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In Hartshorne exercise III.6.7, we are asked to show that if $X = \operatorname{Spec} A$ is an affine Noetherian scheme, and $M, N,$ $A$-modules with $M$ finitely generated, then $$ \operatorname{Ext}^i_A(M, N) \cong \operatorname{Ext}^i_X(\tilde{M}, \tilde{N}). $$ I'm wondering if the $M$ finitely generated hypothesis is even necessary. For since $A$ is Noetherian, sheafifying injective $A$-modules yields injective $\mathcal{O}_X$-modules, so that if $0 \to N \to I^\cdot$ is an injective resolution of $N$ as an $A$-module, one has $$ \operatorname{Ext}^i_A(M, N) = h^i( \operatorname{Hom}_A(M, I^\cdot)) \cong h^i(\operatorname{Hom}_{\mathcal{O}_X}(\tilde{M}, \tilde{I}^\cdot)) = \operatorname{Ext}^i_X(\tilde{M}, \tilde{N}). $$ Is there something wrong with this reasoning?

Legendre
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2024-02-20: There is an important error in your proof (which I should have pointed out earlier): the claim that "sheafifying injective $A$-modules yields injective $\mathcal{O}_X$-modules" isn't true. Take for example $A=\Bbb F_2$ and $M=\Bbb F_2$ - then $M$ is injective, but $\widetilde{M}$ is not injective as a sheaf of abelian groups on $\operatorname{Spec} A=\{pt\}$ because it is not divisible. $\widetilde{M}$ is an injective object in the category of quasi-coherent sheaves, though, but that doesn't help because the various Ext functors here are defined out of $\mathcal{O}_X$-mod.

I suspect the answer is that one actually does need finite generation here for $\operatorname{Ext}^i_X(\widetilde{M},\widetilde{N})\cong \operatorname{Ext}^i_A(M,N)$, but I do not have a counterexample at hand.

2021-12-07: $M$ finitely generated is necessary for $\mathcal{E}xt$, which is the second half of the problem.

To see that $M$ finitely generated is necessary for the second half of the problem, first observe that Hom only commutes with localization when the first argument is finitely presented. For instance, taking $A=N=\Bbb Z$ and $M=\Bbb Z[\frac12]$, $\operatorname{Hom}_A(M,N)=0$ but the Hom sheaf is nonzero over $D(2)\subset\operatorname{Spec} \Bbb Z$, so we get a counterexample involving $\operatorname{Ext}^0$.

KReiser
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  • Thanks! So this shows that the sheafification of $\operatorname{Ext}^i_A(M, N)$ is not necessarily always the Ext sheaf associated to the sheafifications of $M$ and $N$. But will my reasoning above work to show that as abelian groups, $\operatorname{Ext}^i_{\operatorname{Spec} A}(\tilde{M}, \tilde{N}) \cong \operatorname{Ext}^i_{A}(M, N)$ for a noetherian ring $A$ and any $A$-modules $M$ and $N$ (i.e. that the Ext sheaves have the same global sections)? – Legendre Dec 27 '21 at 21:13
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    Oh, sorry, for some reason I assumed you were talking about the sheaves. Your argument is correct for $\operatorname{Ext}$ but fails for $\mathcal{E}xt$, which is the second half of the problem and is stated using the same assumptions by Hartshorne. – KReiser Dec 27 '21 at 23:36