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 $\operatorname{Hilb}^2(X)$?

Comments, references and basically everything is welcome. In particular, partial answers would also be helpful. Thanks in advance.


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.

  • $\begingroup$ 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? $\endgroup$ – James Nov 15 '18 at 13:24
  • $\begingroup$ 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. $\endgroup$ – Sasha Nov 15 '18 at 15:04
  • $\begingroup$ 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. $\endgroup$ – James Nov 15 '18 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.