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This question might be deemed totally unanswerable, unless there is an obvious counterexample. Answers to either effect would be welcome.

Question. Let $X$ be a finite-type scheme over $\mathbb{Z}[[T]]$. For any prime $p \in \mathbb{Z}$, let $X_p$ be the $\mathbb{Z}_p$-scheme obtained by base-changing $X$ along the obvious map $\mathbb{Z}[[T]] \rightarrow \mathbb{Z}_p$. Suppose that $X_p(\mathbb{Z}_p) \neq \emptyset$ for all $p$, do we necessarily have $X(\mathbb{Z}[[T]])\neq\emptyset$?

This question was motivated by the case where $X$ is a projective conic, where the answer is affirmative.

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    $\begingroup$ Won't you get a counterexample by taking $X$ defined over $\mathbb{Z}$ and failing the Hasse principle? $\endgroup$
    – Felipe Voloch
    Oct 23, 2014 at 1:21
  • $\begingroup$ Thank you Felipe, I guess you're absolutely right. I was looking for a "non-isotrivial" counterexample, but I had completely missed this possibility. $\endgroup$
    – Milton
    Oct 23, 2014 at 1:23
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    $\begingroup$ Moreover, when $X$ if of finite type over $\mathbf Z$, the usual Hasse principle (for $X_{\mathbf Q}$) could hold, while the variant for integral points fails. Take for example a non-principal ring of integers $A$ (eg, $A=\mathbf Z[\sqrt -5]$), $L$ a non-trivial invertible $A$-module, and $X=\mathbf V(L)\setminus\{0\}$ (the associated $\mathbf G_m$-torsor). Then $X$ has local integral points, but no $A$-points. (If you insist, take Weil restriction to $\mathbf Z$.) $\endgroup$
    – ACL
    Oct 23, 2014 at 16:15

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