Lemma 51.7.3. Let A be a ring and let J \subset I \subset A be finitely generated ideals. Let i \geq 0 be an integer. Set Z = V(I). If H^ i_ Z(A) is annihilated by J^ n for some n, then H^ i_ Z(M) annihilated by J^ m for some m = m(M) for every finitely presented A-module M such that M_ f is a finite locally free A_ f-module for all f \in I.
Proof. Consider the annihilator \mathfrak a of H^ i_ Z(M). Let \mathfrak p \subset A with \mathfrak p \not\in Z. By assumption there exists an f \in I, f \not\in \mathfrak p and an isomorphism \varphi : A_ f^{\oplus r} \to M_ f of A_ f-modules. Clearing denominators (and using that M is of finite presentation) we find maps
a : A^{\oplus r} \longrightarrow M \quad \text{and}\quad b : M \longrightarrow A^{\oplus r}
with a_ f = f^ N \varphi and b_ f = f^ N \varphi ^{-1} for some N. Moreover we may assume that a \circ b and b \circ a are equal to multiplication by f^{2N}. Thus we see that H^ i_ Z(M) is annihilated by f^{2N}J^ n, i.e., f^{2N}J^ n \subset \mathfrak a.
As U = \mathop{\mathrm{Spec}}(A) \setminus Z is quasi-compact we can find finitely many f_1, \ldots , f_ t and N_1, \ldots , N_ t such that U = \bigcup D(f_ j) and f_ j^{2N_ j}J^ n \subset \mathfrak a. Then V(I) = V(f_1, \ldots , f_ t) and since I is finitely generated we conclude I^ M \subset (f_1, \ldots , f_ t) for some M. All in all we see that J^ m \subset \mathfrak a for m \gg 0, for example m = M (2N_1 + \ldots + 2N_ t) n will do. \square
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