$\mathscr{F}\rightarrow i_*(i^*\mathscr{F})$ is an isomorphism for $i:V(\mathscr{I})\rightarrow X$

Question

This is Remark 7.36 of Görtz/Wedhorn. We have a scheme $X$ and a quasi-coherent $\mathcal{O}_X$-module of finite type $\mathscr{F}$, and we define $\operatorname{Ann}(\mathscr{F})$ as the kernel of the canonical homomorphism of $\mathcal{O}_X$ into the endomorphism sheaf of $\mathscr{F}$ over $\mathcal{O}_X$. Then we assume that $\mathscr{I}$ is a quasi-coherent ideal of $\mathcal{O}_X$, and we thus get an induced closed subscheme $V(\mathscr{I})$ of $X$, whose inclusion into $X$ we denote by $i$. If we assume further that $\mathscr{I}\subseteq \operatorname{Ann}(\mathscr{F})$, then $\mathscr{F}$ is naturally an $\mathcal{O}_X/\mathscr{I}$-module. The authors then say that $\textbf{this}$ shows that the canonical morphism coming from adjuntion $\mathscr{F}\rightarrow i_*(i^*\mathscr{F})$ is an isomorphism, which I don't really get at the moment.

I think that since $i$ is an inclusion of a closed subscheme, we should get isomorphism on stalks, but I don't see how this uses the argument $\textbf{this}$, and also I think my thought is invalid, because otherwise for every closed subscheme we would have this isomorphism. I would be grateful if someone could give me just a little hint.

Answer 1

Hint: work affine locally and describe the map $\mathscr{F}\to i_*i^*\mathscr{F}$ explicitly. What is it? How can you use the fact that $\mathscr{I}\subset \operatorname{Ann}(\mathscr{F})$ with this description?

Full solution:

We work affine-locally. Let $\renewcommand{\Spec}{\operatorname{Spec}}X=\Spec A$, $\mathscr{I}=\widetilde{I}$, and $\mathscr{F}=\widetilde{M}$. On modules, the canonical morphism $\mathscr{F}\to i_*i^*\mathscr{F}$ is $M\to (M\otimes_A A/I)_{A}$, where the subscript indicates that we are considering that $A/I$-module $M\otimes_A A/I$ as an $A$-module. As $M\otimes_A A/I$ is naturally isomorphic to $M/IM$, the assumption that $I\subset \operatorname{Ann}(M)$ means that $IM=0$ and therefore $M\otimes_A A/I=M/IM=M$. So the map $\mathscr{F}\to i_*i^*\mathscr{F}$ is an isomorphism.