# Can discriminant polynomials become perfect powers on hyperplanes?

Let

$$\displaystyle f(x) = a_d x^d + a_{d-1} x^{d-1} + \cdots + a_0.$$

Consider the discriminant of $$f$$, denoted by $$\Delta(f)$$, defined as

$$\displaystyle \Delta(f) = a_d^{2d-2} \prod_{i < j} (\theta_i - \theta_j)^2,$$

where $$\theta_1, \cdots, \theta_d$$ are the roots of $$f(x) = 0$$ (over some algebraic closure, say).

It is well-known that $$\Delta(f)$$ is a homogeneous polynomial of degree $$2d-2$$ in the coefficients $$a_d, \cdots, a_0$$.

We say that a homogeneous polynomial $$F \in \mathbb{C}[x_0, \cdots, x_n]$$ of degree $$m$$ ramifies completely on a hyperplane if there exists a hyperplane $$P$$ in $$\mathbb{P}^n$$ such that $$F |_P$$ is a perfect $$k$$-th power for some $$k > 1$$ dividing $$m$$ (as a polynomial). For example, the cubic polynomial $$F(x,y,z) = x^3 + yz^2$$ ramifies completely on the lines (hyperplanes in $$\mathbb{P}^2$$) $$y = 0, z = 0$$.

For $$d = 2$$, we have that $$\Delta(f) = a_1^2 - 4 a_2 a_0$$ ramifies completely on $$a_2 = 0, a_0 = 0$$. Does this happen for $$d > 2$$? That is, does there exist $$d > 2$$ and a hyperplane $$P \in \mathbb{P}^d$$ such that $$\Delta(f) |_P$$ is a perfect $$k$$-th power, for some $$k | 2d - 2$$?

• I added the algebraic-geometry tag. Hope you don't mind. Nov 7 '18 at 22:27

For any $$d \geq 2$$, there are hyperplanes on which $$\Delta_d$$ ramifies, but for $$d \geq 3$$, it never ramifies completely. I guess there are many proofs of this fact, let me give one based on projective duality.

First note that $$\Delta_d$$ parametrizes polynomials of degree $$d$$ having a multiple root. Let me give a geometric interpretation of $$\Delta_d$$. Let $$X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$$ be the $$d$$-th Vernoese embedding of $$\mathbb{P}^1$$. The equation $$\Delta_d = 0$$ gives in the dual projective space $$\mathbb{P}(S^d \mathbb{C}^2)^*$$ the variety which parametrizes singular hyperplane sections of $$X$$. This variety is known as the projective dual of $$X$$.

For any $$Z \subset \mathbb{P}^N$$, which is irreducible, the reflexivity Theorem tells you that $$(Z^*)^* = Z$$ and for any $$z \in Z_{smooth}$$, the tangency locus of $$z^{\perp}$$ with $$Z^*$$ is identified as a scheme to $$\mathbb{P}(N_{Z/\mathbb{P}^N,z}^*)$$.

Going back to your situation, we have $$X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$$ is smooth, its projective dual $$X^* \subset \mathbb{P}(S^d \mathbb{C}^2)^*$$ is the hypersurface which equation is $$\Delta_d = 0$$. You want to know if there exists $$x \in \mathbb{P}(S^d \mathbb{C}^2)$$ such that $$x^{\perp} \cap X^*$$ is completely non-reduced. This is equivalent to saying that the reduced space underlying the singular locus of $$x^{\perp} \cap X^*$$ is equal to the reduced space underlying $$(x^{\perp} \cap X^*)$$.

Two cases occcur:

1) if $$x \notin X$$, then $$x^{\perp}$$ is not tangent to $$X^*$$ (this is because $$(X^*)^*= X$$). Hence the singular locus of $$x^{\perp} \cap X^*$$ has codimension at least one in $$x^{\perp} \cap X^*$$ and the equation defining $$x^{\perp} \cap X^*$$ can't be a $$k$$-perfect power.

2) if $$x \in X$$, then $$x^{\perp}$$ is tangent to $$X^*$$ (this is again because $$(X^*)^*= X$$). Since $$X$$ is smooth, the reflexivity Theorem ensures that the tangency locus of $$x^{\perp}$$ with $$X^*$$ is scheme-theoretically a hyperplane in $$x^{\perp}$$. As a consequence, the singular locus of $$x^{\perp} \cap X^*$$ is generically scheme-theoretically a hyperplane. Now, if $$x^{\perp} \cap X^*$$ was a $$k$$-perfect power, then the fact that the singular locus of $$x^{\perp} \cap X^*$$ is generically scheme-theoretically a hyperplane in $$x^{\perp}$$ would imply that $$k=2$$ and $$\deg \Delta_d =2$$. This is only possible if $$d=2$$.

As far as non-complete ramification is concerned, the above argument shows that for any $$x \in X$$, the equation of $$x^{\perp} \cap \{\Delta_d = 0\}$$ can be written as $$L^2Q$$ where $$L$$ is a linear factor and $$Q$$ a polynomial of degree $$2d-4$$ which is not a power.

• What you wrote here is really interesting... is it possible to use this to describe the locus, for a given $d$, such that $\Delta_d(f) = \square$? Nov 8 '18 at 14:12
• can't read the square. You want to describe the locus such that $\Delta_d(f) = ???$? Nov 8 '18 at 14:41
• the square is actually just a square, i.e., I want to look for the locus on which $\Delta_d(f)$ is equal to the square of some polynomial Nov 8 '18 at 15:22
• For example, when $d = 4$ the locus contains a countable union of lines. An example of such a line is given by $a_3 = a_1 = 0, a_4 = a_0$, giving rise to polynomials of the shape $a_4 x^4 + a_2 x^2 + a_4$. The discriminant is identically a square on this line. Nov 8 '18 at 15:24
• I don't know if this second question is well-posed. If you take different lines, the system of coordinates vary with the lines you take and this doesn't look obvious how to relate one square to the other. Nov 8 '18 at 16:31