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Let $(R, \mathfrak{m})$ be a noetherian local ring, and let $\hat R$ be its $\mathfrak{m}$-adic completion. Extension of scalars allows one to transform an $R$-scheme into an $\hat R$-scheme. Is this association surjective, up to isomorphism? More precisely, given a finite type scheme $X \to \mathrm{Spec} \ \hat R$, does there exist an $R$-scheme $Y$ and an $\hat R$-isomorphism $$X \cong Y \times_{\mathrm{Spec} \ R} \mathrm{Spec} \ \hat R\ \ ? $$

My interest stems from the case where $R$ is the ring of germs of holomorphic functions at the origin in $\mathbb{C}$. If $t$ is the usual coordinate on $\mathbb{C}$, then we have a canonical isomorphism $\hat R \cong \mathbb{C}[[t]]$, where the latter is the ring of formal power series in $t$. Is every formal family $X \to \mathbb{C}[[t]]$ isomorphic to an analytic family? I suppose it may be relevant that $R$ is henselian in my example (by the implicit function theorem).

Update: Let $K$ be the field of fractions of $R$, and let $\hat K$ be the field of fractions of $\hat R$. Suppose that there exists a $K$-scheme $Y_K$ and a $\hat K$ isomorphism $X_{\hat K} \cong Y_K \times_{\mathrm{Spec} \ K} \mathrm{Spec \ \hat K}$. Can we conclude that there exists an $R$-scheme $Y$ as above, which necessarily has generic fiber $Y_K$? That is, if we already know the generic fiber is algebraic, can we arrange for the whole scheme to be algebraic?

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An elliptic curve over $\widehat{R}$ with $j$-invariant not in $R$ is a counterexample. (Concretely, if the $j$-invariant over $\mathbf{C}[\![t]\!]$ has radius of convergence 0 then it cannot arise from an analytic family of elliptic curves.) –  user30180 Jun 15 '13 at 0:14
    
Nice! This example evaporates if I add an extra hypothesis on the generic fiber of $X$, so I will update my question. –  Xander Faber Jun 15 '13 at 8:31
    
For any noetherian pair $(R,\mathfrak{m})$, one knows: giving a quasicoherent (qc) sheaf $F$ on $S := \mathrm{Spec}(R)$ is the same as giving a qc sheaf $\hat{F}$ on $\hat{S} := \mathrm{Spec}(\hat{R})$, a qc sheaf $F_U$ on $U = S - V(\mathfrak{m})$, and an isomorphism between the two over $U \times_S \hat{S}$. The same comment applies to "quasicoherent sheaves" replaced by "affine/quasiprojective morphisms." –  anon Jun 15 '13 at 8:51
    
Anon - reference? –  Xander Faber Jun 15 '13 at 9:13
    
stacks.math.columbia.edu/tag/05E5 –  anon Jun 15 '13 at 9:25
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This is really just ayanta's example again: even with the condition on the generic fiber, the answer is still "no". Let $R$ be the local ring of $k[s,t]$ at the maximal ideal $\langle s,t\rangle$. Begin with $Y_R$ being $\mathbb{P}^1_R$ -- this is not the final scheme. For the base change $Y_{\widehat{R}}$ of $Y_R$, restrict over the closed subscheme $Z(s) = \text{Spec} \widehat{R}/s\widehat{R}$. Let $D_s \subset \mathbb{P}^1_{Z(s)}$ be a Cartier divisor that is flat over $Z(s)$ of relative degree $4$ and whose corresponding $j$-invariant, $j(D_s)\in \widehat{R}/s\widehat{R}$, is not the pullback of an element in $R/sR$. Now let $X$ be the blowing up of $\mathbb{P}^1_{\widehat{R}}$ along the codimension $2$ closed subscheme $D_s$.

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Great example! If I understand correctly, the fact that $R$ has dimension $>1$ means that you can arrange a descent obstruction without touching the generic fiber of $\mathbb{P}^1_R$. –  Xander Faber Jun 19 '13 at 1:36
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In Jason's example, the obstruction is that even the restriction above $\mathrm{Spec}(\widehat{R}/s\widehat{R})$ does not descend. Now let $U\subset \mathrm{Spec}({R})$ (resp. $\widehat{U}\subset \mathrm{Spec}(\widehat{R})$) be the complement of the closed point. If one assumes that the restriction of $X$ to $\widehat{U}$ descends to $U$, then the answer is yes in many cases. This is discussed in my paper "Un problème de descente", Bull. Soc. Math. France 124 (1996), 559-585.

Of course, in Xander's original case, we have $U=\mathrm{Spec}(K)$.

Unfortunately I have no time right now to elaborate (or even check the statements!). If I remember correctly, there is a descent datum on $X$ relative to $R\to\widehat{R}$, so the problem then becomes an effectivity question.

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@Laurent Many thanks! This article is exactly what I need! –  Xander Faber Jun 19 '13 at 1:32
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