Lemma 91.24.3. Let $\Lambda $ be a ring. Let $X$ be a scheme over $\Lambda $. Then

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda $ on $X$.

Lemma 91.24.3. Let $\Lambda $ be a ring. Let $X$ be a scheme 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 $X$.

**Proof.**
Let $p : X \to \mathop{\mathrm{Spec}}(\Lambda )$ be the structure morphism. Let $q : \mathop{\mathrm{Spec}}(\Lambda ) \to (*, \Lambda )$ be the obvious morphism. By the distinguished triangle of Lemma 91.20.3 it suffices to show that $L_ q = 0$. To see this it suffices to show for $\mathfrak p \in \mathop{\mathrm{Spec}}(\Lambda )$ that

\[ (L_ q)_\mathfrak p = L_{\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda ), \mathfrak p}/\Lambda } = L_{\Lambda _\mathfrak p/\Lambda } \]

(Lemma 91.18.9) is zero which follows from Lemma 91.8.4. $\square$

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