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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?

  • Are $A,B$ implicitly assumed commutative? – YCor May 25 '20 at 23:56
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    This is false even for finite extensions! If $R = \mathbf Z_{(p)}$ and $R^\wedge = \mathbf Z_p$, then any two quadratic extensions of $R$ that split at $p$ become isomorphic to $\mathbf Z_p^2$ after tensoring with $R^\wedge$. For example consider $p = 3$ and $A = R[\sqrt{-2}]$ and $B = R[\sqrt{-5}]$. – R. van Dobben de Bruyn May 26 '20 at 00:52
  • @R.vanDobbendeBruyn thank you, I modified the question a little bit –  May 26 '20 at 02:10
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    Ah, that's a very different question. There are a lot of results about local-global principles, and in many cases it does not hold. For example, there are pairs $E, C$ of genus $1$ curves over $\mathbf Q$ where $E$ has a rational point (hence is an elliptic curve) and $C$ does not, but they become isomorphic over each completion. This is going from $\mathbf Q$ to all $\mathbf Q_p$, not from $\mathbf Z$ to all $\mathbf Z_p$, but spreading out over $\mathbf Z$ and inverting all primes of bad reduction (just to be safe) should give a counterexample to your current question. – R. van Dobben de Bruyn May 26 '20 at 03:06
  • @R.vanDobbendeBruyn The OP insists that $X$ and $Y$ are smooth proper over $R$, so you can't use elliptic curves with bad reduction. – Ariyan Javanpeykar May 27 '20 at 15:10
  • @AriyanJavanpeykar the question was edited after that comment –  May 27 '20 at 15:19
  • @crispr Ow I see. My bad. – Ariyan Javanpeykar May 27 '20 at 16:12
  • And your last edit changed the question into something of which we know very few interesting examples (at least when $R = \mathbf Z$). See for example this question and this one. – R. van Dobben de Bruyn May 27 '20 at 16:27
  • @R.vanDobbendeBruyn I thought maybe one can argue abstractly from the conditions –  May 27 '20 at 16:41
  • My example shows that if you allow $\mathbf Z[1/n]$ for some $n$ there are examples (namely just throw away the bad primes). There seem to be very intricate global properties of $\operatorname{Spec} \mathbf Z$ that make that type of example impossible. It's really quite mysterious. (But maybe over $\mathcal O_K$ it's easier to write something down; I don't know...) – R. van Dobben de Bruyn May 27 '20 at 16:49
  • Ok, there are elliptic curves over number fields with everywhere good reduction. If you can find one that moreover has nontrivial Tate–Shafarevich group, you get your example. These things surely must exist, but I don't know a way to find them except trying a lot of them. – R. van Dobben de Bruyn May 27 '20 at 18:08

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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).

Daniel Litt
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