Let $R$ be a noetherian (commutative) ring. It is a well-known fact that for $R$ regular, $K$-theory of (finitely generated) projective modules and $K$-theory of arbitrary (f.g.) modules agree. Does the *converse* hold, i.e., suppose the natural map


$$K\_i(Proj R-Mod) \rightarrow K\_i(R-Mod)$$


is an isomorphism for all $i \geq 0$ (or just for $i=0$, maybe). Is $R$ regular then?



K-theory and regular rings
I don't know how to answer that question, but suppose we strengthen the hypothesis by making it apply also to all polynomial rings $R[t\_1,\dots,t\_n]$ over $R$, for $n \ge 0$. Then because $K\_i(R-Mod) \to K\_i(R[t]-mod)$ is an isomorphism (Quillen, Theorem 8, Higher Algebraic K-theory:I) for all noetherian rings $R$, it follows that $K\_i(R) \to K\_i(R[t\_1,\dots,t\_n])$ is an isomorphism for all $n$ --- that is called $K\_i$-regularity of $R$. The paper "$K$-regularity, cdh-fibrant Hochschild homology, and a conjecture of Vorst", J. Amer. Math. Soc. 21 (2008), no. 2, 547–561, by Cortiñas, Haesemeyer, and Weibel available at <http://www.math.uiuc.edu/K-theory/0783/chwpost.pdf> shows that $K\_i$-regularity of $R$ for all $i$ implies regularity of $R$ if $R$ is an algebra essentially of finite type over a field of characteristic 0.


