TITLE: Seeking examples or proof: injectivity of Cartan homomorphism for commutative rings? 
QUESTION [5 upvotes]: This question is motivated by some issue raised by David Speyer in this question. 
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 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 $G_0(R)=\mathbb Z$)

REPLY [9 votes]: 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 $\mathbb C$.