Lemma 36.13.9. 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
E is perfect, and
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
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