This question is motivated by this recent [question](http://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 $K_0(R_n)=\mathbb Z^2$.

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