Proof. We will use the equivalence of categories of Cohomology of Spaces, Lemma 69.12.8 without further mention in the proof of the theorem. By Lemma 76.42.6 the functor is fully faithful. Thus we need to prove the functor is essentially surjective.

Consider the collection $\Xi$ of quasi-coherent sheaves of ideals $\mathcal{K} \subset \mathcal{O}_ X$ such that the statement holds for every object $(\mathcal{F}_ n)$ of $\textit{Coh}_{\text{support proper over }A}(X, \mathcal{I})$ annihilated by $\mathcal{K}$. We want to show $(0)$ is in $\Xi$. If not, then since $X$ is Noetherian there exists a maximal quasi-coherent sheaf of ideals $\mathcal{K}$ not in $\Xi$, see Cohomology of Spaces, Lemma 69.13.1. After replacing $X$ by the closed subscheme of $X$ corresponding to $\mathcal{K}$ we may assume that every nonzero $\mathcal{K}$ is in $\Xi$. Let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}_{\text{support proper over }A}(X, \mathcal{I})$. We will show that this object is in the essential image, thereby completing the proof of the theorem.

Apply Chow's lemma (Lemma 76.40.5) to find a proper surjective morphism $f : Y \to X$ which is an isomorphism over a dense open $U \subset X$ such that $Y$ is H-quasi-projective over $A$. Note that $Y$ is a scheme and $f$ representable. Choose an open immersion $j : Y \to Y'$ with $Y'$ projective over $A$, see Morphisms, Lemma 29.43.11. Let $T_ n$ be the scheme theoretic support of $\mathcal{F}_ n$. Note that $|T_ n| = |T_1|$, hence $T_ n$ is proper over $A$ for all $n$ (Morphisms of Spaces, Lemma 67.40.7). Then $f^*\mathcal{F}_ n$ is supported on the closed subscheme $f^{-1}T_ n$ which is proper over $A$ (by Morphisms of Spaces, Lemma 67.40.4 and properness of $f$). In particular, the composition $f^{-1}T_ n \to Y \to Y'$ is closed (Morphisms, Lemma 29.41.7). Let $T'_ n \subset Y'$ be the corresponding closed subscheme; it is contained in the open subscheme $Y$ and equal to $f^{-1}T_ n$ as a closed subscheme of $Y$. Let $\mathcal{F}_ n'$ be the coherent $\mathcal{O}_{Y'}$-module corresponding to $f^*\mathcal{F}_ n$ viewed as a coherent module on $Y'$ via the closed immersion $f^{-1}T_ n = T'_ n \subset Y'$. Then $(\mathcal{F}_ n')$ is an object of $\textit{Coh}(Y', I\mathcal{O}_{Y'})$. By the projective case of Grothendieck's existence theorem (Cohomology of Schemes, Lemma 30.24.3) there exists a coherent $\mathcal{O}_{Y'}$-module $\mathcal{F}'$ and an isomorphism $(\mathcal{F}')^\wedge \cong (\mathcal{F}'_ n)$ in $\textit{Coh}(Y', I\mathcal{O}_{Y'})$. Let $Z' \subset Y'$ be the scheme theoretic support of $\mathcal{F}'$. Since $\mathcal{F}'/I\mathcal{F}' = \mathcal{F}'_1$ we see that $Z' \cap V(I\mathcal{O}_{Y'}) = T'_1$ set-theoretically. The structure morphism $p' : Y' \to \mathop{\mathrm{Spec}}(A)$ is proper, hence $p'(Z' \cap (Y' \setminus Y))$ is closed in $\mathop{\mathrm{Spec}}(A)$. If nonempty, then it would contain a point of $V(I)$ as $I$ is contained in the Jacobson radical of $A$ (Algebra, Lemma 10.96.6). But we've seen above that $Z' \cap (p')^{-1}V(I) = T'_1 \subset Y$ hence we conclude that $Z' \subset Y$. Thus $\mathcal{F}'|_ Y$ is supported on a closed subscheme of $Y$ proper over $A$.

Let $\mathcal{K}$ be the quasi-coherent sheaf of ideals cutting out the reduced complement $X \setminus U$. By Cohomology of Spaces, Lemma 69.20.2 the $\mathcal{O}_ X$-module $\mathcal{H} = f_*\mathcal{F}'$ is coherent and by Lemma 76.42.10 there exists a morphism $\alpha : (\mathcal{F}_ n) \to \mathcal{H}^\wedge$ in the category $\textit{Coh}_{\text{support proper over } A}(X, \mathcal{I})$ whose kernel and cokernel are annihilated by a power of $\mathcal{K}$. Let $Z_0 \subset X$ be the scheme theoretic support of $\mathcal{H}$. It is clear that $|Z_0| \subset f(|Z'|)$. Hence $Z_0 \to \mathop{\mathrm{Spec}}(A)$ is proper (Morphisms of Spaces, Lemma 67.40.7). Thus $\mathcal{H}$ is an object of $\textit{Coh}_{\text{support proper over } A}(\mathcal{O}_ X)$. Since each of the sheaves of ideals $\mathcal{K}^ e$ is an element of $\Xi$ we see that the assumptions of Lemma 76.42.9 are satisfied and we conclude. $\square$

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