Lemma 90.13.2. Let $A \to B$ be a surjective ring map whose kernel $I$ is generated by a regular sequence. Then $L_{B/A}$ is quasi-isomorphic to $I/I^2$.

Proof. This is true if $I = (0)$. If $I = (f)$ is generated by a single nonzerodivisor, then consider the ring map $\mathbf{Z}[x] \to A$ which sends $x$ to $f$. By assumption we have $B = A \otimes _{\mathbf{Z}[x]}^\mathbf {L} \mathbf{Z}$. Thus we obtain $L_{B/A} = I/I^2$ from Lemmas 90.6.2 and 90.13.1.

We prove the general case by induction. Suppose that we have $I = (f_1, \ldots , f_ r)$ where $f_1, \ldots , f_ r$ is a regular sequence. Set $C = A/(f_1, \ldots , f_{r - 1})$. By induction the result is true for $A \to C$ and $C \to B$. We have a distinguished triangle (90.7.0.1)

$L_{C/A} \otimes _ C^\mathbf {L} B \to L_{B/A} \to L_{B/C} \to L_{C/A} \otimes _ C^\mathbf {L} B$

which shows that $L_{B/A}$ has only one nonzero cohomology group which is as described in the lemma by Lemma 90.10.2. $\square$

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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