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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.

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?

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]).

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4 Answers 4

up vote 6 down vote accepted

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.)

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)).

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Even if your question has a negative answer in general, it is related to Artin's approximation theorem (see Publ. IHES, vol 36), which can interpreted as follows. Let S be nice scheme (classically, the spectrum of a field or of a Dedekind domain whose field of functions is a global field, but, by Popescu's desingularisation theorem, any excellent noetherian scheme is eligible). Consider two closed immersions X<-Z->Y of S-schemes of finite type. If X=Y as formal schemes, after completion along Z, then there exist Nisnevich coverings X<-V->Y. So, locally for the Nisnevich topology, X=Y.

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In David S.'s example, I don't see a compatible sytem of isomorphisms $X_n\to Y_n$. Here is a similar example with the required properties.

Let $A=k[[t]]$, let $Y={\rm Spec} A[x]]$ be the affine line over $A$ and $X=Y\setminus \{ y_0 \}$, where $y_0$ is the closed point of $Y$ corresponding to the ideal $(1-tx)A[x]$ (the quotient ring is $A[1/t]=k((t))$). Then the canonical inclusion $X\to Y$ induces clearly a compatible system of isomorphisms $X_n\to Y_n$ for all $n\ge 1$ (they are actually identities), and of course $X$ is not isomorphic to $Y$.

The reason behind is that the point $y_0$ is invisible in all the formal neighborhood of the closed fibler of $Y\to {\rm Spec } A$.

In general, a compatible system of isomorphisms $X_n\to Y_n$ induces an isomorphism between the formal schemes $\hat{X}\to \hat{Y}$. When $X, Y$ are projective, one can use GAGA to show that this isomorphism comes from an isomorphism $X\to Y$. This gives an alternative proof to the use of ${\rm Hom}(X, Y)$.

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In fact, they can be chosen to be compatible; it can be shown from the infinitesimal lifting principle that any trivialization on $k[[t]]/t^n$ can be lifted to one on $k[[t]]/t^{n+1}$. But I like your example as well. –  David Speyer Jan 27 '10 at 16:47
    
I see. Thanks! Probably one can even write explicitly the isomorphisms. The result could also be explained with rigid analytic geometry on $k((t))$. –  Qing Liu Jan 31 '10 at 21:27

I feel like I have seen some theorem saying that this result is true for projective families, and maybe even for proper families. If anyone knows a reference, please post it.

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I don't know a reference but I think it follows from the fact that the hom scheme Hom_A(X,Y) is represented by a scheme -- the iso's between X_n and Y_n correspond to maps A/m^n+1 \to Hom_A(X,Y), and any affine neighborhood Spec B of the image (which is a single point) completes to a map B \to A, i.e. lifting the iso's to a map X \to Y over A. –  David Zureick-Brown Oct 27 '09 at 6:15
    
Actually, in the proper case the hom scheme is an algebraic space and I think you can still make an argument like this work. –  David Zureick-Brown Oct 27 '09 at 6:17
    
This question, and the related "existence" problem (if you have a formal subscheme, does it come from a subscheme ?) was studied by Grothendieck in FGA. See Illusie's survey in the book "FGA explained" (or math.u-psud.fr/~illusie/illusie_trieste.pdf), e.g corollary 4.7 –  Simon Pepin Lehalleur Aug 10 '10 at 0:57

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