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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