Let $p: \mathcal{X} \rightarrow \text{Spec } \mathcal{O}_K$ be a normal proper Artin stack with finite diagonal. A $K$-rational point is by definition a section $x: \text{Spec}(K) \rightarrow \mathcal{X}$ of $p$ over the generic point of $\text{Spec } \mathcal{O}_K$, and an integral point is a section $\bar{x}: \text{Spec } \mathcal{O}_K \rightarrow \mathcal{X}$ of $p$.
Now if $\mathcal{X}$ were to be a proper scheme, then every rational point extends uniquely to an integral point. Indeed, the valuative criterion gives morphisms from every division ring that we can patch up to a morphism from $\text{Spec } \mathcal{O}_K$.
I am currently reading the work "Heights on stacks and a generalized Batyrev–Manin–Malle
conjecture" and there it is claimed that the above statement is no longer true for stacks, but it is still true after a (possibly) ramified extension of $\text{Spec } \mathcal{O}_K$. This leads to my two questions:
- Why is this not true for stacks? Although a specific counterexample would be great, I am also happy with some intuition why this ought to fail for stacks.
- Why is it still true after a (possibly) ramified extension of $\text{Spec } \mathcal{O}_K$? In this case I am mostly interested in a precise argument.
Best Answer
$\DeclareMathOperator{Spec}{Spec}$ For 1. take $\mathcal{X} = B \mu_2$ to be the classifying stack of $\mu_2$ over $\mathbb{Z}$. For any $\mathbb{Z}$-scheme $S$ we have $\mathcal{X}(S) = \mathrm{H}^1(S, \mu_2)$. Thus we have $\mathcal{X}(K) = K^\times/K^{\times 2}$, but this doesn't equal $\mathcal{X}(\mathcal{O}_K)$. For example, if $\mathcal{O}_K$ is a PID then $\mathcal{X}(\mathcal{O}_K) = \mathcal{O}_K^\times/\mathcal{O}_K^{\times2}$.
For 2., this is the valuative criterion for properness of a morphism of stacks (see https://stacks.math.columbia.edu/tag/0CLY, https://stacks.math.columbia.edu/tag/0CQL, and https://stacks.math.columbia.edu/tag/0CL9). Namely, for stacks one only stipulates the existence of an extension of a map from valuation ring after a possible base change of the ring, unlike for the case of schemes where existence of an extension is stipulated for the actual ring.
You can check this property yourself in the above example of $B \mu_2$.