Computation of Hilbert polynomial of two incident lines with nilpotent

Question

It is well known that the ideal $$I=(z^2,yz,xz,y^2-x^2(x+1))$$ occurs in a flat family of twisted cubic, so its Hilbert polynomial is $h^0(\mathcal{O}_{\mathbb P^1}(3t))=3t+1$. The support of the ideal is a plane nodal cubic, and it has a nilpotent element at the singular point.

My question is about the ideal of this form:

1. Does the ideal $J=(z^2,yz,xz,y^2-x^2)$ defines two lines with a nilpotent at $0$?

2. How to compute the Hilbert polynomial of this ideal? (I suspect it is $2t+2$ for some reason but I don't know how to prove this.)

Answer 1

1. Yes. Set-theoretically, $V(J)=V(\sqrt{J})=V(z,yz,xz,y^2-x^2)=V(z,y^2-x^2)$ which is indeed two lines: $V(z,x-y)$ and $V(z,x+y)$. On $D(x)\cap V(J)$, we have that $x$ is invertible, so $z\in J_x$ and therefore $V(J)\cap D(x)$ is reduced. In the local ring at the origin, none of $x,y,z$ are invertible, so $z\notin \mathcal{O}_{V(J),(0,0,0)}$ and $V(J)$ isn't reduced there.

2. If you want a low-tech solution, you can just compute a bunch of terms of the quotient and use that Hilbert polynomials play nicely with exact sequences:

   • $\deg 0: \{1\}, \dim 1$
   • $\deg 1: \{x,y,z\}, \dim 3$
   • $\deg 2: \{x^2,xy\}, \dim 2$
   • $\deg 3: \{x^3,x^2y\}, \dim 2$
   • $\deg 4: \{x^4,x^3y\}, \dim 2$
   • $\deg 5$ and up will be dimension 2 as well: $x^n$ and $x^{n-1}y$ survive in the quotient, but nothing else does - nothing with a $z$ for obvious reasons, and then every monomial $x^ay^b$ is equivalent to either $x^n$ or $x^{n-1}y$ by turning $y^2$ in to $x^2$ a bunch.

So $H_J(t)+2=H_{k[x,y,z]}(t)$, or $H_j(t)=\binom{n+2}{2}-2=\frac{n^2+3n-1}{2}$. If you want to use a bit more technology, you can do the same thing except note that the projective zero locus of $J$ inside $\Bbb P^2$ is two reduced points, which has Hilbert polynomial 2 and then apply the same exact sequence argument.