The evaluation map of Brauer-Manin obstruction is zero?

Question

Let $X$ be a proper $\mathbb{Z}_p$ scheme. The point $x: \mathbb{Z}_p \to X$ induces the map $(A \mapsto A(x))$, $\operatorname{Br}(X) \to \operatorname{Br}(\mathbb{Z}_p)$. Since $\operatorname{Br}(\mathbb{Z}_p) = 0$ we have $A(x)=0$. Now the valuative criterion of properness implies that the natural map $X(\mathbb{Z}_p) \to X(\mathbb{Q}_p)$ is bijective. This would make the Brauer-Manin obstruction vacuous, which isn't true. What went wrong?

Answer 1

There is no contradiction, which you can see by spelling things out a little bit more. If $A \in \operatorname{Br}(X)$, then indeed the evaluation map $\operatorname{ev}_A:X(\mathbb{Q}_p)\to \operatorname{Br}(\mathbb{Q}_p)$ vanishes for the reason you give. However, the Brauer-Manin obstruction is mostly used for Brauer classes $A' \in \operatorname{Br}(X_{\mathbb{Q}_p})$ which come from the generic fiber, so do not necessarily extend to the whole of $X$. The evaluation map for $A'$ no longer factors via $\operatorname{Br}(\mathbb{Z}_p)$, so the contradiction vanishes.