For a Noetherian ring $R$, there seem to be two versions of zeroth K-theory one can associate to it: $K_0(R)$ the Grothendieck group of the exact category of projective modules and $G_0(R)$ the Grothendieck group of the abelian category of finitely-generated modules. There is a map $K_0(R) \rightarrow G_0(R)$ and if $R$ is a regular ring, this map is an isomorphism. What is an example of a (non-regular) ring $R$ such that $K_0(R)$ and $G_0(R)$ are not isomorphic?
(I know that if $R = k[x]/(x^n)$, then $K_0(R), G_0(R)$ are both abstractly isomorphic to $\mathbb{Z}$ but the map above is multiplication by $n$ and hence not an isomorphism. I would an example of $R$ where the two Grothendieck groups are not even abstractly isomorphic).
Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)\cong\mathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $\mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)\to G_0(R_x)\times G_0(R_y)$. We have $M_x=0$ and $M_y\cong R_y$, and $N_x\cong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)\times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.