Lemma 90.24.3. 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 90.24.3. Let $\Lambda $ be a ring. Let $X$ be a scheme over $\Lambda $. Then
where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda $ on $X$.
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 90.20.3 it suffices to show that $L_ q = 0$. To see this it suffices to show for $\mathfrak p \in \mathop{\mathrm{Spec}}(\Lambda )$ that
(Lemma 90.18.9) is zero which follows from Lemma 90.8.4. $\square$
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