## 91.15 Tensor products and the cotangent complex

Let $R$ be a ring and let $A$, $B$ be $R$-algebras. In this section we discuss $L_{A \otimes _ R B/R}$. Most of the information we want is contained in the following diagram

91.15.0.1
$$\label{cotangent-equation-tensor-product} \vcenter { \xymatrix{ L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B) \ar[r] & L_{A \otimes _ R B/B} \ar[r] & E \\ L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B) \ar[r] \ar@{=}[u] & L_{A \otimes _ R B/R} \ar[r] \ar[u] & L_{A \otimes _ R B/A} \ar[u] \\ & L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B) \ar[u] \ar@{=}[r] & L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B) \ar[u] } }$$

Explanation: The middle row is the fundamental triangle (91.7.0.1) for the ring maps $R \to A \to A \otimes _ R B$. The middle column is the fundamental triangle (91.7.0.1) for the ring maps $R \to B \to A \otimes _ R B$. Next, $E$ is an object of $D(A \otimes _ R B)$ which “fits” into the upper right corner, i.e., which turns both the top row and the right column into distinguished triangles. Such an $E$ exists by Derived Categories, Proposition 13.4.23 applied to the lower left square (with $0$ placed in the missing spot). To be more explicit, we could for example define $E$ as the cone (Derived Categories, Definition 13.9.1) of the map of complexes

$L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B) \oplus L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B) \longrightarrow L_{A \otimes _ R B/R}$

and get the two maps with target $E$ by an application of TR3. In the Tor independent case the object $E$ is zero.

Lemma 91.15.1. If $A$ and $B$ are Tor independent $R$-algebras, then the object $E$ in (91.15.0.1) is zero. In this case we have

$L_{A \otimes _ R B/R} = L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B) \oplus L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B)$

which is represented by the complex $L_{A/R} \otimes _ R B \oplus L_{B/R} \otimes _ R A$ of $A \otimes _ R B$-modules.

Proof. The first two statements are immediate from Lemma 91.6.2. The last statement follows as $L_{A/R}$ is a complex of free $A$-modules, hence $L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B)$ is represented by $L_{A/R} \otimes _ A (A \otimes _ R B) = L_{A/R} \otimes _ R B$ $\square$

In general we can say this about the object $E$.

Lemma 91.15.2. Let $R$ be a ring and let $A$, $B$ be $R$-algebras. The object $E$ in (91.15.0.1) satisfies

$H^ i(E) = \left\{ \begin{matrix} 0 & \text{if} & i \geq -1 \\ \text{Tor}_1^ R(A, B) & \text{if} & i = -2 \end{matrix} \right.$

Proof. We use the description of $E$ as the cone on $L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B) \to L_{A \otimes _ R B/A}$. By Lemma 91.13.3 the canonical truncations $\tau _{\geq -2}L_{B/R}$ and $\tau _{\geq -2}L_{A \otimes _ R B/A}$ are computed by the Lichtenbaum-Schlessinger complex (91.13.0.1). These isomorphisms are compatible with functoriality (Remark 91.13.4). Thus in this proof we work with the Lichtenbaum-Schlessinger complexes.

Choose a polynomial algebra $P$ over $R$ and a surjection $P \to B$. Choose generators $f_ t \in P$, $t \in T$ of the kernel of this surjection. Let $Rel \subset F = \bigoplus _{t \in T} P$ be the kernel of the map $F \to P$ which maps the basis vector corresponding to $t$ to $f_ t$. Set $P_ A = A \otimes _ R P$ and $F_ A = A \otimes _ R F = P_ A \otimes _ P F$. Let $Rel_ A$ be the kernel of the map $F_ A \to P_ A$. Using the exact sequence

$0 \to Rel \to F \to P \to B \to 0$

and standard short exact sequences for Tor we obtain an exact sequence

$A \otimes _ R Rel \to Rel_ A \to \text{Tor}_1^ R(A, B) \to 0$

Note that $P_ A \to A \otimes _ R B$ is a surjection whose kernel is generated by the elements $1 \otimes f_ t$ in $P_ A$. Denote $TrivRel_ A \subset Rel_ A$ the $P_ A$-submodule generated by the elements $(\ldots , 1 \otimes f_{t'}, 0, \ldots , 0, - 1 \otimes f_ t \otimes 1, 0, \ldots )$. Since $TrivRel \otimes _ R A \to TrivRel_ A$ is surjective, we find a canonical exact sequence

$A \otimes _ R (Rel/TrivRel) \to Rel_ A/TrivRel_ A \to \text{Tor}_1^ R(A, B) \to 0$

The map of Lichtenbaum-Schlessinger complexes is given by the diagram

$\xymatrix{ Rel_ A/TrivRel_ A \ar[r] & F_ A \otimes _{P_ A} (A \otimes _ R B) \ar[r] & \Omega _{P_ A/A \otimes _ R B} \otimes _{P_ A} (A \otimes _ R B) \\ Rel/TrivRel \ar[r] \ar[u]_{-2} & F \otimes _ P B \ar[r] \ar[u]_{-1} & \Omega _{P/A} \otimes _ P B \ar[u]_0 }$

Note that vertical maps $-1$ and $-0$ induce an isomorphism after applying the functor $A \otimes _ R - = P_ A \otimes _ P -$ to the source and the vertical map $-2$ gives exactly the map whose cokernel is the desired Tor module as we saw above. $\square$

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