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The following is Exercise 15.3 of Görtz-Wedhorn Algebraic Geometry I:

Let $S$ be a Dedekind scheme with function field $K$ and let $f: X\to S$ be a proper morphism of schemes. Then the canonical map $X(S)\to X(K)$ is a bijection.

For example, the map $X(S)\to X(K)$ being injective means ${\rm Spec}\ K\to S$ is an epimorphism of schemes. I'm no idea to prove it, even for $S$ affine. For subjectivity, I can only see (using valuation criterion for being proper) that a morphism ${\rm Spec}\ K\to X$ extends to a morphism ${\rm Spec}\ \mathscr{O}_{S, y}\to X$ ($y$ being the image of ${\rm Spec}\ K\to X\to S$ but don't know how to extends to the whole of $S$.

I'm also wondering if the maps need to respect morphisms to $S$, namely if we need to consider $X_S(S)\to X_S(K)$ instead of $X(S)\to X(K)$.

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  • $\begingroup$ Just to tell that I asked the authors and they really mean $X_S(S)\to X_S(K)$. $\endgroup$ – Lao-tzu Jun 29 at 13:49
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Injectivity is because $X$ is separated. The locus where two morphisms $S \to X$ agree is a closed subscheme and if it contains the generic point, it's everything.

For surjectivity, we can "spread out" the morphism $Spec K \to X$ to a morphism $U \to X$ from an open subset $U \subset S$ and then use the valuative criterion to fill in the finitely many missing point.

Edit: Here are some more details. So spreading it out is the general procedure where if we have an integral ring $R$ with fraction field $K$, a finite type scheme $X$ and a $K$ point of $X$, we can extend it to a $R_{f}$ point because there will only be finitely many denominators. For an easy example, suppose $X = \mathbb A^1$ and $R = \mathbb Z$ and the rational point is $t \to 1/2$. Then, there is an obvious way to think of this as a $\mathbb Z[1/2]$ point.

So given a rational point of $X$, we can extend this to a map from an open subset $U \subset S$ to $X$. Since $S$ is a Dedekind scheme, there are finitely many height one primes in the complement of $U$. Consider the local ring $R_{\mathfrak p}$ at one of these primes which is a DVR.

By the valuative criterion, we can get a $R_{\mathfrak p}$ point of $X$. Again, by spreading out, we can extend this to a map from $V \to X$ for $V$ another open subset of $S$ that contains $\mathfrak p$. On the other hand, the maps from $V \to X$ and $U\to X$ match on their intersection (by the same argument as for injectivity) and so we can extend this to a map $U\cup V \to X$.

Repeating this process for any points still left over, we get a map $S \to X$ that extends our rational map.

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  • $\begingroup$ Thanks! Injectivity agree! For surjectivity, I don't know how to "spread out" and "fill in the finitely many missing point", could you give more details? And do you use $X(S)\to X(K)$ or $X_S(S)\to X_S(K)$? $\endgroup$ – Lao-tzu Jun 20 at 18:02
  • $\begingroup$ OK, I think by "spread out" you mean extending morphisms as in Proposition 10.52 of Görtz-Wedhorn. $\endgroup$ – Lao-tzu Jun 20 at 18:09
  • $\begingroup$ I think it's useful to try and work it out yourself. Anyway, I will try and write more in a hour or two if you are still stuck (busy right now). $\endgroup$ – Asvin Jun 20 at 18:11
  • $\begingroup$ OK, thanks, I will try to work it out myself and tell you if I still stuck. $\endgroup$ – Lao-tzu Jun 20 at 18:12
  • $\begingroup$ I used some results on extending rational morphisms and schematically density to show $X_S(S)\to X_S(K)$ is a bijection, but I still don't know if you are meant $X(S)\to X(K)$. Even so, I'm still wondering how to use the valuative criterion to fill in the finitely many missing point; I can't make it work. It would be grateful if you can also add that to your answer. $\endgroup$ – Lao-tzu Jun 20 at 21:23

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