Reduced K0

Remarks

A finitely generated projective module $P$ has class $0$ in $\tilde{K}_0(R)$ exactly when it is stably free. Indeed, if $[P] = n[R]$ with $n \in \Z$, then $P \oplus R^a \oplus Q \cong R^b \oplus Q$ for some $a, b \ge 0$ with $b - a = n$ and some finitely generated projective $Q$. Adding a module $Q'$ with $Q \oplus Q' \cong R^k$ gives $P \oplus R^{a+k} \cong R^{b+k}$. So $\tilde{K}_0(R) = 0$ exactly when every finitely generated projective $R$-module is stably free.

If $R$ is commutative and nonzero with $\mathrm{Spec}\, R$ connected, for instance an integral domain, the rank gives a split surjection $K_0(R) \to \Z$, and $\tilde{K}_0(R)$ is isomorphic to its kernel. Then $K_0(R) \cong \Z \oplus \tilde{K}_0(R)$.

Example: for a field $F$, $K_0(F) = \Z [F]$, so $\tilde{K}_0(F) = 0$.

Non-example of vanishing: for $R = \Z/6\Z$, $K_0(R) \cong \Z^2$ with $[R] = (1, 1)$, so $\tilde{K}_0(R) \cong \Z$. The generator is the class of $\Z/2\Z$, which is projective but not stably free.