Lemma 92.9.1. 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 92.4.5. 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 92.8.5. 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 92.8.4 and Lemma 92.8.5 show that L_{B/A} = L_{A[x_1, \ldots , x_ n]/A} \otimes B whence the conclusion by Lemma 92.4.7. \square
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