What is an example of an abelian/exact/triangulated category that has Grothendieck group isomorphic to $\mathbb{Z}/k\mathbb{Z}$ for some $k$? The category of finitely generated abelian groups has Grothendieck group $\mathbb{Z}$. I think it is easy to construct categories with Grothendieck group $\mathbb{Z}/2\mathbb{Z}$, for example, if a generator $X$ is isomorphic to $X[1]$. I would like to construct an example that has Grothendieck group $\mathbb{Z}/k\mathbb{Z}$, ideally by localizing a category I already understand.

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    $\begingroup$ This question got a bounty because Tobias Kildetoft drew attention to it at the Pearl Dive. $\endgroup$ – Jyrki Lahtonen Feb 12 at 10:14

For triangulated categories, there are naturally occurring examples coming from stable module categories. For example, let $A=k[x]/(x^n)$ for a field $k$. The stable module category of finitely generated $A$-modules (i.e., the category formed from the module category by factoring by the ideal of maps that factor through projective modules, or this can be constructed as the quotient in the sense of triangulated categories of the bounded derived category $D^b(\text{mod }A)$ by the subcategory of bounded complex of perfect complexes) is a triangulated category whose Grothendieck group is $\mathbb{Z}/n\mathbb{Z}$. This is roughly because the Grothendieck group of the category of finitely generated modules is $\mathbb{Z}$, generated by the class $[k]$ of the one-dimensional module, and the class of the unique indecomposable projective module is $n[k]$.

For abelian categories, I don't think I know a simpler example than the following.

Let $R=k\langle x_0,\dots,x_n\rangle$ be the free algebra over a field $k$ on $n+1$ (noncommuting) generators. Then the category $\text{mod }R$ of finitely presented right modules is abelian, and the quotient by the Serre subcategory of finite dimensional modules has Grothendieck group $\mathbb{Z}/n\mathbb{Z}$.

A proof can be found in Proposition 3.2 of

Ara, Pere, Finitely presented modules over Leavitt algebras., J. Pure Appl. Algebra 191, No. 1-2, 1-21 (2004). ZBL1072.16012,

but the idea is that the Grothendieck group of $\text{mod }R$ is $\mathbb{Z}$, generated by the class $[R]$ of the regular module, and by a theorem of Lewin, every finite dimensional module $M$ has a free resolution $$0\to R^m\to R^s\to M\to0$$ where $m=n\dim(M)+s$, so the class of $[M]$ is a multiple of $n[R]$.

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  • $\begingroup$ Thanks for the examples! Do you know any examples where $R$ is a finitely-generated $k$-algebra? $\endgroup$ – user39598 Dec 13 '18 at 15:14
  • $\begingroup$ @user39598 The $R$ in my answer is a finitely generated $k$-algebra. $\endgroup$ – Jeremy Rickard Dec 13 '18 at 20:14

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