A condition on isolated singularity

Question

Suppose $F: {\mathbb C}^N \to {\mathbb C}$ defines a singularity at the origin (for simplicity one can assume that $F$ is a quasi-homogeneous polynomial). Suppose it is nondegenerate, i.e., $dF(z) = 0$ only at $z = 0$. Let $D_i \subset {\mathbb C}^N$ be the divisor defined by $\partial_i F$.

My question is: is there a natural condition on $F$ to guarantee the following (or can we prove the following under the nondegenerate condition):

For any $I \subsetneq \{1, \ldots, N\}$, $F$ doesn't vanish identically on (any component of) the intersection

$$\bigcap_{i \in I} D_i.$$

Answer 1

Assume that $F$ is quasihomogeneous.

A condition on $F$ would be the following. For each $k$ write $F=x_kG+H_k(x_1,\dots,x_{k-1},x_{k+1},\dots, x_N)$. Then $H_k$ is quasihomogeneous. If for each $k$ the polynomial $H_k$ has an isolated singularity in $\mathbb{C}^{N-1}$ then $F$ does not vanish identically on any irreducible component of $\cap_{i\in I} D_i$ for any $I$ under consideration.

However, if $H_k$ has a nonisolated singularity in $\mathbb{C}^{N-1}$ then $F$ does vanish identically on an irreducible component $\cap_{i \in \{1,\dots, N\}, i\neq k} D_i$.

To prove the first statement, note that it suffices to prove it for any $I$ with $N-1$ elements. Let $I_k$ be the subset where we leave out $k$. Also note that by the Euler relation for quasihomogeneous polynomials we have that the partials of $F$ form a complete intersection. Therefore any irreducible component of $C_k:=\cap_{i\in I_k} D_i$ has dimension one. If $F$ vanishes on an irreducible component of $C_k$ then by the Euler relation for quasihomogeneous polynomials we find that $x_k \partial_k F$ is zero on this particular component. If $\partial_k F$ would be zero then $F$ would have a nonisolated singularity, hence $x_k=0$.

Now the partial of $F$ wrt to $x_i$ for $i\neq k$, when restricted to $x_k=0$, coincide with the partial of $H$ wrt to $x_i$. Since $H$ has an isolated singularity we find that no component of $C_k$ is contained in $x_k=0$. This proves the first statement.

The second statement follows along the same lines.