The Stacks project

Lemma 52.16.3. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Consider the following conditions:

  1. $(\mathcal{F}_ n)$ is in the essential image of the functor (,

  2. $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module,

  3. $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ V$-module for $U \cap Y \subset V \subset U$ open,

  4. $(\mathcal{F}_ n)$ is the completion of the restriction to $U$ of a coherent $\mathcal{O}_ X$-module,

  5. $(\mathcal{F}_ n)$ is the restriction to $U$ of the completion of a coherent $\mathcal{O}_ X$-module,

  6. there exists an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, I\mathcal{O}_ X)$ whose restriction to $U$ is $(\mathcal{F}_ n)$.

Then conditions (1), (2), (3), (4), and (5) are equivalent and imply (6). If $A$ is $I$-adically complete then condition (6) implies the others.

Proof. Parts (1) and (2) are equivalent, because the completion of a coherent $\mathcal{O}_ U$-module $\mathcal{F}$ is by definition the image of $\mathcal{F}$ under the functor ( If $V \subset U$ is an open subscheme containing $U \cap Y$, then we have

\[ \textit{Coh}(V, I\mathcal{O}_ V) = \textit{Coh}(U, I\mathcal{O}_ U) \]

since the category of coherent $\mathcal{O}_ V$-modules supported on $V \cap Y$ is the same as the category of coherent $\mathcal{O}_ U$-modules supported on $U \cap Y$. Thus the completion of a coherent $\mathcal{O}_ V$-module is an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Having said this the equivalence of (2), (3), (4), and (5) holds because the functors $\textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ U) \to \textit{Coh}(\mathcal{O}_ V)$ are essentially surjective. See Properties, Lemma 28.22.5.

It is always the case that (5) implies (6). Assume $A$ is $I$-adically complete. Then any object of $\textit{Coh}(X, I\mathcal{O}_ X)$ corresponds to a finite $A$-module by Cohomology of Schemes, Lemma 30.23.1. Thus we see that (6) implies (5) in this case. $\square$

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