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 $\operatorname{Hilb}^2(X)$?
Comments, references and basically everything is welcome. In particular, partial answers would also be helpful. Thanks in advance.
 A: 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.
