# Hilbert scheme of points and passing curves

It is well known that through five points on a projective plane you can pass a conic.

Q. What happens when points collide ?

More precisely: if I consider a more simple question of two points and passing a line through them, then if I take a point in a Hilbert scheme of two points $Hilb^2$ I can pass exactly one line when this points collide.

Q Now, can I draw a conic (maybe singular) if I choose a point in $Hilb^5$ of projective plane? Or may be I need some different version of moduli space of points? The symmetric powers not work at all.

• "Now, can I draw a unique conic (maybe singular) if I choos a point in $\text{Hilb}^5$ of projective plane." No, you cannot find a unique conic, even in the nice case that the point parameterizes $5$ distinct, reduced points. For instance, consider the case when $4$ of the $5$ points are collinear. – Jason Starr Dec 23 '17 at 16:45
• So maybe the question should be rephrased as - given a general point of the diagonal, is there a unique conic passing through it ? – aginensky Dec 23 '17 at 16:57

## 1 Answer

If you take any zero-dimensional subscheme Z of P^2 of length 5, the only way two conics can both contain it is if they have a common component, which must then be a line L, hence they must be of the form L+L' where L' is another line.

If Z is contained in a line L, then a conic C contains Z iff it contains L, so the parameter space of conics containing Z is 2-dimensional.

If Z is not contained in a line, but it contains a subscheme W of length 4 which is a subscheme of a line L, then there is a one-parameter family of conics containing it; they are of the form L+L' where L' is any line containing p (the only point in P^2 where W is not Z, or the support of the kernel of O_Z\to O_W.

If the length of L\cap Z is at most 3, then the length of L'\cap Z must be at least 2 - and as you pointed out, this means the line L' is unique.