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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 $R$?

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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 $f(a,b)=g(a,b)=0$.

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  • $\begingroup$ Could you explain why is it true the assertion in the second paragraph? Thanks $\endgroup$ Jul 22, 2021 at 16:19
  • $\begingroup$ @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 $\mathbb{P}^1(R)\to \mathbb{P}^1(K)$. $\endgroup$ Jul 24, 2021 at 7:28

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