4
$\begingroup$

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 ?

$\endgroup$
2
  • $\begingroup$ 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)? $\endgroup$ Jun 14 '20 at 20:09
  • $\begingroup$ @JeremyRickard Yes $\endgroup$
    – M. Cousto
    Jun 14 '20 at 20:14
8
$\begingroup$

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 $$P_\bullet=\dots\longrightarrow0\longrightarrow P\otimes_\mathbb{Z}P \stackrel{0}{\longrightarrow}Q\otimes_\mathbb{Z}P\longrightarrow0\longrightarrow\dots$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.