Let $R$ be the ring of integers in a number field. Let $X$ and $Y$ be smooth and proper schemes over $R$. For a maximal ideal $\mathfrak{m}\subset R$ denote the completion of the localization at $\mathfrak{m}$ by $R^\wedge\_{\mathfrak{m}}$. Assume that for every $\mathfrak{m}$ there is an $R^\wedge\_{\mathfrak{m}}$-isomorphism $X\otimes R^\wedge\_{\mathfrak{m}}\approx Y\otimes R^\wedge\_{\mathfrak{m}}$. Are the generic fibers of $X$ and $Y$ isomorphic?



Varieties with everywhere good reduction that are isomorphic over every completion have isomorphic generic fibers
Here's an explicit example. Let $R=\mathbb{Z}[\sqrt{2}]$, let $X=\mathbb{P}^1\_R$, and let $Y$ be the smooth projective conic defined by the equation $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\sqrt{2})xz=0.$$
I claim this conic has good reduction everywhere; as smooth conics over the ring of integers of a $p$-adic field are always split, this implies that $X$ and $Y$ are everywhere locally isomorphic. On the other hand, I claim that $Y$ has no rational points over $\mathbb{Q}[\sqrt{2}]$, whereas $X$ evidently does.


To see these claims, note that away from $2$, $Y$ can be obtained from the conic $$x^2+y^2+(3-2\sqrt{2})z^2=0$$ via the projective transformation $$x\mapsto x+y, y\mapsto y+z, z\mapsto x+z.$$ This immediately implies that $Y$ has good reduction everywhere away from the prime above $2$, and one checks directly that $Y$ has good reduction at this prime. But this simpler conic evidently has no rational points, because $3-2\sqrt{2}$ is totally positive. Thus the proof is complete.


Let me briefly indicate how one can see such examples must exist from "pure thought." Suppose one decides to look for conics with this property; these are classified by $2$-torsion elements of the Brauer group of the ground field. The fact that these conics must be smooth and proper over the ring of integers of the ground field means that their local invariants must be zero at all finite places; hence we need a number field with at least $2$ real places to have a chance. But any such number field will do; we can simply take $X$ to be $\mathbb{P}^1$ and $Y$ to be a conic ramified at exactly two real places. All I did is make this reasoning explicit in the case of $\mathbb{Q}[\sqrt{2}]$.


One final remark; R. van Dobben de Bruyn suggests looking for elliptic curves with everywhere good reduction and non-trivial Tate-Shafarevich group; this is simply the "genus zero" version of that idea, which is a bit easier (since we understand the Brauer group much better than the Tate-Shafarevich group).


