This question is motivated by this recent [question](https://mathoverflow.net/questions/27406/free-resolution-dimension). 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](http://www.math.rutgers.edu/%7Eweibel/Kbook.html)) 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](http://www.math.ku.edu/%7Emandal/talks/bottPP1.pdf)):
$$\tilde K\_0(R\_n) \cong \tilde KO\_0(S^n) $$
the [reduced $K$-groups of real vector bundles](https://math.ucr.edu/home/baez/octonions/node10.html), 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 $n=1,3,7$](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-64/issue-3.P1/On-the-parallelizability-of-the-spheres/bams/1183522319.full).


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.



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


