Lemma 59.82.6. Let $(A, I)$ be a henselian pair. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. For any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z)$.

Proof. Recall that the spectrum of any ring is a spectral space, see Algebra, Lemma 10.26.2. By More on Algebra, Lemma 15.11.16 we see that $\overline{\{ x\} } \cap Z$ is connected for every $x \in X$. By Lemma 59.82.4 we see that the statement is true for sheaves on $X_{Zar}$. For any finite morphism $X' \to X$ we have $X' = \mathop{\mathrm{Spec}}(A')$ and $Z \times _ X X' = \mathop{\mathrm{Spec}}(A'/IA')$ with $(A', IA')$ a henselian pair, see More on Algebra, Lemma 15.11.8 and we get the same statement for sheaves on $(X')_{Zar}$. Thus we can apply Lemma 59.82.2 to conclude. $\square$

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