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?

  • $\begingroup$ I would guess that when $R$ is not regular then the map of $K_0$ groups is not surjective: that, for example, the image does not contain the element determined by $k=R/\frak m$ when $m$ is any maximal ideal such that the local ring is not regular. It would be enough to prove this when $R$ is local. Maybe it can be done using $Tor(k,k)$ somehow? Does anyone know? $\endgroup$ Dec 1, 2011 at 22:28
  • $\begingroup$ @Tom: for $R=k[[x,y]]/(x^2-y^3)$ the map between $K_0$ is an isomorphism. $\endgroup$ Dec 1, 2011 at 22:51
  • $\begingroup$ Oh yes, I can see that. I was being very naive: no surprise. $\endgroup$ Dec 1, 2011 at 23:19

1 Answer 1


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.


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