K0 of a ring

Remarks

Every element of $K_0(R)$ has the form $[P] - [R^n]$ for some finitely generated projective $P$ and some $n \ge 0$. Two modules satisfy $[P] = [Q]$ in $K_0(R)$ exactly when $P \oplus R^n \cong Q \oplus R^n$ for some $n \ge 0$. Equivalently, $K_0(R)$ can be built from idempotent matrices over $R$, since each finitely generated projective module is $R^n e$ for an idempotent $e \in M_n(R)$. A ring homomorphism $R \to S$ induces $K_0(R) \to K_0(S)$ through $P \mapsto S \otimes_R P$.

Example: for a field $F$, every finitely generated module is a vector space $F^n$, and $F^n \cong F^m$ only when $n = m$. So $\mathrm{Proj}(F) \cong \N$ and $K_0(F) \cong \Z$ through the dimension. For $R = \Z/6\Z \cong \Z/2\Z \times \Z/3\Z$, one gets $K_0(R) \cong K_0(\Z/2\Z) \times K_0(\Z/3\Z) \cong \Z^2$.

Non-example: the Grothendieck group of all finitely generated modules under direct sum is a different group. For $R = \Z$ it contains $[\Z/2\Z]$, which is not of the form $[P] - [Q]$ with $P$, $Q$ projective, because projective $\Z$-modules are free and have no torsion. Using short exact sequences as relations instead gives the group $G_0(R)$ of G-theory.