Lemma 91.5.7. Let $A \to B$ be a ring map. Let $\pi$, $\mathcal{O}$, $\underline{B}$ be as in (91.4.0.1). We have

$L\pi _!(\mathcal{O}) = L\pi _!(\underline{B}) = B \quad \text{and}\quad L_{B/A} = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = L\pi _!(\Omega _{\mathcal{O}/A})$

in $D(\textit{Ab})$.

Proof. This is just an application of Lemma 91.5.6 (and the first equality on the right is Lemma 91.4.3). $\square$

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