# Punctual Hilbert scheme of four points

I am looking at $\text{Hilb}^4(\mathbb{C}^2)$, which is the Hilbert scheme of four points on $\mathbb{C}^2$. In particular, I am just looking at four points collided (at the origin), and want to know what all the non-isomorphic ideal representations are. For example, $I = \langle x, y^4 \rangle$ has colength 4, thus is an element in $\text{Hilb}^4(\mathbb{C}^2)$. I'm curious to know what are the non-isomorphic ideals like this of colength four. Thanks for the help.

## 1 Answer

The $T=(\mathbb C^\times)^2$-action on $\textrm{Hilb}^n\mathbb C^2$ (lifted from the natural one on $\mathbb C^2$) has, as fixed points, the monomial ideals, which correspond to (one-dimensional) partitions of $n$. Hence if you can list the $p(n)$ partitions (Young tableaux if you prefer!) of $n$ and you write down the corresponding monomial ideals, you are done. In the case of $n=4$, you get: $$(x,y^4),(y^4,x),(x^2,y^2),(x^3,xy,y^2),(x^2,xy,y^3).$$

NB. (In case you wonder how to find the generators in general.) Represent the polynomial ring $\mathbb C[x,y]$ by listing all possible monomials in $x$ and $y$, along two axes. A Young tableaux shapes for you a "staircase" (called Hilbert staircase), and since the Young tableaux (or partition) is the complement of the corresponding ideal, you can read the generators of the ideal "under the stairs"!

• Thanks for the help @Brenin! I was hoping you could clarify a few things. I see what you are talking about with the tableaux, but if you do this for the n=3 case, I seem to be getting $\langle x, y^3 \rangle$, $\langle x^3, y\rangle$ (isomorphic), but then $\langle x^2, y^2\rangle$, but this has colength 4, not 3. I'm somehow missing the $xy$ term. For this one, my tableaux looks like two vertical blocks and two horizontal (sharing one block). Can you help me see what I'm doing wrong? Thanks again for the help. Apr 7 '15 at 14:00
• Yes, the third fixed point for $n=3$ corresponds to $I=(x^2,xy,y^2)$.Think of the box in the corner as $1\in\mathbb C[x,y]$, and the two other boxes as $x$ and $y$. All the rest is the ideal $I$. Can you see it now? (It's a pity I cannot draw here!) Again, if you draw the tableau, the generators of the corresponding ideal are the monomials which define the shape of the staircase. Apr 7 '15 at 14:54
• Yes, I was misinterpreting where the "x's" and "y's" went but I see that they go in the Tableaux (duh). And yeah as you said, the Tableaux is the compliment of the ideal. That was also misunderstood by me but I see it now. Thanks so much! Apr 8 '15 at 20:37
• My pleasure. Have fun with boxes :) Apr 8 '15 at 23:24