# Homogeneous polynomials which ramify completely on a hypersurface

Let $$F \in \mathbb{Z}[x_0, \cdots, x_n]$$ be a homogeneous polynomial. Let $$V \subset \mathbb{P}^n(\mathbb{C})$$ be a hypersurface (defined over $$\mathbb{Q}$$ say), given by a homogeneous polynomial $$G(x_0, \cdots, x_n)$$ say.

We say that $$F$$ ramifies completely on $$V$$ if there exists a positive integer $$r > 1$$ and polynomials $$S,H$$ such that $$F(x_0, \cdots, x_n) = S^r + GH$$. In other words, the image of $$F$$ with respect to the natural map $$\mathbb{Z}[x_0, \cdots, x_n] \rightarrow \mathbb{Z}[x_0, \cdots, x_n)/I(V)$$ is a perfect $$r$$-th power for some $$r > 1$$.

Unfortunately, assuming that $$F$$ is non-singular is not enough the exclude the possibility that $$F$$ ramifies on some hypersurface. Indeed, let $$G(x_1, \cdots, x_n)$$ be a non-singular polynomial of degree $$d$$ with respect to the variables $$x_1, \cdots, x_n$$, and put

$$\displaystyle F(x_0, \cdots, x_n) = x_0^d + G(x_1, \cdots, x_n).$$

It can easily be checked that $$F$$ is non-singular, and $$F$$ ramifies completely on the hypersurface given by $$G = 0$$.

Is there a way to classify those $$F$$ which ramifies completely on low degree hypersurfaces, namely hypersurfaces of degrees up to three?

Prop : Let $$S$$ and $$V$$ are two smooth hypersurfaces in $$\mathbb{P}^n$$ with $$n \geq 2$$. Assume that $$\deg S \neq \deg V$$, then $$V \cap S$$ is singular at most in a finite number of points.
Proof : Assume that $$v = \deg V < \deg S = s$$ and assume that the singular locus of $$V \cap S$$ contains a curve, say $$C$$. Since $$V$$ and $$S$$ are smooth hypersurfaces, we have: $$C \subset \{x \in V \, \textrm{such that} \, T_{V,x} = T_{S,x} \}.$$ Reformulated : we have $$T_{V,C} = T_{S,C}$$, so that $$N_{V/\mathbb{P}^n,C}(-v) = N_{S/\mathbb{P}^n,C}(-v)$$. Since $$N_{V/\mathbb{P}^n,C} = \mathcal{O}_C$$ and $$N_{S/\mathbb{P}^n,C}(-v) = \mathcal{O}_C(s-v)$$ is ample (because $$s>v$$), we get a contradiction.
In the case $$s=v$$, it is less obvious to say something, as the example in your question shows.