# What are the inverses in the Grothendieck group of a module category?

Let $$R$$ be a unital commutative ring and $$\mathcal{C}$$ be the category of finitely generated $$R$$-modules. The Grothendieck group $$K_{0}(\mathcal{C})$$ is the free abelian group generated by the isomorphism classes $$[M]$$ of modules in $$\mathcal{C}$$ modded out by the relation $$[N] - [M] - [O]$$ whenever there exits a short exact sequence of $$R$$-module homomorphisms of the form $$0 \rightarrow M \xrightarrow{f} N \xrightarrow{g} O \rightarrow 0$$ The isomorphism class of the zero module $$[0]$$ is the identity in this group.

Now let $$[M]$$ be in $$K_{0}(\mathcal{C})$$ and so it must have an inverse, $$[N]$$, such that $$[0] = [M] + [N]$$ but this then implies there is an exact sequence $$0 \rightarrow M \xrightarrow{f} 0 \xrightarrow{g} N \rightarrow 0$$ and $$g$$ being surjective then implies $$N = 0$$.

Given a module (M) what is the inverse of ([M]) in $$K_{0}(\mathcal{C})$$?

• Isn't it just $-[M]$? – Orat Mar 16 at 16:17

Not every element of $$K_0(C)$$ has the form $$[N]$$. Indeed, as you said, $$K_0(C)$$ is a certain quotient of the free abelian group on the set of isomorphism classes of modules, so an element of $$K_0(C)$$ can be represented by any formal $$\mathbb{Z}$$-linear combination of modules. So, the inverse of $$[M]$$ is just the formal $$\mathbb{Z}$$-linear combination $$(-1)\cdot[M]$$.