# Constructing affine hypersurfaces with one singularity

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).

• This was implicit in the comments to your other question. The associated weighted hypersurface is smooth if and only if the affine hypersurface you're interested in has an isolated singularity at the origin. Sep 30 '10 at 22:04
• @mdeland; I want to know when the affine hypersurface has such a singularity. I hope I made that clear this time in my question. Sep 30 '10 at 22:11
• Yes, you did make it clear - I'm just telling you that the answer to your question is the same as the answer to whether or not the weighted projective hypersurface is smooth. Sep 30 '10 at 22:15
• @mdeland: I understand that they are equivalent. I don't understand how this answers my question; you've just rephrased it to "what are conditions on $S$ and/or $H$ such that the weighted projective space is smooth". Having coprime weights that divide the degree is not enough: I can assign degree 2 to $x$ and 1 to $y$ in example 1, and $2|6$, but the singular locus of $x^3$ is not the origin. Sep 30 '10 at 22:30

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.

• His condition that $S$ lies in a hyperplane is equivalent to the existence of a $\mathbb C^*$ action for which each monomial is of the same degree. The weights are given by the normal vector to $H$. For instance, in example 3, $(6,0,0).(1,2,3)=(1,1,1).(1,2,3)=\cdots=6$, so the exponents lie in the hyperplane $a + 2b + 3c = 6$.
– JRG
Sep 30 '10 at 22:49