# Valuative criterion for properness for general integral domains

Consider the following assertion:

Let $$R$$ be an integral domain with fraction field $$K$$. Let $$X$$ be a scheme and $$X \to \operatorname{Spec}R$$ a proper morphism. Then the natural map $$X(R)\to X (K)$$ is a bijection.

As usual, I use the notation $$X(R):=\text{Hom}(\operatorname{Spec}R,X)$$.

I know the assertion is true for Dedekind domains. But is it true for more general integral domains?

Are there counterexamples, for example, for $$X$$ the projective line over $$R$$?

One easy counterexample for the case $$X=\mathbb{P}^1$$ is when $$R=K[x,y]$$, $$K$$ any field. Suppose we have a point $$[f(x,y):g(x,y)]\in \mathbb{P}^1(K(x,y))$$, where we can suppose that $$f(x,y)$$ and $$g(x,y)$$ do not have factors in common, using that $$K[x,y]$$ is a UFD. This representation is unique (up to non-zero constants in $$K$$).
This point lifts then to a point in $$\mathbb{P}^1(K[x,y])$$ if and only if the ideal generated by $$f(x,y)$$ and $$g(x,y)$$ is the total ideal, something that will happen "rarely" (if the field is "big", e.g. algebraically closed); for example, it cannot be true if there is an $$(a,b)\in K^2$$ such that $$f(a,b)=g(a,b)=0$$.
• @Nulhomologous WIth notation as above, for example $[x: y]$ defines a point $\mathbb{P}^1(K)$ since $x$ is a unit in $K$ but not in the image of the map $\mathbb{P}^1(R)\to \mathbb{P}^1(K)$. Jul 24, 2021 at 7:28