The dimension of the punctual Hilbert scheme Let $H$ be the punctual Hilbert scheme of $3$ points in $\mathbb A^3$, over the complex numbers.
Then $H$ can be described in the following equivalent ways:


*

*set-theoretically, $H$ is the set of subschemes $Z\subset \mathbb A^3$ of length $3$ concentrated at the origin;

*$H=\textrm{Hilb}^3(\textrm{Spec }R)$, with $R=\mathbb C[x,y,z]/(x,y,z)^3$;

*$H=p^{-1}(3[P])$, where $p:\textrm{Hilb}^3(\mathbb A^3)\to \textrm{Sym}^3(\mathbb A^3)$ is the Hilbert-Chow morphism and $P\in\mathbb A^3$ is a closed point.


I am struggling to find the dimension of $H$ and I would like some advise on how to compute it. 
On the one hand, there are only two non-isomorphic structures on a triple point in $\mathbb A^3$, but on the other hand the dimension of $H$ seems to be bigger than the expected dimension $3\cdot\dim R=0$. I am not able to compute the dimension of the fiber over $3[P]$, as it lies in the diagonal and I do not really know what to expect.
Thanks for any help.
 A: I am sure there are references online for these kinds of computations: for instance, you could look at the papers of Iarrobino, who has written a lot about these schemes. Anyway, here is one way to calculate what you want; there may be a better one.
Let $A= \mathbf C[x,y,z]$.
We want to find ideals $I$ supported at $0$ (so contained in $\mathfrak m = (x,y,z)$) such that $A/I$ is a vector space of dimension 3. The first observation is that $I$ must contain $\mathfrak m^3$, for otherwise we can easily find 4 linearly independent elements in the quotient. So we are really looking for ideals $\tilde{I}$ contained in $\mathfrak m / \mathfrak m^3$, and the condition that $\operatorname{dim} A/I =3 $ translates to  $\operatorname{dim} \mathfrak m/\tilde{I} =2 $. 
Let $L$ be the vector subspace in $\mathfrak m / \mathfrak m^3$ spanned by the linear forms. Since $L$ has has dimension 3, it must have nonzero intersection with $\tilde{I}$, otherwise the quotient $\mathfrak m/\tilde{I}$ would have dimension at least 3. On the other hand, we cannot have $L \subset \tilde{I}$, since $L$ generates $\mathfrak m$ as an ideal.
So $\operatorname{dim}( L \cap \tilde{I}) = 1 \text{ or } 2$. 
If it is 1, then the dimensions tell us $\tilde{I}$ must contain all of $\mathfrak m^2 / \mathfrak m^3$, so $\tilde{I}$ is  determined by its intersection with $L$. 
If it is 2, then (exercise) we calculate that $(\tilde{I} \cap L)^2$ has dimension 5, hence again by dimensions is all of the degree-2 part of $\tilde{I}$; so again in this case, $\tilde{I}$ is determined by its intersection with $L$. 
So the upshot of all this is that the possible choices of the ideal $I$ are exactly the possible choices of a linear subspace of dimension 1 or 2 in the 3-dimensional vector space $L$. So the Hilbert scheme is isomorphic to $G(1,3) \amalg G(2,3) = \mathbf P^2 \amalg \mathbf P^2$.
By the way, note that $PGL(3)$ acts transitively on each component, giving the two possible scheme structure on a triple point that you mention.
