Lemma 36.3.5. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. All the functors in the diagram

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[rr]_{(06VT)} & & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[ld]^{R\Gamma (X, -)} \\ & D(A) \ar[lu]^{\widetilde{\ \ }} }$

are equivalences of triangulated categories. Moreover, for $E$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ we have $H^0(X, E) = H^0(X, H^0(E))$.

Proof. The functor $R\Gamma (X, -)$ gives a functor $D(\mathcal{O}_ X) \to D(A)$ and hence by restriction a functor

36.3.5.1
$$\label{perfect-equation-back} R\Gamma (X, -) : D_\mathit{QCoh}(\mathcal{O}_ X) \longrightarrow D(A).$$

We will show this functor is quasi-inverse to (36.3.0.1) via the equivalence between quasi-coherent modules on $X$ and the category of $A$-modules.

Elucidation. Denote $(Y, \mathcal{O}_ Y)$ the one point space with sheaf of rings given by $A$. Denote $\pi : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ the obvious morphism of ringed spaces. Then $R\Gamma (X, -)$ can be identified with $R\pi _*$ and the functor (36.3.0.1) via the equivalence $\textit{Mod}(\mathcal{O}_ Y) = \text{Mod}_ A = \mathit{QCoh}(\mathcal{O}_ X)$ can be identified with $L\pi ^* = \pi ^* = \widetilde{\ }$ (see Modules, Lemma 17.10.5 and Schemes, Lemmas 26.7.1 and 26.7.5). Thus the functors

$\xymatrix{ D(A) \ar@<1ex>[r] & D_\mathit{QCoh}(\mathcal{O}_ X) \ar@<1ex>[l] }$

are adjoint (by Cohomology, Lemma 20.28.1). In particular we obtain canonical adjunction mappings

$a : \widetilde{R\Gamma (X, E)} \longrightarrow E$

for $E$ in $D(\mathcal{O}_ X)$ and

$b : M^\bullet \longrightarrow R\Gamma (X, \widetilde{M^\bullet })$

for $M^\bullet$ a complex of $A$-modules.

Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. We may apply Lemma 36.3.4 to the functor $F(-) = \Gamma (X, -)$ with $N = 1$ by Cohomology of Schemes, Lemma 30.2.2. Hence

$H^0(R\Gamma (X, E)) = H^0(R\Gamma (X, \tau _{\geq 0}E)) = \Gamma (X, H^0(E))$

(the last equality by definition of the canonical truncation). Using this we will show that the adjunction mappings $a$ and $b$ induce isomorphisms $H^0(a)$ and $H^0(b)$. Thus $a$ and $b$ are quasi-isomorphisms (as the statement is invariant under shifts) and the lemma is proved.

In both cases we use that $\widetilde{\ }$ is an exact functor (Schemes, Lemma 26.5.4). Namely, this implies that

$H^0\left(\widetilde{R\Gamma (X, E)}\right) = \widetilde{H^0(R\Gamma (X, E))} = \widetilde{\Gamma (X, H^0(E))}$

which is equal to $H^0(E)$ because $H^0(E)$ is quasi-coherent. Thus $H^0(a)$ is an isomorphism. For the other direction we have

$H^0(R\Gamma (X, \widetilde{M^\bullet })) = \Gamma (X, H^0(\widetilde{M^\bullet })) = \Gamma (X, \widetilde{H^0(M^\bullet )}) = H^0(M^\bullet )$

which proves that $H^0(b)$ is an isomorphism. $\square$

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