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

  • 1
    $\begingroup$ Isn't it just $-[M]$? $\endgroup$ – 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 $(-1)\cdot[M]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.