# Zero-dimensional Hilbert scheme of two points and Hilbert-Chow morphism

I would like to understand what happens to the Hilbert scheme of two points on a scheme if the scheme is zero-dimensional. The background for this question is just general curiosity, Hilbert schemes were discussed a lecture I attended but no-one every talked about this case (only about curves and surfaces).

The following should, hopefully, serve as minimal example. If there are issues with it, let me know.
Let $$X$$ be the subscheme of $$\mathbb A^1$$ defined by the ideal $$\langle x^k\rangle$$ for some $$k$$. We can consider the symmetric product $$\operatorname{Sym}^2(X)$$ and it is easy to write down equations for this by computing the ring of invariants. I did so and got a point of multiplicity $$\frac{k(k+1)}{2}$$ in $$\operatorname{Sym}^2(X)$$.
Question 1: Is there, besides the calculations, a reason to expect this number?
From my lecture I know that for a scheme $$Y$$ we have the Hilbert-Chow morphism $$H:\operatorname{Hilb}^2(Y)\to \operatorname{Sym}^2(Y),\quad C\mapsto \sum\limits_{P\in Y}\operatorname{length}_P(C)\cdot P,$$ and I have at least a rough idea of what it does. In particular, I understand that points being mapped to points of the form $$2P\in\operatorname{Sym}^2(Y)$$ can be understood as points together with a tangent direction. Returning to the zero-dimensional situation, $$\operatorname{Hilb}^2(X)$$ should be zero-dimensional, too. But then, how can we talk a tangent direction at a point?
Question 2: Do we have a Hilbert-Chow morphism for zero-dimensional $$X$$ and if so, how can we understand it?
Finally, I am also curious about the multiplicity of $$\operatorname{Hilb}^2(X)$$. Is there a way to calculate it, just as I did for $$\operatorname{Sym}^2(X)$$?
Question 3: What is the multiplicity of $$\operatorname{Hilb}^2(X)$$?

First, note, that if $$Y \subset X$$ is a closed embedding, then $$Hilb^m(Y) \subset Hilb^m(X)$$ is the closed subscheme, that parameterizes the subschemes of $$X$$ contained in $$Y$$. This is easy to check by comparing the functors represented by the two schemes.

Now, if $$X = \mathbb{A}^1$$ and $$Y = (x^k)$$, the $$k$$-th neighborhood of $$0 \in \mathbb{A}^1$$, we see that $$Hilb^m(Y)$$ is a subscheme of $$Hilb^m(X) = Hilb^m(\mathbb{A}^1) = \mathbb{A}^m$$. Note that this already implies that the Hilbert-Chow morphism for $$Y$$ is an isomorphism onto its image (since this holds for $$X$$).

Let me also describe explicitly $$Hilb^m(Y)$$ in the case $$m = 2$$, $$k = 4$$ (other cases are analogous). We have $$Hilb^2(\mathbb{A}^1) = \mathbb{A}^2$$, and the universal subscheme can be described as $$Z = \{ t^2 = at + b \} \subset \mathbb{A}^2 \times \mathbb{A}^1,$$ where $$a,b$$ are the coordinates on $$\mathbb{A}^2$$ and $$t$$ is the coordinate on $$\mathbb{A}^1$$. Therefore, the algebra of functions on $$Z$$ can be written as $$k[Z] = \{ f(a,b) + t g(a,b) \},$$ where $$f$$ and $$g$$ are polynomials, and the multiplication is defined by $$t^2 = at + b$$. Now we should impose the equation $$t^4 = 0$$, defining the subscheme $$Y \subset \mathbb{A}^1$$. Using the multiplication rule, one computes $$t^4 = (a^3 + 2ab)t + (a^2b +b^2),$$ hence $$t^4 = 0$$ is equivalent to the equations $$a^3 + 2ab = a^2b + b^2 = 0$$ defining $$Hilb^2(Y) \subset \mathbb{A}^2$$.

EDIT. Let me be a bit more detailed about the deduction of the equations. Let $$Z \subset Hilb^m(X) \times X$$ be the universal subscheme. Then every function $$\varphi \in k[X]$$ defines a function on $$Z$$, and hence a section, say $$v(\varphi)$$ of the tautological bundle $$V := p_*\mathcal{O}_Z$$, where $$p \colon Z \to Hilb^m(X)$$ is the projection. Now, if $$Y \subset X$$ is a subscheme defined by equations $$\varphi_1,\dots,\varphi_n$$, then $$Hilb^m(Y) \subset Hilb^m(X)$$ is just the zero locus of the sections $$v(\varphi_1)$$, \dots, $$v(\varphi_n)$$. This can be again deduced by comparing the functors represented by $$Hilb^m(Y)$$ and by this zero locus inside $$Hilb^m(X)$$.

In the case of $$X = \mathbb{A}^1$$ and $$\varphi = x^k$$, we have $$v(\varphi) = (a^3 + 2ab)t + (a^2b +b^2),$$ considered as a section of the bundle $$V$$, which is trivial of rank 2 with basis $$t$$, 1. Thus we get the above equations.

• Thank you very much for your answer. I am struggling a little with the universal subscheme and how you got the equations out of it. Could you give a little more details on why you consider this universal subscheme and why the result gives defining equations for the $Hilb^2$? Also you were saying that the Hilbert-Chow morphism is an isomorphism in this case but $Sym^2$ admits a point of multiplicity $10$ ($k=4$) and the local equations you are giving yield a point of multiplicity $6$. So how does this go together?
– user526015
Nov 15 '18 at 13:24
• Right, a wasn't correct saying it is an isomorphism, the right statement is that it is an isomorphism onto its image. I also added an explanation about equations, please check. Nov 15 '18 at 15:04
• I will need some time to work through this. Thank you very much, I will try to understand this and then accept your answer or ask further questions.
– user526015
Nov 15 '18 at 17:54