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?

  • $\begingroup$ 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. $\endgroup$ Aug 14, 2012 at 13:59

1 Answer 1


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 $Hilb_P$.

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    $\begingroup$ 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. $\endgroup$ Aug 14, 2012 at 13:02
  • $\begingroup$ @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? $\endgroup$ Aug 14, 2012 at 13:27
  • $\begingroup$ @Starr: I have modified the question a bit to ensure we always have reduced schemes. $\endgroup$ Aug 14, 2012 at 13:31
  • $\begingroup$ After the modification, Sasha's answer is correct. $\endgroup$ Aug 14, 2012 at 13:59
  • $\begingroup$ @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. $\endgroup$
    – Sasha
    Aug 14, 2012 at 14:10

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