Lemma 91.9.1. If $A \to B$ is a smooth ring map, then $L_{B/A} = \Omega _{B/A}[0]$.

**Proof.**
We have the agreement in cohomological degree $0$ by Lemma 91.4.5. Thus it suffices to prove the other cohomology groups are zero. It suffices to prove this locally on $\mathop{\mathrm{Spec}}(B)$ as $L_{B_ g/A} = (L_{B/A})_ g$ for $g \in B$ by Lemma 91.8.5. Thus we may assume that $A \to B$ is standard smooth (Algebra, Lemma 10.137.10), i.e., that we can factor $A \to B$ as $A \to A[x_1, \ldots , x_ n] \to B$ with $A[x_1, \ldots , x_ n] \to B$ étale. In this case Lemmas 91.8.4 and Lemma 91.8.5 show that $L_{B/A} = L_{A[x_1, \ldots , x_ n]/A} \otimes B$ whence the conclusion by Lemma 91.4.7.
$\square$

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