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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 $k | 2d - 2$?

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    $\begingroup$ I added the algebraic-geometry tag. Hope you don't mind. $\endgroup$ – Libli Nov 7 '18 at 22:27
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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.

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  • $\begingroup$ What you wrote here is really interesting... is it possible to use this to describe the locus, for a given $d$, such that $\Delta_d(f) = \square$? $\endgroup$ – Stanley Yao Xiao Nov 8 '18 at 14:12
  • $\begingroup$ can't read the square. You want to describe the locus such that $\Delta_d(f) = ???$? $\endgroup$ – Libli Nov 8 '18 at 14:41
  • $\begingroup$ 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 $\endgroup$ – Stanley Yao Xiao Nov 8 '18 at 15:22
  • $\begingroup$ 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. $\endgroup$ – Stanley Yao Xiao Nov 8 '18 at 15:24
  • $\begingroup$ 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. $\endgroup$ – Libli Nov 8 '18 at 16:31

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