A ring such that all projectives are stably free but not all projectives are free?

Question

This question is motivated by this recent question. Suppose $R$ is commutative, Noetherian ring and $M$ a finitely generated $R$-module. Let $FD(M)$ and $PD(M)$ be the shortest length of free and projective resolutions (with all modules f.g.) of $M$ respectively.

We will be interested in two conditions on $R$:

(1) For all f.g. $M$, $PD(M)<\infty$ if and only if $FD(M)<\infty$.

(2) For all f.g. $M$, $PD(M)=FD(M)$.

It is not hard to see that one is equivalent to (3): all f.g. projective modules are stably free. Also, (2) is equivalent to (4): all f.g projectives are free. It is natural to ask whether (1) and (2) are equivalent. I don't think they are, but can't find a counter-example. Thus:

Can we find a commutative Noetherian ring which satisfies (3) but not (4)?

Some thoughts: (3) is equivalent to the Grothendick group of projective $K_0(R)=\mathbb Z$. Well-known class of rings satisfying (3): local rings or polynomial rings over fields.

Of course, there are well known rings which fails (4): coordinate rings of $n$-spheres $R_n=\mathbb R[x_0,\cdots,x_n]/(x_0^2+\cdots+x_n^2-1)$ for $n$ even. Unfortunately, for those rings we have $K_0(R_n)=\mathbb Z^2$. (except for $n=8r+6$ which do provide examples, see below)

UPDATE: it seems to me that Tyler's answer suggests $R_5$ may work, but one needs to check some details.

UPDATE 2: I think one can put together a more or less complete solution, based on Tyler and Torsten's answers, the comments and some digging online.

Claim: $R_n$ satisfies (3) but not (4) if $n=5,6$ or $n=8r+k$ with $r>0$, $k\in \{3,5,6,7\}$

Proof: Let $S^n$ be the $n$-sphere and $C(S^n)$ be the ring of continuous functions.

To show that $R_n$ satisfies (3), it suffices to show $K_0(R_n)=\mathbb Z$ (Lemma 2.1, Chapter 2 of Weibel's K-book) or the reduced group $\tilde K_0(R_n)=0$. But one has isomorphism (known to Swan, see for example 5.7 of this paper): $$\tilde K_0(R_n) \cong \tilde KO_0(S^n) $$ the reduced $K$-groups of real vector bundles, and it is well-known that $\tilde KO_0(S^n) =0$ if $n\equiv 3,5,6,7 (\text{mod} 8)$.

It remains to show our choices of $R_n$ fail (4). Let $T_n$ be the kernel of the surjection $R_n^{\oplus n+1} \to R$ defined by the column of all the $x_i$s. $R$ is projective, so the sequence splits, thus $T_n$ is stably free.

On the other hand, $R_n$ embeds in $C(S^n)$ and tensoring with $C(S^n)$ turns $T_n$ into the tangent bundle of $S^n$. If $T_n$ is free, said bundle has to be trivial. But $S_n$ has trivial tangent bundle if and only if $n=1,3,7$.

While it is not surprising that topological machinery is helpful, I would still be interested in seeing a pure algebraic example, so if you have one, please post.

Answer 1

The more canonical example probably is the standard universal example for such a question. So, let $R_n=k[x_i,y_i]/\sum x_iy_i=1$ where $k$ is any field and there are $2n$ variables. By localization one easily checks that $K_0(R_n)=\mathbb{Z}$ for any $n$. But the projective module given by the presentation, $$0\to R_n\stackrel{(x_i,\ldots,x_n)}{\to} R_n^n\to P\to 0$$ is clearly stably free but not free if $n\geq 3$. An algebraic proof (given by myself and Madhav Nori) can be found in the article of Swan (Annals of Math studues, vol 113, pp 432-522).

