This question is motived by this recent [question](https://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free/).


$K\_{0}(R)=\mathbb{Z}$ is often used as a euphemism for saying that every finitely generated projective module is stably free; however, there are some subtleties involved. 


The statement that every finitely generated projective module is stably free is equivalent to saying that $K\_{0}(R)$ is generated by the isomorphism class of $R$. To show that the two are not the same, consider $R = M\_{2}(\mathbb{C})$, $2\times 2$ matrices over the complex numbers. Devissage tells us that $K\_{0}(R)=\mathbb{Z}$, with the unique simple module as a generator. However, every stably free projective has to have even length as an $R$ module, as length($R$)=2, and thus the unique simple module (projective as $R$ is semisimple) is not stably free.


Are there any commutative examples of this phenomena? More precisely, is there a commutative ring with $K\_{0}(R)=\mathbb{Z}$, but with f.g. projectives which aren't stably free? A commutative noetherian one? 



$K_{0}(R) =\mathbb{Z}$ but some f.g. projective not stably free?
If $R$ is a commutative ring with $K\_{0}(R)=\mathbb{Z}$, then $\mathop{\rm Spec} R$ is connected, because otherwise $R$ would split as a product, and $K\_{0}(R)$ would contain a copy of $\mathbb{Z} \oplus \mathbb{Z}$. Hence there is a well defined surjective rank homomorphism $K\_{0}(R) \to \mathbb{Z}$, which must then be an isomorphism. Since $R$ has rank 1, it follows that the class of $R$ generates. This implies that every projective module is stably free.


