POTW Ext Functor on Noetherian Affine Schemes

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Let $M$ and $N$ be finitely generated modules over a Noetherian ring $R$. For every $R$-module $S$ let $\tilde{S}$ be the associated sheaf on the affine scheme $X = \operatorname{Spec}R$. Show that for all $q \ge 0$, the associated sheaf of the $R$-module $\operatorname{Ext}^q_R(M,N)$ is the Ext sheaf $\mathscr{E}xt^q_{\mathcal{O}_X}(\tilde{M},\tilde{N})$.

 

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Spoiler

Since $M$ is finite over Noetherian ring $R$, there is a finitely generated free resolution $\cdots \to L_1 \to L_0 \to M \to 0$ for $M$. Applying the tilde functor yields a resolution of $\tilde{M}$ by finite locally free $\mathcal{O}_X$-modules $\cdots \to \tilde{L}_1 \to \tilde{L}_0 \to \tilde{M} \to 0$. Hence $\mathscr{E}xt^q_{\mathcal{O}_X}(\tilde{M}, \tilde{N})$ is the $q$-th cohomology of the complex $\mathscr{H}om_{\mathcal{O}_X}(\tilde{L}^\cdot, \tilde{N})$. There is an natural isomorphism of complexes $\mathscr{H}om_{\mathcal{O}_X}(\tilde{L}^\cdot, \tilde{N}) = \widetilde{\text{Hom}_{\mathcal{O}_X}(L^\cdot, N)}$ and cohomology commutes with the tilde functor; furthermore, the $q$th cohomology of $\text{Hom}_{\mathcal{O}_X}(L^\cdot, N)$ is $\text{Ext}^q_R(M,N)$. Therefore $\mathscr{E}xt^q_{\mathcal{O}_X}(\tilde{M}, \tilde{N}) \cong \widetilde{\text{Ext}^q_R(M,N)}$, as desired.

 

[mathwonk]

Using the Grothendieck definition of cohomology by injective resolutions, the fact that Tilda of an injective R module is an injective O_X module, that Tilda commutes with kernels and cokernels, and is a fully faithful equivalence from R modules to O_X modules, this seems to reduce to the case of "sheaf Hom", i.e. sheaf Ext^0, where it seems to follow from the fact that on finitely generated modules (actually we seem only to need M finitely generated), Hom commutes with localization.

I.e. to compute Ext, take an injective R module resolution J(*) of N, and apply Hom(M,_), to get a complex C = Hom(M,J(*)), whose cohomology is Ext*(M,N). Applying Tilda to C, gives a complex whose cohomology is Tilda of Ext, since Tilda preserves kernels and cokernels. But since Hom commutes with localization for finitely generated M, applying Tilda to C is the same as first applying Tilda to J, getting an injective O_X module resolution of NTilda, and then applying sheafHom(MTilda,_), to get the complex whose cohomology is sheafExt. Hence sheafExt is Tilda of Ext in this case.