Lemma 36.13.8. Let $X$ be a quasi-compact and quasi-separated scheme. Let $f \in \Gamma (X, \mathcal{O}_ X)$. For any morphism $\alpha : E \to E'$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ such that

1. $E$ is perfect, and

2. $E'$ is supported on $T = V(f)$

there exists an $n \geq 0$ such that $f^ n \alpha = 0$.

Proof. We have Mayer-Vietoris for morphisms in the derived category, see Cohomology, Lemma 20.33.3. Thus if $X = U \cup V$ and the result of the lemma holds for $f|_ U$, $f|_ V$, and $f|_{U \cap V}$, then the result holds for $f$. Thus it suffices to prove the lemma when $X$ is affine, see Cohomology of Schemes, Lemma 30.4.1.

Let $X = \mathop{\mathrm{Spec}}(A)$. Then $f \in A$. We will use the equivalence $D(A) = D_\mathit{QCoh}(X)$ of Lemma 36.3.5 without further mention. Represent $E$ by a finite complex of finite projective $A$-modules $P^\bullet$. This is possible by Lemma 36.10.7. Let $t$ be the largest integer such that $P^ t$ is nonzero. The distinguished triangle

$P^ t[-t] \to P^\bullet \to \sigma _{\leq t - 1}P^\bullet \to P^ t[-t + 1]$

shows that by induction on the length of the complex $P^\bullet$ we can reduce to the case where $P^\bullet$ has a single nonzero term. This and the shift functor reduces us to the case where $P^\bullet$ consists of a single finite projective $A$-module $P$ in degree $0$. Represent $E'$ by a complex $M^\bullet$ of $A$-modules. Then $\alpha$ corresponds to a map $P \to H^0(M^\bullet )$. Since the module $H^0(M^\bullet )$ is supported on $V(f)$ by assumption (2) we see that every element of $H^0(M^\bullet )$ is annihilated by a power of $f$. Since $P$ is a finite $A$-module the map $f^ n\alpha : P \to H^0(M^\bullet )$ is zero for some $n$ as desired. $\square$

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).