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

$\mathcal{G}_ n = \widetilde{M/I^ nM}.$

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

$(\mathcal{G}_ n|_ U) \longrightarrow (\mathcal{F}_ n)$

is injective in the abelian category $\textit{Coh}(U, I\mathcal{O}_ U)$.

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

$M_ n = \mathop{\mathrm{Im}}(M \longrightarrow H^0(U, \mathcal{F}_ n))$

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

$(M_ n)_ f \subset H^0(U, \mathcal{F}_ n)_ f = H^0(D(f), \mathcal{F}_ n)$

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

$\mathop{\mathrm{lim}}\nolimits (M/I^ nM)_ f = \mathop{\mathrm{lim}}\nolimits H^0(D(f), \mathcal{G}_ n) \to \mathop{\mathrm{lim}}\nolimits H^0(D(f), \mathcal{F}_ n)$

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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