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.
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2$\begingroup$ That group is known as $G_0$ and the map you mention is an object of study. Many things can happen. Just take an introductory book on algebraic K-theory. $\endgroup$– Fernando MuroCommented Aug 18, 2012 at 1:31
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3 Answers
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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.
answered Aug 18, 2012 at 11:28
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2$\begingroup$ Isn't the Cartan homomorphism the map induced on the version of K-theory that also splits short exact sequences? from the way the question is phrased I would guess the question is about the group-completion K-theory, which only formally add direct-sum-inverses but do not splits exact sequences. If this is indeed the intention, this homomorphism is almost never surjective. This comment applies (and even more relevant) to all other answers as well. $\endgroup$ Commented Apr 2 at 12:51
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The map is an isomorphism iff every fg module has finite projective dimension iff (in the commutative case) R is regular.
answered Aug 17, 2012 at 19:20
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2$\begingroup$ The ring $R$ has to be noetherian for this to be true. Also, I think W. Politarczyk is mainly interested in the non-commutative case (group rings mostly are). $\endgroup$ Commented Aug 17, 2012 at 20:44
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$\begingroup$ I forgot to mention it, but for me $R$ is noetherian and I am mostly interested in the noncommutative case. $\endgroup$ Commented Aug 18, 2012 at 8:10
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Let $R = \mathbb Z_4$ (by which I mean integers modulo $4$). Then the abelian group $K_0(R)$ is freely generated by $[\mathbb Z_4]$ and $G(R)$ is freely generated by $[\mathbb Z_2]$, and we have $[\mathbb Z_4]=2 [\mathbb Z_2]$ in $G(R)$. Therefore, the canonical map $K_0(R) \to G(R)$ is an embedding of index $2$.