In this question I'd like to examine some properties of universally closed morphisms. The question is self-contained. It can also be seen as a follow-up to this question.

Let $R$ be a discrete valuation ring, $X$ and $S$ two $R$-flat and universally closed $R$-schemes.

Let $f : X\to S$ be a universally closed morphism. Assume the special fiber of $f$ is an isomorphism.

  • If $X$ and $S$ are reduced, is $f$ an isomorphism?
  • If $X$ and $S$ have reduced fibers, is $f$ an isomorphism?

Without either of the reducedness assumptions, the answer is "no", as shown by the example provided here.

My expectation is that the answer is still "no" to both questions, but an example seems subtle to produce.

Remarks In these remarks we further assume $f$ is flat, to see what more can be said in this case. After all, should the questions have positive answer, $f$ would have to be flat. The remarks are listed increasing the strength of the additional assumptions, to see what can be said in each ever-more-special case.

  • From flatness we deduce $f$ is surjective, since generizations lift along $f$ (by flatness), and every point in $S$ specializes to a point in the special fiber (by universal closedness of $S$, if $s\in S$ is contained in the generic fiber, it maps to the generic point in $\text{Spec}(R)$. This specializes to the closed point in $\text{Spec}(R)$, and specializations lift along $S\to \text{Spec}(R)$ by universal closedness, so there is some $s_0\in S$ contained in the special fiber, to which $s$ specializes).

  • If $f$ is, in addition, separated and locally of finite type, then it is also proper, since $f$ is quasi-compact by universal closedness. It is therefore an isomorphism (it is quasi-finite, hence finite, hence finite flat of degree $1$). The question is interesting only if $f$ is not locally of finite type.

  • To summarize, the case when $f$ is flat and separated is already unanswered but one has access to more info on $f$. A non-separated example, flat or not, to answer the question in the negative, would be already what I'm looking for. A flat separated example, if any, would be the best way to settle this puzzle, and I expect it is very subtle to find.

  • $\begingroup$ I keep trying to find nonseparated examples, but they do not quite work. $\endgroup$ Jul 10, 2021 at 21:37
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    $\begingroup$ What if you glue two copies of $\mathbf P^1_{\mathbf Z_p}$ away from a copy of $\operatorname{Spec} \mathbf Z_p$, and then map to two copies of $\mathbb P^1_{\mathbf Z_p}$ glued away from a point in the special fiber? $\endgroup$
    – Will Sawin
    Jul 11, 2021 at 0:39
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    $\begingroup$ Why is $f$ flat? $\endgroup$
    – Johan
    Jul 12, 2021 at 16:46

1 Answer 1


Let $R$ be $\mathbb Z_p$ (or any other dvr).

Let $S$ be obtained by gluing two copies of $\mathbb P^1_{\mathbb Z_p}$ away from the $0$-point in the special fiber, i.e. away from the vanishing locus of the ideal $(p,x)$ in local coordinates away from $\infty$.

Let $X$ be obtained by gluing two copies of $\mathbb P^1_{\mathbb Z_p}$ away from the $0$-section, i.e. away from the vanishing locus of the ideal $x$.

These are obtained by gluing two flat and universally closed schemes along an open set, hence flat and universally closed.

Then we can map $X \to S$ by mapping each copy of $\mathbb P^1_{\mathbb Z_p}$ to a different copy of $\mathbb P^1_{\mathbb Z_p}$, which is compatible with the gluing since we glue along a larger open set in $S$ than in $X$.

This map is an isomorphism in the special fiber since the constructions of $X$ and $S$, restricted to the special fiber, are identical.

But it is manifestly not an isomorphism, e.g. because it makes a non-separated generic fiber to a separated one.

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    $\begingroup$ In the post title and in the previous question, the OP says that the morphism should be universally closed (the OP does not say this in the body of this post). Your morphism is not closed. $\endgroup$ Jul 11, 2021 at 18:37
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    $\begingroup$ @JKR It would have to be a pretty significant adjustment to make it not locally of finite presentation. I'm not sure how that could work... $\endgroup$
    – Will Sawin
    Jul 11, 2021 at 22:00

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