Lemma 15.85.3. Consider a cocartesian diagram of rings
If $B$ is flat over $A$, then the canonical map $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B' \to \mathop{N\! L}\nolimits _{B'/A'}$ is a quasi-isomorphism. If in addition $\mathop{N\! L}\nolimits _{B/A}$ has tor-amplitude in $[-1, 0]$ then $\mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B' \to \mathop{N\! L}\nolimits _{B'/A'}$ is a quasi-isomorphism too.
Proof. Choose a presentation $\alpha : P \to B$ as in Algebra, Section 10.134. Let $I = \mathop{\mathrm{Ker}}(\alpha )$. Set $P' = P \otimes _ A A'$ and denote $\alpha ' : P' \to B'$ the corresponding presentation of $B'$ over $A'$. As $B$ is flat over $A$ we see that $I' = \mathop{\mathrm{Ker}}(\alpha ')$ is equal to $I \otimes _ A A'$. Hence
We have $\Omega _{P'/A'} = \Omega _{P/A} \otimes _ A A'$ because both sides have the same basis. It follows that $\Omega _{P'/A'} \otimes _{P'} B' = \Omega _{P/A} \otimes _ P B \otimes _ B B'$. This proves that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ B B' \to \mathop{N\! L}\nolimits (\alpha ')$ is an isomorphism of complexes and hence the first statement holds.
We have
as a complex of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see Lemma 15.66.2. If this holds, then $\mathop{N\! L}\nolimits (\alpha ) \otimes _ B^\mathbf {L} B' = \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B'$ and we get the second statement. $\square$
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