Lemma 30.23.11. Let $X$ be a Noetherian scheme. Let $\mathcal{I}, \mathcal{J} \subset \mathcal{O}_ X$ be quasi-coherent sheaves of ideals. If $V(\mathcal{I}) = V(\mathcal{J})$ is the same closed subset of $X$, then $\textit{Coh}(X, \mathcal{I})$ and $\textit{Coh}(X, \mathcal{J})$ are equivalent.

Proof. First, assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I, J \subset A$ be the ideals corresponding to $\mathcal{I}, \mathcal{J}$. Then $V(I) = V(J)$ implies we have $I^ c \subset J$ and $J^ d \subset I$ for some $c, d \geq 1$ by elementary properties of the Zariski topology (see Algebra, Section 10.17 and Lemma 10.32.5). Hence the $I$-adic and $J$-adic completions of $A$ agree, see Algebra, Lemma 10.96.9. Thus the equivalence follows from Lemma 30.23.1 in this case.

In general, using what we said above and the fact that $X$ is quasi-compact, to choose $c, d \geq 1$ such that $\mathcal{I}^ c \subset \mathcal{J}$ and $\mathcal{J}^ d \subset \mathcal{I}$. Then given an object $(\mathcal{F}_ n)$ in $\textit{Coh}(X, \mathcal{I})$ we claim that the inverse system

$(\mathcal{F}_{cn}/\mathcal{J}^ n\mathcal{F}_{cn})$

is in $\textit{Coh}(X, \mathcal{J})$. This may be checked on the members of an affine covering; we omit the details. In the same manner we can construct an object of $\textit{Coh}(X, \mathcal{I})$ starting with an object of $\textit{Coh}(X, \mathcal{J})$. We omit the verification that these constructions define mutually quasi-inverse functors. $\square$

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