Does there exist a simple example of a commutative noetherian local ring $R$ such that $K'_0(R) = K_0(\mbox{Mod-}R)$ (by $\mbox{Mod-}R$ I mean the abelian category of finitely generated $R$-modules) is not isomorphic to $\mathbb Z$?

  • $\begingroup$ Do you mean $R$ to be commutative? $\endgroup$
    – YCor
    Apr 24, 2020 at 15:50
  • $\begingroup$ $\mathbb Q[X,Y]/(XY)$ localized at the origin. Or maybe you wanted a domain. $\endgroup$ Apr 24, 2020 at 16:02
  • $\begingroup$ $\Bbb{Q}[[t^2,t^3]]$ should work. $\endgroup$
    – abx
    Apr 24, 2020 at 16:10
  • $\begingroup$ Does it? It's clear that the group is generated by the free module of rank one and the residue field. But the latter is trivial in the group. (There is a principal ideal of codimension $2$ and another of codimension $3$.) $\endgroup$ Apr 24, 2020 at 16:19
  • $\begingroup$ @TomGoodwillie How one can compute K' of your example? $\endgroup$ Apr 24, 2020 at 16:24

1 Answer 1


This is just to flesh out @Tom Goodwillie example.

For any reasonable scheme $X$ and an open set $U$, one has a natural exact sequence,

$$K_0(X-U)\to K_0(X)\to K_0(U)\to 0.$$

Taking $X=\operatorname{Spec} (\mathbb{Q}[x,y]/xy)_{(x,y)}$ (or a number of similar examples) and $U$ the punctured spectrum, we note that the punctured spectrum is two points and thus $K_0(U)=\mathbb{Z}^2$. The kernel is generated by the closed point, but going mod $x+y, x+y^2$, one can easily see that 2 and 3 times the closed point is zero in $K_0(X)$. So, we get $K_0(X)=\mathbb{Z}^2$.


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.