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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 $G_0(R)=\mathbb Z$)

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Why is it called the Cartan homomorphism? I've always followed the usual mathematical practice of calling something after someone who had nothing to do with it -- in this case, calling it "the Poincar\'e morphism", since it's related to K-theoretic Poincar\'e duality. – Allen Knutson Jun 11 '10 at 1:23
@Allen: I don't know the origin, but I added some references. – Hailong Dao Jun 11 '10 at 5:25
Bass, "Algebraic K-theory", Ch.9, S2, p.453 also calls it Cartan homomorphism (he considers both $K_0\to G_0$ and $K_1\to G_1$). – Victor Protsak Jun 11 '10 at 5:45
In the case where $R$ is the path algebra of a quiver, $G_0$ and $K_0$ are finite free $\mathbb{Z}$-modules of the same rank and each comes with a natural basis: the simples and the indecomposable projectives. Let $C$ be the matrix of the map $c$, in that basis. Then $C^{-1} + C^{-T}$ is the Cartan matrix of the quiver. I think (but have not seen anyone explicitly state) that this is the origin of the name. – David Speyer Jun 11 '10 at 11:53
Here $C^{-T}$ is shorthand for $(C^{-1})^T=(C^T)^{-1}$. – David Speyer Jun 11 '10 at 11:54
up vote 7 down vote accepted

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 $\mathbb C$.

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Dear Angelo: why is $G_0(R)=\mathbb Z$? Thanks. – Hailong Dao Jun 11 '10 at 7:11
Hint: Consider the short exact sequence coming from multiplication by $t^2$ on $R$ and conclude that finite length modules may be ignored. – Wilberd van der Kallen Jun 11 '10 at 8:23
Since $R$ is a 1-dimensional domain, it is generated by the class of $R$ and the classes of all $R/m$, where $m$ is a maximal ideal in $R$, so it is sufficient to show that the last are 0. Since every maximal ideal of $R$ is the restriction of one in $S := \mathbb C[t]$, and every maximal ideal in $S$ is principal, every $R/m$ is the cokernel of a homorphism $S \to S$, so it is 0 in K-theory. – Angelo Jun 11 '10 at 10:24
This computation looks related to the computation that the Picard group does not inject into the Weil divisor group in this case. As the Picard group does inject into the Weil divisor group for $R$ normal, one might conjecture that $G_0$ injects into $K_0$ for normal rings. Is this true? – David Speyer Jun 11 '10 at 11:59
Cool! This also gives an example David was looking for in the other question $G_0(R)=\mathbb Z$ but there are many projectives which are not stably free. – Hailong Dao Jun 11 '10 at 16:56

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