Question

In algebraic K-theory one defines $K_0(R)$ as the result of application of the Grothendieck construction to the semigroup of isomorphism classes of left f.g. projective $R$-modules. But we can also consider the category of left f.g. $R$-modules and apply the same construction to obtain a group (let's call it $G(R)$). What can we say about the canonical homomorphism $$K_0(R) \to G(R)$$ induced by the inclusion. Is it onto? or maybe is it an isomorphism? I am mostly interested in the case when $R$ is a group ring of a finitely presented group.

Answer 1

Typically the homomorphism here $K_0 \rightarrow G_0$ fails to be surjective. This shows up in a wide range of examples involving group algebras of finite groups (over fields of characteristic dividing the group order), restricted enveloping algebras of modular Lie algebras, etc. The homomorphism itself is often called the Cartan homomorphism: in some of his early work, Elie Cartan studied the regular representation of a finite dimensional algebra ("hypercomplex system"). See for example Chapter IX.2 of the pioneering 1968 Benjamin lecture notes Algebraic K-Theory by Hyman Bass.

The Grothendieck group formalism was popularized in modular representation theory of finite groups by Serre in his lecture notes (later translated into English as a volume in the Springer GTM series). That's one of many sources for concrete examples showing why the Cartan map is usually not surjective; Curtis-Reiner is another source. There are also many easy examples in the books and papers on representations of algebraic groups and their Lie algebras in prime characteristic, with analogues in characteristic 0 for quantum groups at a root of unity.

I'd emphasize that there are many different situations involving rings and finitely generated projectives, but only in fairly simple cases is the Cartan map likely to be surjective.

Answer 2

The map is an isomorphism iff every fg module has finite projective dimension iff (in the commutative case) R is regular.