10
$\begingroup$

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 ?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jack Huizenga
    Commented Apr 28, 2011 at 19:13
  • $\begingroup$ What are curvilinear ideals? $\endgroup$
    – Sasha
    Commented Apr 29, 2011 at 12:09
  • $\begingroup$ Geometrically, a curvilinear ideal is an ideal corresponding to a scheme that is locally contained in a smooth curve. $\endgroup$
    – Jack Huizenga
    Commented Apr 29, 2011 at 20:03

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$
    – quim
    Commented May 2, 2011 at 17:07
  • $\begingroup$ Indeed I have.. I deleted the answer referred to. $\endgroup$
    – MartinG
    Commented May 2, 2011 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .