• If $f : X \to Y$ is proper, then specializations lift along $f$, and uniquely.

(This means, if $R$ is a discrete valuation ring with fraction field $K$ and I choose a factorization $\text{Spec}K \to \text{Spec} R \to Y$ through $X$, I get a unique factorization of $\text{Spec} R \to Y$ through $X$ inducing it. )

  • If $f : Z \to Y$ is flat, then generizations lift (Stacks Project 040F).

(Let $k$ be the residue field of the DVR $R$ from before. Subtlety: I have to allow $k \subseteq k'$ an extension and choose a factorization $\text{Spec}k' \to \text{Spec} k \in \text{Spec} R \to Y$ through $X$. Then I get a DVR $R'$ with residue field $k'$ and a factorization of $\text{Spec}R' \to \text{Spec} R \to Y$ through $X$. This is ''going down.'')

  • Is there a condition on a scheme/map so that generizations lift uniquely? You have to exclude the trivial generization, which seems ad hoc. I'm not sure how to make this precise. Here is an example to highlight the difference:

Consider $S = \mathbb{A}^1$ and $T = (xy = 0) \subseteq \mathbb{A}^2$ over a base field $\text{Spec} L$. Both are flat over the base automatically. Yet for $S$, there's only one nontrivial generization of any closed point, the generic point. For $T$, the generic points of each component both generize the point $x = y = 0$.

Generizations happen within an open neighborhood of a point, so it seems like I'm asking for the topological space to be locally irreducible.

(Due diligence statement: I tried a handful of different searches to find something like this question. Hensel's Lemma is related. The phrase "generalizations lift" yields too many google responses, so my apologies if I've missed something.)

edit: Strong conditions on dimensions of points to be generized seem necessary. If we require such lifting for codimension-one points, then ``local irreducibility'' seems right. On $\mathbb{P}^2$, there should be infinitely many curves specializing through each point given by lines $\mathbb{P}^1$ through the origin. This may indicate that the expectation is unreasonable or uninteresting, but I'd still like to hear if the notion has any utility with some dimensional restrictions.


1 Answer 1


Your condition is equivalent to stating that, for any map $\operatorname{Spec} R \to X \times_Y X$, if the special point factors through the diagonal $\Delta: X \to X \times_Y X$, then the whole map factors through the diagonal.

A sufficient condition is clearly that the diagonal is an open immersion, because if the special point is in an open set than the whole local ring will be. This is one of the conditions for a morphism to be unramified, so being unramified is also a sufficient condition.

I'm not sure, but I think it's probably not a necessary condition.

  • $\begingroup$ If X \to Y is quasiseparated, the diagonal is compact and is open if and only if stable under generization (stacks 0542). If X \to Y is further locally of finite type, unramified and open diagonal are equivalent 02GE. I'm fine with these minor finiteness hypotheses, and I think yours is the answer. $\endgroup$
    – Leo Herr
    Oct 7, 2021 at 17:26

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