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I would be very grateful for any comment or a reference on the following question.

Let $Hilb_0^n({\mathbb C}^2)$ be the Hilbert scheme of n points in ${\mathbb C}^2$ concentrated, set theoretically, at the origin. Let $X$ be the locus of $Hilb_0^n({\mathbb C}^2)$ formed by curvilinear ideals.

Is the complement of $X$ in $Hilb_0^n({\mathbb C}^2)$ a divisor or it has codimension > 1 ?

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Taking n=3 shows that it is at least possible that the locus you describe has codimension >1. Of course, it is also empty if we take n=2. I don't know the answer off hand for all n. –  Jack Huizenga Apr 28 '11 at 19:13
    
What are curvilinear ideals? –  Sasha Apr 29 '11 at 12:09
    
Geometrically, a curvilinear ideal is an ideal corresponding to a scheme that is locally contained in a smooth curve. –  Jack Huizenga Apr 29 '11 at 20:03

1 Answer 1

It follows from Briançon (Inventiones Math 41 (1977), no. 1, 45–89) Theorem III.3.1 that the codimension is 2 for n=3, and 1 for n>3.

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He computes the dimension of the strata corresponding to different Hilbert functions in terms of "staircases" of initial ideals. For n>3 there is always the staircase corresponding to $(y^2,xy,y^{n-1}$ giving dimension $n-2$. The deformation is similar to that given by Martin, but I don't recall the details (and as Jack pointed out, Martin's has some problem). –  quim May 2 '11 at 17:07
    
Indeed I have.. I deleted the answer referred to. –  MartinG May 2 '11 at 20:42

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