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What is the simplest example of a domain $R$ which is regular (in particular Noetherian) and factorial which admits a finitely generated projective module that is not free?

In fact I'll be at least somewhat happy with any example, since I can't think of one at the moment.

Some brief comments: $R$ needs to have Krull dimension greater than one or else it is a PID. The module in question needs to have rank greater than one, because the hypotheses force the Picard group to be equal to the divisor class group and the divisor class group to be trivial. And famously, by work of Quillen and Suslin, one cannot take $R$ to be a polynomial ring over a field. Oh yes, and of course $R$ can't be local (or even semilocal, I suppose). I'm already out of ideas...

P.S.: If you can get an easier example by removing the hypothesis of finite generation, I'd be interested in that as well.

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I don't think you'll find any non-finitely generated examples; see Theorem 4.3 of projecteuclid.org/DPubS/Repository/1.0/… –  Manny Reyes Feb 4 '11 at 22:15
@Manny: thanks very much for your comment. As far as I'm concerned, it's informative enough to be worthy of leaving as answer. (Note that Corollary 4.5 answers my P.S. even more squarely than the theorem. It also suggests that I guessed wrong at what "finitely many primes minimal above zero" means.) –  Pete L. Clark Feb 5 '11 at 1:01
Dear Pete, I asked a very close question here: mathoverflow.net/questions/27424/…. I think the answers there which are regular (for example, Mohan Kumar's answer) also answer your question, since any stably free ideal is free, see Eisenbud "Commutative Algebra...", Lemma 19.18. –  Hailong Dao Feb 5 '11 at 5:25
@Hailong: I agree with you. Since I hadn't upvoted your question, I can only conclude that I never really read it. (That has now been remedied on both accounts.) But two different questions with the same answer is fine with me (I have given essentially the same answer to many questions...). –  Pete L. Clark Feb 5 '11 at 8:07
@Pete: thank you! Juts to be clear, I do not mind at all that they have similar answers. I thought it would be useful to point out the similarity. I learned quite a few things from the answers to your question! –  Hailong Dao Feb 8 '11 at 2:04
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6 Answers

up vote 13 down vote accepted

Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz-1)$. Map $R^3$ to $R$ by $(f,g,h)\mapsto xf+yg+zh$. Let $P$ be the kernel of this map.

$P$ is the universal example of a rank 2 projective module over a $k$-algebra that becomes free after adding a free rank-one direct summand. (That is, if $A$ is any other $k$-algebra with such a projective module $Q$, then there is a map from $R$ to $A$ such that $Q=P\otimes_RA$.) One could argue that this makes it the simplest example. In particular, Hugh Thomas's example arises from this example in this way.

To see that $P$ is not free, use the fact that Hugh Thomas's example is not free. Alternatively, invoke the much more general result of Mohan Kumar and Nori, which says that if $R=k[x_1,...,x_n,a_1,...a_n]/(\sum x_ia_i-1)$, then the kernel of the map defined by the $1\times n$ matrix $(x_1^{m_1},\ldots x_n^{m_n})$ cannot be free unless $m_1\ldots m_n$ is divisible by $(n-1)!$.

If you want an example where the UFD property is obvious, Mohan Kumar's paper "On Stably Free Modules" gives a family of examples over rings of the form $A_f$ where the rings $A$ are polynomial rings over fields. These examples all have the property that they become free after adding a free rank-one module.

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Among many nice answers, I think this one is the nicest. Thanks very much. –  Pete L. Clark Feb 5 '11 at 8:05
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The simplest example I can think of is $\mathbb C[x_1, x_2, x_3, x_4, x_5]/(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - 1)$. The proof I am going to give is almost certainly an overkill, a simpler approach should be possible. Call $X$ the spectrum of the algebra above. This is the ring of regular functions of the complement of a smooth projective quadric of dimension 3 in a smooth projective quadric of dimension 4. By standard results on Chow groups of quadrics, the Chow group of $X$ has rank 1 in dimension 2; hence the K-theory rings, which is rationally isomorphic to the Chow ring, has rank at least 2, and this implies that there must be vector bundles on $X$ which are non-trivial. These correspond to non-free projective modules.

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Dear Angelo, is your ring factorial? I confess that it is not clear to me. –  Georges Elencwajg Feb 5 '11 at 13:26
Dear Georges, I think it is. One can use Nagata' strick and induction to prove it. –  Hailong Dao Feb 8 '11 at 2:06
The Picard group of the projective quadric is generated by a hyperplane section, so the complement is factorial. –  Angelo Feb 8 '11 at 7:50
Thank you, Hailong and Angelo. –  Georges Elencwajg Feb 8 '11 at 21:57
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If Pete or someone else is still interested despite the fine answers already given, here is an analysis of what might be the simplest situation. Let $k$ be a field of characteristic $\neq 2$ and define $A=k[X,Y,Z]/(X^2+Y^2+Z^2-1)=k[x,y,z]$ where $X,Y,Z$ are indeterminates.

Proposition 1 (Nagata) The ring $A$ is a UFD iff $-1$ is not a square in $k$.

