# Five points, two tangent spaces

An exact comparison of two Artinian quotients with Hilbert function (1, 3, 1).

2026-10-17 · Study note · exact algebra calculation

> Prepared with an assistant from the cited example and exact seed calculations. This record supplies the starting computation for a proposed project; it is not a new theorem or an independently completed project.

## Recorded starting point

My saved A–M reading concerns Artinian structure, nilpotents and local rings. The length-five comparison was selected and computed with an assistant while planning this site; it is not a recorded personal discovery.

## A distinction to keep explicit

The two quotients share a graded Hilbert function, but the 21/15 computation concerns full embedded deformations. Restricting to homogeneous degree-zero maps gives 5/5. A complete personal derivation of the second-order obstruction is still a task ahead.

## The objects

Work over the rational numbers. A finite quotient S/I is a finite-dimensional vector space; its dimension is the length of the corresponding zero-dimensional scheme. Here both ideals are homogeneous and supported at the origin.

$$
S=\mathbb Q[x,y,z],\qquad A_j=S/I_j,
$$

$$
\begin{aligned}I_1&=(x^2,xy,xz,y^2,yz,z^3),\\ I_2&=(x^2,xy,xz,y^2,z^2).\end{aligned}
$$

The standard monomials are those not divisible by any monomial generator of I. They give an explicit vector-space basis of S/I. This gives the same graded dimension in the two cases, but different multiplication.

$$
\begin{aligned}\mathcal B_1&=(1,x,y,z,z^2),\\\mathcal B_2&=(1,x,y,z,yz),\\H_{A_1}(t)&=H_{A_2}(t)=1+3t+t^2.\end{aligned}
$$

## What the Hilbert function forgets

Among products of positive-degree basis elements, only z·z survives in A₁; only y·z survives in A₂. In particular, the socle—elements killed by the maximal ideal—already distinguishes them.

$$
\operatorname{Soc}(A_1)=\langle x,y,z^2\rangle,\qquad\operatorname{Soc}(A_2)=\langle x,yz\rangle.
$$

Thus an equality of Hilbert functions does not imply an isomorphism of algebras. The next question asks what happens to their embedded deformations inside affine three-space.

## From first-order deformations to Hom

Use the dual-number ring D = Q[ε]/(ε²). A flat first-order deformation of the quotient can be written locally by lifting each f in I to f + εg in an ideal over D. Reducing g modulo I defines a map φ: I → S/I; changing the lift does not change its residue. Multiplication by S forces φ to be S-linear. With this sign convention the tangent vector is φ; using f − εg reverses its sign.

$$
T_{[I]}\operatorname{Hilb}^5(\mathbb A^3_{\mathbb Q})\simeq \operatorname{Hom}_S(I,S/I).
$$

This is the tangent space to the full Hilbert scheme, allowing all embedded first-order deformations. It is not just the degree-zero Hom space, and not just deformations preserving a chosen graded presentation.

## A finite system of exact equations

Write I = (m₁,…,mᵣ). An S-linear map is determined by r vectors φ(mᵢ) in the five-dimensional quotient. For every pair of monomial generators, their least common multiple gives a relation. These pair relations generate the relation module, so imposing them suffices.

$$
\frac{\operatorname{lcm}(m_i,m_j)}{m_i}\varphi(m_i)-\frac{\operatorname{lcm}(m_i,m_j)}{m_j}\varphi(m_j)=0\quad\text{in }S/I.
$$

Expand each expression in the standard-monomial basis. Exact rational row reduction gives 30 unknown coefficients with rank 9 for I₁, and 25 unknown coefficients with rank 10 for I₂.

$$
\dim_{\mathbb Q}\operatorname{Hom}_S(I_1,A_1)=21,\qquad\dim_{\mathbb Q}\operatorname{Hom}_S(I_2,A_2)=15.
$$

See the website for the interactive calculation; exact seed scripts and results are provided alongside it.

## The research step still ahead

A tangent dimension alone does not construct a second-order lift. The next task is to choose actual tangent vectors, lift the relations over Q[ε]/(ε³), and identify the quadratic compatibility equations. A dimension jump is a starting observation, not a proof that an arbitrarily chosen direction is obstructed.

A useful project would delimit a family with a fixed basis, compute its rank-jump locus, and then examine liftability along that locus. Published smoothness criteria for monomial points must be checked before making a novelty claim.

## Sources

- [Jelisiejew, Ramkumar & Sammartano — The Hilbert scheme of points on a threefold: broken Gorenstein structures and linkage](https://arxiv.org/html/2409.17009v2)