# When is $K_0(i)$ an injection?

Suppose that $\mathcal A$ and $\mathcal B$ are two abelian categories such that $\mathcal A$ is a full subcategory of $\mathcal B$. If $i: \mathcal A\rightarrow\mathcal B$ is the inclusion functor, when is the induced map $K_0(i): K_0(\mathcal A)\rightarrow K_0(\mathcal B)$ injective? In the situation I am working in, we can assume that $\mathcal A$ is a category in which there are finitely many simple objects and every object has finite length.

Of course, we can use Bass' Resolution Theorem, provided every object of $\mathcal B$ has a resolution by objects in $\mathcal A$ and $\mathcal A$ contains the kernels of its morphisms which are epimorphisms in $\mathcal B$ (see, for example, [$\textit{Algebraic K-Theory and its Applications}$, Rosenberg, thm. 3.1.13]).

Indeed. It seems that situations in which this happens are perhaps not very common. From Auslander and Reiten's paper $\textit{Grothendieck Groups of Algebras and Orders}$, we obtain that the the sequence (proposition 2.4)

$$0\rightarrow K_0(\textbf{mod}\mathcal C/\mathcal D)\rightarrow K_0(\textbf{mod}\mathcal C)\rightarrow K_0(\textbf{mod}\mathcal D)\rightarrow 0$$

is split exact and under the assumption that $(\mathcal C, \mathcal D)$ is a $\textit{coherent pair}$ (And $\mathcal D$ is required to be a skeletally small additive $R$-category) such that either $\textbf{mod}(\mathcal D)$ is regular (ie. every object has finite projective dimension) or every object in $\mathcal D$ has finite length. Here, $\textbf{mod}(\mathcal D)$ is the category of finitely presented contravariant $R$-linear functors.

Let me describe the situation I am most interested in. We are working with the following assumptions about our ground ring:

(1) $R$ is a local Cohen-Macaulay ring of finite Cohen-Macaulay type (ie. $R$ admits only finitely many indecomposable MCM $R$-modules)

(2) $R$ is Henselian

(3) $R$ admits a dualizing module.

Our relevant categories are $\mathcal C = \textbf{mod}(\text{MCM}\hspace{.1 cm}R)$ and $\mathcal D = \textbf{mod}(\text{proj}\hspace{.1 cm}R)$ (this is a coherent pair). In this case, the restriction functor $r: \mathcal C\rightarrow\mathcal D$ is such that $\text{ker}(r)$ is naturally equivalent to $\textbf{mod}(\mathcal C/\mathcal D)$. What I would like to know is when $K_0(i): K_0(\text{ker}(r))\rightarrow K_0(\mathcal C)$ is an injection. So our question becomes, when is $\mathcal D = \textbf{mod}(\text{proj}\hspace{.1 cm}R)$ when do each of its objects have finite length or when is it regular?

By [$\textit{K-groups for Rings of Finite Cohen-Macaulay Type}$, H. Holm, Observation (5.3)] $\textbf{mod}(\text{proj}\hspace{.1 cm} R)$ is equivalent $\text{modfg}\hspace{.1 cm} R$, so in either of those cases, the ring $R$ happens to be regular or Artinian. It would be of interest to know if there are situations in in which $K_0(i)$ is an injection other than when $R$ is regular or Artinian.

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I don't think there's much you can say in general. It is very easy for the induced map to be trivial because $K_0(B)$ is trivial, e.g. consider the inclusion of f.g. abelian groups into abelian groups. –  Qiaochu Yuan Aug 29 '13 at 2:22
I realized it was much too general. I edited my post to include specifics. –  Zach Flores Aug 29 '13 at 3:33