Lemma 59.82.8. Let X be a scheme with affine diagonal which can be covered by n + 1 affine opens. Let Z \subset X be a closed subscheme. Let \mathcal{A} be a torsion sheaf of rings on X_{\acute{e}tale} and let \mathcal{I} be an injective sheaf of \mathcal{A}-modules on X_{\acute{e}tale}. Then H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) = 0 for q > n.
Proof. We will prove this by induction on n. If n = 0, then X is affine. Say X = \mathop{\mathrm{Spec}}(A) and Z = \mathop{\mathrm{Spec}}(A/I). Let A^ h be the filtered colimit of étale A-algebras B such that A/I \to B/IB is an isomorphism. Then (A^ h, IA^ h) is a henselian pair and A/I = A^ h/IA^ h, see More on Algebra, Lemma 15.12.1 and its proof. Set X^ h = \mathop{\mathrm{Spec}}(A^ h). By Theorem 59.82.7 we see that
H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) = H^ q_{\acute{e}tale}(X^ h, \mathcal{I}|_{X^ h})
By Theorem 59.51.3 we have
H^ q_{\acute{e}tale}(X^ h, \mathcal{I}|_{X^ h}) = \mathop{\mathrm{colim}}\nolimits _{A \to B} H^ q_{\acute{e}tale}(\mathop{\mathrm{Spec}}(B), \mathcal{I}|_{\mathop{\mathrm{Spec}}(B)})
where the colimit is over the A-algebras B as above. Since the morphisms \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A) are étale, the restriction \mathcal{I}|_{\mathop{\mathrm{Spec}}(B)} is an injective sheaf of \mathcal{A}|_{\mathop{\mathrm{Spec}}(B)}-modules (Cohomology on Sites, Lemma 21.7.1). Thus the cohomology groups on the right are zero and we get the result in this case.
Induction step. We can use Mayer-Vietoris to do the induction step. Namely, suppose that X = U \cup V where U is a union of n affine opens and V is affine. Then, using that the diagonal of X is affine, we see that U \cap V is the union of n affine opens. Mayer-Vietoris gives an exact sequence
H^{q - 1}_{\acute{e}tale}(U \cap V \cap Z, \mathcal{I}|_ Z) \to H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) \to H^ q_{\acute{e}tale}(U \cap Z, \mathcal{I}|_ Z) \oplus H^ q_{\acute{e}tale}(V \cap Z, \mathcal{I}|_ Z)
and by our induction hypothesis we obtain vanishing for q > n as desired. \square
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