# Do K-equivalent rings have isomorphic Nil-Terms?

Let $R,S$ be rings and $f: R \rightarrow S$ a ring homomorphism such that $f$ induces an isomorphism on the $K$-theory of the rings. The map $f$ also induces a ring homomorphism $f[t]: R[t] \rightarrow S[t]$, which presumably does not have to induce an isomorphism on K-theory. Equivalently, $f$ induces a map $Nil_i(R) \rightarrow Nil_i(S)$, which presumably does not have to be an isomorphism either. However, I have not been able to find an example of such rings. So I am looking for two rings $R,S$ and a ring homomorphism $f: R \rightarrow S$ such that $K_{\star}(f)$ is an isomorphism, but $K_{\star}(f[t])$ is not an isomorphism. Does anybody know anything about this?

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The authors exhibit a ring (actually an algebra over a field of characteristic 0) for which $K_{*}(R)= K_{*}(R[t])$ but $K_{*}(R)\neq K_{*}(R[t,x])$, so take just $S=R[t]$.