MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let ${Hilb_{P,Q}}_{red}$ be the reduced scheme associated to the Hilbert flag scheme parametrizing all pairs $(C,X)$ with $C \subset X \subset \mathbb{P}^3$, where $C$ is a curve and $X$ a degree $d$ surface in $\mathbb{P}^3$. When is the natural projection map $$pr_1:{Hilb_{P,Q}}_{red} \to pr_1({Hilb_{P,Q}}_{red}) \subset {Hilb_P}_{red}$$ a flat morphism?

share|cite|improve this question
After your modification, Sasha's answer is complete and correct. You are not going to find some flatness criterion that does not directly prove that the Hilbert function is constant. One of the main results on constancy of the Hilbert function is the work of Gruson, Lazarsfeld and Peskine. I suggest you start there. – Jason Starr Aug 14 '12 at 13:59

The fiber of the map over a curve $C$ is just $P(H^0(P^3,I_C(d)))$. So, the sufficient and necessary condition is that $\dim H^0(P^3,I_C(d))$ is constant on $Hilb_P$.

share|cite|improve this answer
Your answer is only correct if $\text{Hilb}_P$ is reduced. In fact I believe Mumford's famous example of an everwhere nonreduced component of the Hilbert scheme of smooth, embedded, space curves precisely exploits the fact that the flag Hilbert scheme (for curves in a cubic surface) is not flat over the Hilbert scheme of space curves. – Jason Starr Aug 14 '12 at 13:02
@Sasha: The main motivation of the question was to know when $h^0(I_C(d))$ is constant. So could you suggest some other criterion? – Naga Venkata Aug 14 '12 at 13:27
@Starr: I have modified the question a bit to ensure we always have reduced schemes. – Naga Venkata Aug 14 '12 at 13:31
After the modification, Sasha's answer is correct. – Jason Starr Aug 14 '12 at 13:59
@Jason: Of course, I was a bit careless, what is true is that the flag Hilberts scheme is isomorphic to the projectivization of $(pr_1)_*I_Z(d)$, where $Z \subset Hilb_P\times P^3$ is the universal family. If $Hilb_P$ is reduced then the condition I wrote is equivalent to saying that $(pr_1)_*I_Z(d)$ is a vector bundle. But if it is not reduced then of course it is not. – Sasha Aug 14 '12 at 14:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.