Lemma 90.8.1. Let $A \to A' \to B$ be ring maps such that $B = B \otimes _ A^\mathbf {L} A'$. Then $L_{B/A} = L_{B/A'}$ in $D(B)$.

Proof. According to the discussion above (i.e., using Lemma 90.6.1) and Lemma 90.4.3 we have to show that the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A} \otimes _ P B$ on $\mathcal{C}_{B/A}$ is the pullback of the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A'} \otimes _ P B$. The pullback functor $g^{-1}$ is given by precomposing with the functor $u : \mathcal{C}_{B/A} \to \mathcal{C}_{B/A'}$, $(P \to B) \mapsto (P \otimes _ A A' \to B)$. Thus we have to show that

$\Omega _{P/A} \otimes _ P B = \Omega _{P \otimes _ A A'/A'} \otimes _{(P \otimes _ A A')} B$

By Algebra, Lemma 10.131.12 the right hand side is equal to

$(\Omega _{P/A} \otimes _ A A') \otimes _{(P \otimes _ A A')} B$

Since $P$ is a polynomial algebra over $A$ the module $\Omega _{P/A}$ is free and the equality is obvious. $\square$

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