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
$E$ 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
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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