Let $X$ be a Noetherian scheme and let $f:X\to Y=\operatorname{Spec} A $ a morphism onto an affine scheme.

Proposition III.8.5 in Hartshorne book says that for any quasi-coherent sheaf $\mathcal F$ on $X$ we have:

$$R^if_\ast(\mathcal F)=H^i(X,\mathcal F)^\sim$$

Here $R^if_\ast$ is the $i$-th right derived functor associated to $f_\ast$.

My question is probably very stupid but I'm not able to answer by myself:

$H^i(X,\mathcal F)$ is actually a $\mathcal O_X(X)$-module and not an $A$-module; so how can we construct the sheaf $H^i(X,\mathcal F)^\sim$ on $Y$? I mean in order to use the "$\sim$ construction" on affine schemes we need to start with an $A$-module.

  • $\begingroup$ Any $\mathcal{O}_X(X)$ module is also an $A$-module using $f$. $\endgroup$ – Mohan Sep 28 '17 at 15:30
  • $\begingroup$ Can you be more precise? I mean let $a\in A$ and let $M$ be a $\mathcal O_X(X)$-module. How do you define $a\cdot m$ for $m\in M$? $\endgroup$ – ByContradiction Sep 28 '17 at 15:37
  • 4
    $\begingroup$ Given a morphism $f:X\to Y$, you always have a morphism of rings, $f^*:\mathcal{O}_Y(Y)\to\mathcal{O}_X(X)$, thus any $\mathcal{O}_X(X)$ module is a module over $\mathcal{O}_Y(Y)$. $\endgroup$ – Mohan Sep 28 '17 at 16:21

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