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

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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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## 2 Answers

In

MR2657430 (2011g:19003) Cortiñas, G.; Haesemeyer, C.; Walker, Mark E.; Weibel, C. Bass' NK groups and cdh-fibrant Hochschild homology. Invent. Math. 181 (2010), no. 2, 421–448.

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 $S=R[t]$.

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If $f:R\to S$ in map of finitely generated commutative $k$-algebras for some fields $k$ (maybe you should take $k$ of characteristic 0) and $K_{\ast}(f)$ induces an isomorphism than $K_{\ast}(f[t])$ induces an isomorphism! moreover $i:R\to R[t]$ induces an isomorphism in $K$-theory.

May be i missed somme condition on $k$ but this property is called "homotopy invariance of algebraic $K$-theroy.

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Algebraic K-theory is not homotopy invariant – Fernando Muro Oct 6 2011 at 15:58
Hi Fernando, If you take the category of smooth schemes over a noetherian scheme then $K$-theory is actually $\mathbb{A}^{1}$-invariant!! That is a general statement! – Ilias Oct 6 2011 at 16:35

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If $R$ is a regular ring, then the map $R \to R[t]$ induces an isomorphism on $K$-theory. – Ulrich Pennig Oct 6 2011 at 16:58
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