Lemma 91.26.4. Let $\Lambda $ be a ring. Let $X$ be an algebraic space over $\Lambda $. Then

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda $ on $X_{\acute{e}tale}$.

Lemma 91.26.4. Let $\Lambda $ be a ring. Let $X$ be an algebraic space over $\Lambda $. Then

\[ L_{X/\mathop{\mathrm{Spec}}(\Lambda )} = L_{\mathcal{O}_ X/\underline{\Lambda }} \]

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda $ on $X_{\acute{e}tale}$.

**Proof.**
Let $p : X \to \mathop{\mathrm{Spec}}(\Lambda )$ be the structure morphism. Let $q : \mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}\to (*, \Lambda )$ be the obvious morphism. By the distinguished triangle of Lemma 91.22.3 it suffices to show that $L_ q = 0$. To see this it suffices to show (Properties of Spaces, Theorem 65.19.12) for a geometric point $\overline{t} : \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(\Lambda )$ that

\[ (L_ q)_{\overline{t}} = L_{\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}/\Lambda } \]

(Lemma 91.18.9) is zero. Since $\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}$ is a strict henselization of a local ring of $\Lambda $ (Properties of Spaces, Lemma 65.22.1) this follows from Lemma 91.8.4. $\square$

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