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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 $K_{0}(\mathcal{C})$?

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    $\begingroup$ Isn't it just $-[M]$? $\endgroup$ – Orat Mar 16 at 16:17
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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 $(-1)\cdot[M]$.

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