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
By Theorem 59.51.3 we have
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
and by our induction hypothesis we obtain vanishing for $q > n$ as desired. $\square$
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