We continue the discussion started in Section 52.16. This section can be skipped on a first reading.
Lemma 52.17.1. In Situation 52.16.1. Let $(\mathcal{F}_ n) \to (\mathcal{F}'_ n)$ be a morphism of $\textit{Coh}(U, I\mathcal{O}_ U)$ whose kernel and cokernel are annihilated by a power of $I$. Then
$(\mathcal{F}_ n)$ extends to $X$ if and only if $(\mathcal{F}'_ n)$ extends to $X$, and
$(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module if and only if $(\mathcal{F}'_ n)$ is.
Proof. Part (2) follows immediately from Cohomology of Schemes, Lemma 30.23.6. To see part (1), we first use Lemma 52.16.6 to reduce to the case where $A$ is $I$-adically complete. However, in that case (1) reduces to (2) by Lemma 52.16.3. $\square$
The following two lemmas where originally used in the proof of Lemma 52.16.10. We keep them here for the reader who is interested to know what intermediate results one can obtain.
Lemma 52.17.2. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. If the inverse system $H^0(U, \mathcal{F}_ n)$ has Mittag-Leffler, then the canonical maps are surjective for all $n$ where $M$ is as in (52.16.6.1).
\[ \widetilde{M/I^ nM}|_ U \to \mathcal{F}_ n \]
Proof. Surjectivity may be checked on the stalk at some point $y \in Y \setminus Z$. If $y$ corresponds to the prime $\mathfrak q \subset A$, then we can choose $f \in \mathfrak a$, $f \not\in \mathfrak q$. Then it suffices to show
is surjective as $D(f)$ is affine (equality holds by Properties, Lemma 28.17.1). Since we have the Mittag-Leffler property, we find that
for some $m \geq n$. Using the long exact sequence of cohomology we see that
Since $U = X \setminus V(\mathfrak a)$ this $H^1$ is $\mathfrak a$-power torsion. Hence after inverting $f$ the cokernel becomes zero. $\square$
Lemma 52.17.3. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $M$ be as in (52.16.6.1). Set If the limit topology on $M$ agrees with the $I$-adic topology, then $\mathcal{G}_ n|_ U$ is a coherent $\mathcal{O}_ U$-module and the map of inverse systems is injective in the abelian category $\textit{Coh}(U, I\mathcal{O}_ U)$.
\[ \mathcal{G}_ n = \widetilde{M/I^ nM}. \]
\[ (\mathcal{G}_ n|_ U) \longrightarrow (\mathcal{F}_ n) \]
Proof. Observe that $\mathcal{G}_ n$ is a quasi-coherent $\mathcal{O}_ X$-module annihilated by $I^ n$ and that $\mathcal{G}_{n + 1}/I^ n\mathcal{G}_{n + 1} = \mathcal{G}_ n$. Consider
The assumption says that the inverse systems $(M_ n)$ and $(M/I^ nM)$ are isomorphic as pro-objects of $\text{Mod}_ A$. Pick $f \in \mathfrak a$ so $D(f) \subset U$ is an affine open. Then we have
Equality holds by Properties, Lemma 28.17.1. Thus $\widetilde{M_ n}|_ U \to \mathcal{F}_ n$ is injective. It follows that $\widetilde{M_ n}|_ U$ is a coherent module (Cohomology of Schemes, Lemma 30.9.3). Since $M \to M/I^ nM$ is surjective and factors as $M_{n'} \to M/I^ nM$ for some $n' \geq n$ we find that $\mathcal{G}_ n|_ U$ is coherent as the quotient of a coherent module. Combined with the initical remarks of the proof we conclude that $(\mathcal{G}_ n|_ U)$ indeed forms an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Finally, to show the injectivity of the map it suffices to show that
is injective, see Cohomology of Schemes, Lemmas 30.23.2 and 30.23.1. The injectivity of $\mathop{\mathrm{lim}}\nolimits (M_ n)_ f \to \mathop{\mathrm{lim}}\nolimits H^0(D(f), \mathcal{F}_ n)$ is clear (see above) and by our remark on pro-systems we have $\mathop{\mathrm{lim}}\nolimits (M_ n)_ f = \mathop{\mathrm{lim}}\nolimits (M/I^ nM)_ f$. This finishes the proof. $\square$
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