Find description of $\mathscr H_0 \subset Hilb^3(\mathbb C^2)$ I am trying to understand the Hilbert scheme of points in $\mathbb C^2$ and I have very naive questions about it. 
Consider $\mathscr H_0 \subset Hilb^3(\mathbb C^2)$ the subscheme of lenght $3$, supported at the origin, or equivalenty all the ideals $I \subset \mathbb C[x,y]$ with $\dim_{\mathbb C} \mathbb C[x,y]/I = 3$ and $Z(I) = (0,0)$.
If we look at $A = \mathbb C[x,y]/I$, it's a $\mathbb C$-algebra of dimension $3$. There is a linear term $l \in A$ and if $l^2 \neq 0$ then $A = \mathbb C \{1,l,l^2 \}$.
Else, there is another linear form $l' \in A$, but two linears form have the same $\mathbb C$-span as ${x,y}$, i.e $A = \mathbb C\{1,x,y\}$ or equivalently $I = (x,y)^2$.
In the first case, $I = (l^3,l')$ where $l,l'$ form a basis of the linear form in $x,y$. Of course it does not change anything if we scale $l$ by $\lambda \in \mathbb C$. So really, if $Y = \mathbb P^1 \times \mathbb P^1 \backslash \Delta$ and, I believe the map $\phi : Y \to \mathscr H_0, (l,l') \mapsto (l^3,l')$ should be almost an isomorphism except that $(x,y)^2$ is not in the image. 
Now Eisenbud and Harris, in the book "Geometry of Schemes", page 63, claim that $\mathscr H_0$ can be embedded as a cubic cone in $\mathbb P^3$ with vertex = $(x,y)^2$. I have no ideas why. Can you help me ? 
I also have another problem. When I try to draw $p \in \mathscr H_0$ I just drawn 3 points on the same line $l$ which will collapse at $(0,0)$. I know point does not always arise this way but I like this way of thinking about them. However, for the ideal $(l^3,l')$ I only need $l$ for make this picture. So where does appear this auxiliary direction $l'$ on this picture ? 
Thanks in advance ! 
 A: Edit (April 3, 2017)
My answer yesterday gave an incomplete list of the required ideals.
I have now corrected my answer by completing that list and I apologise for my preceding answer.
a) Here is the complete, irredundant  list of all distinct ideals of codimension three  $I\subset \mathbb C[x,y]$ such that $V(I)$ supported at the origin: 
$$I_{a,b}=(x^3, y+ax+bx^2) \;  \operatorname {where}\: a,b \in \mathbb  C  \\J_c=(y^3,x+cy^2)  \;  \operatorname {where}\: c\in \mathbb  C  \\ \mathfrak m^2=(x,y)^2$$ Note carefully:
i) $V(I_{a,b})\subset V(y+ax+bx^2)$ so that $I_{a,b}$ defines a subscheme of the parabola  $y+ax+bx^2=0$ if $b\neq0$.
ii)  $V(I_{a,0})\subset V(y+ax)$ so that $I_{a,0}$ defines a subscheme of the line  $y+ax=0$ if $b=0$.
iii) $V(J_c)\subset V(x+cy^2)$ so that $J_c$ defines a subscheme of the parabola $x+cy^2=0$ if $c\neq 0$
iv) $V(J_0)\subset V(x)$ so that $J_0$ defines a subscheme of the $y$-axis $x=0$ if $c=0$.
v) However $V(\mathfrak m^2)$ is not a subscheme of any  smooth curve, since its Zariski tangent space has dimension $2$. 
b) The set $\mathcal H$ is probably a closed subscheme of the Hilbert scheme $Hilb ^3(\mathbb A^2_\mathbb C)$ but I haven't checked it.
The set of $I_{a,b} \;(a,b \in \mathbb  C) $ would then be a dense open subset of $\mathcal H$ isomorphic to $\mathbb A^2_\mathbb C$, and thus $\mathcal H$ would be a rational surface with $\mathfrak m^2$ as its only singularity: this might be the cubic cone alluded to by Eisenbud-Harris in their Exercise II-13, page 63.
I find that the authors do a great disservice to beginners by discouraging them with such an absurdly difficult exercise for which they don't have the necessary tools.
Indeed the Hilbert scheme, the natural technique for solving such an exercise,  is only mentioned two hundred pages later (on page 262 !) and even there they only give the barest introduction to that theory.   
