1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Just to tell that I asked the authors and they really mean $X_S(S)\to X_S(K)$. $\endgroup$
    – Lao-tzu
    Commented Jun 29, 2020 at 13:49

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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
    Commented Jun 20, 2020 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
    Commented Jun 20, 2020 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
    Commented Jun 20, 2020 at 18:11
  • $\begingroup$ OK, thanks, I will try to work it out myself and tell you if I still stuck. $\endgroup$
    – Lao-tzu
    Commented Jun 20, 2020 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
    Commented Jun 20, 2020 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.