Let $H$ denote the irreducible component of $\text{Hilb}^{3t+1}\mathbb{P}^3$ whose general member corresponds to a non-singular twisted cubic. Let $C$ be a subscheme lying in the boundary of $H$ and assume it lies in a surface $S \subseteq \mathbb{P}^3$.

Then why is it possible that we can find families $C_R, S_R \subseteq \mathbb{P}^3_R$ over a DVR $R$ with fraction field $K$ such that

1) $C_R \subseteq S_R$

2) The generic fiber $C_K$ is a non-singular twisted cubic

3) $C \subseteq S$ are the closed fibers of the family.

The authors in "Hilbert Scheme Compactification of the Space of Twisted Cubics": https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/Hilbertscheme.pdf make the claim on page 4 (pg 763), line 7 of the proof. Although they are studying embedded points, this claim seems to be something more general about flat limits?

More generally, is it true that if something on the boundary of my component in a Hilbert scheme lied in a hypersurface, then I could find a family over a DVR like above?

  • $\begingroup$ If I understand correctly the question (not sure), this is the valuative criterion of specialisation, see EGA II, Proposition 7.1.4. This tells you that there exists a DVR $R$ and a morphism $\operatorname{Spec}(R)\rightarrow H $ mapping the generic point to the generic point of $H$, and the special point to $[C]\in H$. Then you just pull back the universal subscheme over $H$ to $\operatorname{Spec}(R) $. $\endgroup$
    – abx
    Dec 1, 2018 at 5:18
  • $\begingroup$ Hi thanks for the response. I understand how to lift the curve and surface separately, but my confusion is how could we lift them together i.e. we have a containment in the generic fiber, $C_K \subseteq S_K$. This seems to imply that there's a non-singular twisted cubic that lies in a surface of degree $d$, if some degenerated twisted cubic lies in a surface of degree $d$. $\endgroup$
    – Randomness
    Dec 1, 2018 at 5:40
  • $\begingroup$ I think this would follow from knowing that the Hilbert function is constant in the family of degenerations of twisted cubic. Do you know if this holds? $\endgroup$
    – naf
    Dec 1, 2018 at 5:45
  • $\begingroup$ Hi. I don't think the Hilbert function is constant as they are proving that the degenerate curve has an extra cubic in its ideal (twisted cubic is cut out by three quadrics). I also think they use this family of surfaces to show that the degenerate curve doesn't lie in a plane. $\endgroup$
    – Randomness
    Dec 1, 2018 at 7:11

1 Answer 1


As @abx and @ulrich explain in the comments, the original question is equivalent to a question about constancy of Hilbert functions for the universal family restricted over the irreducible component $H$. The Hilbert function is constant on $H$, as I explain below. I believe the OP is confused because, at this point in the article of Piene and Schlessinger, they have not proved constancy of the Hilbert function. Nor do they need this. For instance, Piene and Schlessinger point out that for the purposes of their proof, it is fine to replace a hyperplane that contains the curve by a quadric hypersurface that contains the curve (in fact, as follows from constancy of the Hilbert function, there is no hyperplane containing a curve parameterized by $H$). So my advice to the OP for reading the article is: just read the remainder of the proof and then come back to this issue after a complete read-through.

Anyway, the Hilbert function is constant. Denote by $p(t)$ the Hilbert polynomial $p(t)=3t+1$. On the Hilbert scheme $\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$, the natural action of $\textbf{PGL}_{4,k}$ on $\mathbb{P}^3_k$ induces an action on $\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$. Denote by $H_0$ the unique open orbit. Denote by $H$ the closure of $H_0$ in $\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$. Denote the restriction of the universal closed subscheme over $H$ by $$Z_H\subset H\times_{\text{Spec}\ k}\mathbb{P}^3_k.$$

Claim. Every geometric fiber $Z_t$ of the projection $Z_H\to H$ has Hilbert function, $$ h_{Z_t}:\mathbb{Z}_{\geq 0} \to \mathbb{Z}_{\geq 0}, \ \ h_{Z_t}(d) := h^0(\mathbb{P}^3_k,\mathcal{O}(d)) - h^0(\mathbb{P}^3_k,\mathcal{I}_{Z_t}(d)), $$ equal to $h_{Z_t}(d) = 3d+1$.

Proof. Probably the fastest way to prove this is to use the stratification of $H$ according to the "type" of $Z_t$. Some version of this is contained in Joe Harris's monograph.

MR0685427 (84g:14024)
Harris, Joe
Curves in projective space.
With the collaboration of David Eisenbud.
Séminaire de Mathématiques Supérieures, 85.
Presses de l'Université de Montréal, Montreal, Que., 1982.
138 pp. ISBN: 2-7606-0603-1

I have a vague recollection that there is a small mistake in Harris's description of the orbit decomposition. I am more familiar with the honors thesis of Yoon-Ho Alex Lee. The most relevant result is Figure 4.4, p. 47.

Yoon-Ho Alex Lee
The Hilbert scheme of curves in $\mathbb{P}^3$

Since dimensions of cohomology groups are upper semicontinuous, it suffices to prove that $h_{Z_t}(d)$ equals $3d+1$ for $Z_t$ in the two "deepest" strata, XVI and XVII in Lee's notation. With respect to homogeneous coordinates $[ s,t,u,v ]$, the ideal for XVI is $\langle u^2,ut,uv,v^3\rangle$, and the ideal for XVII is $\langle u^2,uv,v^2 \rangle$. It is straightforward in each of these cases to compute that $h_{Z_t}(d)$ equals $3d+1$. QED


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