# Examples of noetherian local rings $R$ such that $K'_0(R)$ is not isomorphic to $\mathbb Z$

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 $$\mathbb Z$$?

• Do you mean $R$ to be commutative?
– YCor
Apr 24, 2020 at 15:50
• $\mathbb Q[X,Y]/(XY)$ localized at the origin. Or maybe you wanted a domain. Apr 24, 2020 at 16:02
• $\Bbb{Q}[[t^2,t^3]]$ should work.
– abx
Apr 24, 2020 at 16:10
• 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 $3$.) Apr 24, 2020 at 16:19
• @TomGoodwillie How one can compute K' of your example? Apr 24, 2020 at 16:24

## 1 Answer

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 $$K_0(X)=\mathbb{Z}^2$$.