# Does the Grothendieck ring inject to the Grothendieck group?

To a Noetherian scheme $X$, we can associate a Grothendieck ring of locally free coherent sheaves, and a Grothendieck group of coherent sheaves, with a natural map from the former to the latter. Is this map always injective?

Here's the question in more detail:

First consider the free abelian group on generators $[M]$ where the $M$ are coherent sheaves on $X$. Now for every exact sequence $0\to M'\to M \to M''\to 0$ of coherent sheaves, we mod out by an "exact relation" $[M'] - [M] + [M'']$, and call the result $K_{coh}(X)$ the Grothendieck group of $X$. Intuitively, we're "taking apart" $M$ by declaring it to be "made up" of $M'$ and $M''$, without remembering "how they're put together".

Now repeat the same construction with coherent sheaves replaced by locally free coherent sheaves (essentially vector bundles). Since tensoring with locally frees is exact, this group is a ring under the operation of tensoring, and also acts on $K_{coh}(X)$ by tensoring. Call this ring $K_{lfc}(X)$, the Grothendieck ring of $X$.

There is an inclusion map from the free abelian group on locally free coherent sheaves to the free abelian group on coherent sheaves, and it takes the exact relations to exact relations.

Hence we have a natural map $\phi: K_{lfc}(X)\to K_{coh}(X)$ of $K_{lfc}(X)$-modules. Is this map always injective?

This amounts to asking if the exactness relations for $K_{coh}$ can be added up in some way that cancels to leave a relation involving only locally free sheaves that could not have been obtained from exact sequences on locally free coherent sheaves alone.

This yields a vague "moral" interpretation of the question, which is what makes it interesting to me:

Does "taking apart" coherent sheaves allow us to "take apart" locally free coherent sheaves in new ways?

Cases where I know $\phi$ is injective:

1) If $X$ is a regular separated Noetherian scheme, e.g. a smooth variety, then $\phi$ is an isomorphism. See this question by Ariyan and its answers.

2) If $X=Spec(R)$ where $R$ is a Noetherian local domain, we can define the rank of a coherent $R$-module as its dimension when localized to the quotient field. We can extend this by linearity to the whole Grothendieck group (it's zero on exact relations, by dimension counting) to get a map $rank: K_{lfc} \to \mathbb{Z}$. But since $R$ is local, this induces an isomorphism $rank: K_{lfc} \to \mathbb{Z}$, and since it factors through $K_{coh}$, the map $\phi$ must be injective.

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Dear Andrew, the answer is no, see Angelo's answer to my question here:mathoverflow.net/questions/27747/… – Hailong Dao Feb 23 '11 at 8:01
Ah, thank-you!! – Andrew Critch Feb 23 '11 at 18:52