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I am confused about something in Exercise 4.1 of Chapter 3 of Hartshorne. It asks us to prove : Let $f : X \rightarrow Y$ be an affine morphism of Noetherian, separated schemes. Show that for any quasi-coherent sheaf $\mathcal{F}$ on $X$, there are natural isomorphisms $$H^i(X, \mathcal{F}) \cong H^i(Y, f_* \mathcal{F}) \ \ \forall i \geq 0$$

I seem to have a proof without assuming that $f$ is affine or that the schemes are separated (but I need that $X$ is Noetherian). So I know my proof must be bogus, but I can't figure out why.

Proof : Exercise 3.6 of Chapter 3 of Hartshorne says that we can compute cohomology of a quasi-coherent sheaf as the derived functors of $\Gamma(X, \cdot)$, considered a a functor from the category of quasi-coherent sheaves to abelian groups. So let $\mathcal{F} \rightarrow \mathcal{I}^{\cdot}$ be an injective resolution of $\mathcal{F}$ in the category of quasi-coherent sheaves. Then $f_* \mathcal{F} \rightarrow f_* \mathcal{I^{\cdot}}$ is a flasque resolution of quasi-coherent sheaves of $f_* \mathcal{F}$ (pushforward of quasi-coherent is quasi-coherent since $X$ is Noetherian). Moreover, we clearly have $\Gamma(X, \mathcal{I}^{\cdot}) = \Gamma(Y, f_* \mathcal{I}^{\cdot})$. Therefore, we get the same cohomology groups (this is the same method of proof of Lemma 2.10, Chapter 3, of Hartshorne).

What did I do wrong?

Sincerely,

David

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  • $\begingroup$ Try out your proof in the case that $Y$ is affine, but $X$ is not, and recall Serre's cohomological characterization of affine schemes. It cannot work. $\endgroup$ – Martin Brandenburg Apr 11 '12 at 18:05
  • $\begingroup$ How could it happen that X is not affine, but Y is, when f is affine? I am really worried since I still find the argument above correct... But I imagine Hartshorne assumed separated since this gives us a way to prove the above with Cech complex, isn't it? $\endgroup$ – Piotr Pstrągowski Dec 18 '12 at 1:03
  • $\begingroup$ It can't but remember that the OP's supposed argument would work without the affineness assumption. $\endgroup$ – Ioannis Zolas May 11 at 20:14
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When $f$ is affine, then $f_*$ is an exact functor. The reason is that in affine charts this is just some forgetful functor $\mathrm{Mod}(A) \to \mathrm{Mod}(R)$ for some $R$-algebra $A$. Of course, this is needed to get a resolution of $f_* F$ from a resolution of $F$. So the error is: You have forgotten the definition of a resolution.

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