The following two statements are known as Bass conjecture for $K_0$ (resp. $G_0$) :
(1) If $R$ is a finitely generated $\mathbb Z$-algebra ( similarly $X$ is a $\mathbb Z $-scheme of finite type ) , then $G_0(R)$ (resp. $G_0(X) $ ) is a finitely generated abelian group.
(2) If $R$ is a regular, finitely generated $\mathbb Z$-algebra ( similarly $X$ is a regular $\mathbb Z $-scheme of finite type ) , then $K_0(R)$ (resp. $K_0(X) $ ) is a finitely generated abelian group.
My question is : Are these two statements equivalent ?
I think they are, but I can't quite prove it ... I know that if $X$ is a separated, Noetherian, regular scheme then $K_0(X)\cong G_0(X)$, but unfortunately, the statement (1) above doesn't only concern regular schemes...
Please help
Yes, 2) implies 1). Scheme statement can be easily reduced to the affine case, so let me do the affine case. Proof is standard devissage, inducting on the Krull dimension. Since all finitely generated modules can be filtered by $R/P$ for prime ideals $P$, you are reduced to the case of a domain. If $R$ is a domain, then you can find a non-zero $f\in R$ such that $R_f$ is regular. Then you have an exact sequence $G_0(R/f)\to G_0(R)\to G_0(R_f)\to 0$, but the last is finitely generated by 2). $G_0(R/f)$ is finitely generated since $\dim R/f<\dim R$.