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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).