The Stacks project

Lemma 15.58.4. 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 15.57.12). 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 15.58.2 applied to the material inside the brackets. By Lemmas 15.57.5 and 15.57.6 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 12.22.8. 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 15.58.1. 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 15.57.16 and 15.57.17 as well as the previously mentioned lemma. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08YU. Beware of the difference between the letter 'O' and the digit '0'.