac {\partial f_{3}} {\partial x}$$ and examining the components of said determinant (particularly those components where the degree of $$x$$ is less that or equal to $$r$$), and how they go into the determinant formula:

Consider the Sylvester $$S$$ matrix of $$f_{3}(x)$$ and $$\frac {\partial f_{3}(x)}{\partial{x}}$$ (preferably written like this) . The discriminant $$\mathrm{Disc}_x (f_3)$$ is the determinant of said matrix. Consider the sub-matrix made of the last $$r$$ columns of $$S$$. We see from the Leibnitz formula of the determinant (sum of products of permutations) that every product in the determinant sum must take $$r$$ terms from said sub-matrix. However, the terms appearing in the sub-matrix are coefficients of degree less than $$r$$ in $$f_3(x)$$ and $$f_3'(x)$$, and $$0$$. Given that $$f_{3}(x) = y M(x,y) + x^{r+1}J(x)$$, we see that all such coefficients are multiples of $$y$$. In other words, all terms in the sum that gives the determinant are multiples of $$y^r$$. The determinant is a multiple of $$y^r$$, thus completing our proof.

Edit: at the last step of the proof you have to make sure the Sylvester matrix is large enough to be able to select an $$r$$-column sub-matrix. If $$J(x) \neq 0$$ this always happens. However, you can "trick" it if you set $$J(x) = 0$$. That way you can make $$r$$ arbitrarily large and no longer fulfill the discriminant condition. Substituting the variables backwards when $$J(x) = 0$$ to go back to $$f$$ makes it divisible by $$y-z$$, giving us the following:

We should add an additional condition to the original problem: $$f$$ should not be divisible by $$y-z$$ . For example if we have $$f = (y-z)(x^2 - z^2)$$ we can plainly see that the sum of the coefficients of $$\frac{\partial^i f}{\partial x^i}$$ is $$0$$ for all $$i \in \mathbb{N}$$, yet $$\textrm{Disc}_x(f) = 4(y - z) (y z^2 - z^3)$$.