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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.

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My answer to the question you linked to shows that (1) holds for all rings. It is not equivalent to (3) since being stably free means you can direct sum with a finite free module to get a free module. –  Kevin Ventullo Jun 8 '10 at 5:34
    
Kevin: Let R=Z/6Z, M=Z/2Z. M is projective but does not have a finite free resolution. –  Hailong Dao Jun 8 '10 at 5:40
    
Ahh, FD didn't mean what I thought. So in general, does the projective dimension of a f.g. module always mean the shortest resolution by finite projectives, or just projectives? –  Kevin Ventullo Jun 8 '10 at 6:13
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Very nice question! I was asking people in the dept this a couple of weeks ago, and I we couldn't answer it. My interest was the following: by homotopy invariance of K-theory we know that $K_0(k[T_1,\dots,T_n])=0$, and I wondered if this implied Serre's conjecture (even though elementary, direct proofs of Serre's conjecture are now known). Seems not. –  Matthew Morrow Jun 8 '10 at 9:11
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Swan constructed examples of Noetherian rings of dimension $m$ such that all projective modules of rank $\ne m$ are free, for all $m\equiv 2 (\mod 4).$ This may not be quite what you wanted, because the construction is topological and the rings are difficult to describe explicitly, but the conclusion is very strong. See Swan, Richard G. Topological examples of projective modules. Trans. Amer. Math. Soc. 230 (1977), 201--234 MR0448350 –  Victor Protsak Jun 8 '10 at 23:46

5 Answers 5

up vote 9 down vote accepted

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).

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Wow, hi Mohan! Welcome to MO. –  Hailong Dao Jun 9 '10 at 19:35
    
I was looking at this example (due to Raynaud, if I'm not mistaken), but couldn't figure out how to prove that $K_0=\mathbb{Z}.$ Can you, please, sketch "by localization one easily checks" part? –  Victor Protsak Jun 9 '10 at 23:36
    
@Victor: I think Mohan probably refers to this sequence: $G_0(R/xR) \to G_0(R) \to G(R[1/x] \to 0$. One can use this + induction to compute $G_0(R)$. –  Hailong Dao Jun 10 '10 at 0:04
    
Thank you, Hailong! By the way, there is another proof of the theorem of Mohan Kumar and Nori on Swan's homepage, math.uchicago.edu/~swan/MKN.pdf –  Victor Protsak Jun 10 '10 at 4:38

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.

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Thanks Torsten! Looks like we we have a solution. –  Hailong Dao Jun 8 '10 at 8:42

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.

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Hy Tyler, very interesting! For completeness and benefit of other viewers, can you provide some references? –  Hailong Dao Jun 8 '10 at 4:36
    
Tyler -- the ring of continuous functions on the 5-sphere is definitely not Noetherian. –  algori Jun 8 '10 at 4:43
    
But it may still be true that $R_5$ works. –  Hailong Dao Jun 8 '10 at 5:00
    
Hailong -- what is $R_5$? My bet would be the easiest example is an algebraic vector bundle. –  algori Jun 8 '10 at 5:12
    
Namely, let T be the $R_5$ module defined as kernel of the map defined by the column with all the $x_i$s. $R_5$ is a subring of your ring R, and T tensoring with R gives the tangent bundle which shows T is not free. –  Hailong Dao Jun 8 '10 at 5:13

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.

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Simon: there is a stably-free but not free for all $n$ even. But we need to show (3) holds, which I don't think is true. –  Hailong Dao Jun 8 '10 at 8:41
    
Sorry. I clearly didn't read your question sufficiently carefully. –  Simon Wadsley Jun 8 '10 at 10:27
    
Dear Simon, no problem, Weibel's book is really helpful. –  Hailong Dao Jun 8 '10 at 19:03

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.

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Hi Seamus, thanks! The result looks very interesting, do you have a reference for Frank's paper? –  Hailong Dao Jun 9 '10 at 19:16
    
I don't think it's been published yet, but I'll see if I can get an electronic copy for you. Swan's paper is worth having a look at but it is very difficult and long so it's a rather major undertaking! –  Seamus Jun 9 '10 at 19:33
    
The situation for noncommutative rings is substantially different (see McConnell and Robson's book). For example, every projective module of rank at least 2 over the first Weyl algebra is free. But there are (many) nonprincipal left ideals. The proofs aren't too hard. –  Victor Protsak Jun 9 '10 at 23:42

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