# Varieties with everywhere good reduction that are isomorphic over every completion have isomorphic generic fibers

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, 2020 at 23:56
• 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 $B = R[\sqrt{-5}]$. May 26, 2020 at 0:52
• @R.vanDobbendeBruyn thank you, I modified the question a little bit
– user158636
May 26, 2020 at 2:10
• 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. May 26, 2020 at 3:06
• @R.vanDobbendeBruyn The OP insists that $X$ and $Y$ are smooth proper over $R$, so you can't use elliptic curves with bad reduction. May 27, 2020 at 15:10

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 $$\mathbb{Q}[\sqrt{2}]$$.