Constructing affine hypersurfaces with one singularity

Question

This is a followup to my previous poorly-worded question.

Consider a finite collection of points $S \subset \mathbb N^n$ lying in a hyperplane $H$. These points define exponents of a collection of monomials in $\mathbb C[x_1, \cdots, x_n]$, which may be combined with some choice of coefficients to produce a polynomial that is homogeneous under the $\mathbb C^*$ action with weights determined by $H$.

Are there conditions on $S$ and/or $H$ such that by choosing generic coefficients, the resulting polynomial $p$ has singular locus at the origin?

I am happy to assume that the sum of the entries of $\alpha \in S$ (the naive degree) is greater than two, if this is easier.

In case it is still not clear, three examples: #1 is no good, but #2,3 are allowed.

1. If I choose $\{(3,0)\}\subset \mathbb N^2$, then for any non-zero coefficient $a$, the polynomial $p=ax^3$ has singularities away from the origin in $\mathbb C^2$.

2. If I choose $\{(3,0), (0,3)\} \subset \mathbb N^2$, then for any choice of non-zero coefficients $a,b$, the polynomial $p = ax^3 + by^3$ will be singular only at the origin.

3. If I choose $\{(6,0,0), (0,3,0), (0,0,2), (1,1,1))\} \subset \mathbb N^3$, then for non-zero $a,b,c,d$, the polynomial $a x^6 + b y^3 + cz^2 + d x y z$ has singularities only at the origin.

Please note that I am asking about singularities of the affine hypersurface, not the (weighted) projective hypersurface (hence conical singularity).

Answer 1

I guess that to show that if the $\mathbb{Q}$-span of $S$ is not $\mathbb{Q}^n$ then $H$ is automatically singular is a not too hard exercise. I.e., in this case you construct an affine cone over a weighted projective cone. (Since it is almost midnight here, I cannot be bothered to do this..)

If $S$ spans all of $\mathbb{Q}$ then I am not so sure that you always have a $\mathbb{C}^*$-action. A necessary and sufficient condition seems the following: Let $m=|S|$ and let $M$ be the $m\times n$-matrix where the rows are the elements of S. Then the vector consisting of m 1s should be in the column span of M.

If this is the case then $p$ is weighted homogeneous. For such polynomials one has the weighted Euler relation. If $p$ has weighted degree $d$, $w_i$ is the weight of $x_i$, and $p_{x_i}$ is the derivative of $p$ wrt $x_i$. Then $d p=\sum w_ix_ip_{x_i}$.

Hence to find singularities of $p=0$ you only need to consider the partials of $p$. Since you consider only general coefficients, it suffices to determine if there is a monomial that divides $p_{x_i}$. I.e., for each $i$ let $T_i\subset S$ of elements such that the $i$-coordinate is non-zero. Let $f_i$ be the product of coordinates $x_j$ such that for each element in $T_i$ the $j$-th coordinate is non-zero (if $j\neq i$) or at least 2 (if $j=i$.).

For a general choice of coefficients the singular locus of $p$ coincides with $V(f_0,\dots,f_n)$.
I hope you can work out the details.