Remark 36.7.7 (Warning). Let $X$ be a quasi-compact scheme with affine diagonal. Even though we know that $D(\mathit{QCoh}(\mathcal{O}_ X)) = D_\mathit{QCoh}(\mathcal{O}_ X)$ by Proposition 36.7.5 strange things can happen and it is easy to make mistakes with this material. One pitfall is to carelessly assume that this equality means derived functors are the same. For example, suppose we have a quasi-compact open $U \subset X$. Then we can consider the higher right derived functors

$R^ i(\mathit{QCoh})\Gamma (U, -) : \mathit{QCoh}(\mathcal{O}_ X) \to \textit{Ab}$

of the left exact functor $\Gamma (U, -)$. Since this is a universal $\delta$-functor, and since the functors $H^ i(U, -)$ (defined for all abelian sheaves on $X$) restricted to $\mathit{QCoh}(\mathcal{O}_ X)$ form a $\delta$-functor, we obtain canonical tranformations

$t^ i : R^ i(\mathit{QCoh})\Gamma (U, -) \to H^ i(U, -).$

These transformations aren't in general isomorphisms even if $X = \mathop{\mathrm{Spec}}(A)$ is affine! Namely, we have $R^1(\mathit{QCoh})\Gamma (U, \widetilde{I}) = 0$ if $I$ an injective $A$-module by construction of right derived functors and the equivalence of $\mathit{QCoh}(\mathcal{O}_ X)$ and $\text{Mod}_ A$. But Examples, Lemma 109.46.2 shows there exists $A$, $I$, and $U$ such that $H^1(U, \widetilde{I}) \not= 0$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).