Sketch of proof: If $-1=i^2$ for some $i\in k$, the equality $x^2+y^2=(1+z)(1-z)$ shows that $A$ is not a UFD. In the other direction, if $-1$ is not a square, factoriality follows fairly easily from the following theorem of Nagata: if a noetherian domain $A$ has a prime element $t\in A$ such that $A[\frac{1}{t}]$ is a UFD, then $A$ was already a UFD.

Proposition 2 (Serre, Samuel) Consider the module $M=A\partial _x \oplus A\partial _y \oplus A\partial _z$ (the free module of rank 3 over $A$ ) and its quotient $P=M/(x\partial _X+ y\partial _y+ z\partial _z)$ . Then $P$ is projective over $A$. If moreover the field $k$ is formally real, then that module $P$ is not free over $A$.

Comments Formally real means that $-1$ is not a sum of squares in $k$. By Artin-Schreier theory this is equivalent to $k$ being orderable. Projectivity follows from an easy explicit calculation. Non-freeness over $\mathbb R$ follows , by the Serre-Swan correspondence, from the well known theorem in topology that the tangent bundle to the sphere is not trivial . In the general case of a formally real field one resorts to logical trickery of Tarski type to reduce to the case $k=\mathbb R$. Interestingly up to a few years ago, experts assured me that there was no purely algebraic proof in the case of $\mathbb R$. I don't know if one has been found since. Finally, if $k=\mathbb C$ Serre has shown that $P$ is free (however, by Proposition 1, the ring $A$ is not a UFD in the case $k=\mathbb C$ )

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Thanks, Georges, this is a great answer! As I am guessing you know, it is very closely related to a lot of things I have written about in various notes on my webpage, so it speaks deeply to me. –  Pete L. Clark Feb 5 '11 at 14:28
You do not need the Serre-Swan correspondence to prove that $P$ is not free. Simply observe that if it were free, it would have two generators, either one of which defines a nowhere-vanishing vector field on the 2-sphere. (It is a vector field on the two-sphere because it is everywhere orthogonal to the vector $(x,y,z)$, and nowhere vanishing because it is part of a basis, hence part of a basis for the tangent space after evaluating at any point.) –  Steven Landsburg Feb 5 '11 at 23:25
Yes, your argument is simpler, Steve. Thanks. –  Georges Elencwajg Feb 6 '11 at 1:23
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[Expanding my comment to an answer, at Pete's suggestion.]

According to the following paper of Bass, aptly titled Big projective modules are free, you won't find any non-finitely generated examples.


In particular, Theorem 4.3 states that if $R$ has only finitely many minimal primes, has no nontrivial idempotents, and is Noetherian modulo its Jacobson radical, then every nonfinitely generated projective module is free. Of course, this holds in particular if $R$ is a noetherian domain, as in the question asked.

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Let me just comment that (as I mentioned in my class today!), the hypothesis of connectedness is certainly necessary, since otherwise there are cheap examples. Geometrically, they correspond to finding a separation of $\operatorname{Spec} R$ and taking a trivial vector bundle of rank $\alpha$ on one open set of the separation and a trivial vector bundle of rank $\beta \neq \alpha$ on the other. (Here $\alpha$ and $\beta$ can be infinite cardinals, although in my class they were $0$ and $1$). Obviously this gives a nontrivial vector bundle on the whole space, but it's not very impressive. –  Pete L. Clark Feb 5 '11 at 8:13
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What about $R = \mathbb R[x,y,z]/\langle x^2+y^2+z^2-1\rangle$ and $M$ the projective module corresponding to the tangent bundle? This seems to satisfy your criteria except probably for factorialness (for which I have no intuition).

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Sorry, since this doesn't answer the question as posed, maybe I should have left it as a comment. –  Hugh Thomas Feb 4 '11 at 20:44
I think it is factorial: The Picard group of the real projective quadric $x^2-y^2-z^2-t^2$ is generated by hyperplane sections and $x^2-y^2-z^2-1$ is a complement of a hyperplane and thus the whole Picard group is killed off. –  Torsten Ekedahl Feb 5 '11 at 5:36
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Let $A=\mathbb R[x_0,\ldots,x_n]/(x_0^2+\ldots+x_n^2-1)$ be the coordinate ring of $S^n$. Let $P$ be the kernel of the surjection $A^{n+1}\rightarrow A$ defined by $(x_0,\ldots,x_n)$. If $n\not=1,3,7$, then $P$ is Not free. Clearly $P\oplus A \sim A^{n+1}$. For $P$ not free, note that $P\otimes C(S^n,\mathbb R)$ corresponds to tangent bundle of $S^n$, which is not free, unless $n=1,3,7$. This gives examples of projective $A$-modules of rank $=$ dimension $A$ which is stably free but not free. Note that when rank $P>$ dimension $A$, then $P$ is cancellative (Bass' 1964), i.e. $P\oplus A^t\sim Q\oplus A^t$ implies $P\sim Q$. The above example says that Bass result is best possible. Note that Suslin (~1977) has proved that if $A$ is affine algebra over algebraically closed field, then projective $A$ modules of rank = dimension $A$ are also cancellative. Hence, if we replace $\mathbb R$ by $\mathbb C$ in $A$, then $P$ is free.

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