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Let $R$ be a ring (not necessary commutative) and let $P_{\bullet}$ be a perfect $R$-bimodule (chain complex). I will denote the category of perfect right $R$-chain complexes by $\textbf{Perf}(R)$. The endofunctor $-\otimes_{R}P_{\bullet} :\textbf{Perf}(R)\rightarrow \textbf{Perf}(R)$ induces a map in algebraic $K$-theory given by

$K_{\ast}(-\otimes_{R}P_{\bullet}):K_{\ast}(R)\rightarrow K_{\ast}(R)$.

If the class $[P_{\bullet}] \in K_{0}(R)$ is trivial $(=0)$ does it mean that $K_{\ast}(-\otimes_{R}P_{\bullet})$ is a 0 map ?

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  • $\begingroup$ When you write $[P_{\bullet}] \in K_{0}(R)$, do you mean the class of $P_\bullet$ considered as a complex of right $R$-modules (forgetting the left $R$-module structure)? $\endgroup$
    – Jeremy Rickard
    Commented Jun 14, 2020 at 20:09
  • $\begingroup$ @JeremyRickard Yes $\endgroup$
    – M. Cousto
    Commented Jun 14, 2020 at 20:14

1 Answer 1

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No. Let $R=\mathbb{Z}\times\mathbb{Z}$, let $P$ and $Q$ be the projective modules $\mathbb{Z}\times0$ and $0\times\mathbb{Z}$, and let $$P_\bullet=\dots\longrightarrow0\longrightarrow P\otimes_\mathbb{Z}P \stackrel{0}{\longrightarrow}Q\otimes_\mathbb{Z}P\longrightarrow0\longrightarrow\dots$$

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