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.
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.
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
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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