Let (A,m) be a complete local Noetherian ring and let X and Y be two schemes of finite type over A (and flat over A). Let Xn and Yn be the reductions of X and Y mod mn+1.



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> **Question**: Suppose there is a compatible system of isomorphisms between Xn and Yn (for all n). Does there necessarily exist an isomorphism between X and Y over A?
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In other words, suppose the formal schemes \hat{X} and \hat{Y} are isomorphic; are X and Y isomorphic? 


*Remark*: The answer is `no' if we drop flatness (you can just stick an extra component over the generic fiber) or finite type (A[t] vs. A{t} = the completion of A[t]).



Can isomorphisms of schemes be constructed on formal neighborhoods?
No. Let A be k[[t]]. Let X be A^1 \setminus {-1,0,1} and Y be A^1 \setminus {1,t,-1}. In explicit equations, X = Spec k[[t]][x, y]/y(x-1)x(x+1)-1 and Y = Spec k[[t]][x, y]/y(x-1)(x-t)(x+1)-1.


Over k[[t]]/t^{n+1}, the reductions of X and Y are isomorphic because all infinitesimal deformations of a smooth affine scheme are trivial. (See corollary 4.7 in [Hartshorne's notes on deformation theory](http://math.berkeley.edu/~robin/math274root.pdf).) 


However, X and Y are not isomorphic because the two fibers over the general point are not: For any field K, if we have P\_K^1 \setminus {a,b,c,d} for {a,b,c,d} \in P^1(K), then the cross ratio of a,b,c and d is a well defined element of K. In particular, this applies when K=k((t)).


