7
$\begingroup$

I am interested in examples of rings (or triangulated categories) that have zero Grothendieck group but are somehow still interesting. More example, for what rings $R$ is the category of finitely-generated projective $R$-modules have zero Grothendieck group? I recently learned that the category of all modules has zero Grothendieck group: if $M$ is any $R$-module, then $M \oplus M^{\oplus \infty} \cong M^{\oplus \infty}$ and so $M$ is zero in the Grothendieck group. I also know that "infinite sum rings" with the property that $R \oplus R \cong R$ have vanishing Grothendieck group. I would like some "smaller" examples of rings with vanishing Grothendieck group. Also, such an $R$ cannot be commutative ring, since commutative rings have invariant basis property and hence have non-zero Grothendieck group.

It is easy to construct categories that are not split-closed where the Grothendieck group is zero. For example, suppose that some triangulated is generated by an object $A$. Then the subcategory generated by $A \oplus A[1]$ has zero Grothendieck group because the shift acts like $-1$ in the Grothendieck group; however this category is not split-closed since the projective object $A$ is not in the category. So I would like an example of a split-closed category that has zero Grothendieck group.

Maybe an easier but less concrete question: what does the vanishing of the Grothendieck group imply? By a note in a paper of Thomason, $D$ is zero in the Grothendieck group if there exist $A,B,C$ and exact triangles $A \rightarrow B \oplus D \rightarrow C \rightarrow$ and $A \rightarrow B \rightarrow C \rightarrow$ but this is not very enlightening. Does someone have another interpretation?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.