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Given an abelian category C of finite length (i.e., every object has a Jordan-Holder filtration), let $A$ be the set of isomorphism classes of simple objects. there's an obvious map

$\mathbb Z^A \to K_0(C)$

given by sending a vector in $\mathbb Z^A$ (the free abelian group on A) to the obvious class in the Grothendieck group. The map is a surjection, obviously, but what's the best way to show it's an injection?

That is, given some equation in the Grothendieck group

$\sum_i n_i [S_i] = 0$

where $S_i$ is a simple object and $n_i \in \mathbb Z$, how do I show that all the $n_i$ must be zero?

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Show that there is a splitting $K_0(C) \to \mathbb Z^A$. The Jordan-Holder theorems tell you that given any object $M$ in $C$ the multiplicities of the simple objects that occur in a filtration are well defined so send the class $[M]$ to the vector giving these multiplicities. You just need to show that these multiplicities are additive accross short exact sequences.

Then to see that this splits your map $\mathbb Z^A \to K_0(C)$ just show that the composition $\mathbb Z^A \to K_0(C) \to \mathbb Z^A$ is the identity on the basis elements of $\mathbb Z^A$ (the ones that map to the class of a simple object of $C$).

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  • $\begingroup$ Wonderful. Thank, you, Jim. $\endgroup$ – Guest Sep 18 '13 at 4:51

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