Lemma 30.24.3. Let $A$ be Noetherian ring complete with respect to an ideal $I$. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a projective morphism. Let $\mathcal{I} = I\mathcal{O}_ X$. Then the functor (30.23.3.1) is an equivalence.

Proof. We have already seen that (30.23.3.1) is fully faithful in Lemma 30.24.1. Thus it suffices to show that the functor is essentially surjective.

We first show that every object $(\mathcal{F}_ n)$ of $\textit{Coh}(X, \mathcal{I})$ is the quotient of an object in the image of (30.23.3.1). Let $\mathcal{L}$ be an $f$-ample invertible sheaf on $X$. Choose $d_0$ as in Lemma 30.24.2. Choose a $d \geq d_0$ such that $\mathcal{F}_1 \otimes \mathcal{L}^{\otimes d}$ is globally generated by some sections $s_{1, 1}, \ldots , s_{t, 1}$. Since the transition maps of the system

$H^0(X, \mathcal{F}_{n + 1} \otimes \mathcal{L}^{\otimes d}) \longrightarrow H^0(X, \mathcal{F}_ n \otimes \mathcal{L}^{\otimes d})$

are surjective by the vanishing of $H^1$ we can lift $s_{1, 1}, \ldots , s_{t, 1}$ to a compatible system of global sections $s_{1, n}, \ldots , s_{t, n}$ of $\mathcal{F}_ n \otimes \mathcal{L}^{\otimes d}$. These determine a compatible system of maps

$(s_{1, n}, \ldots , s_{t, n}) : (\mathcal{L}^{\otimes -d})^{\oplus t} \longrightarrow \mathcal{F}_ n$

Using Lemma 30.23.3 we deduce that we have a surjective map

$\left((\mathcal{L}^{\otimes -d})^{\oplus t}\right)^\wedge \longrightarrow (\mathcal{F}_ n)$

as desired.

The result of the previous paragraph and the fact that $\textit{Coh}(X, \mathcal{I})$ is abelian (Lemma 30.23.2) implies that every object of $\textit{Coh}(X, \mathcal{I})$ is a cokernel of a map between objects coming from $\textit{Coh}(\mathcal{O}_ X)$. As (30.23.3.1) is fully faithful and exact by Lemmas 30.24.1 and 30.23.4 we conclude. $\square$

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