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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