Answer 2

This is an attempt to complete Tyler's argument. We first note that $KO^0(S^5)=\mathbb Z$ (note this true for all spheres of dimension $\equiv 5,6,7 \bmod 8$). This means that every topological vector bundle on $S^5$ is topologically stably trivial. Let now $E$ be an algebraic vector bundle on $S^n$, i.e., an f.g. projective module over $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$, of rank $k$. As it is topologically stably trivial that means that there are continuous sections $f_1,\dots,f_k$ of some $E\bigoplus R^m$ trivialising it, i.e., form a basis at each fibre. Now, being a trivialisation is an open condition under the sup norm (with respect to some metric on the vector bundle to be precise) so if we can show that the algebraic sections of any vector bundle $F$ are dense in the space of continuous sections we get that it is also algebraically trivial. However, picking a $G$ such that $F\bigoplus G$ is trivial reduces this to showing that $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ is dense in the ring of continuous (real-valued) functions on $S^n$ but this follows from the Stone-Weierstrass theorem.

Addendum: I made a small thinko, I implicitly use that an element of $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ that is invertible as continuous function on the sphere is invertible in the ring which of course is not true (it can be zero on a complex point of the sphere). Hence, it is necessary to consider the localisation where all such functions are inverted. This still gives a Noetherian ring however.

Addendum 1: I confused real and complex K-theory in my initial description which is now fixed. I also (unsurprisingly) found a reference for the arguments in Bochnak, Coste, Roy: Real algebraic geometry, Cap 12.

Answer 3

EDIT: Not an example because I missed the Noetherian hypothesis. Left here for posterity.

Let R be the ring of continous functions on the 5-sphere. Swan's theorem implies that finitely generated projective modules over this ring are the same thing as vector bundles on $S^5$. The tangent bundle of the 5-sphere is not trivial (it is not parallelizable, which is classical). However, the limiting homotopy group $\varinjlim \pi_5 BO(n)$ classifying stable isomorphism classes of vector bundles is trivial, and so every vector bundle on $S^5$ is stably trivial.

Answer 4

Here is an example of a non commutative ring which satisfies (3) but not (4): the integral group ring $\mathbb{Z}[C_\infty \times Q_8]$ where $Q_8$ is the quaternion group of order 8 and $C_\infty$ is the infinite cyclic group. Using the methods developed by Swan in (Projective resolutions over binary polyhedral groups, Journal fur die Reine und Ang. Math. 342 66---172 (1983)), Frank Johnson at UCL has shown:

(i) $K_0(\mathbb{Z}[C_\infty \times Q_8])\cong K_0(\mathbb{Z}[Q_8])\cong \mathbb{Z}$ (which is (3));

(ii) $\mathbb{Z}[C_\infty \times Q_8]$ has infinitely many isomorphically distinct stably free modules of rank 1. So (4) fails quite badly.

The proof involves decomposing the ring as a fibre product and then analyzing the module structure a la chapter 2 of Milnor's Algebraic K-Theory.

Swan's paper gives examples of integral group rings of finite groups with stably free, non-free modules. However there are no finite commutative groups whose integral group rings have non-trivial stably frees. T.Y. Lam's excellent book on Serre's problem has lots of information on this kind of thing.

Answer 5

Example 1.2.2 in Chapter 1 of Weibel's book in progress on K-theory http://www.math.rutgers.edu/~weibel/Kbook.html says that $R_2$ in the notation of your question has a stably free module that is not free.

Intertestingly, further down the page is the "Bass Cancellation Theorem for stably free modules". This says that if $R$ is a commutative Noetherian ring of Krull dimension $d$ then every stably free module of rank $>d$ is free.

Answer 6

Although this question already has some great answers, I thought it might be nice to note that if we drop the commutativity condition then in the case of integral group rings this question has an answer of non-abelian torsion-free polycyclic groups.

In particular, in the paper of Stafford http://www.numdam.org/article/CM_1985__54_1_63_0.pdf, Theorem 2.12 constructs a (right) stably-free non-free module over the integral group ring of any non-abelian torsion-free polycyclic group. We also have that projective modules are stably-free as polycyclic groups are solvable, and hence satisfy the Farrel-Jones conjecture.

Finally, virtually polycyclic groups are the only known examples of Noetherian integral group rings (or group rings in general), so fit this condition quite nicely.