Lemma 92.14.1. Let $A = \mathbf{Z}[x_1, \ldots , x_ n] \to B = \mathbf{Z}$ be the ring map which sends $x_ i$ to $0$ for $i = 1, \ldots , n$. Let $I = (x_1, \ldots , x_ n) \subset A$. Then $L_{B/A}$ is quasi-isomorphic to $I/I^2[1]$.
Proof. There are several ways to prove this. For example one can explicitly construct a resolution of $B$ over $A$ and compute. We will use (92.7.0.1). Namely, consider the distinguished triangle
\[ L_{\mathbf{Z}[x_1, \ldots , x_ n]/\mathbf{Z}} \otimes _{\mathbf{Z}[x_1, \ldots , x_ n]} \mathbf{Z} \to L_{\mathbf{Z}/\mathbf{Z}} \to L_{\mathbf{Z}/\mathbf{Z}[x_1, \ldots , x_ n]}\to L_{\mathbf{Z}[x_1, \ldots , x_ n]/\mathbf{Z}} \otimes _{\mathbf{Z}[x_1, \ldots , x_ n]} \mathbf{Z}[1] \]
The complex $L_{\mathbf{Z}[x_1, \ldots , x_ n]/\mathbf{Z}}$ is quasi-isomorphic to $\Omega _{\mathbf{Z}[x_1, \ldots , x_ n]/\mathbf{Z}}$ by Lemma 92.4.7. The complex $L_{\mathbf{Z}/\mathbf{Z}}$ is zero in $D(\mathbf{Z})$ by Lemma 92.8.4. Thus we see that $L_{B/A}$ has only one nonzero cohomology group which is as described in the lemma by Lemma 92.11.2. $\square$
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