Let $S$ be a scheme and $f : X\to S$ be an $S$-scheme. This question asks for examples of maps of sets $X(S) \to X(S)$ that do not come from an $S$-scheme endomorphism of $X$, but that, roughly, specialize to maps $X_s(\kappa(s))\to X_s(\kappa(s))$ that do come from a $\kappa(s)$-scheme endomorphism of the fiber $X_s$ of $f$ at each closed point $s\in S$, with $\kappa(s)$ the residue field of $S$ at $s$.

More precisely, let $R$ be a strictly henselian complete discrete valuation ring and $s$ the closed (geometric) point in $S=\text{Spec}(R)$ and assume $f$ is proper and smooth.

Suppose there are:

  • a set-theoretic self map $a : X(R)\to X(R)$
  • a set-theoretic self map $a_0 : X_s(\kappa(s))\to X_s(\kappa(s))$
  • an endomorphism of $\kappa(s)$-schemes $\alpha_0 : X_s\to X_s$ such that $a_0$ is induced by $\alpha_0$ on $\kappa(s)$-points
  • calling $\pi$ the natural map $X(R) \to X_s(\kappa(s))$, $a$ and $a_0$ satisfy the condition $$a_0\circ\pi = \pi\circ a$$

Q1: Is there an endomorphism of $S$-schemes $\alpha : X\to X$ such that $a$ is induced by $\alpha$ on $S$-points and $\alpha_0 = \alpha\times_S\text{Spec}(\kappa(s))$?

To give a sense of what the question asks, for general $S$ this would mean whether and when we can interpolate scheme theoretic endomorphisms of the closed fibers of $f$ to an $S$-scheme endomorphism of $f$, and whether the condition that we can do so set-theoretically on $S$-points is enough.

I'd expect the answer to be no. I'm asking for a couple of concrete counterexamples, if any. In other words

Q2: Is there an explicit example of such an $f$ together with the data described before Q1 and with no such $\alpha$?

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    $\begingroup$ When you say "let $S$ be a strictly complete henselian discrete valuation ring", that should be $R$, right? $\endgroup$
    – LSpice
    Jun 22, 2021 at 19:58
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    $\begingroup$ The condition $\pi\circ a=a_0\circ \pi$ (where $\pi:X(R)\to X(\kappa(s))$ is the natural map) is necessary. Did you mean to assume about $a$ and $a_0$? $\endgroup$ Jun 22, 2021 at 20:44
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    $\begingroup$ We can take $X$ to be a curve of genus $\geq 2$, $a_0$ the identity map, $\alpha_0$ the identity, and $a$ any nontrivial permutation of the points reducing to a given one. Because there are infinitely many rational points, $\alpha_0$ is uniquely determined by $a_0$, and then $\alpha_0$ has a unique lift, which is the identity, so any non-identity $a$ works. $\endgroup$
    – Will Sawin
    Jun 22, 2021 at 22:22
  • $\begingroup$ Thanks, this already settles it $\endgroup$
    – user290895
    Jun 24, 2021 at 2:50

1 Answer 1


Choose a smooth projective $X/R$ of positive dimension, and pick a set-theoretic splitting $\varphi:X(\kappa(s))\to X(R)$ of the reduction map $\pi:X(R)\to X(\kappa(s))$. Take $a=\varphi\circ \pi$, and let $a_0$ and $\alpha_0$ be the identity. Then $a$ is constant on residue disks, so is locally constant for the analytic topology. If $a$ were induced by an algebraic morphism $\alpha:X\to X$, then $\alpha$ would have to be locally constant for the Zariski topology (I'm using the fact that the Zariski closure of an analytic neighborhood contains a Zariski neighbohood). This would imply the image of $\alpha$ is finite, which is impossible because $\#Im(a)=\# X(\kappa(s))=\infty$ since $X$ is positive dimensional and $\kappa(s)$ is separably closed.

  • $\begingroup$ Great this is exactly what I was looking for. Thanks $\endgroup$
    – user290895
    Jun 24, 2021 at 2:50

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