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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?

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    $\begingroup$ 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 $m_P (\mathcal C, \mathcal C') \ge r$. $\endgroup$ – Luca Bressan Jul 24 at 17:20
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    $\begingroup$ 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 $\mathrm {Disc}_x (f)$. $\endgroup$ – Luca Bressan Jul 24 at 17:25
  • $\begingroup$ 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? $\endgroup$ – LWW Jul 25 at 2:32
  • $\begingroup$ Bressan // And why the degree of $y-z $ in $\mathrm {Disc}_x (f) $ is the same as intersection multiplicity? $\endgroup$ – LWW Jul 25 at 2:50
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    $\begingroup$ 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 $(1 : 1 : 1)$. $\endgroup$ – Luca Bressan Jul 25 at 7:23
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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) $.

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