Lemma 15.60.5. Let $A \to B \to C$ be ring maps. Let $N^\bullet$ be a complex of $B$-modules and $K^\bullet$ a complex of $C$-modules. The compositions of the functors

$D(A) \xrightarrow {- \otimes _ A^\mathbf {L} N^\bullet } D(B) \xrightarrow {- \otimes _ B^\mathbf {L} K^\bullet } D(C)$

is the functor $- \otimes _ A^\mathbf {L} (N^\bullet \otimes _ B^\mathbf {L} K^\bullet ) : D(A) \to D(C)$. If $M$, $N$, $K$ are modules over $A$, $B$, $C$, then we have

$(M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K = M \otimes _ A^\mathbf {L} (N \otimes _ B^\mathbf {L} K) = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} K)$

in $D(C)$. We also have a canonical isomorphism

$(M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K \longrightarrow (M \otimes _ A^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C)$

using signs. Similar results holds for complexes.

Proof. Choose a K-flat complex $P^\bullet$ of $B$-modules and a quasi-isomorphism $P^\bullet \to N^\bullet$ (Lemma 15.59.10). Let $M^\bullet$ be a K-flat complex of $A$-modules representing an arbitrary object of $D(A)$. Then we see that

$(M^\bullet \otimes _ A^\mathbf {L} P^\bullet ) \otimes _ B^\mathbf {L} K^\bullet \longrightarrow (M^\bullet \otimes _ A^\mathbf {L} N^\bullet ) \otimes _ B^\mathbf {L} K^\bullet$

is an isomorphism by Lemma 15.60.2 applied to the material inside the brackets. By Lemmas 15.59.3 and 15.59.4 the complex

$\text{Tot}(M^\bullet \otimes _ A P^\bullet ) = \text{Tot}((M^\bullet \otimes _ R A) \otimes _ A P^\bullet$

is K-flat as a complex of $B$-modules and it represents the derived tensor product in $D(B)$ by construction. Hence we see that $(M^\bullet \otimes _ A^\mathbf {L} P^\bullet ) \otimes _ B^\mathbf {L} K^\bullet$ is represented by the complex

$\text{Tot}(\text{Tot}(M^\bullet \otimes _ A P^\bullet )\otimes _ B K^\bullet ) = \text{Tot}(M^\bullet \otimes _ A \text{Tot}(P^\bullet \otimes _ B K^\bullet ))$

of $C$-modules. Equality by Homology, Remark 12.18.4. Going back the way we came we see that this is equal to

$M^\bullet \otimes _ A^\mathbf {L} (P^\bullet \otimes _ B^\mathbf {L} K^\bullet ) \longleftarrow M^\bullet \otimes _ A^\mathbf {L} (N^\bullet \otimes _ B^\mathbf {L} K^\bullet )$

The arrow is an isomorphism by definition of the functor $-\otimes _ B^\mathbf {L} K^\bullet$. All of these constructions are functorial in the complex $M^\bullet$ and hence we obtain our isomorphism of functors.

By the above we have the first equality in

$(M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K = M \otimes _ A^\mathbf {L} (N \otimes _ B^\mathbf {L} K) = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} K)$

The second equality follows from the final statement of Lemma 15.60.1. The same thing allows us to write $N \otimes _ B^\mathbf {L} K = (N \otimes _ B^\mathbf {L} C) \otimes _ C^\mathbf {L} K$ and substituting we get

\begin{align*} (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K & = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} ((N \otimes _ B^\mathbf {L} C) \otimes _ C^\mathbf {L} K) \\ & = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (K \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C)) \\ & = ((M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C)) \\ & = (M \otimes _ C^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C) \end{align*}

by Lemmas 15.59.14 and 15.59.15 as well as the previously mentioned lemma. $\square$

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