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This question is motived by this recent question.

$K_{0}(R)=\mathbb{Z}$ 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?

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up vote 10 down vote accepted

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.

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Hah! I've lost a few seconds trying to explain what goes wrong in the noncommutative situation! – Victor Protsak Jun 10 '10 at 5:19
Victor, this is not a race. – Angelo Jun 10 '10 at 5:34
I didn't imply it was, Angelo, in fact, I voted you up immediately, which is why you have an extra point as of now. But have you noticed an uncanny resemblance between our solutions? It's even more amazing, since I've completely rewritten mine twice before posting. – Victor Protsak Jun 10 '10 at 8:30
Yes, our solutions are identical. It is a very natural idea, so it is not so suprising. Sorry if I was brusque, I did not mean to be offensive, but sometimes I find that this reputation business generates a lot of competition, which I don't like. – Angelo Jun 10 '10 at 8:38
And I absolutely agree with you about the reputation business! It's a natural idea, but it took me some effort to come up with a polished presentation. I am proud that it came out nearly the same as the one done by a professional algebraic geometer (I am a representation theorist myself). – Victor Protsak Jun 10 '10 at 8:48

No. Since $K_0(R)=\mathbb{Z},$ the ring $R$ is not a direct product, i.e. $\operatorname{Spec} R$ is connected. Therefore, every projective module has constant rank. The rank function is an isomomorphism between $K_0(R)$ and $\mathbb{Z}$ which maps the free module of rank 1 $R$ into the generator 1 of $\mathbb{Z}$. Therefore, every projective module is stably free.

In your noncommutative example, $M^{\oplus n}\simeq R,$ so the "rank" of $M$ isn't an integer.

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Or if you use uniform rank instead of naive "rank", the rank of $R$ isn't 1. – Victor Protsak Jun 10 '10 at 8:34
You've lost to an angel. Not a big loss. – Wadim Zudilin Jul 5 '10 at 8:14

When $R$ is commutative, $K_0(R)$ is a commutative ring with multiplicative unit the class $[R]$ of $R$. The only ring structure with additive group $\mathbb{Z}$ is the familiar one, so every element of $K_0(R)$ is an integer multiple of $[R]$ - every finitely generated projective $R$-module is stably free.

When $R$ is non-commutative, there is no natural ring structure on $K_0(R)$ which explains how Rishi's phenomenon occurs.

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I'm used to the notation that $K_0(R)$ is the Grothendieck group of finitely generated modules modulo short exact sequences, and $K^0(R)$ is the one generated by locally free modules. Only $K^0(R)$ has a natural ring structure. But maybe this is not standard notation? – David Speyer Jun 10 '10 at 14:16
David, my use of $K_0(R)$ for the Grothendieck group of finitely generated projective modules (modulo direct sums) is completely standard; it is used in Bass's Algebraic K-Theory. – Robin Chapman Jun 10 '10 at 17:33

Under the mild condition that the rank of a free module is an invariant, the free modules form a monoid isomorphic to $\mathbb N$. This induces an injection: $$f: \mathbb Z \to K_0(R) $$

Then every projective is stably free is equivalent to $f$ being an isomorphism.

When $R$ is commutative, $f$ splits (since one can map $R$ to a field, which induces a map $K_0(R) \to \mathbb Z$ such that when compose to $f$ is the identity). So in this case $f$ is an isomorphism iff $K_0(R)\cong \mathbb Z$. It works whenever the map $f$ splits.

In your example, $f$ is the injection $2\mathbb Z \to \mathbb Z$.

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I'm going to disagree with all of the previous answers, and say that there is an example. This will relate to the issue of notation that I raise in my comment to Robin's answer.

Take $R = \mathbb{Z}/p^2$, for $p$ some prime. Every finitely generated $R$-module is a direct sum of modules of the form $\mathbb{Z}/p$ and $\mathbb{Z}/p^2$. From the short exact sequence $0 \to \mathbb{Z}/p \to \mathbb{Z}/p^2 \to \mathbb{Z}/p \to 0$, we get the relation $[\mathbb{Z}/p^2] = 2 [\mathbb{Z}/p]$ in $K_0(R)$. So $K_0(R)$ is generated by $[\mathbb{Z}/p]$. As length is an invariant on $K_0(R)$, there are no further relations. So $K_0(R) \cong \mathbb{Z}$, generated by $[\mathbb{Z}/p]$, and $\mathbb{Z}/p$ is not stably free.

There should be many similar examples whenever $R$ is not generically reduced.

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It depends of what you mean by $K_0$. The natural interpretation seemed to be that of the Grothendieck group of projective finitely generated modules; if you take all finitely generated modules, then of course you are right. – Angelo Jun 10 '10 at 14:33
I think $K_0$ for projectives is standard notation. The Grothendieck group of f.g. modules is often called $G_0$. – Hailong Dao Jun 10 '10 at 15:38
Just to clarify, by $K_{0}(R)$ I was referring to the Grothendieck group of the f.g. projectives, not all f.g. modules. – Rishi Vyas Jun 10 '10 at 17:37
Victor, that reminds me. I've seen this ring turn up in another answer… :-) – Robin Chapman Jun 10 '10 at 20:23
$K_0(R)=K^0(\textrm{Spec}R)$ so the variance make sense both ways. – Ben Wieland Jun 10 '10 at 23:23

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