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?



Homogeneous polynomials which ramify completely on a hypersurface
The following result is completely standard in projective geometry.



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> **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.
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**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.


