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 root 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 $k | 2d - 2$?