I have the following problem. Let $\mathcal{O}$ be a valuation ring and $S=Spec(\mathcal{O})$, denote with $s$ the closed point and with $\eta$ the generic one. Let $X\rightarrow S$ be a proper, flat scheme of relative dimension $n$ and $Z\subset X_s$ be an equidimensional closed subscheme of the special fiber with dimension $d \lt n$. Given a point $t_{\eta}\in X(\eta)$ we know that this extends to a point $t\in X(S)$.

What can we say about subscheme $U\subset X_{\eta}$ of points specializing to $Z$?

Assume that $X\rightarrow S$ is very nice (like $X_{\eta}$ smooth and $X_s$ with semistable singularities) can I bound the dimension of $U$ in terms of the dimension of $Z$?

  • 2
    $\begingroup$ The set $U$ is not constructible in general (so not in particular not a subscheme). From the point of view of formal geometry, it is a closed subset but it is certainly not a Zariski closed subset in general. $\endgroup$ Apr 11, 2013 at 10:54
  • $\begingroup$ @Damian Rössler do you have any reference/counterexample for that? $\endgroup$
    – meti
    Apr 11, 2013 at 13:01

1 Answer 1


Suppose for simplicity that $\mathcal O$ is a henselian (e.g., complete) DVR (otherwise, a point of $X_\eta$ may have several specializations in $X_s$) with field of fractions $K$. Consider $X=\mathbb P^1_S$ parametriezd by a rational function $t$, and let $Z$ be the single point $t=0$ in the closed fiber. The the set $U$ is the open disc $|t|<1$ in the generic fiber $\mathbb P^1_K$ (in more algebraic terms, they correspond to maximal ideals $P(t)K[t]$ with $P(t)\in \mathcal O[t]$ monic and $P(0)\in \mathfrak m$ the maximal ideal of $\mathcal O$). So it is not a scheme, but a rigid analytic subspace.

  • $\begingroup$ It is worth pointing out that this kills the possibility of a naive dimension bound. There is a formal model of $\mathbb{P}^{n,an}_K$ with special fiber $\mathbb{P}^n_{\tilde{K}}$, and the preimage of a closed point is, again, an open ball. It should be true in some generality that $U$ in the question is an open subspace. $\endgroup$ Sep 26, 2013 at 18:44
  • $\begingroup$ @AndrewDudzik: the $U$ is always open as a rigid analytic subspace of $X_\eta^{an}$. $\endgroup$
    – Qing Liu
    Oct 23, 2013 at 21:42

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