Lemma 15.81.1. Let R be a ring. Let K^\bullet be a complex of R-modules. Consider the R-algebra map R[x] \to R which maps x to zero. Then
K^\bullet \otimes _{R[x]}^{\mathbf{L}} R \cong K^\bullet \oplus K^\bullet [1]
in D(R).
Proof. Choose a K-flat resolution P^\bullet \to K^\bullet over R such that P^ n is a flat R-module for all n, see Lemma 15.59.10. Then P^\bullet \otimes _ R R[x] is a K-flat complex of R[x]-modules whose terms are flat R[x]-modules, see Lemma 15.59.3 and Algebra, Lemma 10.39.7. In particular x : P^ n \otimes _ R R[x] \to P^ n \otimes _ R R[x] is injective with cokernel isomorphic to P^ n. Thus
P^\bullet \otimes _ R R[x] \xrightarrow {x} P^\bullet \otimes _ R R[x]
is a double complex of R[x]-modules whose associated total complex is quasi-isomorphic to P^\bullet and hence K^\bullet . Moreover, this associated total complex is a K-flat complex of R[x]-modules for example by Lemma 15.59.4 or by Lemma 15.59.5. Hence
\begin{align*} K^\bullet \otimes _{R[x]}^{\mathbf{L}} R & \cong \text{Tot}(P^\bullet \otimes _ R R[x] \xrightarrow {x} P^\bullet \otimes _ R R[x]) \otimes _{R[x]} R = \text{Tot}(P^\bullet \xrightarrow {0} P^\bullet ) \\ & = P^\bullet \oplus P^\bullet [1] \cong K^\bullet \oplus K^\bullet [1] \end{align*}
as desired. \square
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