Let $d>1, n>1$ and consider the vector space $V=\mathbb{C}[x_0,\ldots,x_n]_d$. Let $\mathscr A \subseteq \mathbb{P}(V)$ be the set of all forms with the property that their zero locus in $\mathbb{P}^n$ has at least $k$ singularities. Let $p \in \mathbb{P}(V)$ be square free such that its zero locus has less than $k$ singularities and all of them are nodes. Is it true that in that case $p$ is not in the Zariski closure of $\mathscr A$?
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2$\begingroup$ Yes, that is true. Consider the closed subscheme of a hypersurface that is defined by the Fitting ideal of the sheaf of relative differentials. The dimension of this closed subscheme is upper semicontinuous, so the parameter point $[X]\in \mathbb{P}(V)$ cannot be in the Zariski closure of the closed locus parameterizing hypersurfaces whose singular scheme has positive dimension. For nodal hypersurfaces, the length of the singular scheme is precisely the number of nodes, whereas for $k$-singular schemes, it is at least $k$. Finally, the length is upper semicontinuous. $\endgroup$– Jason StarrCommented Oct 19, 2015 at 11:39
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$\begingroup$ Uhm, it seems to me that for an isolated singularity the length coincides with the Milnor number, since the Milnor number is defined by taking the quotient of the local ring by the Jacobian ideal, right? $\endgroup$– Francesco PolizziCommented Oct 19, 2015 at 12:57
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Yes, this is true. This answer is essentially an expanded version of Jason Starr's comment, also including a reference.
The result that we need is the following, that can be easily deduced from [G. M. Greuel, C. Lossen, E. Schustin, Introduction to singularities and deformations, Theorem 2.6 Chapter I].
Proposition. Let $f \colon \mathcal{X} \to \Delta$ be a deformation of the affine hypersurface $X=f^{-1}(0)$ over a disk. Assume that $0$ is the only singular point for $X$ and that $X_t$ has only isolated singularities. Then the total Milnor number of the fibres is upper semicontinuous, i.e. for all $t \in \Delta$ we have $$\mu(X, \, 0) \geq \sum_{x \in \textrm{Sing} \, X_t} \mu(X_t, \, x).$$ In other words, the total Milnor number cannot decrease under specialization.
In your case, the total Milnor number of every hypersurface in $\mathscr A$ is at least $k$, whereas the total Milnor number of the hypersurface $\{p=0\}$ is the number of its nodes, hence strictly less than $k$.
Therefore, $\{p=0\}$ is not a specialization of hypersurfaces belonging to $\mathscr A$.
answered Oct 19, 2015 at 12:47