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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.

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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"!

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  • $\begingroup$ 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. $\endgroup$
    – user46348
    Apr 7 '15 at 14:00
  • $\begingroup$ 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. $\endgroup$
    – Brenin
    Apr 7 '15 at 14:54
  • $\begingroup$ 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! $\endgroup$
    – user46348
    Apr 8 '15 at 20:37
  • $\begingroup$ My pleasure. Have fun with boxes :) $\endgroup$
    – Brenin
    Apr 8 '15 at 23:24

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