## 15.85 The naive cotangent complex

In this section we continue the discussion started in Algebra, Section 10.134. We begin with a discussion of base change. The first lemma shows that taking the naive tensor product of the naive cotangent complex with a ring extension isn't quite as naive as one might think.

Lemma 15.85.1. Let $R \to S$ and $S \to S'$ be ring maps. The canonical map $\mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ induces an isomorphism $\tau _{\geq -1}(\mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ in $D(S')$. Similarly, given a presentation $\alpha$ of $S$ over $R$ the canonical map $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ induces an isomorphism $\tau _{\geq -1}(\mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ in $D(S')$.

Proof. Special case of Lemma 15.84.6. $\square$

Lemma 15.85.2. Let $R \to S$ and $R \to R'$ be ring maps. Let $\alpha : P \to S$ be a presentation of $S$ over $R$. Then $\alpha ' : P \otimes _ R R' \to S \otimes _ R R'$ is a presentation of $S' = S \otimes _ R R'$ over $R'$. The canonical map

$NL(\alpha ) \otimes _ S S' \to \mathop{N\! L}\nolimits (\alpha ')$

is an isomorphism on $H^0$ and surjective on $H^{-1}$. In particular, the canonical map

$\mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'}$

is an isomorphism on $H^0$ and surjective on $H^{-1}$.

Proof. Denote $I = \mathop{\mathrm{Ker}}(P \to S)$. Denote $P' = P \otimes _ R R'$ and $I' = \mathop{\mathrm{Ker}}(P' \to S')$. Suppose $P$ is a polynomial algebra on $x_ j$ for $j \in J$. The map displayed in the lemma becomes

$\xymatrix{ \bigoplus _{j \in J} S' \text{d}x_ j \ar[r] & \bigoplus _{j \in J} S' \text{d}x_ j \\ I/I^2 \otimes _ S S' \ar[r] \ar[u] & I'/(I')^2 \ar[u] }$

where the left column is $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ and the right column is $\mathop{N\! L}\nolimits (\alpha ')$. By right exactness of tensor product we see that $I \otimes _ R R' \to I'$ is surjective. Hence the bottom arrow is a surjection. This proves the first statement of the lemma. The statement for $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'}$ follows as these complexes are homotopic to $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ and $\mathop{N\! L}\nolimits (\alpha ')$. $\square$

Lemma 15.85.3. Consider a cocartesian diagram of rings

$\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }$

If $B$ is flat over $A$, then the canonical map $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B' \to \mathop{N\! L}\nolimits _{B'/A'}$ is a quasi-isomorphism. If in addition $\mathop{N\! L}\nolimits _{B/A}$ has tor-amplitude in $[-1, 0]$ then $\mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B' \to \mathop{N\! L}\nolimits _{B'/A'}$ is a quasi-isomorphism too.

Proof. Choose a presentation $\alpha : P \to B$ as in Algebra, Section 10.134. Let $I = \mathop{\mathrm{Ker}}(\alpha )$. Set $P' = P \otimes _ A A'$ and denote $\alpha ' : P' \to B'$ the corresponding presentation of $B'$ over $A'$. As $B$ is flat over $A$ we see that $I' = \mathop{\mathrm{Ker}}(\alpha ')$ is equal to $I \otimes _ A A'$. Hence

$I'/(I')^2 = \mathop{\mathrm{Coker}}(I^2 \otimes _ A A' \to I \otimes _ A A') = I/I^2 \otimes _ A A' = I/I^2 \otimes _ B B'$

We have $\Omega _{P'/A'} = \Omega _{P/A} \otimes _ A A'$ because both sides have the same basis. It follows that $\Omega _{P'/A'} \otimes _{P'} B' = \Omega _{P/A} \otimes _ P B \otimes _ B B'$. This proves that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ B B' \to \mathop{N\! L}\nolimits (\alpha ')$ is an isomorphism of complexes and hence the first statement holds.

We have

$\mathop{N\! L}\nolimits (\alpha ) = I/I^2 \longrightarrow \Omega _{P/A} \otimes _ P B$

as a complex of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see Lemma 15.66.2. If this holds, then $\mathop{N\! L}\nolimits (\alpha ) \otimes _ B^\mathbf {L} B' = \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B'$ and we get the second statement. $\square$

Lemma 15.85.4. Let $A \to B$ be a local complete intersection as in Definition 15.33.2. Then $\mathop{N\! L}\nolimits _{B/A}$ is a perfect object of $D(B)$ with tor amplitude in $[-1, 0]$.

Proof. Write $B = A[x_1, \ldots , x_ n]/I$. Then $\mathop{N\! L}\nolimits _{B/A}$ is represented by the complex

$I/I^2 \longrightarrow \bigoplus B \text{d}x_ i$

of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is finite free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see Lemma 15.66.2. By definition $I$ is a Koszul regular ideal and hence a quasi-regular ideal, see Section 15.32. Thus $I/I^2$ is a finite projective $B$-module (Lemma 15.32.3) and we conclude both that $\mathop{N\! L}\nolimits _{B/A}$ is perfect and that it has tor amplitude in $[-1, 0]$. $\square$

Lemma 15.85.5. Consider a cocartesian diagram of rings

$\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }$

If $A \to B$ and $A' \to B'$ are local complete intersections as in Definition 15.33.2, then the kernel of $H^{-1}(\mathop{N\! L}\nolimits _{B/A} \otimes _ B B') \to H^{-1}(\mathop{N\! L}\nolimits _{B'/A'})$ is a finite projective $B'$-module.

Proof. By Lemma 15.85.4 the complexes $\mathop{N\! L}\nolimits _{B/A}$ and $\mathop{N\! L}\nolimits _{B'/A'}$ are perfect of tor-amplitude in $[-1, 0]$. Combining Lemmas 15.85.1, 15.74.9, and 15.66.13 we have $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B' = \mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B'$ and this complex is also perfect of tor-amplitude in $[-1, 0]$. Choose a distinguished triangle

$C \to \mathop{N\! L}\nolimits _{B/A} \otimes _ B B' \to \mathop{N\! L}\nolimits _{B'/A'} \to C[1]$

in $D(B')$. By Lemmas 15.74.4 and 15.66.5 we conclude that $C$ is perfect with tor-amplitude in $[-1, 1]$. By Lemma 15.85.2 the complex $C$ has only one nonzero cohomology module, namely the module of the lemma sitting in degree $-1$. This module is of finite presentation (Lemma 15.64.4) and flat (Lemma 15.66.6). Hence it is finite projective by Algebra, Lemma 10.78.2. $\square$

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