Lemma 91.14.1. 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 $I/I^2[1]$.

## 91.14 The cotangent complex of a local complete intersection

If $A \to B$ is a local complete intersection map, then $L_{B/A}$ is a perfect complex. The key to proving this is the following lemma.

**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 (91.7.0.1). Namely, consider the distinguished triangle

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 91.4.7. The complex $L_{\mathbf{Z}/\mathbf{Z}}$ is zero in $D(\mathbf{Z})$ by Lemma 91.8.4. Thus we see that $L_{B/A}$ has only one nonzero cohomology group which is as described in the lemma by Lemma 91.11.2. $\square$

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.
$\square$

Lemma 91.14.3. Let $A \to B$ be a surjective ring map whose kernel $I$ is Koszul. Then $L_{B/A}$ is quasi-isomorphic to $I/I^2[1]$.

**Proof.**
Locally on $\mathop{\mathrm{Spec}}(A)$ the ideal $I$ is generated by a Koszul regular sequence, see More on Algebra, Definition 15.32.1. Hence this follows from Lemma 91.6.2.
$\square$

Proposition 91.14.4. Let $A \to B$ be a local complete intersection map. Then $L_{B/A}$ is a perfect complex with tor amplitude in $[-1, 0]$.

**Proof.**
Choose a surjection $P = A[x_1, \ldots , x_ n] \to B$ with kernel $J$. By Lemma 91.11.3 we see that $J/J^2 \to \bigoplus B\text{d}x_ i$ is quasi-isomorphic to $\tau _{\geq -1}L_{B/A}$. Note that $J/J^2$ is finite projective (More on Algebra, Lemma 15.32.3), hence $\tau _{\geq -1}L_{B/A}$ is a perfect complex with tor amplitude in $[-1, 0]$. Thus it suffices to show that $H^ i(L_{B/A}) = 0$ for $i \not\in [-1, 0]$. This follows from (91.7.0.1)

and Lemma 91.14.3 to see that $H^ i(L_{B/P})$ is zero unless $i \in \{ -1, 0\} $. (We also use Lemma 91.4.7 for the term on the left.) $\square$

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