# A finitely generated, locally free module over a domain which is not projective?

This is a followup to a previous question

What is the right definition of the Picard group of a commutative ring?

where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring. I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]:

Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. The following are equivalent:
(i) $M$ is projective.
(ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{r(\mathfrak{p})}$.
(iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$.
(iv) $M$ is locally free in the stronger sense: there exist $f_1,\ldots,f_n \in R$, generating the unit ideal, such that for each $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
(v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M_f$ is a free $R_f$-module.

This answers my previous question, because the rank function of an invertible module is identically one.

In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is not projective. Thus $R$ must be non-Noetherian. The wikipedia article on projective modules contains some nice information, in particular sketching an example of such a module over a Boolean ring. For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives.

I would like to see an example with $R$ an integral domain, if possible. It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.

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unfortunately cyclic modules (as in the boolean example) won't work. –  Martin Brandenburg Feb 2 '10 at 12:32

## 2 Answers

It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below.

Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.)

There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th invariant factor $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$.

Lemma. (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I_n(M))$ is open in $\mathrm{Spec}(R)$.

This openness translates to finite generation.

Proposition. If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.

Corollary. The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective.

Corollary. Over an integral domain $R$, every finitely generated flat $R$-module is projective.

Corollary. A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.

Example. Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it.

Let $S:=\bigoplus_{n=1}^{\infty}\mathbf{F}_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s_1,s_2,\dots)$ of elements of $\mathbf{F}_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.

I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R_{\mathfrak{p}}$-module $I_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I_{\mathfrak{p}}=R_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I_{\mathfrak{p}}$ is a principal ideal of $R_{\mathfrak{p}}$ with trivial annihilator.

It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.

[This answer was reorganized on the recommendation of Pete Clark.]

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Thanks again, Clark. I have voted up and accepted your answer, but to me the main point of the answer is the "No, such examples do not exist over a domain" part, so I think it would be even better if that were displayed more prominently. (Note that, as I said above, wikipedia provides an example with $R$ a Boolean ring.) Do you by any chance want to take a shot at the question of whether any of the standard references contain this fact, and if not why not? –  Pete L. Clark Feb 4 '10 at 5:54
My apologies: I'd (mis)interpreted your question to mean that you were looking for a down-to-earth way of thinking about the difference between projectivity and finite generation plus flatness, so I was trying to describe a principled way of building new counterexamples. I edited the answer as you suggested. I don't think I've ever seen the result on integral domains before (and in the first draft of my answer, I wasn't even sure it was true!), but I don't see any reason for it to be excluded from standard texts; after all, Cartier's proof is quite simple. –  Clark Barwick Feb 4 '10 at 14:21
Is Algebre Homologique the 10th chapter of Algebre or Algebre Commutative? –  Harry Gindi Feb 4 '10 at 14:22
@Everyone: This is a really, really good answer. It deserves more upvotes than the question. –  Pete L. Clark Feb 4 '10 at 14:23
@Harry Gindi: It's the 10th chapter of Algèbre. I fixed the reference to make it clearer. @Pete: Thanks! I'm glad it's interesting! –  Clark Barwick Feb 4 '10 at 14:30

After mulling things over for a while, I realize that assuming the results from Bourbaki which I mentioned in the statement of the question, there is a very straightforward answer to my question. (I mean this as no slight to Clark Barwick's excellent answer, which came instantaneously and contains lots of other valuable information. Rather, I mean that had I thought more carefully I would not have needed to ask the question at all.)

The key is the following simple result:

Lemma: Let $R$ be a[n always commutative] ring, $\mathfrak{p}_1 \subset \mathfrak{p}_2$ prime ideals of $R$, and $M$ a finitely generated locally free (in the weaker sense) $R$-module. Then $r(\mathfrak{p}_1) = r(\mathfrak{p}_2)$.

The proof is obvious, once you realize that localizing at $\mathfrak{p}_1$ is the same as localizing at $\mathfrak{p}_2$ and then localizing at (the ideal in $R_{\mathfrak{p}_2}$ naturally corresponding to) $\mathfrak{p}_1$. (This is the same argument that allows you to see that is enough to require $M_{\mathfrak{p}}$ to be free at every maximal ideal $\mathfrak{p}$.) Also I am using that the rank of a finitely generated free module over a [commutative!] ring is well-determined, as one sees by tensoring to the quotient field of some maximal ideal.

[I had some kind of psychological block coming from a vague memory that the rank function was merely semicontinuous. As far as I can see now, semicontinuity does not come up anywhere in the study of the rank function. I had to see essentially this argument in print -- in Milnor's Introduction to Algebraic K-Theory just this evening -- in order to become unblocked.]

Corollary: Let $R$ be a ring with a unique minimal prime ideal (e.g. an integral domain). Then any finitely generated locally free $R$-module has constant rank function and therefore (by the Bourbaki result above) is projective.

This also illustrates why the most natural counterexamples come from zero-dimensional rings: the rank function is determined by its behavior on the minimal primes, so you might as well look at the zero-dimensional case.

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