Sheaf cohomology and torsion

Question

Let $X=Spec(R)$ be an affine scheme. Let $Y$ be a closed subset of $X$ and denote by $U$ its complement. Assume $U$ is quasicompact. Then $U= \cup_{i=1}^{n} D(f_i)$, where $f_i \in R$. Denote the inclusions by $j: U \to X$ and $i:Y \to X$. Let $F \in D_{qcoh}(X)$, where $D_{qcoh}(X)$ stands for the derived category of complex with quasicoherent cohomology (If you don't like $D_{qcoh}(X)$ replace it with $D^{b}_{qcoh}(X)$). Assume that $j^{\ast} F=0$.

Is the cohomology of $F$ $(f_1,\ldots,f_n)$-torsion? And how to prove it?

A qick way to get convince (not a proof). Assume that $U=D(f)$. Since $j^{\ast} G=0$ where $G$ is a quasicoherent sheaf, $j^{\ast} G=0$ is equivalent to $G(X)_f=0$. which implies that $G(X)$ is $(f)$-torsion.

Thanks

Answer 1

The answer is yes, assuming by $(f_1,\ldots, f_n)$-torsion you mean that each element of each cohomology group of $F$ is killed by some power of this ideal.

There are several ways to see this. The most straightforward is probably to note that the inclusion $j_i\colon D(f_i) \to X$ factors via $j\colon U \to X$ so $j_i^*F \cong F_{f_i} \cong 0$ for each $i=1,\ldots,n$. Your example, the fact that localization commutes with taking cohomology, and that there are finitely many $f_i$ shows that this is enough.