Lemma 91.14.2. Let $A \to B$ be a surjective ring map whose kernel $I$ is generated by a Koszul-regular sequence (for example a regular sequence). Then $L_{B/A}$ is quasi-isomorphic to $I/I^2[1]$.

Proof. Let $f_1, \ldots , f_ r \in I$ be a Koszul regular sequence generating $I$. Consider the ring map $\mathbf{Z}[x_1, \ldots , x_ r] \to A$ sending $x_ i$ to $f_ i$. Since $x_1, \ldots , x_ r$ is a regular sequence in $\mathbf{Z}[x_1, \ldots , x_ r]$ we see that the Koszul complex on $x_1, \ldots , x_ r$ is a free resolution of $\mathbf{Z} = \mathbf{Z}[x_1, \ldots , x_ r]/(x_1, \ldots , x_ r)$ over $\mathbf{Z}[x_1, \ldots , x_ r]$ (see More on Algebra, Lemma 15.30.2). Thus the assumption that $f_1, \ldots , f_ r$ is Koszul regular exactly means that $B = A \otimes _{\mathbf{Z}[x_1, \ldots , x_ r]}^\mathbf {L} \mathbf{Z}$. Hence $L_{B/A} = L_{\mathbf{Z}/\mathbf{Z}[x_1, \ldots , x_ r]} \otimes _\mathbf {Z}^\mathbf {L} B$ by Lemmas 91.6.2 and 91.14.1. $\square$

Comment #5104 by Noah Olander on

I think the method of proof here works immediately for Koszul regular sequences (and is even a little simpler since you don't need the induction and you don't have to argue flat locally in the next lemma). Just compute $L_{B/A}$ for $A = \mathbf{Z} [x_1, \dots , x_r]$, and then if $(f_1, \dots , f_r)$ is a Koszul regular sequence on $C$, then $C / (f_1, \dots , f_r) = B \otimes _A ^{\mathbf{L}} C$.

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