This is a followup to my [previous](https://mathoverflow.net/questions/40621/singular-locus-of-a-homogeneous-polynomial) 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 $H$. 



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



Constructing affine hypersurfaces with one singularity
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.


