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Let $\mathfrak{X}$ be a locally noetherian adic formal scheme over $\text{Spf}(A)$, with $A$ an $I$-adically complete and separated noetherian ring.

Suppose the mod $I$-fiber of $\mathfrak{X}$ is an algebraic $\text{Spec}(A/I)$-scheme equipped with an ample line bundle.

Grothendieck's formal existence theorem implies $\mathfrak{X}$ is algebraizable by some $X$ over $\text{Spec}(A)$, projective, and such that $I$-adic completion gives an equivalence of categories:

$${Coh}(X)\cong {Coh}(\mathfrak{X}).$$

This equivalence implies that every closed formal subscheme $\mathfrak{Z}\subset\mathfrak{X}$ uniquely algebraizes to a closed subscheme $Z\subset X$.

QUESTION. Can we algebraize open formal subschemes of $\mathfrak{X}$ too?

For example by declaring, for any open formal subscheme $\mathfrak{U}\subset\mathfrak{X}$, that its algebraization $U$ is the open complement of the unique closed subscheme $Z$ that algebraizes $\mathfrak{Z} := \mathfrak{X}-\mathfrak{U}$, where this last one is endowed with the reduced induced formal scheme structure.

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  • $\begingroup$ can you fix the typo in the title? $\endgroup$ – YCor Mar 18 '18 at 7:42
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    $\begingroup$ For a formal closed subscheme $(Z_n)_{n\geq 0}$ of the formal projective scheme scheme $(X_n, \mathcal{O}_{X_n}(1))_{n\geq 0}$ over $\text{Spec}(A/I^{n+1})$, the reduced induced scheme structure on $Z_n$ is typically just $Z_0$. So you would be defining $U$ to be the complement in $X$ of the closed fiber $Z_0$. In particular, if $A$ is a complete DVR with $I$ the maximal ideal, the generic fiber of $U$ equals the entire generic fiber of $X$. Is that what you intended? $\endgroup$ – Jason Starr Mar 18 '18 at 10:40
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    $\begingroup$ Yes. Less clear, and one aspect the question is meant to ask, is whether this construction is functorial, in the sense that maps $\mathfrak{U}_1\to\mathfrak{U}_2$ of formal open subschemes of $\mathfrak{X}$, algebraize to $U_1\to U_2$ in a unique way. Likely uniqueness fails, for instance, but even existence is not so clear. $\endgroup$ – user95222 Mar 18 '18 at 16:31
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    $\begingroup$ @AG2. There is a maximal open subscheme $U$ of $X$ whose associated formal scheme equals $\mathfrak{U}$. In that sense, the construction is functorial. However, as mentioned in my previous comment, $U$ is simply the complement in $X$ of $Z_0$. $\endgroup$ – Jason Starr Mar 19 '18 at 10:56

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