Lemma 36.3.5. Let X = \mathop{\mathrm{Spec}}(A) be an affine scheme. All the functors in the diagram
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)).
Lemma 36.3.5. Let X = \mathop{\mathrm{Spec}}(A) be an affine scheme. All the functors in the diagram
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
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
are adjoint (by Cohomology, Lemma 20.28.1). In particular we obtain canonical adjunction mappings
for E in D(\mathcal{O}_ X) and
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
(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
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
which proves that H^0(b) is an isomorphism. \square
Comments (0)