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?



Flat limit (of twisted cubic) contained in surfaces
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 $\mathbb{P}^3$  

<https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/ALee_Hilbertschemes.pdf>


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**


