Disclaimer: This was first asked here on math.stackexchange with no answers.

Let $F,G$ be quasicoherent sheaves of modules on a scheme $X$, then is the sheaf $\mathcal{Ext}^i(F,G)$ equal to the sheafification of the presheaf $$E^i(F,G) : U\mapsto Ext^i_U(F|_U,G|_U)?$$

It seems to me that (maybe modulo some technical points?), this is probably true.

Indeed, let $G\rightarrow I^\bullet$ be a injective resolution for $G$. By Lemma III.6.1 in Hartshorne's Algebraic Geometry, we have that for any open $U\subset X$, $I^\bullet|_U$ is an injective resolution for $G|_U$. (Actually this result seems somewhat suspicious to me, since I have heard multiple times that "localization does not always preserve injectives").

But, assuming Hartshorne's lemma is correct, the cohomology of the complex $Hom_U(F|_U,I^\bullet|_U)$ gives precisely the groups $Ext^i_U(F|_U,G|_U)$.

On the other hand for any inclusion of opens $U\subset V$, localizing yields a morphism of complexes $$Hom_V(F|_V,I^\bullet|_V)\rightarrow Hom_U(F|_U,I^\bullet|_U)$$ and hence a map on cohomology, which is clearly functorial w.r.t. further localization. Thus, for any inclusion of opens $U\subset V$, this gives restriction maps

$$Ext^i_V(F|_V,G|_V)\rightarrow Ext^i_U(F|_U,G|_U)$$ which shows that the rule $U\mapsto Ext^i_U(F|_U,G|_U)$ defines a presheaf.

Now, we want to define a map of presheaves $E^i(F,G)\rightarrow \mathcal{Ext}^i(F,G)$. To do this, we note that we have an equality of complexes $$Hom_U(F|_U,I^\bullet|_U) = \Gamma(U,\mathcal{Hom}_U(F|_U,I^\bullet|_U))$$ If $\partial$ denotes the differential of the complex $\mathcal{Hom}_U(F|_U,I^\bullet|_U)$, and $d$ denotes the differential of the complex $Hom_U(F|_U,I^\bullet|_U)$, then we get a map $$\ker d^i\stackrel{=}{\longrightarrow}\Gamma(\ker\partial^i)\rightarrow\Gamma(\ker\partial^i/\text{im }\partial^{i-1})$$ which kills $\text{im }d^{i-1}$, and hence for every open $U$, we get compatible maps $$Ext^i_U(F|_U,G|_U)\rightarrow\Gamma(U,\mathcal{Ext}^i(F,G))$$ whence a morphism of presheaves $$\varphi : E^i(F,G)\rightarrow\mathcal{Ext}^i(F,G)$$ which seems to induce isomorphisms on open affines (is there anything to check here? must we impose any finiteness conditions like $X$ noetherian or $F$ coherent?).

If the above is correct, then $\varphi$ should induces an isomorphism on stalks, and hence identifies $\mathcal{Ext}^i(F,G)$ with the sheafification of $E^i(F,G)$.

Can someone check this proof and/or confirm the result? References would be appreciated as well. On the one hand it seems like a nice statement to have in mind, and yet on the other hand if it were true, I would have expected to have seen/heard it somewhere.

  • $\begingroup$ One thing to be said about Hartschorne's lemma is that his injectives are injectives in the category of $O_X$ modules and not in $QCoh(X)$. So restricting to smaller open is not localization of a module. $\endgroup$
    – user45150
    Jun 16, 2017 at 19:57
  • $\begingroup$ @user45150 Hmm, are you saying that the restriction (in $Mod(X)$) of a quasicoherent $O_X$-module to an open $U\subset X$ is not necessarily quasicoherent? Do you have an example in mind? $\endgroup$ Jun 16, 2017 at 20:10
  • $\begingroup$ No. I am saying that when Hartschorne says injective sheaf $\mathcal{I}$, $\mathcal{I}$ is not assumed to be quasi-coherent, but only an $O_X$-module $\endgroup$
    – user45150
    Jun 16, 2017 at 20:53
  • $\begingroup$ @user45150 I'm thinking about this again, and I'm confused now: If the restriction of a quasicoherent injective module to $U$ is also quasicoherent, then doesn't Hartshorne's Lemma 6.1 "restrictions of injective objects in $Mod(X)$ is an injective of $Mod(U)$" imply that the restriction of a quasicoherent injective is also quasicoherent injective? $\endgroup$ Jul 7, 2017 at 21:37
  • $\begingroup$ the injective objects are in $Mod(X)$ not in $QCoh(X)$. He never considers "quasicoherent injectives" $\endgroup$
    – user45150
    Jul 8, 2017 at 15:44

1 Answer 1


This is proven in the Stacks Project as pointed out in the comments on math.stackexchange by essentially the same proof you suggested.

  • $\begingroup$ Ah, how embarassing. I had thought for some reason that this question was never resolved, so I didn't think to reread the comments. Oops. $\endgroup$ Jun 16, 2017 at 20:55

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