This question is motivated by some issue raised by David Speyer in [this question](https://mathoverflow.net/questions/27634/k-0r-mathbbz-but-some-f-g-projective-not-stably-free). 


Let $R$ be a ring. Let $K\_0(R)$ and $G\_0(R)$ be the Grothendieck groups of f.g. projective modules and f.g. modules over $R$, respectively (you just kill all relations generated by short exact sequences). There is a natural map, called the *Cartan homomorphism* (see Serre's "Linear reps of finite groups", Chapter 15) $$c: K\_0(R) \to G\_0(R)$$ given by forgetting a module is projective. 


In general, $c$ needs not be injective nor surjective. For non-surjectivity, take $R$ to be a local ring, then $K\_0(R)=\mathbb Z$ but $G\_0(R)$ can be huge (in particular, if $R$ is normal, $\mathbb Z\oplus \text{Cl}(R)$ is a quotient of $G\_0(R)$). Examples of non-injectivity can be found by taking $R$ to be some group rings, as the Cartan matrix is not always invertible, see for example Section 4 of [this paper](http://www.math.temple.edu/%7Elorenz/papers/reps.ps) by Martin Lorenz . But I don't know any commutative example of non-injectivity.


*Is $c$ always injective if $R$ is commutative? How about if $R$ is commutative and Noetherian?*


(If this is true, one can prove the original question quoted above with the assumption $G\_0(R)=\mathbb Z$) 



Seeking examples or proof: injectivity of Cartan homomorphism for commutative rings? 
No, it is not injective in general, unless $R$ is regular notherian. There are many counterexamples; for a simple one you can take the ring $R := \mathbb C[t^2, t^3] \subseteq \mathbb C[t]$, compute that $G\_0(R) = \mathbb Z$, while $K\_0(R)$ maps onto the Picard group of $R$, which is the additive group $\mathbb C$.


