# A question about Dedekind schemes and proper morphisms

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 $$X(S)\to X(K)$$.

• Just to tell that I asked the authors and they really mean $X_S(S)\to X_S(K)$. – Lao-tzu Jun 29 at 13:49

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.

• 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 $X_S(S)\to X_S(K)$? – Lao-tzu Jun 20 at 18:02
• OK, I think by "spread out" you mean extending morphisms as in Proposition 10.52 of Görtz-Wedhorn. – Lao-tzu Jun 20 at 18:09
• 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). – Asvin Jun 20 at 18:11
• OK, thanks, I will try to work it out myself and tell you if I still stuck. – Lao-tzu Jun 20 at 18:12
• 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. – Lao-tzu Jun 20 at 21:23