Gluing subschemes of fibers of Hilbert Scheme (in mixed characteristic)

Let $\mathfrak{X}\rightarrow Spec(R)$ be a smooth family of smooth projective varieties over a local 1-dimensional ring of mixed characteristic. Suppose that there are non-empty subschemes (locally closed) $Y_0\subset Hilb_Q(\mathfrak{X}_{(0)}/Frac(R))$ and $y_m\subset Hilb_Q(\mathfrak{X}_{(m)}/k)$ (where $k=R/m$) defined by the same conditions (e.g. locally complete intersection and some cohomological conditions) in characteristic $0$ (generic fiber) and in characteristic $p$ (special fiber).

Is it always true that there is a subscheme $Z\subset Hilb_Q(\mathfrak{X}/Spec(R))$ such that $Z_{(0)}=Y_0$ and $Z_{(m)}=y_m$?

• One of the basic properties of the Hilbert scheme is "properness". So if $R$ is a Dedekind domain (e.g., a discrete valuation ring), then for every given $Y_0$ over the generic point, for every maximal ideal $m\subset R$, there exists one and only one $y_m$ over the closed point that extends $Y_0$. – Jason Starr Jul 9 '15 at 10:15
• Thank you very much for your help. But I cannot see why must the specialization be the $y_m$ ? I have not a good intuition on hilbert schemes but is it impossible for this specialization to parametrize different kind of subvarieties ? – user052715 Jul 9 '15 at 10:43
• "But I cannot see why must the specialization be the $y_m$?" That is my point: for every $Y_0$, there is one and only one $y_m$, and we do not get to choose what it is: "$Y_0$ chooses its $y_m$". – Jason Starr Jul 9 '15 at 10:49
• Thank you again. This is one illustration of my worry: suppose $Y_0$ parametrizes subvarieties with trivial dualizing sheaf, and trivial higher cohomology groups with value in ideal sheaf. Why the unique extension of $Y_0$ to the special fiber should parametrize the same kind of subvarieties? – user052715 Jul 9 '15 at 11:44
• Regarding the dualizing sheaf, if the subvarieties are all smooth and (geometrically) connected, then the corresponding Picard functor is separated. So if the dualizing sheaf for $Y_0$ is trivial, then the same is true for $y_m$. For cohomology vanishing, that is more subtle. However, there are many results in this direction. Sometimes the Hodge Theorem can force vanishing (this leads to one proof of the Kawamata-Viehweg Vanishing Theorem). – Jason Starr Jul 9 '15 at 12:00