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
Explanation: The middle row is the fundamental triangle (92.7.0.1) for the ring maps $R \to A \to A \otimes _ R B$. The middle column is the fundamental triangle (92.7.0.1) 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
and get the two maps with target $E$ by an application of TR3. In the Tor independent case the object $E$ is zero.
Lemma 92.15.1. If $A$ and $B$ are Tor independent $R$-algebras, then the object $E$ in (92.15.0.1) is zero. In this case we have 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.
\[ L_{A \otimes _ R B/R} = L_{A/R} \otimes _ A^\mathbf {L} (A \otimes _ R B) \oplus L_{B/R} \otimes _ B^\mathbf {L} (A \otimes _ R B) \]
Proof. The first two statements are immediate from Lemma 92.6.2. 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 92.15.2. Let $R$ be a ring and let $A$, $B$ be $R$-algebras. The object $E$ in (92.15.0.1) 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 92.13.3 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 (92.13.0.1). These isomorphisms are compatible with functoriality (Remark 92.13.4). 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
and standard short exact sequences for Tor we obtain an exact sequence
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
The map of Lichtenbaum-Schlessinger complexes is given by the diagram
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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