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 $\mathbb Z$?
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$\begingroup$ Do you mean $R$ to be commutative? $\endgroup$– YCorApr 24, 2020 at 15:50
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$\begingroup$ $\mathbb Q[X,Y]/(XY)$ localized at the origin. Or maybe you wanted a domain. $\endgroup$– Tom GoodwillieApr 24, 2020 at 16:02
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$\begingroup$ $\Bbb{Q}[[t^2,t^3]]$ should work. $\endgroup$– abxApr 24, 2020 at 16:10
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$\begingroup$ 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 $3$.) $\endgroup$– Tom GoodwillieApr 24, 2020 at 16:19
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$\begingroup$ @TomGoodwillie How one can compute K' of your example? $\endgroup$– Mykola PochekaiApr 24, 2020 at 16:24
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1 Answer
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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 $K_0(X)=\mathbb{Z}^2$.
