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 $k | 2d - 2$?



Can discriminant polynomials become perfect powers on hyperplanes?
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.


